Math 55a, Fall 2018
December 9, 2018
Contents
1 9/5/18
1.1 Course Mechanics . . . . . . . . . . . .
1.2 Course Content . . . . . . . . . . . . .
1.3 Academic Integrity . . . . . . . . . . .
1.4 Groups . . . . . . . . . . . . . . . . . .
1.5 Examples of Groups . . . . . . . . . .
1.6 Subgroups and Group Homomorphisms
2 9/7/18
2.1 Set Theory Review . . . . . . . . . . .
2.2 Finite and Infinite Products of Groups
2.3 Back to Groups! . . . . . . . . . . . . .
2.4 Equivalence Relations and Partitions .
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3 9/10/18
3.1 Announcements . . . . . . . . . . . . . . .
3.2 Recap + Constructions . . . . . . . . . . .
3.3 Equivalence Relations and Partitions on S
3.4 Groups and Equivalence Classes . . . . . .
3.4.1 Cosets . . . . . . . . . . . . . . . .
3.4.2 Normal Subgroups . . . . . . . . .
3.5 Examples . . . . . . . . . . . . . . . . . .
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4 9/12/18
5 9/14/18
5.1 Three Constructions . . .
5.1.1 Group centers . . .
5.1.2 The Commutator .
5.1.3 Conjugacy Classes
5.2 Free Groups . . . . . . . .
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6 9/17/18
6.1 Rings and Fields . . . . . . . . . .
6.2 New Rings/Fields from Old Ones .
6.3 Power Series . . . . . . . . . . . . .
6.4 Polynomials as Functions . . . . . .
6.5 Vector Spaces over a Field . . . . .
6.6 Basic Notions about Vector Spaces
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7 9/19/18
7.1 More on Power Series . . . . . .
7.2 Characteristic . . . . . . . . . .
7.3 Vector Spaces and Linear Maps
7.4 Bases . . . . . . . . . . . . . . .
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8 9/21/18
8.1 Bases and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Direct Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9 9/24/18
9.1 Dual Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2 Quotient Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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10 9/26/18
10.1 Quotient Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 Dual Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3 Linear Algebra - thinkin’ ’bout matrixez . . . . . . . . . . . . . . . . . .
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11 9/28/18
11.1 Generalized Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.2 Categories and Functors . . . . . . . . . . . . . . . . . . . . . . . . . . .
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12 10/1/18
12.1 Jordan Normal Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.2 Categories and Functors . . . . . . . . . . . . . . . . . . . . . . . . . . .
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13 10/3/18
13.1 Bilinear Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.2 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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14 10/5/18
14.1 “Life Inside an Inner Product Space” . . . . . . . . . . . . . . . . . . . .
14.2 Operators on Inner Product Spaces . . . . . . . . . . . . . . . . . . . . .
14.3 Wrapping Up Categories: Functors . . . . . . . . . . . . . . . . . . . . .
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15 10/10/18
15.1 Self-Adjoint Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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15.2 Orthogonal Transformations . . . . . . . . . . . . . . . . . . . . . . . . .
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16 10/12/18
16.1 Modules over Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.2 Hermitian Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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17 10/15/18
17.1 Billinear Forms (Conclusion) . . . . . . . . . . . . . . . . . . . . . . . . .
17.2 Multilinear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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18 10/17/18
18.1 Tensors! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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19 10/19/18
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20 10/22/18
20.1 Group Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20.2 Burnside’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20.3 Actions of a group on itself . . . . . . . . . . . . . . . . . . . . . . . . . .
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21 10/24/18
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22 10/26/18
22.1 The Symmetric Group . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22.1.1 Normal Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . .
22.2 The Sign Homomorphism and the Alternating Group . . . . . . . . . . .
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23 10/29/18
23.1 Counting Conjugacy Classes . . . . . . . . . . . . . . . . . . . . . . . . .
23.2 The Alternating Group . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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24 10/31/18
24.1 Sylow Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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25 11/2/18
25.1 Sylow Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25.2 Proving Sylow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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26 11/5/18
26.1 Finishing Sylow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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27 11/7/18
27.1 Finite Abelian Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27.2 Representation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27.3 Summer Research . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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32 11/19/18
32.1 Finishing up characters . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32.2 Representation Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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34 11/30/2018
34.1 Induced Representations . . . . . . . . . . . . . . . . . . . . . . . . . . .
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35 12/3/2018
35.1 Frobenius Reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35.2 Some Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35.3 Real Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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36 12/5/2018 Last Class!
36.1 Group Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36.2 Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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1 9/5/18
1.1 Course Mechanics
• We will have lectures 3 times a week, but they’ll be slightly different: Monday and
Wednesday will be normal lectures (as well as this Friday), while Fridays will be
a combination of review, QA sessions (not to say you can’t ask questions during
lecture), and introduction of additional topics not on the syllabus. The course’s
content is so integral to the rest of mathematics that there’s no shortage of things
to talk about. We will also have recitation (i.e. section), hour-long meetings run
by the CAs which will be a little more informal and allow for specific questions,
and office hours for all three instructors. We will wait until Friday (or later), once
enrollment has steadied, to decide on times.
• Throughout this week, Brian will be holding office hours in the Math Lounge to talk
about this class and the math department in general; they are today, 11:45-1:30
and 3:00-5:00; tomorrow, 12:00-1:30 and 3:00-5:00; and Friday, 12:00-2:00.
• Homework consists of weekly problem sets due every Wednesday to be submitted
via Canvas. If you like to handwrite your problem sets, please scan it and/or hand
it directly to one of us (but paper problem sets are easily lost, so be careful!).
4
Otherwise, LATEX is highly preferred: it’s used throughout math and many of the
sciences, so it will be a great help to learn it now.
• We will have 2 hour-long exams. The first will be a few weeks from now and is
in part an assessment to determine whether or not you should be in this course;
it will definitely be before the add-drop deadline. The second will probably be in
November. Don’t worry – they won’t count for much by way of grading. Finally,
we’ll have a take-home final due during reading period.
We will be using Canvas as our course website. Please enroll in this course as soon as
possible if you plan on taking it so that you have access; if you’re auditing this course or
part of the extension school, please give us your email so we can give you access as well.
We have a number of course texts; all but one of them are freely available online
through SpringerLink. Our main text will be Artin’s Algebra; when we do linear algebra,
we will use part of Axler’s Linear Algebra Done Right. There are three additional texts
to be used for reference; see the syllabus.
1.2 Course Content
Math 55 will take a slightly different trajectory this year than in previous years. We
will begin with groups, segue into fields and vector spaces, discuss linear and multilinear
algebra, and finally conclude with representation theory.1
1.3 Academic Integrity
We highly encourage collaboration with your peers and interaction with the instructors.
Please talk to each other and exchange ideas as much as possible. But when it comes
time to write up your solutions, please do so on your own. We also exhort you to keep
track of and list your collaborators for every problem set.
Note. Y’all should declare a math concentration RIGHT NOW. Talk to Cindy on the
third floor.
1.4 Groups
To quote Artin, groups are the most fundamental notion of mathematics. They’ve been
around for about 150 years; take a look at the original 1870 book on groups by Jordan.
In the 20th century, mathematics became abstract; in Jordan’s book, groups were mostly
subsets of permutation groups and linear groups, but now we treat them as abstract
objects. You’ll have to take this definition essentially on faith as we tell you it’s important.
Definition 1.1. A group G is a set S together with a law of composition: a map m :
S × S → S (we assign each ordered pair of elements in S to another element in S) where
we denote (a, b) 7→ ab so that the following axioms hold:
1
Talk to Serina! She likes representation theory!
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i) There exists e ∈ S so that ea = ae = a for all a ∈ S: we call e the identity.2
ii) For every a ∈ S, we have some element b = a−1 ∈ S so that ab = ba = e; in other
words, every element has an inverse.
iii) Most importantly, for any three elements a, b, c ∈ S, we have (ab)c = a(bc); this
axiomatizes associativity. Thus abc is unambiguous notation.
Definition 1.2. If it also holds that ab = ba for all elements a, b ∈ S, we say that G is
abelian.
Definition 1.3. If we drop the second axiom (so inverses don’t have to exist), we get
what’s called a semigroup.
We now descend a bit into the abyss of bad mathematical notation (and you’ll never
get out). Technically, a group is a pair (S, m) of a set and a binary operation, but we
usually write G for the set and just talk about elements of G.
1.5 Examples of Groups
Example 1.4. Number systems (the integers, the reals, the rationals) are sets together
with the operation of addition, so they are all groups. The most fundamental group
in existence is probably (Z, +); we also have (Q, +), (R, +), and (C, +). We also have
perhaps a less familiar number system: (Z/n, +), the integers modulo n.
Example 1.5. We can take “products” of groups. Given G a group, we can define Gn :=
{(a1 , . . . , an ) ∈ G} along with termwise composition. In other words, (a1 , . . . , an )(b1 , . . . , bn ) =
(a1 b1 , . . . , an bn ). So we can talk about Zn , Qn , etc.
Example 1.6. We also have groups where the operation is not addition. Take Q∗ =
(Q \ {0} , ×); now every element has an inverse and our identity is 1. We can do the same
to other number systems to obtain R∗ and C∗ . Among the nonzero complex numbers,
we could also restrict ourselves further to {z ∈ C : |z| = 1}, which is also a group under
multiplication (verify this yourself if you don’t believe it).
Example 1.7. Take a real number x 6= 0; we can define a group “R/x” as ([0, x), +) where
(a, b) 7→ ε and 0 ≤ ε < x so that a + b = nx + ε for some integer n. This is analogous to
modding out Z by an integer. Notice that we could have some abuse of notation here; I
think it will be clearer once we start talking about equivalence classes and stuff soon.
Every group we’ve talked about so far is abelian; in general, groups arising from number
systems are abelian, though not always.
Example 1.8. We can also talk about symmetry groups. The simplest ones to consider
arise from permutations of sets, which are bijective maps from a set to itself.3 For a set
2
We could also use ∃ for “there exists” and ∀ for “for every”, but because I am texing instead of writing
I will just write things out.
3
A map f : S → T is injective if f (a) = f (b) =⇒ a = b; it’s surjective if for all c ∈ T , there is some
a ∈ S so that f (a) = c (so every element of T is hit by f ); it’s bijective if it’s both injective and
surjective.
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S, we have a group G = (Perm(S), ◦). The law of composition is actually composition
here; inverses exist because all permutations are bijections and thus invertible. If S is
finite, the order of G, denoted |G|, will be |S|!. The permutation group of {1, . . . , n} is
denoted Sn (sorry).
Example 1.9. There are more complicated (in some sense) symmetry groups too. We can
consider specific subgroups of the permutations of Rn that also have additional conditions.
For example, we have GLn R, which is the set of n×n invertible matrices with real entries
equipped with matrix multiplication. Similarly, we have GLn C, which acts on complex
spaces. We can then consider some restrictions: we can ask for volume- and orientationpreserving (SLn R), distance-preserving (On R), etc.4 We can also consider these over Q
or even finite groups; SLn (Z/n) is one such studied group.
One more example! Examples are extremely important, and when you see some sort
of statement about groups (and in general other things) it’s helpful to understand this
statement for specific groups.
Example 1.10. We can also consider symmetries of geometric objects, like rotations of
regular n-gons, or all their symmetries. For a square, we have 4 rotations, 2 reflections,
and the various other symmetries we get by composing these. Similarly, we can do this
for 3-dimensional objects like cubes and tetrahedra.
1.6 Subgroups and Group Homomorphisms
You might notice that there are some repeats on the board. For example, the group of
rotations of an n-gon is in some sense “the same” as Z/n. The study of groups is all about
the relationships between groups, and we will make some of these notions precise.
Definition 1.11. Let G be a group. A subgroup H ⊂ G is a subset of G closed under
composition and inversion. More formally: H ⊂ G is a subgroup if and only if
i) e ∈ H,
ii) m : G × G → G restricted to H × H has image contained in H (closed under
composition),
iii) For all h ∈ H, h−1 ∈ H as well (closed under inversion).
These are somewhat redundant since closure under composition and inversion implies
that the identity lies in H. Note also that for H to be a subgroup, it needs to share a
composition law with G.
Now let G, H be any groups; we want to define a “structure-preserving” map G → H.
Definition 1.12. A map ϕ : G → H is a homomorphism if it respects the law of
composition. In other words, for all a, b ∈ G, ϕ(a, b) = ϕ(a)ϕ(b); we get the same thing
4
GL stands for general linear, SL for special linear, and O for orthogonal.
7
whether we do the composition followed by the map, or the map followed by composition.
G×G
We can denote this formally with this commutative diagram:
ϕ×ϕ
mG
G
H ×H
mH
ϕ
H
Another way of defining a subgroup is as a subset H ⊂ G whose inclusion is a homomorphism.5 In that sense, a homomorphism is a generalization of the notion of a
subgroup.
Example 1.13. Suppose m|n for m, n ∈ Z. We then have a homomorphism Z/n → Z/m
where we take every element of n to its value modulo m.
2 9/7/18
2.1 Set Theory Review
Definition 2.1. Let S, T be sets. A map f : S → T is injective if f (a) = f (b) =⇒ a =
b. It is surjective if ∀b ∈ T ∃a ∈ S : f (a) = b. Bijective if both injective and surjective.
Definition 2.2. We say |S| = |T | if there exists a bijection between them.
If there is an injection f : S ,→ T then |S| ≤ |T |.
Theorem 2.3. Schröder - Bernstein Theorem: If |S| ≤ |T | and |T | ≤ |S|, then
|S| = |T |. (proof in Halmos, p. 88)
Ex: We say that S is countably infinite if |S| = |N|. Note that |N| = |Z| = |Q|.
Note: Finite products of sets have the same cardinality as the sets: |S × S| = |S|. (but
not necessarily infinite products!)
Observation: |R| > |Z|.
Claim: if x1 , x2 , x3 , · · · ∈ R is any sequence of of real numbers, then ∃x ∈ R not in the
sequence:
x1 = 0.x11 x12 x13 ...
x2 = 0.x21 x22 x23 ...
..
.
then we can form t = 0.t1 t2 t3 .... ∈
/ {x1 , x2 , · · · }
s.t.ti 6= xii
But: If T = {(x1 , x2 , · · · ) : xi ∈ R} (set of all sequences of real numbers) then |T | = R.
Claim: If S is any set, we can form the power set P (S) the set of all subsets of S. We
claim that |P (S)| > |S|. Proof: Suppose φ : S → P (S) such that φ is a bijection. Let
A = {x ∈ S : x ∈
/ φ(x)}. Then A cannot be in the image of φ by construction, which
contradicts the supposition that φ is a bijection.
5
The inclusion map takes every element of H to itself as an element of G, as you would expect.
8
2.2 Finite and Infinite Products of Groups
Definition 2.4. If G, H are groups, then the product group G × H is the set {(g, h) :
g ∈ G, h ∈ H} with composition rule (g, h)(g 0 , h0 ) = (gg 0 , hh0 ). (Here, g 0 , h0 are just some
other elements of G and H, respectively. I could equivalently write g2 , h2 .)
We can concatenate this process to form products of any finite set of groups G1 × G2 ×
· · · × Gn .
Definition 2.5. Suppose now that we have
Q∞a countably infinite sequence of groups
G1 , G2 , · · ·. I can form the infinite product i=1 Gi with the set {(g1 , g2 , · · · ) : gi ∈ Gi }.
This group is never countable (unlessL
the groups are all trivial).
We can also form the direct sum ∞
i=1 Gi with the set
{(g1 , g2 , · · · ) : gi ∈ Gi and only finitely many gi 6= ei the identity}
This is countable (as long as all the constituent groups are countable).
2.3 Back to Groups!
Let G be a group. If H, H 0 both subgroups of G (which I will denote H, H 0 ⊂ G), then
the intersection H ∩ H 0 is again a subgroup.
Remark. If Σ is a subset of the underlying set of G, then there is a smallest subgroup
Σ ⊂ G containing Σ. We call Σ the subgroup generated by Σ. We form:
\
H = {an1 1 an2 2 · · · ank k ai ∈ Σ ni ∈ Z}
< Σ >≡
Σ⊂H
If Σ = {a} (one element), then a = {an n ∈ Z}. If an 6= e for any n 6= 0, then
< a >∼
= Z. Otherwise, a ∼
= Z/N where N is the smallest integer such that aN = e.6
Definition 2.6. We say that a group is cyclic if it can be generated by a single element.
Example 2.7. Show that SL2 Z can be generated by 2 elements, where
a b
SL2 Z = {
: a, b, c, d ∈ Z ad − bc = 1}
c d
Definition 2.8. Let G, H be groups. A homomorphism φ : G → H is a map that
respects the law of composition of each group. That is, ∀a, b ∈ G we have φ(ab) =
φ(a)φ(b), or equivalently the following diagram commutes:
G×G
φ×φ
mG
G
H ×H
mH
φ
H
where mG , mH are the composition maps for G, H respectively.
6
The symbol ∼
= stands for ”is isomorphic to”, which we will define soon!
9
Definition 2.9. A group homomorphism is injective/surjective if its action on sets
is injective/surjective, and an isomorphism if both. We often denote an injection by
φ : G ,→ H, and we denote the existence of an isomorphism by G ∼
= H.
Definition 2.10. An automorphism is an isomorphism φ : G → G (from the group to
itself).
Definition 2.11. If φ : G → H is any homomorphism, the kernel of φ is defined:
ker(φ) ≡ {g ∈ G : φ(g) = eH }
and ker(φ) is a subgroup of G.
Definition 2.12. The image of φ is
im(φ) ≡ {h ∈ H : φ(g) = h for some g ∈ G}
and (φ) is a subgroup of H.
2.4 Equivalence Relations and Partitions
This is discussed in Halmos, and in Artin Ch. 2.
Definition 2.13. Suppose S is a set. An equivalence relation on S is a relation that
may hold between any given pair of elements of S, denoted by a ∼ b, satisfying the
following axioms:
i) Reflexive: a ∼ a for all a ∈ S
ii) Symmetric: a ∼ b =⇒ b ∼ a
iii) Transitive: a ∼ b and b ∼ c implies a ∼ c.
Remark. An equivalence relation ∼ can be thought of as a subset of S × S of elements
{(a, b) : a ∼ b}.
Definition 2.14. Given a set S and an equivalence relation ∼ on S, we get a partition
of S into equivalence classes. A partition is an expression of S as a disjoint union of
subsets ti Si , and an equivalence relation is a subset [a] ⊂ S such that for all a, a0 ∈ [a]
we have a ∼ a0 .
Remark. Given an equivalence relation, we can define a unique partition. Given a partition, we can define a unique equivalence relation.
Example 2.15. Some trivial examples:
• a ∼ b ⇐⇒ b ∼ a
• a ∼ b for all a, b ∈ S.
Less trivial examples:
10
• Fix n ∈ Z, then a ∼ b ⇐⇒ n|a − b (equivalence mod n).
• If f : S → T is a map of sets, we get an equivalence
relation a ∼ b ⇐⇒ f (a) = f (b).
F
This corresponds to the partition S = t∈(f ) f −1 (t).
We want to think of equivalence classes induced by group homomorphisms, which will
lead us to the idea of a quotient group.
Q: Given a subgroup H ⊂ G, when does there exist a homomorphism φ : G → K such
that ker(φ) = H?
3 9/10/18
3.1 Announcements
• Every Tuesday, there’s this cool event called Math Table which involves undergrads
giving short talks about math they enjoy. Next Math Table is tomorrow, 6pm,
in this room (507); Davis Lazowski, a current senior, will be talking about quiver
representations. Food will be provided!
• Problem 3 will not be graded on problem set 1.
3.2 Recap + Constructions
Proposition. All subgroups of Z are of the form nZ for some n.
Proof. We use the Euclidean algorithm: given H ⊂ Z a subgroup, let n be the smallest
positive integer in H (assuming H is not trivial; if H is trivial, then H = {0} and
n = 0). We claim that H = nZ. We know that H must include all multiples of n and
all multiples of −n. Suppose H 6= nZ; then H ⊃ nZ and there exists some m ∈ H with
n 6 |m. Using the Euclidean algorithm and closure under addition, we can obtain some ε
so that 1 ≤ ε < n and m = an + ε for a ∈ Z; ε ∈ H, but this contradicts our claim that
n was the smallest element in H.
This is relevant because we will use some of it in class: Let G be any group and a ∈ G
an element. We can get a homomorphism ϕa : Z → G sending n 7→ an . ker ϕa is a
subgroup of Z; either ker ϕa = nZ for some n 6= 0 or ker ϕa = {0}. If we have the latter,
the image of ϕa is then isomorphic to Z and we say a has order ∞; in the former case,
the image is isomorphic to Z/n and we say a ∈ G has order n.
Recall that we defined the product of two groups G, H as G × H = {(g, h) : g ∈ G, h ∈
H} with the law of composition (g, h)(g 0 , h0 ) = (gg 0 , hh0 ). A natural question to ask is
whether or not we can express a group as a product of “simpler” groups.
Proposition. Z/6 ∼
= Z/2 × Z/3.
Proof. We exhibit an explicit isomorphism by mapping n 7→ (n (mod 2), n (mod 3)).
There can, moreover, be more than one isomorphism, and it turns out that we can move
between these isomorphisms by composing by another automorphism.
11
Proposition. Z/4 6∼
= Z/2 × Z/2.
Proof. There are several ways to see this; one is to see that Z/4 is cyclic (can be generated
by a single element) while in the product, this is not true. Alternately, Z/2 × Z/2 has all
elements with order 2.
In general, Z/m × Z/n ∼
= Z/mn if and only if (m, n) = 1 (this is the statement of the
Chinese Remainder Theorem).
3.3 Equivalence Relations and Partitions on S
As we said last time, an equivalence relation is a relation among pairs of elements in S
satisfying reflexivity, transitivity, and symmetry. This gives rise to a partition of S by
saying that any two equivalent elements should be placed in the same equivalence class.
Given any map S → T , we have an associated equivalence relation
a
` (or, equivalently,
0
0
−1
partition) given by s ∼ s if f (s) = f (s ). We get a partition S = t∈im f f (t), where
we define f −1 (U ) = {s ∈ S : f (s) ∈ U } for U ⊂ T . We use f −1 (t) as shorthand for
f −1 ({t}). Saying that a map is injective is equivalent to saying that this equivalence
relation is the trivial one (every element is equivalent only to itself), or that the partition
we get is trivial. If you want to see proofs, go see Halmos.
3.4 Groups and Equivalence Classes
3.4.1 Cosets
Let G, H be groups and ϕ : G → H any homomorphism. We can generate an equivalence
relation as follows: let K = ker ϕ = {g ∈ G : ϕ(g) = eH }. This is one equivalence class
in our partition of G, and the fibers of ϕ are cosets of K.
Definition 3.1. Let K ⊂ G be a subgroup. A left coset of K in G is a subset of the
form aK = {ak : k ∈ K}.
Cosets of K ⊂ G also produce a partition corresponding to an equivalence relation by
setting a ∼ b ⇐⇒ b = ah for some h ∈ K. We should check that this is indeed an
equivalence relation.
• a ∼ a is trivially true since a = ae and e ∈ K by definition.
• a ∼ b ⇐⇒ b ∼ a is true since, if b = ah for H ∈ K, then a = bh−1 and we know
h−1 ∈ K because K is closed under inversion.
• a ∼ b, b ∼ c =⇒ a ∼ c: we know b = ah and c = bh0 for h, h0 ∈ K; thus c = a(hh0 )
and K is closed under multiplication, so a ∼ c.
`
Therefore, given a subgroup K ⊂ G, we get a partition G = aK; this set of equivalence
classes is denoted G/K . Right now, we have no group structure and this is just a set.
However, notice that we do have a set map G → G, b 7→ ab for each a. It’s definitely not a
homomorphism, but it is a bijection because we can define the inverse (left multiplication
12
by a−1 ). In particular, under this set map every subgroup K 7→ aK, the coset. This
shows that all the cosets are bijective to on another and in particular have the same
cardinality, |K|. So we actually have written G as a disjoint union of subsets, all of which
have the same cardinality. As a consequence, we have
Theorem 3.2. If G is finite, for K ⊂ G, |G| = |K| · [G : K] (where [G : K] is the
cardinality of |G/K |).
We call [G : K] the index of K ⊂ G.
Corollary 3.3. If G is any finite group and K ⊂ G a subgroup, we know that |K|||G|.
Corollary 3.4. In particular, if a ∈ G, recall that we can produce a subgroup by considering the image of ϕa : Z → G, n 7→ an . This image is a subgroup, and its size divides
the order of G. Therefore, the order of any element a ∈ G (the smallest positive power
of a so that an = e) is also a divisor of |G|.
Corollary 3.5. Any group of prime order p must be cyclic – isomorphic to Z/p – because
any nontrivial element has order p and its powers therefore generate the group.
If we have groups of composite order, this is not necessarily true; we saw above that
we have at least 2 groups of order 4. A good question to think about is finding the
isomorphism classes of groups of a certain order. It’s a hard problem that gets more
complicated as n increases, as expected.
3.4.2 Normal Subgroups
Now suppose that ϕ : G → H is any homomorphism with kernel K ⊂ G. A priori, we
only know that the kernel is a subgroup. We saw that the fibers of ϕ are precisely the
cosets of K in G. More concretely, if b = ϕ(a), then ϕ−1 (b) = aK. Why are defining
left cosets instead of right cosets, though? It turns out that we would have to be a little
more careful about order of composition, but otherwise everything would be the same.
So ϕ−1 (b) = Ka as well. Therefore, in order for K to be the kernel of a homomorphism,
it must have aK = Ka for all a (its left and right cosets coincide); in other words,
aKa−1 = K. So if we multiply an element of k on the left by some element a and on the
right by a−1 , we get back another element of k.
Definition 3.6. A subgroup K ⊂ G so that aKa−1 = K for all a ∈ G is a normal
subgroup.
Next time, we will see that all normal subgroups can be written as kernels of homomorphisms, or that we can quotient out by said normal subgroup.
3.5 Examples
Example 3.7. Let S3 be the permutation group on 3 letters; it can also be realized as the
symmetries of an equilateral triangle by visualizing symmetries as acting on the vertices.
It’s not hard to see that every permutation of 3 vertices can be realized by a symmetry
of the triangle. What does this group look like?
We have three types of elements:
13
• e, the identity; it does nothing. This has order 1.
• 2 rotations which rotate by 2π/3 and 4π/3 and correspond to the cycle permutations
(123) and (132).7 These have order 3.
• 3 reflections which transpose two elements, which we can write as (12), (13), and
(23).8 These haver order 2.
What are the subgroups of S3 ?
• The trivial subgroup {e};
• The cyclic group generated by rotations, {e, (123), (132)} (isomorphic to Z/3);
• The subgroups of order 2 each consisting of a single transposition, {e, (12)}, etc.,
which are isomorphic to Z/2
• All of S3 .
Which ones are normal? The trivial group and the entire group are clearly normal (e
commutes with everything, and the group is closed under multiplication). It turns out
that the subgroup of rotations is also normal, but not the transpositions. Intuitively, if
we rotated the triangle 120◦ , reflected, and rotated back, we would get another reflection,
but it would be around another axis. For example, (123) ◦ (23) ◦ (132) = (13).
Another way of saying this is that we have a map S3 → Z/2 with kernel Z/3.
4 9/12/18
Notation Screw-up Alert: The group of symmetries of a regular n-gon is denoted
Dn , not D2n .
Joe’s OH: Friday, 1:00 PM - 3:00 PM, SC 339 (inside Math Library)
Normal Subgroups
Definition 4.1. A subgroup H ⊂ G is normal if aHa−1 = H for all a ∈ G. This is
equivalent to the statement aH = Ha.
If we have a law of composition, we can also compose subsets! Let A, B ⊂ G, then
AB = {ab : a ∈ A, b ∈ B}. In particular, if A = {a}, then we write aB (left coset) and
Ba (right coset).
Theorem 4.2. Given a subgroup H ⊂ G there exists a surjective homomorphism φ :
G → K with kernel H if and only if H is a normal subgroup. In this case, we call
K = G/H the “quotient”.
7
These parenthesized triples are known as cycle notation; for a permutation σ, we can write a permutation as a series of (1σ(1)σ 2 (1) . . . ), denoting where σ applied repeatedly sends 1. If we have elements
not hit by σ n (1), we can write other series for them. For example, (123) is the permutation where
σ(1) = 2, σ(2) = 3, and σ(3) = 1.
8
Omitting elements means that the permutation fixes them.
14
Proof. Suppose there is a surjective homomorphism φ : G K with kernel H. We want
to show that aHa−1 = H.
aHa−1 = {aha−1 : h ∈ H, a ∈ G}
φ(aha−1 ) = φ(a)φ(h)φ(a−1 ) = φ(a)eK φ(a−1 ) = φ(a)(φ(a))−1 = eK
So everything in aHa−1 is in the kernel of φ, from which we can conclude aHa−1 = H
and thus that H is normal. (!)
Now, suppose H is a normal subgroup. Take K = {cosets aH of H in G}. Define
the law of composition on K by (aH)(bH) = abH (note that this depends on H being
normal). The identity element eK = eH = H. We define the map φ : G → K by
φ(a) = aH. Note that φ(a) = H if and only if a ∈ H. That is, ker(φ) = H.
Correspondence Theorem
Let f : G → K = G/H be a surjective homomorphism of groups. Then we have a
bijection:
{subgroups of K} ⇐⇒ {subgroups J ⊂ G : J ⊃ H}
That is, L ⊂ K is a subgroup if and only if f −1 (L) ⊂ G where f −1 (L) = {a ∈ G : f (a) =
L}.
Examples
Consider S3 = Perm({1, 2, 3}) (which is also equal to the symmetries of an equilateral
triangle) (see section 1.5 in Artin for notation). We have three kinds of elements
e → identity
(123), (132) → rotations
(12), (23), (13) → reflections
We have subgroups:
{e,
← Normal
 (123), (132)}

{(e, (12)}
← Not Normal
{e, (13)}


{e, (23)}
Definition 4.3. We say that a sequence of groups and homomorphisms:
···
An
φn
An+1
φn+1
An+2
···
is exact if for all n ker(φn+1 ) = (φn ).
In the case of S3 , we have the following exact sequence:
{e}
Z/3
S3
15
Z/2
{e}
Remark. For a sequence of length 5 (often called a short exact sequence):
{e}
A
α
B
β
C
{e}
being exact is equivalent to saying that α is injective, and β is surjective. This then
implies that A ⊂ B is normal, and C ∼
= B/A.
Q: Is there only one group that fits into the above exact sequence, given all the other
groups? That is, given that the following is an exact sequence:
{e}
Z/3
G
Z/2
{e}
Is the group G unique? No! We could take G = Z/6 or G = S3 , for example!
Definition 4.4. A group is called simple if it has no normal subgroups.
Simple groups are the “atomic objects” of groups - so can we classify all the simple
groups in a sort of “periodic table of groups”?
[I have intentionally left out the part where Joe explained a homework problem]
Free Groups
First we will look at the free group on 2 generators F2 , calling the generators a, b. As a set
F2 = {‘words’ in a, b, a−1 , b−1 }. That is, elements like an1 bn2 an3 · · · bnN or bm1 am2 · · · amM .
We ‘reduce’ a word by combining everything that can be combined a la a3 a4 b3 = a7 b3 .
We define the composition by juxtaposition - that is, ab2 a · b7 a3 b3 a7 = ab2 ab7 a3 b3 a7 .
Remark. If G is a group with generators α, β, we get a surjective homomorphism F2 G
that sends a 7→ α and b 7→ β.
5 9/14/18
Today, we’ll end our discussion of groups. On Monday, we’ll start linear algebra.
5.1 Three Constructions
5.1.1 Group centers
Let G be any group.
Definition 5.1. Z ⊂ G is the center of G and is defined as {a ∈ G : ab = ba∀b ∈ G}; in
other words, the elements that commute with everything else in the group. In particular,
they commute with each other, so Z is an abelian subgroup of G.
Proposition. Z ⊂ G is a normal subgroup.
16
Proof. One way to see this is to note that ab = ba =⇒ bab−1 = a. Thus conjugating
an element of Z by an arbitrary element of G just returns the same element of X and
gZg −1 = Z for all g ∈ G.
This argument works for any abelian group, actually, and this shows that every abelian
group is normal as well.
Thus we have an exact sequence
{e} ⊂ Z
H = G/Z
G
{e}
Admittedly this might not tell us very much about the group: for example, if Z is trivial.
Example 5.2. Z(Sn ) = {e} for n > 2.
Example 5.3.
(
SLn R =
Z/2
if n even
.
{e} if n odd
Example 5.4. Z(GLn ) is the group of multiples of the identity.
If we have a subgroup H ⊂ G, it is not necessarily true that Z(H) ⊂ Z(G). For
example, if we take the subgroup generated by a single cycle in Sn , it’s abelian and thus
its own center.
5.1.2 The Commutator
The center essentially expresses how “abelian” a group is; what if we want to express
how not-abelian it is? Let G be any group and a, b ∈ G.
Definition 5.5. The commutator of a and b is denoted [a, b] and is aba−1 b−1 .
Notice that the commutator is the identity if and only if a and b commute. We can
now consider the subgroup generated by all commutators.
Definition 5.6. Let C(G) = h{[a, b] : a, b ∈ G}i; this is called the commutator subgroup.
C(G) is in fact normal, which you can verify directly by checking conjugation or through
the following:
Note. We say that H ⊂ G is characteristic if for all ϕ ∈ Aut(G), ϕ(H) = H. In
other words, every automorphism carries the subgroup to itself. In particular, every
characteristic subgroup is normal. The commutator subgroup is one such group since
ϕ([a, b]) = [ϕ(a), ϕ(b)].
Example 5.7. What is C(Sn )? If we commuted two of the transpositions, we would get
one of the rotations (one way of seeing this is through the sign homomorphism 9 ). It
turns out that the commutator subgroup of Sn is An , the set of even permutations of An .
9
Every permutation can be expressed as a product of transpositions, and it turns out that no matter
how you write this product, the parity of the number of transpositions is constant, so we get a map
Sn → {±1}. This is in fact a homomorphism and is known as the sign homomorphism.
17
Example 5.8. What is C(GLn R)? Well, it’s definitely a subgroup of SLn R, and it turns
out that SLn R is the commutator subgroup of GLn R.
The commutator subgroup is normal, so we get an exact sequence
0
C(G)
H = G/C(G)
G
0.
Notice that because all commutators have been set to e in H, H is in fact abelian.
Definition 5.9. We call H the abelianization of G.10
5.1.3 Conjugacy Classes
It turns out that conjugates of elements share many properties with the elements themselves, like order.
Observation. Say that a ∼ b when b = σaσ −1 for some σ ∈ G; this defines an equivalence relation and therefore partitions G into what we call conjugacy classes.
Definition 5.10. The conjugacy class of an element a ∈ G is the equivalence class of a
under this equivalence relation.
Conjugacy classes are by no means subgroups, except for the one corresponding to the
identity. Notice that the identity is equivalent only to itself since aea−1 = e for all a. By
a similar argument, if an element is in the center of a group its conjugacy class contains
itself. But if a group is non-abelian, it’s much harder to figure out what the conjugacy
classes are, or even how many there are.
5.2 Free Groups
Let a1 , . . . , an be formal variables.
Definition 5.11. The free group on n generators Fn consists of the set of reduced words
of the form ani11 · ani22 · · · anikk where ni 6= 0 for any i and il 6= il+1 for all l. In other words,
we don’t repeat symbols. Note that exponents can in fact be negative because of closure
under inversions. Its law of composition is juxtaposition, or concatenation, of words,
followed by reduction, or cancellation: if two symbols adjacent to each other are the
same, we add their exponents; if we end up with a 0 exponent, we remove the symbol.
This turns out to be a group with identity element the empty word e (k = 0).
Given any group G and any finite collection of elements b1 , . . . , bn ∈ G, we have a unique
map Fn → G sending ai 7→ bi . In particular, if G is finitely generated by b1 , . . . , bn , then
our map Fn → G is in fact a surjection.
Suppose K = ker(ϕ) where ϕ : Fn → G.11
10
We’ll figure out sometime later why Abel has his name on such an important group property when
“group theory” wasn’t even really a thing in his time.
11
The subgroup of a finitely generated group is not necessarily finitely generated.
18
Definition 5.12. If K is finitely generated, then we say that G is finitely presented.
Say that K = hc1 , . . . , cl i with ci ∈ Fn . G is generated by the bi with relations c1 , . . . , cl .
Example 5.13. Consider D4 , the symmetries of the square. It is generated by r, the π2
rotation, and s, the reflection along the horizontal axis of the square. We have a map
F2 → D4 which sends the generators of F2 to r, s respectively. What is the kernel of this
map? It includes r4 , s2 , and srsr, and is in fact generated by these (it can be generated
by other relations too, and this is just one such example).
Given this, we say that D4 has two generators with relations r4 = s2 = rsrs. This
specifies D4 completely.
In many situations, the group presentation is a good way to view groups (and in others
not so much). One big problem in group theory is the “word problem:” if we have a
finite collection of symbols and a finite collection of relations, when is the group we get
trivial? This problem is still open.
6 9/17/18
6.1 Rings and Fields
Definition 6.1. A commutative ring is a set S with two laws of composition +, ×,
satisfying the following axioms:
i) (S, +) is an abelian group with identity element denoted by 0.
ii) There exists an identity element 1 ∈ S such that 1a = a1 = a for all a ∈ S.
iii) (ab)c = a(bc) for all a, b, c ∈ S (× is being suppressed here).
iv) ab = ba for all a, b ∈ S.
v) a(b + c) = ab + ac for all a, b, c ∈ S.
There are a lot of useful rings that are not commutative, and we will see some in this
class. But we will tend to focus on commutative rings.
Definition 6.2. A field is a ring such that ∀a 6= 0 there exists b such that ab = ba = 1.
That is, all nonzero elements of a field have inverses.
Note that as a consequence of the field axioms, 1 6= 0 and ab = 0 =⇒ a = 0 or b = 0.
Definition 6.3. A ring/field homomorphism is a map φ : S → T that respects both
laws of composition. That is:
φ(a + b) = φ(a) + φ(b)
φ(1S ) = 1T
φ(ab) = φ(a)φ(b)
Proposition. If φ : S → T is a field homomorphism, then it is injective.
19
Proof. Say φ(a) = 0 but a 6= 0. Then we have some b such that ab = 1, so φ(ab) = 1.
But φ(ab) = φ(a)φ(b) = 0 × φ(b) = 0. This is a contradiction.
Example 6.4. Rings: Z, Z/p (p prime)
Fields Q, R, C
Figure 1: We first learn about fields in the context of counting systems. Here, the Count
holds up 4 fingers, representing the fact that Z/4 is not a real field, just like he
is not a real count.
6.2 New Rings/Fields from Old Ones
Given a field k, we can form the ring of polynomials of x:
k[x] ≡ {a0 + a1 x + a2 x2 + · · · + an xn | ai ∈ k, n ∈ N}
(1)
Here, x is what we call a formal variable. It isn’t an element of anything12 , and we can
equivalently think of these polynomials as the sequences of their coefficients. Note that
this is not a field. We can define the field of rational functions of x:
p
x
p
∼
⇐⇒ py = qx
(2)
k(x) ≡ { | p, q ∈ k[x], q 6= 0}
q
q
y
These notions generalize to any number of variables, k[x, y], k(x, y), k[x1 , · · · , xn ], k(x1 , · · · , xn ).
12
As you might be suspecting, there are well-defined ways of associating functions on a field to elements
of the polynomial ring.
20
6.3 Power Series
Definition 6.5. Let k be a field. We define the ring of formal power series in x as:
(∞
)
X
k[[x]] ≡
an x n | an ∈ k
(3)
n=0
where we add and multiply as we would polynomials.
Definition 6.6. Let k be a field. We define the field of Laurent series in x as:
k((x)) ≡ an xn + an xn+1 + · · · | ai ∈ k, n ∈ Z
(4)
6.4 Polynomials as Functions
Given k a field, I can look at polynomials k[x]. If we think about these polynomials as
functions on k, then for a given polynomial f (x) there may not be an x ∈ k such that
f (x) = 0.
For instance, for k = Q the polynomial x2 − 2 has no roots. But we can form the field:
√
√
(5)
Q( 2) ≡ {a + b 2 | a, b ∈ Q}
Note
√ that this is a valid field (treat
2”. Note that R(i) = C.
√
2 as you normally would). We call this “Q adjoined
6.5 Vector Spaces over a Field
Definition 6.7. Fix a field k. A vector space over k is a set V with with two operations:
i) + : V × V → V (addition)
ii) k × V → V (scalar multiplication)
satisfying the following axioms:
i) (V, +) is an abelian group (identity element 0)
ii) 1v = v for all v ∈ V .
iii) b(av) = bav for all a, b ∈ k and v ∈ V .
iv) (a + b)v = av + bv for all a, b ∈ k and v ∈ V .
v) a(v + w) = av + aw for all a ∈ k and v, w ∈ V .
Definition 6.8. Given V /k (V a vector space over k), a subset W ⊆ V that is closed
under the vector space operations is called a subspace (that is W + W ⊆ W and
kW ⊆ W ).
21
Example 6.9.
•
k n ≡ {(k1 , k2 , · · · , kn ) | ki ∈ k}
(6)
k is a subspace of k n for n > 0.
• We can consider k[x] as a vector space over k by forgetting about the multiplication
on k[x]. We can consider k[[x]] as a vector space over k as well.
• Question: Are k[x] and k[[x]] isomorphic as vector spaces?
• If S is any set, the set of maps S → k is a vector space over k.
• Consider set of maps from R → R as a vector space over R. We have a subspace
of continuous maps from R → R. A subspace of that is differentiable maps from
R → R.
6.6 Basic Notions about Vector Spaces
Definition 6.10. Let V be a vector space over k, and let v1 , · · · , vn ∈ V . The span of
v1 , · · · , vn is the smallest subspace of V containing v1 , · · · , vn . The span is expressible as
{c1 v1 + · · · + cn vn | ci ∈ k}.
Definition 6.11. We say that v1 , · · · , vn ∈ V span V if their span is equal to V .
Definition 6.12. We say that v1 , · · · , vn are linearly independent if
c1 v1 + · · · cn vn = 0 ⇐⇒ c1 = · · · = cn = 0
(7)
Equivalently, given v1 , · · · , vn we get a map φ : k n → V that sends (c1 , · · · , cn ) 7→
c1 v1 + · · · + cn vn . We say that v1 , · · · , vn span V if φ is surjective and v1 , · · · , vn are
linearly independent if φ is injective.
Definition 6.13. If v1 , · · · , vn spans V and is linearly independent, we say it is a basis
for V .
Example 6.14. For k[x], we have a basis {1, x, x2 , · · · }.
Question: Does there exist a basis for k[[x]]? If so, what is it?
7 9/19/18
Announcement: please speak up during lecture! If you’re not understanding, say something; don’t be afraid to ask questions, and it’s better to clear up any confusion now
rather than later.
22
7.1 More on Power Series
Recall that a field (S, +, ×) is a set S with these two operations so that (S, +) is an
abelian group with identity 0; (S \ {0}, ×) is an abelian group with identity 1; and these
operations are compatible through the distributive law. If we don’t require existence of
multiplicative inverses, we get a commutative ring. If multiplication is not commutative,
we just get a ring.
In general, the study of rings is a much deeper subject than that of fields. From rings,
we are able to come up with the concept of primes, unique factorization, etc. – take 123
if you want to find more.
Remember that last time, we defined k[[x]] as the ring of formal power series that is,
{a0 + a1 x + a2 x2 + · · · : ai ∈ k}. Here, these ai are any elements in a field. We use
formal because if we’re doing analysis, say, we would rather deal with power series to
converge, but arbitrary fields don’t necessarily have norms, so we have no such notion of
“convergence.” This is a commutative ring, but not a field. However, we claim that
Proposition. a0 + a1 x + · · · has a multiplicative inverse if and only if a0 6= 0.
P i
Proof.
We
solve
a
system
of
equations.
Take
some
other
power
series
bi x so that
P
P
ai xi bi xi = 1. Multiplying this out, we know that a0 b0 = 1, that a1 b0 + a0 b1 = 0,
and so on. Thus b0 = a10 and b1 = −aa10b0 , etc. At each stage, we can solve for the next bi
as long as a0 is nonzero. Conversely, if a0 = 0, it’s clear we have no possible value for b0 ,
so our power series has no multiplicative inverse.
If we enlarge our ring a little, though, we can end up with a field; as we discussed, this
is the field k((x)) of Laurent series; namely, power series of the form an xn +an+1 xn+1 +· · ·
when n ∈ Z. If we allow fractional powers of x, there are many things we could ask for:
only bounded denominators, all possible expressions, etc.
7.2 Characteristic
Given a field k, we always have the canonical homomorphism of rings ϕ : Z → k. This
simply sends 1 → 1k (i.e. the number 1 to the multiplicative identity of k). Does this
have a kernel?
If it doesn’t, then we get an injection, and most of the fields we’ll use are of this form.
For example, Q, R, C are like this, as are their polynomial rings. If it does, then
Proposition. The kernel must be pZ for some prime p.
Proof. If ker ϕ = nZ for composite n, we can write n = ab for a, b < n. We also know that
n is the smallest positive integer for which ϕ(n) = 0. Now we know that ϕ(n) = ϕ(a)ϕ(b);
thus either ϕ(a) or ϕ(b) is 0 because fields have no zero divisors, contradiction.
Definition 7.1. We say that k has characteristic p if the kernel of ϕ is pZ.
So far, the only example we’ve seen of fields of characteristic p is Z/p, but eventually
you will learn how to construct finite fields.
Theorem 7.2. For all positive n and all primes p, there exists a unique field with pn
elements, and these are all the finite fields.
23
7.3 Vector Spaces and Linear Maps
Definition 7.3. As we said last class, a vector space V over a field k is a set with 2
operations: addition, V × V → V ; and scalar multiplication, k × V → V .
For example, 0 is a vector space; so is k, and in general k n for any integer positive
integer n.
Definition 7.4. Let V, W be vector spaces over k. A homomorphism of vector spaces,
which we call a linear map, ϕ : V → W is any map that respects the vector space
structure on V . That is, ϕ(v) + ϕ(w) = ϕ(v + w), and λϕ(v) = ϕ(λv), for all λ ∈ k and
v, w ∈ V .
We will be dealing with linear maps for essentially the rest of the semester!
Proposition. Notice that for V, W vector spaces over k, we can consider the set of all
homomorphisms V → W , which we denote Hom(V, W ). This is itself a vector space over
k.
Proof. Suppose ϕ, ψ ∈ Hom(V, W ). We define addition and multiplication pointwise;
that is, (ϕ + ψ)(v) = ϕ(v) + ψ(v), and (λϕ)(v) = λ · ϕ(v). It is easy to verify that these
sums and products are themselves linear maps.
What is the dimension of Hom(V, W )? It’s the product of the dimensions of V and W ;
we’ll do this in class soon.
How much of the vector space’s structure depend on the field? Suppose k 0 ⊂ k is a
subfield. For example, we can consider R ⊂ C. If we have a vector space V over k, we
get another vector space over the smaller field k 0 by what we call “restriction of scalars.”
After all, k 0 × V ⊂ k × V , so we can just restrict our multiplication map. In particular,
k is a vector space over k 0 .
Example 7.5. C is a vector space over R; we can think about it as the set of {a + bi :
a, b ∈ R}, and C ∼
= R2 .
What happens to Hom(V, W ) if we go over a different field? For example, how do linear
maps from V → W over C relate to those over R? We have an inclusion HomC (V, W ) (
HomR (V, W ); intuitively, there are more restrictions on a linear map over C than over R.
Long story short, fields matter.
7.4 Bases
Suppose V is a vector space over k and let Σ ⊂ V be any subset.
Definition 7.6. We say that Σ spans V if every element in V can be written as a linear
combination of elements of Σ. We say that P
Σ is linearly independent if no nontrivial
linear combination of elements of Σ is 0 (i.e.
ci vi = 0 =⇒ ci = 0 for vi ∈ Σ). If Σ is
both spanning and linearly independent, then we say that Σ is a basis of V .
24
Example 7.7. More concretely, let V = k n . A basis for V is the set of elements ei , where
ei = (0, . . . , 0, 1, 0, . . . , 0) (1 in the ith place, 0 at all others), and is the one you probably
saw in high school.
Remark. Other bases exist; for example, in k 2 , we could consider the basis {(1, 1), (1, −1)}.
Note that if k has characteristic 2, this is not a basis.13
Given a space V overP
k and some vectors v1 , . . . , vn ∈ V , we get a linear map ϕ : k n → V
where (c1 , . . . , cn ) 7→
ci vi . To say that the set spans V is equivalent to ϕ being
surjective; saying that the set is linearly independent is equivalent to ϕ being injective.
So ϕ being an isomorphism is exactly the condition for v1 , . . . , vn being a basis and we
can think about a basis as moving back to k n when dealing with abstract vector spaces.
Definition 7.8. V is finite-dimensional if it has a finite spanning subset.
Observation. If v1 , . . . , vn is any spanning set, then some subset of v1 , . . . , vn is a basis.
That is, every spanning subset contains a basis by essentially throwing out vectors.
Proof. If v1 , .P
. . , vn are independent, we’re done. If not, there is some nontrivial linear
combination
ci vi = 0, with some ci 6= 0. WLOG say cn 6= 0; then we can write vn
in terms of the remaining vs and v1 , . . . , vn−1 span V as well. Repeat this process until
our remaining set is independent; we then have a basis. This argument depends a lot on
the finiteness of the set; otherwise, we would be hard-pressed to claim that the process
“finishes” without using the all-powerful Axiom of Choice.
Proposition. Suppose V is finite-dimensional. Then any two bases for V have the same
number of elements, and we call the size of this set the dimension. Thus our notion of
“dimension” is well-defined.
Lemma 7.9. IF v1 , . . . , vm and w1 , . . . , wn are bases for V , for some i, v1 , . . . , vm−1 , wi
is again a basis.
8 9/21/18
Recall that last time, we defined a linear map ϕ : V → W as a map so that ϕ(v + w) =
ϕ(v) + ϕ(w) and ϕ(λv) = λϕ(v). We also mentioned that for a field k and a subfield
k 0 ⊂ k, a vector space over k is also one over k 0 by restricting scalar multiplication; the
endomorphisms of the vector space differ depending on which field it’s over.
Example 8.1. For example, we could take k 0 = F ⊂ k = C and let V = C. Here,
HomC (V, V ) = C since every map is just some kind of scalar multiplication. On the other
hand, if we look at HomR (V, V ), this space is much larger. We now have two basis vectors
for V (usually 1 and i) that we can send anywhere. Thus HomR (V, V ) = M2,2 (R), the
set of 2 × 2 matrices with entries in R.
13
Characteristic 2 f**ks everything up.
25
8.1 Bases and Dimension
Suppose V is a finite dimensional vector space over k.
Observation. If v1 , . . . , vm are linearly independent vectors in V , then we can complete
it to a basis v1 , . . . , vn . Likewise, if v1 , . . . , vm are a spanning set, then some subset
v1 , . . . , vm is a basis.
Theorem 8.2. if v1 , . . . , vm and w1 , . . . , wn are two bases for V , then m = n.
Proof. We claim that there exists some index 1 ≤ j ≤ n so that v1 , . . . , vm−1 , wj is a basis.
We know that v1 , . . . , vm−1 don’t span V (otherwise, vm would be a linear combination
of the other vi , contradicting linear independence). Thus for some j, wj is not
P in the
span of v1 , . . . , vm−1 . On the other hand if we use all the vi we can write wj = ai vi for
am 6= 0. We can thus write vm in terms of wj and v1 , . . . , vm−1 :
vm =
1
wj − a1 v1 − · · · − am−1 vm−1
am
so wj , v1 , . . . , vm−1 is both spanning and linearly independent, making it a basis. Repeating this process, we can replace vm−1 another w, etc. Notice that we never replace with
the same w twice since otherwise our chosen replacement would be in the span of our
m − 1-length set; when we finish, we get some subset w1 , . . . , wm of w1 , . . . , wn that’s a
basis. Therefore, m = n.
Definition 8.3. We say that the dimension of V is the cardinality of any basis.
Given a basis v1 , . . . , vn ∈ V , we get a map ϕ : k n → V given by (a1 , . . . , an ) 7→
a1 v1 + · · · + an vn . Linear independence is equivalent to injectivity of ϕ, and spanning
set is equivalent to surjectivity. If v1 , . . . , vn is a basis, it’s equivalent to an isomorphism
kn ∼
= V . That is, up to isomorphism there is only one n-dimensional vector space over
k, and we can say it’s k n . But there are lots of different identifications, since choosing a
basis of V chooses your isomorphism k n → V . Sometimes, when considering linear maps
on vector spaces, it’s useful to invoke this identity.
Definition 8.4. We denote the m × n matrices with entries in k as Mm,n .
Thinking about them as tuples in k, we can identify Mn,m with k mn . We can identify
this space also with Hom(k m , k n ). That is, given V, W vector spaces with dimensions m
V
W
and n respectively, we can choose bases for V and W to get a map k m , k n :
km
kn
More concretely, if we have some V, W and choose bases, given an m × n matrix (aij ), we
can define a map ϕ where
X
X
X
ci vi 7→
ci ai1 w1 , . . . ,
ci ain wn
which we can easily verify to be linear. Likewise, given a map ϕ, we can choose bases
and determine ϕ(vi ) in terms of the basis of w; this gives us a matrix of coefficients.
26
Definition 8.5. A matrix A = (aij ) is called the matrix representative
P of ϕ with respect
to the bases v1 , . . . , vm and w1 , . . . , vn if it defines the map ϕ : vi 7→
aij wj .
In particular, dim Hom(V, W ) = dim Mn,m = mn = dim V dim W .
Now, what if we had chosen a different basis for V , our target space. Call the new
basis, v10 , . . . , vn0 . We can relate the new basis to the old by writing the new basis vectors
as linear combinations of the old. What happens to the matrix representation of a given
transformation in this new basis? If we write
vi0 = a1i vi + a2i v2 + · · · + ami vm
we have an m × m matrix aij ; we call this the change of basis matrix P .14 Given
ϕ : V → W , A the matrix representative of ϕ with respect to the bases {vi } and {wi },
and A0 the matrix representative with respect to {vi0 } and {wi }, then A0 = AP . Similarly,
if w10 , . . . , wn0 is another basis for W , with change of basis matrix Q, then our new matrix
representative B = QA. Combining this, the matrix representative of ϕ with respect to
our prime bases is QAP . We can see that these change of basis matrices are invertible
because we can also construct an inverse map taking {vi0 } → {vi }.
The whole point of linear algebra is to avoid choices of matrices, vectors, and bases
whenever possible. We want to express as much as we can in a coordinate-free language –
instead of choosing an identification with k n , we can work with the abstract vector space.
8.2 Direct Sums
Let V, W be vector spaces over k.
Definition 8.6. The direct sum V ⊗ W = {(v, w) : v ∈ V, w ∈ W }.
L
We can extend this to finite collections of vector spaces to construct ni=1 vi = {(v1 , . . . , vn ) :
vi ∈ Vi }. However, if we have an infinite collection Vα with α ∈ I some index set. We
now have two possible constructions: either
M
Vα = {(· · · , vα , · · · )α∈I : vα ∈ I, almost all vα = 0},
a∈I
which we call the direct sum, or
Y
Vα = {(· · · , vα , · · · )α∈I : vα ∈ I}.
a∈I
Yes, I can be uncountable; for example, functions R → R can be thought of as a product
of R copies of R.
L
Q
Example 8.7. If we compare
n∈Z k vs.
n∈Z k, we can identify the first to k[x]
(polynomials), and the second with k[[x]] (power series).
14
If any of you explicitly write one of these on your homework, I’m going to cry. And take off points.
Just know that you can write linear maps in terms of any basis.
27
Putting all this infinite-dimensionality behind us, we now focus on the finite case.
Suppose we have a collection of subspaces V1 , . . . , Wn of a fixed vector space W .
P
Definition 8.8. We say that the
W
are
independent
if
ci wi = 0 with vi ∈ k, wi ∈ Wi
α
P
implies ci = 0∀i. Alternately,
wi = 0 =⇒ wi = 0∀i. Notice that this is parallel with
the definition of linear independence of vectors.
Thus, independence of the Wα implies that they are pairwise disjoint.
Definition 8.9. Likewise, we say that they span W if every w ∈ W is a linear combination
of the elements of the Wi .
L
Given
any
W
,
.
.
.
,
W
⊂
W
,
we
have
a
map
ϕ
:
Wi → W where (w1 , . . . , wn ) 7→
1
n
P
wi . The Wi are independent if and only
Lif ϕ is injective; they span if and only if ϕ is
surjective; and if both, we say that W =
Wi .
Say V, W are finite-dimensional vector spaces over k with ϕ : V → W a linear map.
We can associate with this map two subspaces: ker ϕ and im ϕ. The dimension of the
image is called the rank of ϕ. Axler calls the dimension of the kernel the nullity, which
is less standard notation.
Theorem 8.10. dim ker ϕ + dim im ϕ = dim V .
Proof. Begin by choosing a basis for ker ϕ (it’s a subspace of a finite-dimensional space),
say u1 , . . . , um . We can extend it to a basis of V , adding the vectors v1 , . . . , vn−m (where
dim V = n). We claim that ϕ(v1 ), . . . , ϕ(vn−m
must span the
P ) is a basis
P for im ϕ. They P
entire image, since wePcan write every v = ai ui + bi vi and ϕ(v)
P = ϕ( bi vi ), which
by linearity
is just bi ϕ(vi ). If they were linearly dependent, ϕ( ai vi ) = 0 for ai 6= 0
P
and
ai vi ∈ ker ϕ, which contradicts that u1 , . . . , um , v1 , . . . , vn−m is a basis.
Corollary 8.11. If ϕ : V → W is a linear map of rank r, there exist basesfor
I
that the matrix representative of ϕ with respect to these bases has the form
0
V,W so
0
.
0
9 9/24/18
Today we’ll be discussing dual spaces, quotient spaces, and linear operators, which can
be found in chapter 3 of Axler; on Friday, we’ll introduce the language of categories and
functors.
9.1 Dual Vector Spaces
Definition 9.1. Suppose V is a vector space over a field k. Recall that we saw that
Hom(V, W ) was a vector space. Now, in particular we’re interested in studying Hom(V, k),
which is the set of linear functions V → k. This is the dual vector space to V , denoted
V ∗ ..
28
Example 9.2. If V = k n , then V ∗ ∼
well: an n-tuple (a1 , . . . , an ) represents a
= k n as P
function ϕa : V → k where (x1 , . . . , xn ) 7→
ai xi . Because every finite-dimensional
n
vector space is isomorphic to k for some n, then, why do we even need the notion of
dual space when V ∼
= V ∗ in the finite-dimensional case?
Definition 9.3. Suppose we have a map ϕ : V → W . We naturally get a map W ∗ → V ∗ .
Given some linear function l : W → k, we have a map l ◦ ϕ : V → k. This defines a map
ϕ∗ that goes in the opposite direction, and we call it the transpose of ϕ.
You’ve probably seen the word transpose in other settings, like the transpose of the
matrix; it turns out that if we take a basis, using the identification with the dual of V
we gave above, the matrix representations of ϕ and its transpose are in fact transposes
of each other.
Observation. We have a natural map V → (V ∗ )∗ . In general, we don’t have a natural
map V → V ∗ , since the isomorphisms we do get require us to take a basis.
Given a vector v ∈ V , we want a corresponding linear function on V ∗ , and we can
do so by evaluation at v. Given a linear function l ∈ V ∗ , define ϕ(l) = l(v). We can
easily verify that ϕ is in fact linear. This is in fact a canonical isomorphism when V is
finite-dimensional, but not necessarily so when V is infinite dimensional. For example,
suppose V is the space of polynomials over k. A linear function on this vector space is
the map taking a polynomial to its nth coefficient, and thus all finite possible sums of
these are also linear functions. Thus every element of V maps to the linear functional
corresponding to
P a “dot product.” However, we don’t hit every linear function with this
identification:
ai is not a linear combination of the ai because it’s an infinite sum, but
it’s a valid linear functional regardless.
In general, if
function ϕb :
P(b0 , b1 , . . . , ) is any infinite sequence, we have a linear
(a0 , a1 , . . . ) 7→
ai bi . In general, we have an inclusion of k[[x]] ,→ V ∗ : the dual space is
really big! In particular, if we apply this process again, (V ∗ )∗ ⊃ W ∗ strictly contains V .
We can now think about relationships between properties of ϕ and ϕ∗ . For example,
when is ϕ∗ injective?
Proposition. ϕ is surjective if and only if ϕ∗ is surjective. Likewise, ϕ is injective if
and only if ϕ∗ is surjective.
Proof. Suppose ϕ : V → W is surjective. Then, any nonzero linear function l : W → k
will map to some nonzero l ◦ϕ (l(w) 6= 0 =⇒ ϕ(v) = w for some v) and ϕ∗ is an injective
map.
Suppose ϕ is not surjective. Then im ϕ ( W . Then there exists some l ∈ W ∗ so that
l 6= 0 and l|im ϕ = 0; then ϕ∗ (l) = l ◦ ϕ = 0.
Proof of the other statement left as an exercise.
9.2 Quotient Spaces
Suppose V is a finite-dimensional vector space over k and U ⊂ K is a subspace. Notice
that both U and V are groups under addition (forgetting scalar multiplication), so we can
29
form the quotient group V /U = {cosets v + U }. Notice that we don’t have to worry about
normality or anything because U, V are abelian. It turns out that this quotient also has
the structure of a vector space; we define scalar multiplication as λ(v + U ) = λv + U . We
can easily verify that this is well-defined.
Suppose we have V ⊂ W a subspace.
Definition 9.4. The annihilator of V Ann(V ) is the set of linear functionals on W that
kill V .
Proposition.
W ∗/Ann(V )
= V ∗.
(8)
Proof. Given the inclusion ι : V ,→ W , the kernel of ι∗ is just Ann(V ) since ι is injective
and ι ◦ ϕ = ϕ|V , so it’s 0 exactly when ϕ kills all of V . We in fact have the more general
fact that given a surjective map ϕ : V W , W ∼
= V /ker ϕ.
Recall that given ϕ : V → W , we can choose bases v1 , . . . , vm and w1 , . . . , wn so that
ϕ(vi ) = wi for i = 1, . . . , r and ϕ(vi ) =0 for i > r where r is the rank of ϕ; equivalently,
Ir 0
ϕ has a matrix representation
. Suppose now that we have a map ϕ : V → V ,
0 0
but we can’t choose separate bases for the source and the target. What can we say
about ϕ instead? In other words, how do we classify the conjugacy classes of A, a matrix
representation of ϕ?
Definition 9.5. A subspace U ⊂ V is invariant for ϕ if ϕ(U ) ⊂ U . That is, the subspace
is carried into itself.
Example 9.6. A one-dimensional subspace is invariant under a map if it’s carried into
a multiple of itself.
Definition 9.7. A vector v ∈ V is an eigenvector for v if ϕ(v) = λv for some λ. λ is
called the eigenvalue associated to v.
If we could find a basis for V consisting entirely of eigenvalues of ϕ, then ϕ would have
a diagonal matrix representation looking like


λ1 0 · · · 0
 0 λ2 · · · 0 


 ..
.. . .
..  .
.
. .
.
0 0 · · · λn
We can’t achieve this diagonalization all the time, but we can get close (for some definition
of close). For that, we’ll have to return to our study of fields.
Definition 9.8. We say that a field k is algebraically closed if every non-constant polynomial f (x) ∈ k[x] has a root in k; that is, there is some α ∈ k so that f (α) = 0.
Moreover, any polynomial f ∈ k[x] can be written as a scalar multiple of a product of
linear polynomials λ(x − λ1 ) · · · (x − λn ).
30
The fundamental theorem of algebra states that C is algebraically closed. Most proofs
of this involve (crucial) non-algebraic properties of C, though.
Theorem 9.9. If k is any field, then there exists an algebraically closed field k with an
inclusion k ,→ k where k is generated by adjoining the roots of f ∈ k[x]. We call k the
algebraic closure of k.
√
The
√ basic construction here is something we’ve discussed before: we defined Q( 2) =
{a 2 + b : a, b ∈ Q} with the√expected rules for addition and multiplication, which was
a field containing Q but also 2. Take 123 if you want to learn more.
10 9/26/18
10.1 Quotient Vector Spaces
Definition 10.1. Let U ⊆ V be a subspace. We define
V /U ≡ {a + U : a ∈ V }
λ(a + U ) = λa + U
(9)
(10)
That is, the vector space of cosets.
If f : V → W is any linear map, then U ⊂ ker(f ) ⇐⇒ f factors as f˜ ◦ π, that is
f
V
W
f˜
π
V /U
Another basic property is that
dim(V /U ) = dim(V ) − dim(U )
(11)
We call dim(V /U ) the codimension of U in V .
Example 10.2. Let V = k[x] and U = xk[x]. Then the codimension of U in V is 1.
10.2 Dual Vector Spaces
Definition 10.3. Let V be a vector space over a field k. The dual vector space V ∗ is
defined as
V ∗ ≡ Hom(V, k) = {linear maps f : V → k}
(12)
Suppose e1 , · · · , en is a basis for V (finite dimensional). Then a linear map from V → k
is defined by what it does to these basis vectors. We can build a basis for V ∗ by defining
maps xj such that
(
1 j=i
xj (ei ) =
(13)
0 j 6= i
The set of linear maps x1 , · · · , xn is then a basis for V ∗ .
31
Definition 10.4. Given a linear map ϕ : V → W , we define the transpose of ϕ as
t
ϕ : W ∗ → V ∗ such that α 7→ α ◦ ϕ.
We also have a natural map V → (V ∗ )∗ . We do this by “evaluation”
ev : V → (V ∗ )∗
v 7→ evv : V ∗ → k
(14)
(15)
where evv (φ) = φ(v). If V is finite dimensional, this is an isomorphism. In general
though, it is just an inclusion.
Remark. If ϕ : V → W is any map, V, W finite dimensional, then t (t ϕ) = ϕ
tβ
β
If I have a pair of maps: U α V
W then I can form W ∗
V∗
and I have t (β ◦ α) =t α ◦t β. That is, transposition commutes with composition.
tα
U∗
Definition 10.5. Let U ⊂ V be a subspace. Then the annihilator of U , denoted
Ann(U ) ⊂ V ∗ is defined as
Ann(U ) ≡ {l ∈ V ∗ : l|U = 0}
(16)
Observe that Ann(U ) = (V /U )∗ .
10.3 Linear Algebra - thinkin’ ’bout matrixez
Definition 10.6. Let ϕ : V → W be a linear map. The rank of ϕ is the dimension of
(ϕ).
Definition 10.7. A linear operator on V is a linear map ϕ : V → V .
In the case of a linear operator, we are constrained to choosing the same basis on each
side. The key difference here is that linear operators can be composed with themselves
repeatedly - that is, we have a notion of ϕn ≡ ϕ ◦ ϕ ◦ · · · ◦ ϕ.
{z
}
|
n times
Consider the vector space Hom(V, V ). We have a law of composition! Given S, T ∈
Hom(V, V ) we can form the “product” ST ≡ S ◦ T . With this law of composition,
Hom(V, V ) is a ring (non-commutative).
Suppose I have a pair of maps ϕ1 : V1 → W1 and ϕ2 : V2 → W2 . Then I can form the
map ϕ1 ⊕ ϕ2 : V1 ⊕ V2 → W1 ⊕ W2 . Now, if V = V1 ⊕ V2 , W1 = V1 and W2 = V2 , then
ϕ1 ⊕ ϕ2 : V → V .
So if we choose a basis for V such that v1 , · · · , vm ∈ V1 and vm+1 , · · · , vn ∈ V2 , then
the matrix representative of ϕ would be block-diagonal:


ϕ1
0


0
ϕ2
32
More generally, assume only that ϕ(V1 ) ⊆ V1 . We can choose a basis v1 , · · · , vm for V1
and extend this to a basis for V . The matrix of ϕ will then be block upper-triangular


ϕ|V1
∗


0
∗
Our question is, given ϕ : V → V can we find U ⊂ V such that ϕ(U ) ⊆ U ? Or, even
better, can we find a decomposition V = U ⊕ U 0 such that ϕ(U ) ⊆ U and ϕ(U 0 ) ⊆ U 0 .
Lemma 10.8. If k is algebraically closed15 and ϕ : V → V is any linear operator on
a finite dimensional vector space, then there exists v ∈ V , v 6= 0 and λ ∈ k such that
ϕ(v) = λv. This implies the existence of an invariant subspace (namely, the one spanned
by v).
We call v an eigenvector of ϕ, with eigenvalue λ.
Proof. Say dim(V ) = n. Choose any vector v ∈ V (v 6= 0). Look at v, ϕ(v), ϕ2 (v), · · · , ϕn (v).
We have n + 1 vectors here, so this must be a linearly dependent list!
a0 v + a1 ϕ(v) + · · · + an ϕn (v) = 0
(17)
Now consider the operator a0 + a1 ϕ + · · · + an ϕn . We know that this sends v 7→ 0, so
this operator has a nontrivial kernel. Forgetting for a moment what ϕ means, this is a
polynomial in k[ϕ]. Since k is algebraically closed, this polynomial factors:
a0 + a1 ϕ + · · · + an ϕn = c(ϕ − λ1 ) · · · (ϕ − λn )
(18)
Each of these factors is an operator, and we can consider this multiplication as composition. Then the linear operator c(ϕ − λ1 ) · · · (ϕ − λn ) : V → V has a nontrivial kernel.
This implies that ϕ − λi has a nontrivial kernel, for some i. This implies that there exists
a w ∈ V such that ϕ(w) = λi w.
Corollary 10.9. Given ϕ : V → V over an algebraically closed field k, there exists a
basis v1 , · · · , vn for V such that the matrix representation of ϕ is upper-triangular.
11 9/28/18
11.1 Generalized Eigenvectors
Definition 11.1. An endomorphism is a map from a vector space to itself. An automorphism is an isomorphism from a vector space to itself. Sometimes, we denote the spaces
of endomorphisms and automorphisms as End(V ) and Aut(V ), respectively.
15
Algebraically closed means that every polynomial of positive degree with coefficients in the field has
at least one root, or equivalently, every polynomial factors as a product of linear factors. The classic
example is C.
33
Definition 11.2. Given a map f : U → V , we define coker f = V /im f ; this essentially
measures how surjective the map is.
Notice that coker f = ker(t f )∗ .
Proposition. Suppose V is finite dimensional over an algebraically closed field k. Then
every f : V → V has an eigenvector.
Corollary 11.3. Given f : V → V , there exists a basis e1 , . . . , en for V so that the matrix
representative of f with respect to this basis is upper triangular; that is, f (he1 , . . . , ei i) ⊂
he1 , . . . , ei i for all i.
Proof. Begin with some eigenvector e1 for f , which exists by our proposition, with eigenvalue λ1 . Now we proceed by induction. Because f carries e1 to its own subspace, we
get an induced map f : V /e1 → V /e1 . By induction, we have a basis e2 , . . . , en so that
f (he2 , . . . , ei i) ⊂ he2 , . . . , ei i for all i. Then choose ei ∈ V to be any representative of the
coset represented by ei ∈ V /ei . Observe that the basis e1 , . . . , en also satisfies the upper
triangular condition.
Suppose we have some f : V → V represented by an upper triangular matrix M with
diagonal entries λ1 , . . . , λn .
Proposition. M is nonsingular – that is, f is an isomorphism – if and only if all the
diagonal entries are nonzero. (This does not require k to be algebraically closed).
Proof. If λi = 0, then f (he1 , . . . , ei i) ⊂ f (he1 , . . . , ei−1 i), so f is not an isomorphism.
If λi 6= 0 for all i, then we can see that f is surjective: f (e1 ) = λ1 e1 and λ1 6= 0, so
e1 ∈ im(f ); f (e2 ) = λ2 e2 + a1 e1 for some a1 ; λ2 6= 0, so e2 ∈ im(f ) as well. We can
continue in this way until we see that im f contains all the basis vectors.
Corollary 11.4. the following are equivalent:
i) λ is an eigenvalue for f ;
ii) f − λ has a kernel;
iii) λ = λi for some diagonal entry λi .
Therefore, the eigenvalues of f are the diagonal entries of an upper triangular representation of f .
In particular, we can never have more than n eigenvalues if dim V = n. If e1 , . . . , ei are
eigenvectors for f with distinct eigenvalues, then they are linearly independent.
Proof. Suppose e1 , . . . , el is a minimal dependentP
set (that is, e1 , . . . , el are linearly dependent
but no proper subset of them are). So
ai ei = 0; applying f , we have that
P
λi ai ei = 0 as well. Multiply our first equation by λ1 e1 and subtract; we then get
a linear dependence relation between e1 , . . . , el (this relation is not trivial because the
eigenvalues are distinct).
34
Let f : V → V with V finite-dimensional. An alternate definition of eigenvector was
that v ∈ ker(f − λ) for some λ ∈ k; we can extend this to
Definition 11.5. We say that v is a generalized eigenvector for f if v ∈ ker(f − λ)m for
some m > 0.
For example, consider the map f : (x, y) 7→ (x + y, y). This has only one eigenvector,
but notice that (f − I)2 = 0.
11.2 Categories and Functors
Categories give a way of talking about mathematics in a way that crosses different areas
and gives an idea of what constructions might be useful. It’s based on the observation
that there are some aspects of math are repeated over and over again in different contexts.
Definition 11.6. A category C consists of a collection of objects Ob(C); for every A, B ∈
Ob(C), a set Hom(A, B); and a map Hom(B, C) × Hom(A, B) → Hom(A, C) called
composition so that:
i) (f ◦ g) ◦ h = f ◦ (g ◦ h) for all f, g, h in the appropriate sets of morphisms;
ii) for all A, we have a map idA ∈ Hom(A, A);
iii) for all f ∈ Hom(A, B), f ◦ idA = idB ◦f = f .
Example 11.7. For example, sets form a category where the arrows are set maps. Likewise, we have a category of abelian groups where the arrows are group homomorphisms;
and vector spaces over a field k with linear maps.
Example 11.8. We could also have a category of pointed sets, where objects are sets
with a distinguished element (S, x) and morphisms require us to send the distinguished
elements to each other.
Example 11.9. We also have a category of nested pairs (U, V ) with U ⊂ V ; maps require
the subspaces to also be sent to each other.
Let C be any category and A, B ∈ Ob(C). We want to define an object A × B ∈ Ob(C):
Definition 11.10. The product A × B is defined to be an object Z ∈ Ob(C) together
with projection maps πA : Z → A, πB : Z → B such that for all T ∈ Ob(C) and maps
α : T → A, β : T → B we get a unique map T → Z.16
16
The product doesn’t exist in every category, be careful.
35
12 10/1/18
12.1 Jordan Normal Form
Definition 12.1. Let T : V → V and v ∈ V . We say that v is in the generalized kernel
of T if T n v = 0 for some n. We say that v is a generalized eigenvector with eigenvalue λ
if v ∈ ker(T − λI)n for some n.
Proposition. Suppose k is algebraically closed
L and let Vλ be the generalized kernel of
T − λI. Then Vλ is a subspace of V and V = λ Vλ . Observe that T (Vλ ) = Vλ for all λ.
Proof. Let mλ = dim Vλ . Since k is algebraically closed, we can write T as an upper
triangular matrix in some basis of V . Let the diagonal entries in this representation be
λ1 , . . . , λn . We claimPthat mλ is the number of times that λ appears in the diagonal.
This will imply that
mλ = dim V .
Reorder the bases so that the first mλ entries on the diagonal are λ. If we take T − λ,
then, this becomes a block upper triangular matrix with the first block having 0s on the
diagonal andthe second
being upper triangular. Therefore, for some n the matrix for
0
∗
and has no zeroes on the diagonal of the UT section. Therefore,
(T − λI)n is
0 UT
ker(T − λI)n has dimension mλ . Now it just suffices to show that the Vλ are independent.
Suppose we have v1 , . . . , vl with vi ∈ Vi and v1 + · · · + vl = 0. We can see that vi = 0
for all i from the same argument we used to show that eigenvectors are independent.
Therefore, we are done.
Definition 12.2. The characteristic polynomial p(x) of T is defined to be
Y
(x − λ)mλ .
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λ
Notice that p(T ) = 0 from the above, since T commutes with itself and multiples of the
identity.
L
If V =
Vλ , then T |Vλ − λ is nilpotent. Moreover, we can write our matrix as block
diagonal where every block is upper triangular with constant diagonal. Can we simplify
these blocks even further? To do so, consider T0 = T |Vλ ; we know that (T0 − λI)mλ = 0.
We know T0 − λ is nilpotent and we want to find a simple representation of nilpotent
maps.
Then say T : V → V is nilpotent.
Definition 12.3. The exponent of T is the smallest q > 0 so that T q = 0.
Proposition. There exists v ∈ V so that T q−1 v = 0; we then have a set A = hv, T v, . . . , T q−1 vi
and there exists B ⊂ V so that A, B are preserved under T and V = A ⊕ B.
If we then apply this proposition to B and iterate, we end up with a basis v1 , . . . , vn
so that T vi = vi−1 and T v1 = 0. Then the matrix representation has 1s directly above
the diagonal and 0s everywhere else. Now, combining this with our note that T0 − λ is
nilpotent, we have
36
Theorem 12.4. Given any T : V → V , there exist bases for V so that the matrix
representation of T is block diagonal: each block has λi along the diagonal and 1 directly
above the diagonal, and the size of the block is the dimension of the generalized eigenspace
corresponding to λi . These are called Jordan blocks and this form is known as Jordan
normal form.
12.2 Categories and Functors
Recall that a category C is a collection of objects Ob(C) along with a set of morphisms
between all pairs of objects and an associative composition map; every object has an
identity arrow to itself that acts as we expect the identity to.
Example 12.5. Sets, groups, abelian groups, rings, fields, and vector spaces over given
fields are all categories.
Produts: consider any two sets A, B. We can form their familiar Cartesian product
A × B = {(a, b), a ∈ A, b ∈ B}. Can we describe this construction categorically? Namely,
can we characterize A × B using *only* the data we have in the category of sets?
The answer is yes, we can. Observe that A × B has the following property: given an
arbitrary set T , to give a map of sets T → A × B is the same as to give a map of sets
T → A and T → B. More precisely, these two pieces of data are completely equivalent,
given that a certain diagram must be made to commute.
Now, note that in this seemingly obscure and pointless definition, we made no reference
whatsoever to elements of A or B - we only used maps. So, this generalizes to arbitrary
categories! Thus:
Definition 12.6. Let C be any category ,and A, B ∈ ( Ob)C. A product of A and B is
an object Z and maps πA : Z → A and πB : Z → B satisfying the following condition:
for any T ∈ ( Ob)(C) and maps φA : T → A and φB : T → B, there exists a unique map
ψ : T → Z such that πA ◦ ψ = φA and πB ◦ ψ = φB .
It is important to remember that the maps πA : Z → A and πB : Z → B are part of
the data of a product of two objects.
Example 12.7. In the category of sets, the product of A and B is the usual Cartesian
product Z = A × B, and πA : Z → A is the usual projection (a, b) 7→ a; ditto for πB .
Example 12.8. In the category of groups, the product of G and H is the usual product
of groups Z = G × H (with the usual group structure: (g, h) · (g 0 , h0 ) = (gg 0 , hh0 )), where
πA and πB are the same as in the previous example.
A similar (in fact dual, in a way that can be made precise) concept is that of coproduct
or sum in a category. As previously, let’s`
play around with sets first. Given sets A and B,
we can form their disjoint union Z = A B. What similar property does this set have?
Well, last time we considered maps T → Z, so since we wrote the word “dual”, let’s
consider maps Z → T . Well, rejoice, because we observe that to define a map Z → T is
the same as to define a map A → T and B → T ! Again, this observation lives entirely in
the category of sets (i.e. makes no reference to elements of sets, it only uses the maps),
so we can generalize this to arbitrary categories.
37
Definition 12.9. Let C be a category, A, B ∈ Ob(C) be objects in C. Then the sum
or coproduct of A and B is defined to be an object Z ∈ Ob(C) equipped with maps
iA : A → Z and iB : B → Z satisfying the following property: for any T ∈ Ob(C) and
maps φA : A → T and φB : B → T , there exists a unique map ψ : Z → T making the
natural diagram commute, i.e. satisfying φA = ψ ◦ iA and ditto for B.
Time for examples. We just saw how these notions work in Sets - in fact, that’s what we
were basing our generalizations on. Now consider the category Vectk of finite-dimensional
vector spaces over k. Do we have products and coproducts here?
Example 12.10. Let V1 , V2 ∈ Ob(( V ect)k ) be finite-dimensional vector spaces over
k. Then V1 ⊕ V2 , equipped with the natural projections π1 : V1 ⊕ V2 → V1 (given by
(v1 , v2 ) 7→ v1 ) and ditto for π2 : V1 ⊕ V2 → V2 , is a product of V1 and V2 in the category
of finite-dimensional vector spaces over k! (Exercise: check this.)
Here comes something slightly confusing: observe that we also have natural maps
i1 : V1 → V1 ⊕ V2 , given by v1 7→ (v1 , 0); ditto for i2 : V2 → V1 ⊕ V2 . Is it possible that
this makes V1 ⊕ V2 , equipped with these maps, a coproduct (or sum) of V1 and V2 ? The
answer is yes! (Another exercise: check this.)
Q
Now, if we have infinite products and sums,
Ai satisfies the condition of being a
product and ⊕Ai the coproduct because we can’t take infinite linear combinations. So
in the category of vector spaces, products are direct products and coproducts are direct
sums. When we have finitely many factors, these coincide.
13 10/3/18
“Multilinear algebra. It’s like linear algebra, but more of it!”
-Brian Warner, 2016, in response to a question from Prof. Noam Elkies
13.1 Bilinear Forms
Definition 13.1. Let V be a vector space over k. A bilinear form b is a map b :
V × V → k that is linear in each variable. That is, for all u, v, w ∈ V we have
b(λv, w) = λb(v, w)
b(u + v, w) = b(u, w) + b(v, w)
b(v, λw) = λb(v, w)
b(v, w + u) = b(v, w) + b(v, u)
(20)
(21)
(22)
(23)
Keep in mind that a bilinear form is not a linear map from V ×V → k, i.e. b(λ(v, w)) =
b(λv, λw) = λ2 b(v, w) 6= λb(v, w).
Definition 13.2. Let b : V × V → k be a bilinear form. We say that b is symmetric if
b(v, w) = b(w, v) for all w, v ∈ V . We say it is skew symmetric if b(v, w) = −b(w, v)
for all w, v ∈ V .
38
An example is the bilinear form b : k 2 × k 2 → k that maps ((x, y), (u, v)) 7→ xv − yu,
this is skew symmetric.
Suppose we have a bilinear map b : V × V → k. We get a map ϕb : V → V ∗ by taking
v 7→ b(v, ), where b(v, ) : V → k by taking w 7→ b(v, w). If this map is an isomorphism,
then we say that b is non-degenerate.
link to important content 10/10 will be on final exam should memorize this content
click on this link
Definition 13.3. The rank of a bilinear form b, denoted rank(b) is defined as the rank
of ϕb .
Fix V /k. Let B(V ) ≡ {bilinear forms on V }. This is again a vector space. What is
dim B? Choose a basis e1 , · · · , en for V to specify b. It is sufficient to specify b(ei , ej ) for
all i, j ∈ {1, · · · , n}. This is because of bilinearity:
n
n
n
X
X
X
ai e i ,
bj ej ) =
ai bj b(ei , ej )
b(
i=1
j=1
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i,j=1
For instance, we could define ba,b (ea , eb ) = 1 and ba, b(ei , ej ) = 0 for all (i, j) 6= (a, b).
These ba,b form a basis for the vector space B(V ).
We can conclude that dim B(V ) = (dim(V ))2 .
2
Note that the
V2set of symmetric bilinear forms S (V ) and the set of skew-symmetric
(V ) are both vector subspaces of B(V ) (you should prove this!).
bilinear forms
Given a bilinear form b : V × V → k, if we set
aij = b(ei , ej )
(25)
Then A ≡ (aij ) is the matrix representation of b with respect to the basis e1 , · · · , en for
V . Conversely, a matrix defines a bilinear form in this way. That is,
X
X
(v, w) = (
xi e i ,
yj ej )
=⇒ b(v, w) = xAt y
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“If you take anything away from this course, in some sense it
should be that taking a basis is a last resort.”
-Prof. Joe Harris, 2018
So let’s do this without bases! If we’re given a bilinear form b : V × V → k, we get a map
ϕb : V → V ∗ that sends v 7→ b(v, ). In defining ϕb , we used the fact that b is linear in
the second variable. The fact that b is linear in the first variable tells us that ϕb is itself
linear.
So we have a linear map ϕ∗ : B(V ) → Hom(V, V ∗ ) that sends b 7→ ϕ∗ . Note that ϕ∗ is
also linear! We claim that this is a natural17 isomorphism.
Proof. Given ϕ : V → V ∗ define the bilinear form (v, w) 7→ ϕ(v)(w). This is an inverse18
of ϕ∗ . (Check this!)
17
18
We are still avoiding defining the word “natural”. Hang in there!
Exhibiting an inverse (and showing that it is in fact an inverse) is equivalent to proving that a map is
an isomorphism. This is because a map of sets is invertible if and only if it is bijective.
39
So dim B(V ) = dim Hom(V, V ∗ ) = (dim V )(dim V ∗ ) = (dim V )2 .
Definition 13.4. Suppose V is a vector space over k with a bilinear form b : V × V → k,
and suppose S ⊂ V is some vector subspace. We say that the orthogonal complement
of S, denoted S ⊥ ⊂ V , is defined as
S ⊥ ≡ {v ∈ V | b(v, w) = 0 ∀w ∈ S}
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S ⊥ is a vector subspace. Another way to phrase this is that S ⊥ = Ann(ϕb (S)).
P
In the case of Rn , the dot product b(v, w) = i vi wi is a bilinear form. Then V = S⊕S ⊥
in this case, because you can’t have a vector that is in both S and S ⊥ other than 0, and
by definition there are no vectors not in either S and S ⊥ . But in general you can have
bilinear forms that have some vector subspace intersect with its orthogonal complement.
For example:
0 1
: k2 × k2 → k
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−1 0
((x, y), (uv)) 7→ xv − yu
(29)
Look at the subspace S spanned by (1, 1). Then S ⊥ = S. So an orthogonal complement
is not always a complement in the set-theoretic sense.
40
13.2 Inner Product Spaces
Definition 13.5. Let V be a vector space over R. We say a bilinear form b : V × V → R
is an inner product if
i) b is symmetric.
ii) For all v ∈ V , b(v, v) ≥ 0, and b(v, v) = 0 if and only if v = 0. This is called being
positive-definite.
Note that this definition makes use of an ordering on our field, that is, a sense in which
some elements are “larger” than other elements. What would happen if we were to try
this over C? We would want to do this because C is algebraically closed, which is really
nice.
Remark. We cannot defined an inner product over C. Let V be a vector space over C,
and let b be a bilinear form. Then b(iv, iv) = i2 b(v, v) = −b(v, v). So we lose positivity,
so we cannot hope to impose the same condition.
So how do we solve this? We fiddle with definitions!
Definition 13.6. Let V be a vector space over C. We define a Hermitian form as a
map h : V × V → C which is linear in the first variable, and conjugate19 linear in the
second variable. That is,
h(λv, w) = λh(v, w)
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h(v, λw) = λh(v, w)
h(v + u, w) = h(v, w) + h(u, w)
h(v, w + u) = h(v, w) + h(v, u)
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(32)
(33)
14 10/5/18
14.1 “Life Inside an Inner Product Space”
Definition 14.1. An inner product space is a vector space V over R with a bilinear
form , : V × V → R such that
• u, v = v, u for all u, v ∈ V .
• u, u ≥ 0 with equality iff u = 0
Definition 14.2. Given an inner product, a norm kvk of v ∈ V is defined as
√
v, v.
Definition 14.3. We say v, w ∈ V are orthogonal if v, w = 0.
Note: by bilinearity, if v, w are orthogonal then kv − wk2 = v − w, v − w = v, v −
w, w + 2v, w = kwk2 + kvk2 .
19
If a complex number is z = a + ib, then its “conjugate” is z = a − ib.
41
Definition 14.4. In general, define the angle between two vectors < (v, w) ≡ arccos
Note that this definition warrants a proof that v, w ≤ kvk kwk.
v,w
kvkkwk
Definition 14.5. Let V be an inner product space, and let S ⊂ V be any subspace. The
orthogonal complement of S, denoted S ⊥ , is defined as
S ⊥ ≡ {v ∈ V | v, w = 0 ∀w ∈ S}
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Suppose that dim V == n and dim S = m. Then dim S ⊥ ≥ n − m, but S ∩ S ⊥ = {0},
following from the positive-definiteness and non-degeneracy of an inner product. So we
can write V = S ⊕ S ⊥ .
Recall that any bilinear form b on a vector space V gives rise to a map ϕb : V → V ∗ that
sends v 7→ v, . We say that the bilinear form is non-degenerate if ϕb is an isomorphism.
Proposition. If V is finite dimensional and the bilinear form b is positive-definite, then
b is non-degenerate.
Proof. Positive-definiteness implies that ϕb is injective. Since V is finite dimensional, this
implies that ϕb is an isomorphism, which means that b is non-degenerate.
Theorem 14.6. Let u, v ∈ V be vectors in an inner product space. Then
u, v ≤ kuk kvk
(35)
Proof. Since this inequality is unaffected by scaling, we can assume WLOG that kuk = 1.
Let S = span(u) ⊂ V . We have V = S ⊕ S ⊥ . Let p : V → S be the projection on the
first factor, explicitly p(v) = v, uu. We can write for any v ∈ V that v = v, uu + w where
w ∈ S ⊥ . WE can then take
kvk2 = kp(v)k2 + kwk2 ≥ kp(v)k2 = v, w2
(36)
Definition 14.7. Let V be a finite dimensional inner product space20 . A basis v1 , · · · , vn
for V is called orthonormal if
(
1 j=i
vi , vj =
(37)
0 j 6= i
Note that an orthonormal basis for V defines an isomorphism V ∼
= Rn , whereby vi 7→ êi
where ê1 = (1, 0, · · · , 0), ê2 = (0, 1, 0, · · · , 0), etc., that preserves the inner product. In
this sense, any inner product space with an orthonormal basis is just Rn for some n.
Lemma 14.8. Let V be a finite dimensional inner product space. Then there exists an
orthonormal basis for V .
20
Remember, when we say inner product space it has to be over R. We will treat “Hermitian product
spaces” - i.e. the analogue over C, as a separate kind of object.
42
.
Proof. We will prove this two ways. First, by induction on the dimension of V .
v
Choose any v ∈ V such that v 6= 0. Then kvk > 0. Set v1 = kvk
. Note that kv1 k = 1.
⊥
⊥
We can write V = S ⊕ S where S = span(v1 ). S is again an inner product space with
the same inner product. By induction, there exists an orthonormal basis for S ⊥ , call it
v2 , · · · , vn . Observe that v1 , v2 , · · · , vn is an orthonormal basis for V .
Now, we will prove this by constructing an orthonormal basis with the Gram-Schmidt
process.
w1
. Then take
Start with an arbitrary basis for V , call it w1 , · · · , wn . Set v1 = kw
1k
w2 −w2 ,v1 v1
w2 − w2 , v1 v1 , note that this is orthogonal to v1 . Then we can define v2 = kw2 −w2 ,v1 v1 k . In
general, we take
P
wj − j−1
k=1 wj , vk vk vj = (38)
P
j−1
w
−
w
,
v
v
j
k=1 j k k The set v1 , · · · , vn is an orthonormal basis for V .
Note that we then have that every finite-dimensional inner product space is isomorphic
(as an inner product space, not just as a vector space) to Rn where n is the dimension of
said inner product space.
14.2 Operators on Inner Product Spaces
We have two special classes of operators on inner product spaces.
Definition 14.9. Let V be an inner product space. We say a linear operator A : V → V
is orthogonal if it respects the inner product. That is,
Av, Aw = v, w
∀v, w ∈ V
(39)
Definition 14.10. Let V be an inner product space and ϕ : V → V a linear operator on
V . Fix v ∈ V , and consider the linear function on V that sends w 7→ v, ϕw. This is the
inner product of w with some unique vector u (since the inner product is non-degenerate).
Set ϕ∗ (v) = u, i.e.
ϕ∗ : V → V
v, ϕw = ϕ∗ v, w
(40)
(41)
In other words, the inner product induces an isomorphism V ∼
= V ∗ so that the following
ϕ
V
V
∗
t
square commutes
∼
∼
=
= We call ϕ , the composition of ϕ with the isomorphism
tϕ
V∗
above, the adjoint of ϕ.
V∗
Definition 14.11. We say an operator A : V → V is self-adjoint if A∗ = A.
That is, our two special classes of operators on inner product spaces are orthogonal
operators and self-adjoint operators.
43
14.3 Wrapping Up Categories: Functors
Categories consist of objects, morphisms, and a law of composition for morphisms. Some
examples that we have seen are Sets, Groups, Abelian Groups, and Vectk . Some
new examples are Pairs of Sets, where the objects are pairs of a set and a subset and
the morphisms are set maps that carry the distinguished subsets into each other. We
can consider the analogous category for vector spaces, where distinguished subspaces are
carried into each other.
When we define a type of object, we like to know how the objects talk to each other.
Definition 14.12. Let C, D be categories. A covariant functor F : C → D consists of
the following data
• A map Ob(C) → Ob(D) denoted A 7→ F A.
• For all pairs of objects A, B ∈ C, a map HomC (A, B) → HomD (F A, F B).
Satisfying the following conditions:
• For any A, B, C ∈ C the following diagram commutes:
HomC (B, C) × HomC (A, B)
◦C
F
HomC (A, C)
F
HomD (F B, F C) × HomD (F A, F C)
◦D
HomD (F A, F C)
A contravariant functor has the same definition, except that the association of morphisms is backwards: HomC (A, B) → HomD (F B, F A) in both the data and the conditions.
A wonderful example of a general kind of functor is the forgetful functor. Take a
category where the objects have underlying sets, like Groups or Vector Spaces. Then
“bang your head up against the wall” and forget about all the extra structure, and you
can consider everything going on in this category only on the level of sets.
Example 14.13. We can define a contravariant functor from Vectk → Vectk that maps
V 7→ V ∗ . The map on morphisms takes ϕ 7→t ϕ.
Let U ⊂ V be a subspace. We can define a quotient space V /U . We claim that
this defines a functor from Pairs of Vect → Vect by U ⊂ V 7→ V /U . This acts
on morphisms by pre-composing with the quotient maps (recall that we stipulate that
morphisms in the first category must send subspaces to subspaces).
15 10/10/18
15.1 Self-Adjoint Operators
Let V be a finite-dimensional vector space over R with an inner product h·, ·i. We have
two special types of operators.
44
T is orthogonal if hT v, T wi = hv, wi (and T is always injective and thus invertible),
and if e1 , . . . , en is an orthonormal basis for V , then T ei also form an orthonormal basis.
Moreover, if M is the matrix representation for T with respect to the orthonormal basis,
then M · t M = I.
T is self-adjoint if for all v, w ∈ V , then hT v, wi = hv, T wi. To interpret this, think
of hv, T wi as a linear function on V so that l : w 7→ hv, T wi. But as we saw, every
linear functional on V corresponds to an inner product, so l(w) = hu, wi for some u ∈ V
and here this u = T v. Note that self-adjoint operators are not necessarily invertible; for
example, 0 is self-adjoint.
In general, we can define the adjoint T ∗ : V → V as hT ∗ v, wi = hv, T wi; we can also
think about this by realizing that t T is a map V ∗ → V ∗ and if we use our inner product
identification of V ∗ with V , then we get a map T ∗ : V → V . In terms of an orthonormal
basis, the adjoint is just the transpose.
In terms of matrices on an orthonormal basis, T is orthogonal if t M = M −1 and
self-adjoint if t M = M .
Observation. If T : V → V is self adjoint and T (S) ⊂ S for some subspace S, then
T (S ⊥ ) ⊂ S ⊥ .
Proof. Suppose v ∈ S ⊥ . We know that hv, wi = 0 for all w ∈ S. Moreover T w ∈ S. So
hv, T wi = hT v, wi = 0 for all w ∈ S, so for all v, T v ∈ S ⊥ and we’re done.
Proposition. T 2 is positive (that is, hT 2 v, vi ≥ 0).
Proof.
hT 2 v, vi = hT (T v), vi = hT v, T vi ≥ 0
by definition of inner product, so we’re done.
Corollary 15.1. For all a ∈ R, a > 0, T 2 + a is invertible.
Proof. Again consider an inner product:
h(T 2 + a)v, vi = hT 2 v, vi + ahv, vi
and for v 6= 0, we have ahv, vi > 0, so h(T 2 + a)v, vi > 0 and T 2 + a cannot send v to 0.
This gives us injectivity and therefore invertibility.
More generally, if p(x) = x2 + ax + b and p has no real roots, then this is equivalent to
saying that p(x) > 0 for all x and a2 < 4b.
Proposition. Given such a quadratic p with no real roots, p(T ) is invertible.
Proof. If we show that hp(T )v, vi > 0 for v 6= 0, then p(T ) must be invertible. Well,
expanding this out, we have
hp(T )v, vi = hT 2 v, vi + ahT v, vi + bhv, vi
45
and we know that hT 2 v, vi ≥ 0 and hv, vi ≥ 0; however, we need some bound on the
middle term. By Cauchy-Schwarz, ahT v, vi ≥ −a·kT vk·kvk. Moreover, hT 2 v, vi = kT vk2
and bhv, vi = b kvk2 . The polynomial q(x) = x2 − ax + b is also always positive; so
kT vk
2
2
2
hp(T )v, vi ≥ kT vk − a · kT vk kvk + b kvk = kvk · q
>0
kvk
when v 6= 0, as desired.
Theorem 15.2. Spectral Theorem. If T : V → V is self-adjoint, then T has a real eigenvector. Therefore, we can find an eigenbasis for T ; in other words, T is diagonalizable
with real entries.
Proof. Take some v 6= 0 ∈ V and consider v, T v, T v , . . . , T n v. These are linearly dependent and so for some ai ,
(T n + an−1 T n−1 + · · · + a0 )(v) = 0.
Now, if we were working over the complex numbers, we would just factor this. We can’t
in the reals, but we can factor it into linear and quadratic factors. So
Y
Y
(T − λi ) (T 2 + ai T + bi )v = 0
where each T 2 + ai T + bi is irreducible. At least one of these must have a kernel. But it
can’t be any of the quadratic factors because we just showed that those are invertible, so
a linear factor must have a kernel and it must be an eigenvector.
Now that we know we have a real eigenvector v with eigenvalue λ, let S = hvi ⊂ V . We
saw before that S ⊥ is invariant under T . By induction, because S ⊥ is an inner product
space with one less dimension, it has an eigenbasis too. We also have that eigenvectors
with distinct eigenvalues are orthogonal: hT v, wi = hv, T wi, so λhv, wi = µhv, wi for
λ 6= µ; then we must have hv, wi = 0. Hence, in terms of an orthonormal basis, T will be
diagonal.
15.2 Orthogonal Transformations
Suppose now that T is orthogonal. In the one-dimensional case, T is multiplication by
a scalar, and therefore T = ±I. If dim V = 2, T is either rotations or reflections. In
particular, we’ll see that the set of orthogonal transformations form a group. We call this
group O(V, B) ⊂ GL(V ) where B is the inner product. We saw that up to isomorphism
our only inner product space is Rn with the dot product, so this is often just written O(n)
or OR (n). In the case n = 2, one way to say this is that we have a subgroup consisting
solely of rotations. This can be thought of as S 1 , R/Z, or SO(n). What’s the quotient by
this subgroup? It’s just Z/2 since the subgroup has index 2. Higher dimensional spaces
are more complicated and thus more interesting.
Theorem
L 15.3. If T is orthogonal, then we have subspaces Vi ⊂ V so that dim Vi = 1, 2,
V = i Vi , and T (Vi ) = Vi . If dim Vi = 1, then T |Vi = ±I; if dim Vi = 2, then T |Vi is
either a rotation or a reflection.
46
Notice that this gives us a very nice geometric way to think about individual transformations as reflections and rotations on individual subspaces, but no information about
the composition of orthogonal transformations.
16 10/12/18
16.1 Modules over Rings
“Nothing is true. Everything is permitted.”
- Ezio Harris, 1503
Remember that rings need not be commutative, but that we will often assume they are
unless specified otherwise. Our classic examples of rings are Z, Z/n, k[x], k[x1 , · · · , xn ].
Definition 16.1. A module M over a ring R is a set with two operations:
• + : M × M → M , addition, giving (M, +) the structure of an abelian group.
• × : R × M → M satisfying λ(v + w) = λv + λw and (µ + λ)v = µv + λv (scalar
multiplication).
• (λµ)m = λ(µm), and etc. for associativity.
Definition 16.2. Let Σ ⊂ M . We say Σ is a spanning set if every element of M is a
linear combination of elements of Σ. That is, the map ϕ : RΣ → M is surjective.
We say Σ is linearly independent if ϕ is injective.
We say Σ is a basis if ϕ is an isomorphism, and thus M ∼
= RΣ . In this case, we say
that M is a free module.
We say that M is finitely generatedif there exists Σ ⊂ M that is finite and spans
(i.e. it is a quotient of a free module).
Now we begin to learn the horrible truth.
Observation. A spanning set of a module need not contain a basis.
For instance, let R = Z and M = Z/n. Since nx = 0 for all x ∈ Z/n, so ϕ : ZΣ → Z/n
cannot be injective.
Observation. An independent set may not be a subset of a basis.
Observation. A submodule of a finitely generated module need not be finitely generated.
For instance, let R = k[x1 , x2 , · · · ] (polynomials in infinitely many variables) and let
M = R (every ring is a module over itself). So M is finitely generated, namely by 1R .
Take the submodule M 0 consisting of the polynomials with no constant term. M 0 is not
finitely generated, because there is nothing in R that can get xk out of any xj for j 6= k.
Definition 16.3. Fix a ring R and let M, N be modules over R. A homomorphism
of R-modules is a map ϕ : M → N such that
47
• ϕ is a homomorphism of abelian groups.
• ϕ(λm) = λϕ(m). for all λ ∈ R and all m ∈ M .
Observation. HomR (M, N ) has the structure of an R-module, wherein (ϕ + ψ)(v) =
ϕv + ψv and (λϕ)(v) = λϕ(v).
(Ayyyy it’s page 55 look at that)
Note that HomR (Rm , Rn ) ∼
= Rmn .
Observation. But we can have modules M, N 6= 0 such that Hom(M, N ) = 0.
For example, let R = Z and M = /n, N =. There are no nonzero module homomorphisms between M and N . Additionally, if R = k[x] and M = k, where multiplication is
defined as
(a0 + a1 x + · · · )λ = a0 λ
then Hom(k, k[x]) = 0.
Given a module M over R we can set M ∗ = Hom(M, R) and ϕ : M → N with t ϕ : N ∗ →
M ∗ . But we don’t necessarily have M ∼
= (M ∗ )∗ as we did in vector spaces.
16.2 Hermitian Forms
We are now leaving the harsh world of modules and returning to the safe and warm
embrace of vector spaces. But this time, things are more complex.
Definition 16.4. We assume that V is a vector space over . A Hermitian form H on
V is a map H : V × V → that satisfies:
• H(λv, w) = λH(v, w)
• H(v, λw) = λ̄H(v, w)
• H(v + u, w) = H(v, w) + H(u, w) and H(v, w + u) = H(v, w) + H(v, u).
• H(v, w) = H(w, v). (Conjugate-symmetric.)
This is sometimes called sesquilinear.
The reason we define this is that a bilinear map cannot be positive-definite.
Definition 16.5. We say H is positive-definite if H(v, v) > 0 for all v 6= 0, and
H(v, v) = 0 implies v = 0. (Note that we require H(v, v) be real, which is contained in
the “> 0” statement.)
Example 16.6. Let V = {C ∞ functions fromS 1 →} then we have a Hermitian product
given by
Z
f, g ≡
f (t)g(t)dt
(42)
S1
48
Note. Let B : V × V → k be a bilinear form. This gives rise to a linear map ϕ : V → V ∗
taking v 7→ B(v, ). This does not work for Hermitian forms! It does provide a welldefined map if we send v 7→ H( , v), but this map is not linear!
Definition 16.7. Given a positive-definite Hermitian form H on V , we say a basis
e1 , · · · , en for V is orthonormal if
H(ei , ej ) = δij
(43)
Where δij = 0 if i 6= j and 1 if i = j is a useful shorthand.
Theorem 16.8. An orthonormal basis exists. (This proof is the same as in the case of
inner products.)
Definition 16.9. A Hermitian inner product is a positive-definite, conjugate symmetric Hermitian form.
∼n
Corollary 16.10. Given a Hermitian inner product on V , we have an
P isomorphism V =
of Hermitian inner product spaces where H is associated to z, w = i,j zi wj (the regular
dot-product on n ).
Definition 16.11. Let V be a Hermitian inner product space. We say an operator
T : V → V is unitary if H(T v, T w) = H(v, w) for all v, w ∈ V .
Observation. The set of all unitary operators T : V → V forms a group under composition, denoted U (V, H). We often write U (n) where n = dim V , since all finite-dimensional
Hermitian inner product spaces are isomorphic.
Note that U (1) ∼
= S 1 . What about U (2)? Try to give a description of U (2)!
Definition 16.12. We say an operator T : V → V is self-adjoint if H(T v, w) =
H(v, T w) for all v, w ∈ V .
Theorem 16.13. Self-adjoint operators on a finite-dimensional Hermitian inner product
space are diagonalizable with real eigenvalues. (This follows from the real case, and the
proof is the same.)
Note. I can view a complex vector space of dimension n, as a real vector space of dimension
2n:
m
×V
V
m|×V
×V
(V, m|×V ) is a real vector space with dimension 2n.
49
17 10/15/18
17.1 Billinear Forms (Conclusion)
Suppose V is a finite-dimensional vector space over K and B : V ×V → K is a symmetric,
nondegenerate bilinear form. How can we classify such B?
Given a vector space V and a symmetric bilinera form, we can start by finding any
v ∈ V so that B(v, v) 6= 0 and look at hvi⊥ .
Proposition. Over C, there exists a unique nondegenerate symmetric bilinear form so
that there exists a basis e1 , . . . , en for V with B(ei , ej ) = δij .
p
Proposition. Over R, we can’t necessarily always scale to 1 because B(v, v) may not
exist, but we can scale to ±1. Then there exists a basis e1 , . . . , en for V so that B(ei , ej ) =
±δij , and every such B has the form
k
n
X
X
X
X
B(
xi e i ,
yi ei ) =
xi y i −
xi yi .
i=1
i=k+1
Thus there are n + 1 such bilinear forms, and we say that B has the signature (k, l).
In general, if B : V × V → K is a symmetric nondegenerate bilinear form with signature
(k, l), the group of automorphisms of V preserving Bis usually written O(k, l).
Over Q, this is difficult to classify. The case n = 3 was done first by Gauss (the theory
of quadratic forms).
What about, say, the skew-symmetric case? Suppose that char k 6= 2. It turns out
that we can still normalize a basis, but we have to modify our procedure a little because
B(v, v) = 0. Take any vector e1 ∈ V . In the symmetric case, we would pass to the
orthogonal complement, but we can’t here. Instead, choose e2 so that B(e1 , e2 ) 6= 0.
Write V = he1 , e2 i ⊕ he1 , e2 i⊥ and repeat on this orthogonal complement. Notice here
that we’re constructing our basis two vectors at a time, instead of one. Also, on a onedimensional vector space, skew-symmetric forms just take everything to 0, so dim V must
be enve, and we have a basis e1 , . . . , e2n so that
B(
X
ai e i ,
X
bj e j ) =
n
X
a2i−1 b2i − a2i b2i−1 .
i=1
So B(ei , ej ) = 1 if i = 2k − 1, j = 2k, −1 if j = 2k − 1, i = 2k, and 0 otherwise. In
other words, we can reorder the basis to get e1 , . . . , en , f1 , . . . , fn so that B(ei , ej ) = 0
and B(fi , fj ) = 0 for all i, j, while B(ei , fj ) = δij .
In other words, there exists a unique, nondegenerate skew-symmetric bilinear form on
any even-dimensional vector space. The group of automorphisms respecting this bilinear
form is called the symplectic group and is sometimes denoted Sp(2n, k).
Notice that in this construction, we didn’t need algebraic closure. In fact, we didn’t
even need char k 6= 2. Everything does go through, but we just have to change the
definition of skew-symmetric to require B(v, v) = 0 (since we always have B(v, w) =
−B(w, v)) (and in characteristic not 2, this condition implies B(v, w) = −B(w, v)).
50
17.2 Multilinear Algebra
We begin with the tensor product, and we can get other constructions by adding more
structure.
Let V, W be finite dimensional vector spaces over K. The tensor product is a vector
space V ⊗ W along with a bilinear map V × W → V ⊗ W . We now give three definitions.
Definition 17.1. Choose bases e1 , . . . , em and f1 , . . . , fn are bases for V, W respectively.
Then V ⊗ W is the vector space with basis the collection of symbols ei ⊗ fj for all i, j, and
our bilinear map simply takes ei , fj 7→ ei ⊗ fj (extend by linearity). Notice that V × W
has dimension m + n, while by definition the tensor product has dimension mn.
This definition shows that the tensor exists, gives us an explicit formula for the dimension, and is pretty concrete: but it looks like it depends on the choice of basis.
Definition 17.2. Start with a vector space U with basis {v ⊗ w|v ∈ V, w ∈ W }. Notice
that this space is much larger than V ⊗W : in most cases, it has an uncountable basis. We
now impose additional conditions. Consider the subspace R ⊂ U spanned by elements
with the following forms:
(λv) ⊗ w − λ(v ⊗ w)
v ⊗ (λw) − λ(v ⊗ w)
(u + v) ⊗ w − u ⊗ w − v ⊗ w
v ⊗ (w + z) − v ⊗ w − v ⊗ z
and set all of these to 0, thus imposing bilinearity, by defining
V ⊗ W = U/R.
We define the map V × W → V ⊗ W as ϕ(v, w) = v ⊗ w + R.
This definition does show that the tensor exists and is not dependent on choice of basis,
but we don’t know anything about the dimension, etc.
Definition 17.3. Finally, we give the least clear definition of the three by stating the
universal property of the tensor product. Given V, W , we want a new vector space V ⊗W
and a bilinera map V × W → V ⊗ W . In some sense, the tensor product is the “freest”
possible space we can get that’s still “bilinear,” and this is encoded in the following: given
a bilinear map V × W , there is a unique linear f : V ⊗ W → U so that b = f ◦ ϕ.
V ×W
ϕ
V ⊗W
b
f
U
Now this tells us that the tensor is unique by universal property.
51
This definition clearly tells us what characterizes the tensor product, and tells us that
it’s unique if it exists. But we don’t know that such an object exists purely from this
definition. Usually, texts then go back to another definition and show that it satisfies the
universal property.
From these we can extrapolate several properties.
i) First, V ⊗ W = W ⊗ V ; this follows from all three.
ii) Second, (V ⊗ W )∗ = V ∗ ⊗ W ∗ ; this is most easily shown using universal property.
iii) Third, (U ⊕ V ) ⊗ W = (U ⊗ W ) ⊕ (V ⊗ W ); follows from all three.
Something to think about: Hom(V, W ) = V ∗ ⊗ W .
18 10/17/18
18.1 Tensors!
A useful reference for this section is Appendix B of Noam Elkies’ notes on Representation
Theory.
Definition 18.1. Let V, W be finite dimensional vector spaces over k. The tensor
product is a vector space V ⊗ W together with a bilinear map ϕ : V × W → V ⊗ W
such that for all bilinear maps b : V × W → T there exists a unique map f : V ⊗ W → T
that makes the following diagram commute:
b
V ×W
T
∃!f
ϕ
V ⊗W
We can construct V ⊗ W explicitly as a quotient U/R, where
U ≡ span{v ⊗ w | v ∈ V, w ∈ W }
(44)
R ≡ span{λ(v ⊗ w) − λv ⊗ w, λ(v ⊗ w) − v ⊗ λw, (u + v) ⊗ w − u ⊗ w − v ⊗ w, (45)
u ⊗ (w + z) − u ⊗ w − u ⊗ z | v, u ∈ V, w, z ∈ W, λ ∈ k}
(46)
The vector space U/R together with the bilinear map ϕ defined as:
ϕ
V ×W
U
U/R
has the universal property of the tensor product.
b
V ×W
ϕ
U
U/R
52
T
(oops not enough time for this diagram!)
Suppose V, W are finite dimensional vector spaces. Then we have a bilinear map
ϕ : V × W → V ⊗ W . Note that ϕ is not linear. So while the image of ϕ spans V ⊗ W ,
it is not equal to V ⊗ W (it is not even a linear subspace).
Example 18.2. Say V, W are 2-dimensional vector spaces, with bases u, v and w, z,
respectively. Then every element of V ⊗ W is expressible uniquely in the form:
a(u ⊗ w) + b(u ⊗ z) + c(v ⊗ w) + d(v ⊗ z)
for some a, b, c, d ∈ k. Let m = xu + yv ∈ V and n = sw + tz ∈ W . Then
m ⊗ n = xs(u ⊗ w) + xt(u ⊗ z) + ys(v ⊗ w) + yt(v ⊗ z)
Note that these have to satisfy ad − bc = 0 - we call tensors like this primitives or pure
tensors since they can be expressed as the tensor product of two elements m ⊗ n. An
example of a non-pure tensor would be something like
u⊗z+v⊗w+v⊗z
Call ϕ(m, n) = m ⊗ n ∈ V ⊗ W , a pure tensor. Every element of V ⊗ W is a linear
combination of pure tensors. Qustion: how many pure tensors are needed to express an
arbitrary element of V ⊗ W ?
Definition 18.3. Say α ∈ V ⊗ W has rank r if it is the sum of r pure tensors (but no
fewer).
Definition 18.4. Let V1 , · · · , Vl be vector spaces over k. We say a map m : V1 ×· · ·×Vl →
T is multilinear if it is linear in each variable separately.
Definition 18.5. We can define V1 ⊗ · · · ⊗ Vl via the universal property for multilinear
maps:
V1 × · · · × Vl m T
ϕ
∃! f
V1 ⊗ · · · ⊗ Vl
We can also define this as a quotient in the same manner as before.
We have some basic equalities:
V ⊗W =W ⊗V
(U ⊕ V ) ⊗ W = (U ⊗ W ) ⊕ (V ⊗ W )
(U ⊗ V ) ⊗ W = U ⊗ (V ⊗ W ) = U ⊗ V ⊗ W
(47)
(48)
(49)
In the special case of:
V ⊗ · · · ⊗ V ≡ V ⊗n
|
{z
}
(50)
V ⊗k × V ⊗l → V ⊗(k+l)
(51)
n times
we have a bilinear map
53
L
⊗n
Definition 18.6. The tensor algebra of a vector space V is R ≡ ∞
. This is a
n=0 V
⊗k
⊗l
non-commutative ring with multiplication defined by the map V × V → V ⊗(k+l) .
If V, W are any vector spaces, we can take Hom(V, W ) = V ∗ ⊗ W .
Proof. We have a map b : V ∗ × W → Hom(V, W ) that takes (f, w) 7→ ϕ : V → W
such that ϕ(v) = l(v)w. By the universal property we must have a unique linear map
f : V ∗ ⊗ W → Hom(V, W ). Note that since all homomorphisms V → W can be made
out of the b(f, w), f must be surjective. It follows by equality of dimension that f is an
isomorphism.
Note. Hom(V, W ) = V ∗ ⊗ W = W ⊗ V ∗ = (W ∗ )∗ ⊗ V ∗ = Hom(W ∗ , V ∗ ). This is just the
transpose!
Suppose V is a finite dimensional vector space. We have the associated vector space
B(V ) ≡ {bilinear forms V × V → k}. We claim that B(V ) = V ∗ ⊗ V ∗
Proof. We have a natural map V ∗ × V ∗ → B(V ) that takes (l, m) 7→ bl,m where
bl,m (v, w) = l(v)m(w). We then have a map V ∗ ⊗ V ∗ → B(V ) by the universal property,
and since all bilinear maps can be written as a linear combination of the bl,m this map is
surjective. Equality of dimension then implies that the map is an isomorphism.
So B(V ) = V ∗ ⊗ V ∗ . For any V , we have a natural linear map ϕ : V ⊗ V → V ⊗ V
called an involution that exchanges the order of the variables: ϕ(v1 ⊗ v2 ) = v2 ⊗ v1 .
Note that ϕ2 is the identity, so we get a decomposition of V ⊗ V into eigenspaces for ϕ.
In particular we have:
+1 eigenspace = {symmetric elements v1 ⊗ v2 = v2 ⊗ v1 }
−1 eigenspace = {va ⊗ vb = −vb ⊗ va }
and V ⊗ V is the direct sum of these two eigenspaces.
In particular, the +1-eigenspace in V ∗ ⊗ V ∗ is the space of symmetric bilinear forms
on V , and the −1-eigenspace is the space of skew-symmetric bilinear forms on V .
Definition 18.7.
Sym2 V ≡ {x ∈ V ⊗ V | ϕ(x) = x}
∧2 V ≡ {x ∈ V ⊗ V | ϕ(x) = −x}
(52)
(53)
Actually, we could also define Sym2 V as a quotient V ⊗ V /v ⊗ w − w ⊗ v | v, w ∈ V .
A third way is via the universal property
V ×V
b symmetric
T
∃! f
Sym2 V
These three definitions all agree, as long as char(k) 6= 2.
54
19 10/19/18
Mikayel is live-texing for the first time today. Russian writer Issac Babel has two great
stories titled ”My First Fee” and ”My First Goose”; hopefully something will come out
of this as well.
Recall from last time that given V, W finite dimensional vector spaces over k, we have
a canonical isomorphism
Hom(V, W ) ' V ∗ ⊗ W.
Let’s describe this isomorphism more concretely by choosing bases. Let e1 , . . . , em be
a basis for V ; then we have a natural choice of basis for V ∗ , namely e∗1 , . . . , e∗m , where
e∗i (ej ) = δi,j . Let f1 , . . . , fn be a basis for W .
Now, what does the above isomorphism look like? Well, given basis element e∗i ⊗ fj on
the RHS, it maps to the homomorphism φ : V → W given by v 7→ e∗i (v) · fj . Note that
this map has rank 1: indeed, all vectors map to a multiple of fj . Thus, rank 1 tensors
correspond to rank 1 operators (note that this isn’t tautological - we have independent
definitions of ranks for both sides). This turns out to hold for any rank.
It is also convenient to describe the matrix representation of φ as above, with respect
to the chosen bases of V and W . It is of the utmost simplicity - it has a 1 in the (j, i)-th
entry and zeroes elsewhere.
We may also understand the isomorphism Hom(V, W ) ' V ∗ ⊗ W in terms of the
universal property of the latter: we have a bilinear map V ∗ × W → Hom(V, W ), given by
(l, w) 7→ (v 7→ l(v) · w). Observe that this assignment is bilinear; this implies that, by the
universal property of tensor products, there exists a unique map V ∗ ⊗ W → Hom(V, W )
that makes the natural triangle commute.
Now, it is clear that the original map V ∗ × W → Hom(V, W ) is surjective - indeed,
any matrix is a sum of matrices with one non-zero entry in an obvious way. This implies
that the linear map V ∗ ⊗ W → Hom(V, W ) is surjective, and, because the dimensions of
both sides are equal and equal n · m, is an isomorphism, as desired.
Let’s discuss the functoriality of tensor powers. Fix an arbitrary positive integer n,
and consider the assignment V 7→ V ⊗n := V ⊗ V ⊗ . . . ⊗ V (n times). Let’s see that this
naturally is a functor Vectk → Vectk , where Vectk is the category of finite-dimensional
k-vector spaces and linear maps between them.
Indeed, given a linear map f : V → W , there is an induced map V ⊗n → W ⊗n , defined
by v1 ⊗ . . . ⊗ vn 7→ f (v1 ) ⊗ . . . ⊗ f (vn ). It is quite straightforward to check that this is
well-defined and linear, and also that this does indeed complete defining the above functor
Vectk → Vectk . Note that one could conserve paper by inducing the map V ⊗n → W ⊗n
via universal property - try it, it is a fun (and neverending as you do more and more
math) game!
Anyways, let’s move on to symmetric powers. Let V be a finite-dimensional vector
space over k as before; recall that V ⊗n was defined to ”represent” (or be the universal
recipient of) n-multilinear maps from V × . . . × V (n times).
Definition 19.1. Symn V is defined to represent (or be the universal recipient of) nmultilinear and symmetric maps from V × . . . × V (n times). As before, this means that
we’re given an n-multilinear map V × . . . × V → Symn V , and for any vector space W
55
equipped with an n-multilinear map V × . . . × V → W , there exists a unique linear map
Symn V → W that makes the obvious diagram (that I’m too slow to draw) commute.
If any confusion arises, consult the definition of the tensor powers.
Warning: This does not complete the definition of Symn V - indeed, our intentions
may be of the highest orders, but we still have to prove that there exists a vector space
Symn V equipped with a multilinear map V × . . . × V → Symn V that satisfies this
universal property. Of course, if we find any construction, it will be unique up to unique
isomorphism (make sure you understand what the ”unique” in ”unique isomorphism”
means!).
Very well, here are a few constructions. The first one is:
Definition 19.2. We define Symn V to be the subspace of V ⊗n spanned by the symmetric
tensors (think through this if you’re confused - there’s only one thing it may mean). More
succinctly, the symmetric group on n elements Sn acts on V ⊗n by permuting the terms
(e.g. the non-trivial element of S2 acts on v1 ⊗ v2 ∈ V ⊗2 by v2 ⊗ v1 ), and we define
Symn V := (V ⊗n )Sn , where for a group G acting on a set (or vector space) W we define
W G to be the fixed-point set (or subspace), i.e. all w ∈ W s.t. g · w = w for all g ∈ G.
One checks that this Symn V thus defined satisfies the desired universal property. It
is, in fact, quite straightforward - the main ”functoriality” (whatever that means) is
subsumed in the universal property of V ⊗n , and it remains to pick out the subspaces
arising from the symmetricity condition, which turns out to precisely be the required
symmetricity of the multilinear map V × . . . V → W .
Construction number 2:
Definition 19.3. Symn V := V ⊗n /W , where W ⊂ V ⊗n is the subspace spanned by all
elements of the form v1 ⊗ . . . ⊗ vn − vσ(1) ⊗ . . . ⊗ vσ(n) for all σ ∈ Sn .
The meaning of this construction is clear - we’re enforcing symmetricity on our tensors.
This is precisely what we did when we defined V ⊗n - we took a free vector space and
enforced bilinearity et cetera by taking a quotient.
Now, how do we see that these definitions agree? For this, we need char k = 0; I was
too slow to write this down live, so I’ll post a short note discussing this over the weekend.
Symmetric algebra:
Definition
19.4. For V a finite-dimensional vector space over k, we define Sym• V :=
L∞
n
n=0 Sym V . This is a graded k-algebra.
What is a graded k-algebra?
Definition 19.5. A graded k-algebra is a k-algebra (i.e. a ring R equipped withLa structure map k → R) equipped with a grading (i.e. a direct sum decomposition R = ∞
n=0 Rn
for R0 , R1 , . . . finite-dimensional k-vector spaces) that respects the algebra structure (i.e.
Ri · Rj ⊂ Ri+j for all i, j ≥ 0).
56
Prototypical example: polynomial algebra k[x1 , . . . , xn ], the grading being the standard
grading by degree: x + y has degree 1, x3 y 4 has degree 7, et cetera. It is clear that the
axioms are satisfied.
Now, why is Sym• V a graded k-algebra? Well, if under life-threatening circumstances
you’re forced to somehow multiply v1 ⊗ v2 and v3 + v4 , you will certainly go with v1 ⊗ v2 ·
(v3 +v4 ) = v1 ⊗v2 ⊗v3 +v1 ⊗v2 ⊗v4 ! Note that another natural choice was (v3 +v4 )⊗v1 ⊗v2 ,
but the results are the same because we’re working with Sym. One easily checks that the
axioms for being a graded ring are satisfied.
Let V be a finite-dimensional vector space, and consider Sym• V ∗ . How can we interpret
this with respect to V ? Well, the only natural interpretation is that Sym• V ∗ is the
algebra of polynomial functions on V ! Indeed, what is (v1∗ )2 + v2∗ if not the ”polynomial
on V ” that ”squares v1 ” and ”adds v2 ”? Of course, this is somewhat strange: what
even is a polynomial function on a vector space? Well, we essentially define it to be
Sym• V ∗ : indeed, that is the only natural candidate. Because of this one sometimes
writes k[V ] = Sym• V ∗ , by analogy with the usual polynomial algebra.
Now, a reconciliation with the usual description of polynomials is as follows: let
v1 , . . . , vn be a basis of V . Then one can prove that the map k[v1 , . . . , vn ] → Sym• V ∗ ,
given by vi 7→ vi∗ and uniquely extending to a map from the whole k[v1 , . . . , vn ], is an
isomorphism. Thus, Sym• V ∗ is indeed ”the” ”polynomial algebra” on V ; it also has the
advantage that it is defined canonically, i.e. with no choices (in particular, of a basis)
made.
Let V be a finite-dimensional vector space. Recall that B(V ) is the vector space of
bilinear maps V × V → k; now we know that B(V ) ' V ∗ ⊗ V ∗ . We say that b ∈ B(V ) is
symmetric if b(v1 , v2 ) = b(v2 , v1 ); equivalently, if under the above identification B(V ) '
V ∗ ⊗ V ∗ , b lies in Sym2 V ∗ ⊂ V ∗ ⊗ V ∗ .
Now, we have a complementary cousin of symmetric bilinear forms: anti-symmetric
bilinear forms. What do they correspond to inside V ∗ ⊗ V ∗ ?
This motivates the exterior powers and algebra:
Definition 19.6. Let V be a finite-dimensional vector space over k. Then Λk V :=
V ⊗n /W , where W is the subspace spanned by v1 ⊗ . . . ⊗ vi ⊗ . . . ⊗ vj ⊗ . . . ⊗ vn + v1 ⊗
. . . vj ⊗ . . . ⊗ vi ⊗ . . . ⊗ vn . This expression simply expresses the objective to make our
functions anti-symmetric, i.e. flip sign as we exchange two variables.
We say that Λk V is the k-th exterior power of V .
As before, one could also construct Λk V as a subspace:
Definition 19.7. For v a finite-dimensional vector space over k, we define Λk V V ⊗k to
be subspace spanned by elements ”that swap sign whenever we swap two vectors”, i.e.
elements like v1 ⊗ v2 − v2 ⊗ v1 ; if this isn’t clear, compare with the analogous discussion
of the various definitions of Symn V . As with Sym, there is a succinct way of saying the
above: S n acts on V ⊗n , and we define Λn V to be the subspace ”on which Sn acts via the
character sign : Sn → Z/2Z: it consists of all elements v ∈ V ⊗n for which every σ ∈ Sn
satisfies σ(v) = (−1)sign(σ) (v).
The equivalence of these constructions also hinges on the characteristic of k being zero.
I will include a discussion of which one is the correct notion in characteristic p later.
57
There is a very important and concrete description of Λk V : it is the vector space of
all expressions of the form v1 ∧ . . . ∧ vk , subject to bilinearity and anti-symmetricity, i.e.
v2 ∧ v1 = −v1 ∧ v2 , et cetera.
As before, we define:
Definition 19.8. Let V be a finite-dimensional vector space. WeLdefine the exterior
k
algebra of V to be the graded skew-commutative k-algebra Λ• V := ∞
k=0 Λ V .
What is a graded skew-commutative k-algebra?
Definition 19.9. A graded skew-commutative k-algebra is a graded k-algebra R such
that for any x ∈ Ri and y ∈ Rj , xy = (−1)ij yx.
Now, what is the algebra structure on Λ• V ? As before, to multiply, say, v1 ∧v2 by v3 +v4
(in that order - we no longer have commutativity), we will just get v1 ∧v2 ∧v3 +v1 ∧v2 ∧v4 .
The axioms for being a graded skew-commutative k-algebra are easily checked.
Observation: given x = v1 ∧ . . . ∧ vn ∈Λn V , if vi = vj and the characteristic of k isn’t
2, then x = 0. Indeed - if we swap vi and vj , on one hand we get the same x, and on
the other hand we get −x, hence x = −x and thus x = 0. This implies that unlike
Sym• V , Λ• V is ”small” - the graded pieces after the n-th are all zero; in particular, it is
finite-dimensional over k.
Observe that by the same token, if n = dim V , then Λn V is 1-dimensional. Indeed, let
v1 , . . . , vn be a basis of V ; then it is straightforward to see that all elements of Λn V equal
to one of the form a · v1 ∧ . . . ∧ vn for some a ∈ k. In particular, it is one-dimensional.
This allows for a wonder (that, apparently, is often asked of Math 55 graduates at
finance company interviews): a canonical definition of the determinant!
Let f : V → W be any (linear) map. It is easy to see that there is an induced map
k
λ f : λk V → λk W for any k, defined precisely as for the symmetric algebra. Thus, we
every fixed positive integer n we get a functor Vectk → Vectk , mapping V to Λn V and f
to Λn f .
Now, the determinant: let f : V → V be any (linear) endomorphism. Let n be the
dimension of V , and consider the induced map Λn f : Λn V → Λn V . This is a linear
endomorphism of a 1-dimensional vector space, hence it must be multiplication by a
constant! More precisely, for any 1-dimensional vector space W , there is a canonical
isomorphism End(W ) ' k.
So, given T , we get a scalar. What is that scalar? It is precisely the determinant! Of
course, one must check that this coincides with the usual definition of the determinant;
however, we have (amazingly) defined the determinant basis-free, and thus turned the
gigantic expression for the determinant into a pretty endomorphism of the n-th exterior
power of V . Note that we now also have a 1-line proof that det(T1 T2 ) = det(T1 ) det(T2 ):
indeed, this is precisely how multiplication-by-scalar maps compose! Thus, we found
the fundamental explanations for two mysterious phenomena: the independence of the
determinant of an endomorphism (originally defined by a huge formula) of choice of
basis, and the multiplicativity of the determinant (normally proved by very concrete
computation).
Now that we’re doing this, let’s dispel one final miracle: the independence of trace of
choice of basis. But that’s on the homework..
58
20 10/22/18
20.1 Group Actions
Definition 20.1. We say that a group G acts on a set S if we have a homomorphism
ρ : G → Perm(S). In other words, we associate with every g a map S → S, and we say
that s 7→ gs under this map. This map has an inverse, namely that s 7→ g −1 s. There is
also an identity map where s 7→ s, given by g = e. We make this a homomorphism by
ensuring that (hg)s = h(gs).
Definition 20.2. An action is faithful if ρ is injective.
Definition 20.3. The orbit of s under G to be Os = {t ∈ S : gs = t} (for some s).
Notice that s ∈ Os , that t ∈ Os ⇐⇒ s ∈ Ot , and if t ∈ Os , u ∈ Ot , then u ∈ Os as
well. That is, orbit inclusion is an equivalence relation, and we have a partition of S into
its orbits. If our homomorphism ρ is the trivial one, then every element is its own orbit;
if the entire set is one orbit, then we say that our action is transitive. Notice that we can
break up any group action into a disjoint union of transitive ones (just by considering
individual orbits), so it’s most important to study transitive group actions.
Definition 20.4. The stabilizer of s is the subgroup H = {g ∈ G : gs = s}; it is often
denoted stab(s).
Definition 20.5. The fixed points of g is the set {s ∈ S : gs = s}, denotd S g .
Observe that if s0 = gs, then stab(s0 ) = g · stab(s) · g −1 . After all, g · stab(s) · g −1 s0 =
gs = s0 , and we get the opposite inclusion through s = g −1 s0 . So elements in the same
orbit have conjugate stabilizers.
Definition 20.6. Given H ⊂ G a subgroup, we have a set G/H ; among the cosets is H
itself, and we sometimes write [H] for H the coset. G acts on G/H in the obvious way,
taking g[aH] = [gaH]. This must be transitive, and stab([H]) = H.
Now suppose G acts on any set S, and for s ∈ S let H = stab(s). Then we get a
bijection ε : G/H → Os taking [aH] 7→ as. This construction gives us the building block
for any group action. The group action on each orbit corresponds to an orbit on some
kind of quotient, so every group action can be built up as a disjoint union of actions of
this form.
Now suppose that |G| and |S| are finite; then we can talk
Pabout orders and |G| =
|Os | · | stab(s)|. Also, since orbits give a partition of S, |S| = |Os |.
20.2 Burnside’s Lemma
Let G be a group acting on a set S so that |G|, |S| < ∞. We introduce a correspondence
Σ = {(g, s) ∈ G × S : gs = s} (a correspondence here being a subset of a product). There
are two ways of computing |Σ|. First, since Σ ⊂ G × S, we can
P consider the projections
to G and S. If we consider the projection to G, then |Σ| = g |S g |; if we consider the
59
P
projection to S, then |Σ| = s | stab(s)|. But summing over all the stabilizers also allows
us to sum over all the orbits; that is, we could instead consider
X X
| stab(s)|
orbits O⊂S s∈O
but we just stated that all the stabilizers of elements in the same orbit are conjugate
|G|
under G, so they must have the same cardinality, which is |O|
. So we actually get
X
|O| ·
orbits O⊂S
Lemma 20.7. (Burnside’s)
X
X
|G|
=
|G|.
|O| orbits
|S g | = |G| · number of orbits.
g∈G
We’re not going to do a lot of examples here, but you should consult Artin.
20.3 Actions of a group on itself
What can this actually tell us about the structure of a group? Well, one question we
can ask is how G acts on itself. Well, we already have a law of composition: g acts
on h by sending it to gh, and it turns out that this is a group action. Well, this isn’t
super interesting. The stabilizer of every element is the identity, the fixed points of every
nontrivial element are also the identity, and the orbit of every nontrivial element is the
entire group. But at least it’s a faithful action: we have an injective map G → Perm(G).
This implies the following:
Theorem 20.8. (Cayley) Every finite group is isomorphic to a subgroup of Perm(G) =
S|G| .
Well, this still doesn’t tell us very much about groups because usually S|G| is extremely
large compared to G. But at least we know we have something.
There are, however, other ways for groups to act on themselves. Namely, G can act
by conjugation instead of left translation (and it turns out that in general conjugation
is more interesting than left translation, not just in this group action case). That is,
−1
h 7→ ghg . We showed some time ago on the problem sets that conjugation was indeed
a group homomorphism. Unlike the first case, we have nontrivial (!!) orbits. The orbits
are conjugacy classes of G, and the stabilizer of an element h is the set of elements that
commute with h ;we call this the centralizer of h and denote it Z(h). Notice also that
the intersection of all the centralizers is Z(G) and that the kernel of the homomorphism
induced by this group action is Z(G), so this group action is faithful if and only if the
center is trivial.
P
From this, we see that the conjugacy classes of G partition G, so |G| =
C⊂G |C|
where the C are the conjugacy classes. Moreover, the identity’s conjugacy class is only
itself, so there is at least one term in this sum equal to 1. Combining this with the fact
that the order of every conjugacy class divides |G|, we can come up with some nice results
with only these two facts. For example,
60
Theorem 20.9. If |G| = p2 , then G must be abelian.
Lemma 20.10. Z(G) 6= {e}.
P
Proof. Well, |G| = p2 and |G| = C⊂G |C|. Because |C|||G| for all G, then |C| = 1, p, p2 .
Therefore, there have to be a multiple of p 1s, and we know there is at least one 1, so there
exist at least p − 1 other conjugacy classes (besides the identity) consisting of a single
element. But to say that an element’s conjugacy class is itself means that it commutes
with every other element; thus all these elements are in Z(G). In particular, this implies
that |Z(G)| ≥ p.
We now prove the theorem.
Proof. Let g ∈ G be any element; we claim that g ∈ Z(G). Consider Z(g). If g ∈
/ Z(G),
Z(g) must strictly contain Z(G). Then |Z(g)| > p, so Z(g) = G and g ∈ Z(G) actually.
Therefore, every element is in the center and G is abelian. Thus either G ∼
= Z/p2 or
Z/p × Z/p.
Proposition. There exist exactly 5 groups of order 8 (abelian or otherwise).
We’ve already seen the abelian ones: Z/8, Z/2 × Z/4, and Z/2 × Z/2 × Z/2. We also saw
D4 . Finally, there is the quaternion group, which is the set {±1, ±i, ±j, ±k} with 1 the
identity, −1 acting by flipping the sign of every element, and other compositions
i2 = j 2 = k 2 = (−i)2 = (−j)2 = (−k)2 = −1
ij = k, jk = i, ki = j.
21 10/24/18
Recall our definition of group actions from last time: a group G acts on a set S if,
equivalently,
i) There is a map G×S → S that satisfies certain conditions (unitarity, associativity),
or
ii) A group homomorphism G → Perm(S).
Also recall the barrage of definitions associated with this construction: given an action
of G on S, we defined
i) stab(s) = {g ∈ G : gs = s}
ii) Os = {t ∈ S : t = gs for some g ∈ G}
Here’s a very fun application of these ideas: we will classify the finite subgroups of the
(very infinite) group SO3 .
Recall that given V an n-dimensional real vector space with inner product, we defined
O(V ) = {T ∈ GL(V ) : hT v, T wi = hv, wi, ∀v, w ∈ V }. SO(V ) is the subgroup of O(V )
consisting of endomorphisms with determinant 1 (note that all endomorphisms lying in
O(V ) must have determinant 1 or −1. Thus, O(V ) is ”not connected” (this can be made
precise), and SO(V ) is the ”connected piece containing the identity matrix” (ditto)).
61
Theorem 21.1. Let V be a real vector space with an inner product. For any T ∈ O(V ),
we have a T -invariant direct sum decomposition V = ⊕α Vα satisfying dim Vα ∈ {1, 2}.
Concretely, if dim Vα = 1, then T |Vα = ±1; if dim Vα = 2, then T |Vα is a rotation in
that plane.
Now let’s specialize to V a 3-dimensional real inner product space. By the theorem,
given T ∈ SO(3) there are two possibilities: V = V1 ⊕ V2 ⊕ V3 , with T acting as ±1 on
each, or V = V1 ⊕ V2 , with T acting as rotation on V1 and ±1 on V2 . However, we haven’t
imposed the condition that the determinant of T equals 1 yet: doing so tells us that in
the first case, T must be either be the identity or act as −1 on precisely two of the vector
spaces.
Now, given some Σ ⊂ R3 , we may look at its group of ”rotational symmetries”: {T ∈
SO(3) : T (Σ) = Σ}. This group could be infinite - e.g. consider a 1-dimensional circle
around the origin in some plane; then all rotations around the axis perpendicular to that
plane will keep this circle invariant.
Thus, a relation is emerging between geometric objects and subgroups of SO(3); we
will consider some examples, figure out when the group of rotational symmetries of an
object is finite, and use this to classify the finite subgroups of SO(3).
Example 21.2. Let Σ be a regular N -gon, placed in a plane in R3 containing the origin.
What are the rotational symmetries of this? There are the cyclic rotations; there are also
180o degree rotations around certain axes. One shows this group is in fact isomorphic to
Dn .
Example 21.3. Let Σ be a cone on a regular N -gon in a plane. It is possible to show
that the group of rotational symmetries is cyclic of order N .
Example 21.4. We could also take cubes, tetrahedrons, and octahedrons; for a cube
(placed beautifully around the origin in the natural manner) the symmetry group is S 4 (I
think this was a homework problem :); similarly, we get some finite groups by considering
the rotational symmetries of the others.
Theorem 21.5. This is a complete list! Thus, cyclic groups, dihedral groups, and whatever we’ll get from tetrahedra, cubes and icosahedra.
Proof. The key observation is that given any nontrivial element T ∈ SO(3), T must be
a rotation around some axis. With some thought, this implies that there are two unit
vectors fixed by T . We (and Artin) will call these the poles of T .
Now let G ⊂ SO(3) be finite. Let P be the union of all poles of all elements of G, i.e.
T = {v ∈ R3 : ||v|| = 1, gv = v for some g ∈ G, g 6= 1}.
Now, suppose v is a pole of some T ∈ G, and take any S ∈ G. Observe that Sv is a
pole of ST S −1 . Proof: direct calculation. This implies that our group G acts on P ; this
action will be key to understanding the group.
At this point, it is instructive to draw a tetrahedron or some other Platonic solid, list
its rotational symmetries and poles, and see what action you get. For the tetrahedron, we
find that there are three types of poles, and the action of the group of the tetrahedron’s
symmetries has three orbits.
62
But let us return to the proof. Let p ∈ P be a pole of some element of the group.
Consider stabG (p) , the stabilizer of p in G. Then stabG (p) ' Z/rp for some 1 < rp ∈ N.
Proof: transformation fixes p ⇒ transformation must be rotation around that axis, which
implies the claim.
Now we will count. Consider Σ = {(g, p) : g ∈ G, g 6= e, g(p) = p} ⊂ G × P . As a
subset of G × P , Σ comes equipped with projections to G and P .
The map Σ → G is precisely 2-to-1: indeed, this is a restatement of the fact that every
g ∈ G has precisely 2 poles.
Now consider Σ → P . What are the fibers of this map? For a p ∈ P , the size of its
preimage is rp − 1.
Thus, letting N be the cardinality of G, we have that
X
2N − 2 =
(rp − 1).
p∈P
Let’s simplify the RHS by summing over the set of orbits of G on P . Then we have
X X
2N − 2 =
(rp − 1).
orbits Oi p∈Oi
The point of doing this is that poles in the same orbit have equal rp ’s, i.e. gp1 = p2 ⇒
rp1 = rp2 . Indeed, this is true for all group actions on all sets - the stabilizer subgroups
of p1 and p2 are conjugate, hence have equal order.
Thus, our equality simplifies to
X
2N − 2 =
|Oi |(rp − 1).
orbits Oi
Let O1 , . . . , Ok be the set of orbits of G acting on P . For each i, let ri = rp for any
p ∈ Oi . Thus, we have
k
X
2N − 2 =
|Oi |(ri − 1).
i=1
However, recall orbit-stabilizer: cardinality of orbit times cardinality of stabilizer equals
cardinality of G, i.e. N . Applying this here gives us
k
X
1
.
2N − 2 =
N 1−
ri
i=1
Rejoice, for the skies are clearing - the right-hand side looks suspiciously big, for we’re
summing summands of size at least N2 , k times, and this is what will allow us to deduce
the values of ri . For convenience, rewrite this as
k X
1
2
1−
2−
=
.
N
ri
i=1
Now, the left-hand side is less than 2, and the right-hand side is at least k2 , which
implies that k ≤ 3. The rest of the proof will be very concrete: we will consider the
various possibilities.
63
i) k = 1: impossible, as LHS ≥ 1, RHS < 1.
ii) k = 2: We have that 2 − N2 = 1 − r11 + 1 − r12 , i.e. N2 = r11 + r12 . Since each ri divides
N , the only solution is N = r1 = r2 . In this case we have two poles, p and p0 , each
fixed under all of G. This implies that G = stabG (p) = Z/N Z.
This concludes the case k = 2.
iii) k = 3: 2 −
2
N
=3−
1
r1
−
1
r2
−
1
.
r3
Assume r1 ≤ r2 ≤ r3 .
Claim: r1 = 2. Proof: if r1 > 2, then r2 > 2 and r3 > 2, at which point LHS < 2
and RHS ≥ 2.
a) r2 = 2. Then r3 = N2 . Thus, we have a pair of poles forming an orbit. By
prolonged inspection, this gives us the dihedral group.
Remaining cases: k = 3, r1 = 2, r2 ≥ 3.
b) If r2 ≥ 4, then
1
r1
+
1
r2
+
1
r3
≤ 1, and we have a contradiction.
Thus, we may assume r2 = 3. Remaining cases: r1 = 2, r2 = 3 and r3 = 3, 4
or 5.
c) r3 = 3: tetrahedron!
d) r3 = 4: cube!
e) r3 = 5: icosahedron!
It’s starting to get spooky in here!
22 10/26/18
22.1 The Symmetric Group
The symmetric group is very important: it’s like the “ambient space” where all finite
groups live. We defined Sn as the group of permutations of any set with n elements, but
usually we just consider the set {1, . . . , n}. If we want to write an element of the symmetric group, we can just indicate σ(i) for every i. But this doesn’t convey information
about σ very efficiently. So instead, we’ll use the following notation:
Begin by choosing any i ∈ {1, . . . , n} and consider the sequence i, σ(i), σ 2 (i), . . . , which
must eventually return to i because σ must have finite order. Suppose σ k (i) = i for the
smallest k where this occurs; then we indicate this cycle by (i, σ(i), . . . , σ k−1 (i)) (note:
k = 1 is possible, since σ(i) = i is possible; then we just write (i), or ignore it to indicate
σ fixes i). Next, choose any other index not in this set, and repeat. Eventually, we
stop because there are only finitely many elements being permuted. Each of these cycles
is disjoint. By convention, we tend to pick our i, j, . . . in increasing order. The cycles
correspond to the orbits of σ. For example, the permutation 123456 7→ 432516 can be
written in cycle notation as (145)(23)(6) or just (145)(23).
64
Definition 22.1. A k-cycle σ ∈ Sn is a permutation of the form (a1 · · · ak ); in other
words, σ permutes k elements cyclically and fixes the rest.
Definition 22.2. We say two k-cycles σ, τ are disjoint if the sets of elements they cycle
are disjoint.
Note that two disjoint cycles must commute because each acts as the identity on
elements they don’t cycle. Thus cycle notation tells us how to write σ as a product of
disjoint cycles; because those commute, this expression is unique up to order of the cycles.
What does conjugation do to these elements? Suppose σ is a k-cycle (and if we can describe conjugation on k-cycles, clearly we can describe it for any permutation) (a1 · · · ak ).
Let τ ∈ Sn be any permutation.
Proposition. τ στ −1 = (τ (a1 ) · · · τ (ak )). In other words, we get another k-cycle, and it’s
just the cycle on the images of the ai .
Proof. Follow the elements around: τ (ai ) 7→ ai 7→ ai+1 7→ τ (ai+1 ). Any other element
gets fixed by σ and τ τ −1 is just the identity, so the others are fixed.
Corollary 22.3. All k-cycles are conjugate. More generally, any permutations are conjugate if and only if they have the same cycle lengths, so the conjugacy classes of Sn
correspond to partitions of n.
Example 22.4. When n = 3, the partition 1 + 1 + 1 corresponds to the identity; 1 + 2
to the transpositions; and 3 to the 3-cycles. The sizes of our conjugacy classes are 1, 3,
and 2.
Example 22.5. For n = 4, the partition 1+1+1+1 corresponds to the identity; 1+1+2
to the transpositions; 1 + 3 to the 3-cycles; and 4 to the 4-cycles; 2 + 2 to products of 2
disjoint transpositions. The sizes of our conjugacy classes are 1, 6, 8, 6, and 3.
Example 22.6. We’ll do the n = 5 case as well. I’ll start writing a table for these.
partition
description
1+1+1+1+1
identity
1+1+1+2
transpositions
1+1+3
3-cycles
1+4
4-cycles
5
5-cycles
1+2+2
2 transpositions
2+3
a 2-cycle and 3-cycle
size
1
10
20
30
24
15
20
Notice that we can recursively get the sizes of the conjugacy classes.
We can take an interlude here to talk about the partition function p(n). We’ve seen
that p(1) = 1, p(2) = 2, p(3) = 3, p(4) = 5, p(5) = 7, and also p(6) = 11. It doesn’t
have a closed form; it grows faster than any polynomial, but slower than any exponential.
Many of its properties were elucidated by Ramanujan. It’s very fascinating!
65
22.1.1 Normal Subgroups
What are the normal subgroups of the symmetric group? To do so, we use the class
equation. Recall that H ⊂ G is normal when aHa−1 = H for all a; in particular, H is
closed under conjugation and is thus a union of conjugacy classes. Well, we’ve written
out the sizes of these conjugacy classes for S3 , S4 , and S5 . Artin says that we’ve written
the class equations, which simply tell us what the conjugacy class sizes are. For example,
the class equation for S5 is
120 = 1 + 10 + 20 + 30 + 24 + 15 + 20.
To find normal subgroups, we note that we have additional constraints: e ∈ H, and
|H| | |G|. Also, we cannot include the conjugacy class of all the transpositions, which
has size 10. So it turns out that the only possible unions we could get are 1 + 24 + 15
and 1 + 24 + 15 + 20. Now we can sit and stare. It turns out that there is no normal
subgroup of the former form because it’s not a group; of the latter, we have 2 conjugacy
classes of size 20, but only the conjugacy class corresponding to 3 + 1 + 1 works.
22.2 The Sign Homomorphism and the Alternating Group
Consider the n-dimensional vector space with basis the elements of our n-element set;
every permutation thus extends uniquely to an isomorphism of this vector space. So to
every σ, we associate an element of GL(n); this gives us an inclusion Sn ,→ GL(n). We
can also take the determinant; now, because each σ has finite order, its corresponding
linear map must have determinant a root of unity, which is ±1. How can we compute the
sign? Well, ∧n is spanned by the single vector e1 ∧ e2 ∧ · · · ∧ en , and note that switching
any two of these reverses the sign. So the transpositions have sign −1.
Definition 22.7. We say that sgn(σ) = 1 if σ is a product of an even number of transpositions, −1 if odd.
The well-definedness of this construction follows from the way we defined the homomorphism.
So, a k-cycle has sign (−1)k−1 . Moreover, if σ has cycle lengths k1 , . . . , kl , then its sign
is (−1)n−l .
Definition 22.8. The alternating group An is ker sgn.
What are the conjugacy classes in An ? Suppose C ⊂ Sn is a conjugacy class. We have
basically three possibilities.
Proposition. Either C is odd and C ∩ An = ∅, or C is even and C ⊂ An ; then either C
is a conjugacy class in An , or it splits into conjugacy classes in An .
Example 22.9. In the S5 case, A5 consists of the identity, the 5-cycles, the pairs of
transpositions, and the 3-cycles. In A5 , the 3-cycles still form a conjugacy class. However,
the 5-cycles split. So our conjugacy classes have sizes 1, 20, 15, 12, 12. We can now ask
what the normal subgroups are. It turns out that there are no nontrivial subgroups of
A5 . So A5 is simple.
66
23 10/29/18
Note: recall that last time, we talked briefly about p(n), the number
of ways of writing
Q∞
1
n as a sum of positive integers. The generating function for p(n) is k=1 1−t
k.
23.1 Counting Conjugacy Classes
Recall that in Sn , the conjugacy classes correspondP
to cycle lengths and thus partitions of
n. There is another way of writing a partition: as
ibi for some sequence b1 , b2 , . . . ; the
bi ’s count how many of the aj = i. Here it clearly matters what order the bi are in, and
an unspecified number of them can be 0. How large are these conjugacy classes, then?
Using this notation, we can write an explicit formula. Let b = {b1 , b2 , . . . , }. Note first
that there are (n − 1)! n-cycles in Sn . To express some σ ∈ Cb , we have to decompose
{1, . . . , n} into b1 subsets of length 1, b2 subsets of length 2, etc. We can do this in
n!
1!b1 2!b2 3!b3
···
ways. Meanwhile, for each subset of size k, there are (k − 1)! ways to determine what the
k-cycle looks like. Therefore, we have
Q
n! (i − 1)!bi
Q b
.
i! i
There’s a lot of cancellation: it eventually turns into
n!
Q b .
ii
But we’re not done! We’ve overcounted because we could permute the bk k-cycles and
still end up with the same permutation, and need to account for this. Therefore, the size
of Cb is
n!
Q i
.
(ib · bi !)
In particular, note that if we sum over all possible b, we should get n!. Can you prove
this directly?
23.2 The Alternating Group
Let’s go back to the alternating group. Recall that
P we noted that for some conjugacy
class Cb ⊂ Sn , one of three things can happen. If
bi (i − 1) is odd, then Cb isn’t in the
alternating group. If it’s even, either Cb is still a conjugacy class in An , or it splits into
several conjugacy classes – in fact, exactly two.
Recall that any class in G is an orbit of the action of G on itself by conjugation. If G
acts transitively on S and H ⊂ G is a subgroup of index 2, then G = H ∪ Ha. We see
then that for all s ∈ S, S = Gs = Hs ∪ Has. Thus H has at most 2 orbits.
Therefore, because Sn acts transitively on Cb by conjugation and An is a subgroup of
index 2, An has at most 2 orbits when restricted to Cb .
67
Proposition. Cb does not split into conjugacy classes whenever an element σ ∈ Cb is
fixed by conjugation by an odd permutation.
Proof. Suppose there existed τ so that τ στ −1 = σ? Then whenever we have any µ ∈ Cb ,
we know µ = ασα−1 for some α ∈ Sn , but also that µ = (ατ )σ(ατ )−1 ; one of α and ατ
must be even, so then any two elements of Cb are conjugate by an even transposition.
Proposition. Cb splits into two conjugacy classes if and only if the cycle lengths of σ
are all odd and distinct.
Proof. We can write this as b2i = 0 and b2i+1 ≤ 1. Suppose this is not true and σ has
two cycles of length 2i + 1, c and d. Then we claim that there is an odd permutation
that fixes σ under conjugation: the permutation (c1 d1 )(c2 d2 ) · · · (c2i+1 d2i+1 ). After all,
conjugating by this permutation switches c and d, σ cycles them, and the permutation
switches them back, so we get σ. (If we have an even cycle, conjugate by this even cycle,
since it’s an odd permutation.)
Example 23.1. In S5 , the 5-cycles split into two conjugacy classes: those of (12345) and
(12354). The other conjugacy classes either don’t exist or don’t split.
As we said last time, A5 is a simple group and has no proper normal ssubgroups; we
see this once we write out the sizes of the conjugacy classes.
Theorem 23.2. An is simple for n ≥ 5.
Proof. We do this by induction on n, assuming A5 and A6 are simple (which you will
prove on the homework). It’s sufficient to show that for N ⊂ An normal and nontrivial,
that N contains a 3-cycle, since the following lemma will tell us that N = An .
Lemma 23.3. An is generated by the 3-cycles.
Proof. Induct on n. Our base case, n = 3, is easily verified. Suppose An−1 is generated
by its 3-cycles; then if σ ∈ An has a fixed point, it’s generated by some product of 3cycles. So suppose σ doesn’t have a fixed point. Choose any i and say σ(i) = j. Now
take σ and compose it with (jik) for any k; this fixes i, so the permutation (jik)σ is
generated by 3-cycles. The inverse of (jik) is itself a 3-cycle, (kij), so σ is also generated
by 3-cycles.
Observe that if x ∈ N , then x−1 ∈ N . Moreover, gxg −1 ı ∈ N for g ∈ An because N is
normal, and so gxg −1 x−1 ∈ N as well. Let L = ord(x). Choose any prime p so p|l and
replace x with xl/p . In other words, we can assume x has prime order. But the order of
an element is the LCM of its cycle lengths; thus, all the cycles of x are either 1 or p. If
there are any 1s, then x has a fixed point and x ∈ An−1 . Then reduce to N ⊂ An−1 and
by induction this is just An−1 , so certainly contains a 3-cycle. Otherwise, x is a product
of disjoint p-cycles. We’re out of time, but this proceeds similarly.
68
24 10/31/18
Theorem 24.1. An is simple for all n ≥ 5.
Proof. (By induction on n). We have to show N ⊂ An normal. {e} =
6 N =⇒ N = An .
It suffices to show that N contains a 3-cycle. If ∃σ ∈ N with a fixed point (say σ(n) = n)
then N ∩ An−1 is normal in An−1 , σ ∈ N ∩ An−1 6= {e} =⇒ N ∩ An−1 = An−1 =⇒ N
contains a 3-cycle.
We know that N 6= {e}, so ∃x ∈ N s.t. x 6= e. Then we also have x−1 ∈ N and
gxg −1 ∈ N for all g ∈ An . Additionally, gxg −1 x−1 ∈ N for all g ∈ An .
We can assume that the order of x is a prime number l. (Otherwise we could just
replace x with a power of x that has prime order .) Then x consists of the product of
disjoint l-cycles x = (123...l)(other l cycles).
i) l ≥ 5. In this case, consider the commutator of x by (432).
(432)(123...l)(234)(...) = (245)
ii) l = 3. Then x = (123)(456)(...), and we can consider the commutator of x with
(432).
iii) l = 2. Then x must be a product of at least 3 transpositions x = (12)(34)(56)(...).
Take the commutator of x with (531) to get (153)(246), which reduces to case ii.
24.1 Sylow Theorems
Goal: Classify groups of a given order n = G. We know that for any x ∈ G, order(x)|n.
Is the converse true? That is, if l|n must there be an element of order l? No! Because
then there would be an element of order n and the group would be cyclic. .
For any subgroup H ⊂ G, we know that H|n. Is the converse true? That is, if l|n
must there be a subgroup of order l? .
Fix a prime p. We write G = n = pe m such that m is not divisible by p. We say that
a subgroup H ⊂ G of order pe is a Sylow p-subgroup of G.
Theorem 24.2.
i) There exists a Sylow p-subgroup for every p.
ii) If H ⊂ G is a Sylow p-subgroup and K ⊂ G is any subgroup with order a power of
p. Then there exists a conjugate H 0 of H such that K ⊂ H 0 (that is, H 0 = gHg −1
for some g). If we take K itself to be a Sylow p-subgroup, this says that all Sylow
p-subgroups are conjugate.
iii) Let s denote the number of Sylow p-subgroups in G. Then s|n and s = 1 mod p.
Example 24.3. Classify all groups of order 15.
Suppose G = 15, then there exists a Sylow group H of order 3 and a Sylow group K of
order 5. Let s3 be the number of Sylow 3-subgroups, then s3 |5 and s3 = 1 mod 3. So
69
s3 = 1, which implies that H is normal! Similarly, s5 |3 and s5 = 1 mod 5, which implies
that s5 = 1 and thus that K is normal!
Using the lemma below, we can conclude that G ∼
=H ×K ∼
= /3 × /5 ∼
= /15. So there
is only one group of order 15.
Lemma 24.4. Let G be a finite group of order n. Suppose H, K ⊂ G are both normal
subgroups of orders a, b respectively. Assume that ab = n and that a and b are relatively
prime. Then G = H × K.
Proof. We have a map H × K → G which is (h, k) 7→ hk. This is an injection, and since
ab = n it must be a surjection as well. Note that in this specific case this map is in fact
a homomorphism, so G ∼
= H × K.
Example 24.5. Classify all groups of order 21.
There exist Sylow 3- and 7- subgroups H and K. Note that s3 is either 1 or 7, and s7 = 1.
If s3 = 1, G = /21 by the same argument as the previous example. If s3 = 7, let x be a
generator of K =7 and y be a generator of H =3 . Then x7 = y 3 = e in G. Every element
of G is uniquely expressible as xa y b where 0 ≤ a ≤ 6 and 0 ≤ b ≤ 2.
Since K is normal, yxy −1 = xα for some α ∈ {1, ..., 6}. Note that α will determine the
group law. What is α? We can show by the order of y that α = 1, 2 or 4.
25 11/2/18
25.1 Sylow Theorems
Recall that G was a finite group with order n = pe · m, where p - m.
Definition 25.1. A Sylow p-subgroup is a subgroup of order pe .
Theorem 25.2.
i) Sylow p-subgroups exist for all p||G|.
ii) If H ⊂ G is a Sylow p-subgroup and K ⊂ G any other p-group, then there exists a
conjugate H 0 of H containing K. In particular, all Sylow p-subgroups are conjugate
to each other. Moreover, if there is exactly one Sylow p-subgroup for a certain p,
it’s normal.
iii) The number of Sylow p-subgroups, say s, satisfies s|m and s ≡ 1 (mod p).
Corollary 25.3. If p is a prime dividing |G|, then there exists an element of of order p.
Proof. Let H ⊂ G be a Sylow p-subgroup and choose y ∈ H so y 6= e. Since |H| = pe ,
k−1
then ord(y) = pk for some k. Thus y p
has order p and is our desired element.
If G has order a prime power to begin with, then the Sylow theorems tell us absolutely
nothing because there’s exactly one Sylow p-subgroup, the entire one. Thus p-groups are
the hardest to classify, and the number of them grows dramatically with e.
70
Example 25.4. Consider the case p = 2.
n Groups
1
2
3
4
5
1
2
5
14
51
Last time, we were looking at groups of order 21. From the discussion below, we had
determined that the nonabelian groups had group law determined by the value of yxy −1
when x has order 7 and y 3. After all, the group is the set of expressions of the form
xα y β , and we can figure out how to multiply when we set a value for yx. Since the Sylow
7-group, say K, is normal, then yxy −1 ∈ K. Equivalently, yx = xα y for some α. Let ϕ
2
be conjugation by y. We know ϕ(x) = xα ; applying ϕ more times, ϕ2 (x) = y 2 xy −2 = xα
3
and ϕ3 (x) = y 3 xy −3 = xα (where all exponents are mod 7). However, y 3 = e. Thus
ϕ3 (x) = x, and we want to know when α3 ≡ (mod 7). This is satisfied by α = 1, 2, 4.
We now need to check whether or not we produce the same group with different values
of α.
If α = 1, then x, y commute and the group is abelian. If α = 2, 4, we in fact get the
same group. Since we chose generators of our groups, it matters what y we chose (i.e.
either y or y 2 ), and it turns out that the two groups we get are indeed related by the
isomorphism y 7→ y 2 .
So we now know that there exist at most 1 nonabelian group of order 21. We now need
to verify that this group actually exists. Our possible group has the relations x7 = y 3 = e,
yx = x2 y. To do this, we can find this as a subgroup of S7 .
25.2 Proving Sylow
To carry out this proof, we’ll need some preliminaries. Let G be any group and H ⊂ G a
subgroup. A question to ask is, what is the largest subgroup of G so that H is normal?
Definition 25.5. The normalizer of H, N (H), is the largest subgroup K ⊂ G so that
H ⊂ K and H is normal in K. This has a concrete description: N (H) = {g ∈ G :
gHg −1 = H}. This is a subgroup of G and by “largest,” we mean that N (H) contains
all other subgroups that satisfy this property.
Example 25.6. Let G = S3 ; then for H a 3-cycle, its normalizer is all of G. For H a
transposition, the normalizer is just H. So for a normal subgroup, the normalizer is the
entire group.
Notice that the normalizer can detect how “close” H comes to being normal. If H is
normal, N (H) = G; is H is not normal at all, its normalizer is itself.
G acts on itself by conjugation, so G acts on the set of H’s conjugate subgroups. If
the group is normal, this set has one element. By definition, this action is transitive.
Therefore, by orbit-stabilizer, |OH | · |N (H)| = |G|. In other words, the number of subgroups conjugate to H is |G/N (H)|. This already proves the first part of the third Sylow
theorem.
71
We now prove two lemmas. It turns out that the proof of the Sylow theorems is
essentially looking at various group actions and seeing what we can deduce about the
structure of G.
Lemma 25.7. Given n = pe · m with (p, m) = 1, then p - pne .
Proof. Just write out the product:
pe −1
Y n−k
n
n(n − 1) · · · (n − pe + 1)
=
.
=
e−k
pe (pe − 1) · · · (2)(1)
p
pe
k=0
In this product, we note that the highest power of p dividing each of n − k and pe − k is
the highest power of p dividing k; hence p does not divide this factor and thus not the
product.
Lemma 25.8. Let U ⊂ G be any subset and consider the action of G on P(G) by left
multiplication. The stabilizer of [U ] ∈ P(G), stab([U ]), has cardinality dividing |U |.
Proof. Let H = stab([U ]). U itself must be a union of H-orbits because H acts on U
by left-multiplication and preserves U . However, multiplication on the left has no fixed
points, so every orbit has size |H| and |H|||U |.
We now prove the first Sylow theorem.
Proof. Let S be the set of all subsets of G with size pe ; we claim that one of these
` is in fact
a subgroup. Consider the action of G on S by left multiplication. Write S =
orbits O
for this action. By the first lemma, p - |S|. Thus there must be an orbit Ou for this action
so that p - |Ou |. Therefore, pe | stab(Ou ). But by the second lemma, | stab(Ou )|||U | = pe ,
so stab(Ou ) has cardinality pe . Thus the stabilizer is a Sylow p-subgroup and we have
shown existence.
26 11/5/18
Today, we’ll finish the Sylow theorems and talk a little about finite abelian groups.
On Wednesday, we will talk a little about summer “research” opportunities and begin
representation theory.
26.1 Finishing Sylow
Recall the two lemmas we stated last time:
Lemma 26.1. If n = pe · m with p - m, then p -
n
pe
.
Lemma 26.2. Let G be a finite group and consider its action on P(G). Let U ⊂ G be
any subset of G and H = stab([U ]). Then |H| | |U |.
Let’s also go over the proof of the first Sylow theorem.
72
Theorem 26.3. Sylow p-subgroups exist.
Proof. Consider the action of G on the set S of subsets of G of cardinality pe . |S| = pne
P
and by the first lemma p - |S|, so |S| = |O| over the orbits of S, and there exists some
subset U so that O[U ] is not divisible by p. Let H = stab([U ]). We know that |H||pe .
Moreover, |H| · |O[U ] | = |G| = pe m. Thus pe | |H|, and |H| = pe . We have found our
desired Sylow p-subgroup.
We now prove the second Sylow theorem.
Theorem 26.4. If H ⊂ G is any Sylow p-subgroup and K ⊂ G any p-subgroup, then
there exists a conjugate H 0 of H with K ⊂ H 0 . In particular, all the Sylow p-subgroups
are conjugate.
Proof. We construct a set C on which G acts so that p - |C|, the action is transitive,
and there exists c ∈ G so that stab(c) = H. For example, the set of left cosets of
H ⊂ G satisfies these properties: |C| = |G|/|H| = m and p - m; the action is clearly
transitive; and the stabilizer of H is itself (the identity coset). Now restrict our action to
the subgroup K ⊂ G (since p - C).
By the fixed point theorem, there exists a fixed point c0 ∈ C so that K ⊂ stab(c0 ). By
transitivity, c0 = gc for some g ∈ G. Thus K ⊂ stab(c0 ) = g · stab(c) · g −1 , but that’s just
gHg −1 .
Finally, let’s prove the third theorem.
Theorem 26.5. Let s = sp be the number of Sylow p-subgroups of G (we use sp when
we need to specify what p we’re talking about). Then s | m and s ≡ 1 (mod p).
Proof. Consider the action of G on the Sylow p-subgroups by conjugation. By the previous theorem, ths action is transitive. In particular, if H ⊂ G is any Sylow p-subgroup,
the stabilizer is the set {g|gHg −1 = H} = N (H) and we know H ⊂ N (H) so N (H) has
cardinality at least pe . Moreover, s = |G|/|N (H)| by orbit-stabilizer. In particular, since
N (H) has cardinality a multiple of pe , so s | m.
Now consider the action of G on the set of Sylow p-subgroups. For H ⊂ G any Sylow
p-subgroups, restrict the action to H. We have one orbit of size 1, and we claim that this
is the only orbit of size 1, with all other orbits having size divisible by p. Suppose that
we had another fixed point H 0 fixed under conjugation by H. Then H ⊂ N (H 0 ). But
|N (H 0 )| | |G|, so H will also be a Sylow p-subgroup in N (H 0 ). Likewise for H 0 , so both
H, H 0 are Sylow p-subgroups of N (H 0 ). By the second Sylow theorem, they’re conjugate
in N (H 0 ). But H 0 is normal in N (H 0 ) (one would hope), so H 0 = H.
Therefore, H has one orbit of size 1 and all the orbits have size dividing p. So s ≡ 1
(mod p).
Example 26.6. One more example: classifying groups of order 12. We know there’s
a subgroup K of order 3 and s3 = 1, 4; likewise, there’s a subgroup H of order 4 with
s2 = 1, 3. At least one of H, K must be normal: suppose K is not normal (so there are
4 Sylow 3-subgroups). Then we have K1 , . . . , K4 ∼
= Z/3 and these are pairwise disjoint.
73
So we’ve accounted for 9 elements of our group, all of which have order 3. None of these
can be in groups of order 4 (except the identity), so there is exactly one group of order
4 and thus is normal.
If both H, K are normal, then G is abelian and is either Z/4 × Z/3 or Z/2 × Z/2 × Z/3 (the
former is the cyclic group of order 12, and the latter is Z/2 × Z/6). If K is not normal
but H is, consider the action of G on the Ki . Consider an element of one of the Ki ;
conjugation by this element permutes the 4 Ki by fixing K1 and permuting the others.
Thus G is A4 . When K is normal but H is not, this becomes trickier because H can
be either cyclic or the Klein four group. If H = Z/4, consider the action of an element
x ∈ H on K by conjugation. K = {e, y, y 2 }. Then xyx−1 = y 2 , so G is generated by x, y
with x4 = y 3 = e and xy = y 2 x; we don’t know yet that this group necessarily exists.
If H = Z/2 × Z/2, then G is generated by x, y, z with x2 = z 2 = y 3 = e and xz = zx.
Moreover, zy = yz but xy = y 2 x. If these groups exist, there are 5 groups of order 12 up
to isomorphism.
One of these new groups is dihedral and the other is actually a new group.
27 11/7/18
Today, we’ll talk a little about finite abelian groups before moving on to representation
theory.
27.1 Finite Abelian Groups
Often, properties of rings are reflected in the properties of modules over them. If we
consider specifically the ring Z, which is quite special (it’s what’s called a principal ideal
domain), we have the following:
Theorem 27.1. Any finitely generated
L abelian group is a direct sum of cyclic groups. In
particular, if G is finite, then G = i Z/ni .
We won’t prove this, leaving it instead for your study of modules.
pn
Observe that
us a subgroup of
Q for a prime p, setting Gp = {g ∈ G : ∃n|g = e} gives
n
G and G = p Gp . Gp is some direct sum of cyclic groups of order p for some n (but we
don’t know what). This isn’t canonical, but it’s basically as canonical a representation
as we’re going to get.
Given an abelian group G, consider Hom(G, Z). This “sees” the infinite part of G, not
the finite part.
Definition 27.2. For G a finite abelian group, the dual group is Ĝ = Hom(G, C∗ ) where
C∗ is the group of nonzero complex numbers with multiplication. A ψ ∈ Ĝ is the character
of a G.
Note again that Ĝ must be a finite abelian group. We can give it a group law by
pointwise multiplication, and each of the finitely many generators of G can only take
values in the |G|th roots of unity, so Ĝ is finite.
74
Example 27.3. For example, if G = Z/n, a character is determined completely by where
1 ∈ G is sent; thus Ĝ ' Z/n as well. So for any finite abelian group, G ' Ĝ but not
canonically.
27.2 Representation Theory
First, some history. In the 19th century, people first began working on groups. In the
second half of the century, they really explored the structure of groups and the idea
became used in other fields. But they didn’t have our current definition of a group. To
those mathematicians, a “group” was a subset of GL(n) closed under multiplication and
inversion. This changed in the 20th century. As people began to work with these objects
more and more, they realized that the same groups were arising in different contexts (for
example, isomorphic matrix groups in different dimensions). This led to a reorganization
of the theory, due to Emmy Noether; from him our modern definition of a group arose.
In effect, the 19th century problem of classifying groups broke into two parts. Either
they could classify abstract groups (which we’ve been beginning to do), or describe how
an abstract group G could be thought of as a subgroup of GL(n).
We’re going to generalize this a little bit to describe all homomorphisms G → GL(n),
and this is what’s known as representation theory. Note that in this class, we will only
talk about representations of finite groups, but this extends to all kinds of groups –
countably infinite groups like Q, continuous groups like S 1 , etc.
Definition 27.4. A representation of a group G is a vector space V on which G acts
linearly. So we have a map G × V → V so that the map V → V induced by each g is
linear. This way, the action will respect the vector space structure.
Recall that if S were a set equipped with a group action, this group action could also
be thought of as a homomorphism G → Perm(S). We can make an analogous definition
for representations of G: a homomorphism G → GL(V ). It’s actually much better to
consider arbitrary homomorphisms rather than embeddings, as we’ll see.
Our representations will usually be over C, but occasionally not: we can also look at
R and fields of positive characteristic. The latter are called modular representations and
are much trickier.
Definition 27.5. A homomorphism ϕ of two representations of G, V → W , is a linear
map V → W that respects the action of the group. In other words, ϕ(gv) = gϕ(v).
Well, now this makes our Hom terminology a little confusing, so let’s clarify. When we
say Hom(V, W ), we mean all linear maps V → W ; if we want to talk about homomorphisms of a representation of G, we denote this by HomG (V, W ) and call it the space of
G-linear maps.
As with other objects, we can make new representations out of ones we already know.
Given V, W representations and ϕ : V → W a G-linear map, ker ϕ is itself a representation
of G since it’s closed under the action of G; likewise Im(ϕ) is a representation. In fact,
coker(ϕ) is also a representation (though it’s a little harder to show). After all, given
some element g ∈ G, it fixes Im(ϕ), so by the universal property of the quotient, we get
75
an induced map on the quotient coker(ϕ) → coker(ϕ). In fancy language, the category
of representations is abelian.
We can thus define a subrepresentation.
Definition 27.6. A subrepresentation of V a representation of V is a subspace W so
that gW = W for all g ∈ G.
Definition 27.7. A representation V is irreducible if it has no nontrivial proper subrepresentations.
27.3 Summer Research
If you want to stay at Harvard and do more math, there are a number of programs that
will fund you to stay over the summer and work with faculty. Ask Cindy Jimenez, the
undergraduate coordinator for the math department: she knows everything there is about
what we are going to say. The program we know are PRISE, HCRP, Herschel-Smith, and
Mellon-Mays. For each, you make a proposal for your research (in conjunction with a
professor). PRISE has a smaller stipend, but provides you with housing and some food;
the others have larger stipends and allow you to possibly travel. To motivate you to
stay during the summer, the math department runs tutorials over the summer, which are
courses taught by graduate students on a topic of their interest. The tutorials are small,
the environment is more informal, and you’ll get to know the graduate student. The math
department will pay you to take the summer tutorials! If you’re not doing research under
one of these programs, you can also CA a summer school course (the students definitely
pay good money) and do research with a professor.
Well, all these programs profess to fund research, but in math undergraduate research
can be interpreted very liberally. For example, over the summer at Harvard you can
pursue some independent work, not taught in a course and learned at your own pace
without lectures, that isn’t necessarily original.
Outside Harvard, there are a number of REUs (research experiences for undergraduates). These are funded typically by agencies like the NSF and AMS. A math department
will set up a program for undergraduates, including room and board, with some kind of
coordinated project that is intended to lead to original results. They vary a lot in topic
(and also quality). Some famous ones are Emory, Duluth, SMALL, and Twin Cities
(among others, ask Cindy).
28 11/9/18
Let G be a finite group.
Definition 28.1. A representation of G is a vector space V over C on which G acts; i.e.,
a pair (V, ρ) with ρ : G → GL V ).
Definition 28.2. A homomorphism of representations ϕ : V → W is a linear map
respecting the action of g:
76
ϕ
V
g
V
W
g
ϕ
W
Definition 28.3. A subspace W ⊂ V invariant under G is called a subrepresentation.
Definition 28.4. V is irreducible if it has no nontrivial subrepresentations.
Given two representations V, W of G, we can define the representation V ⊕ W where
g(v, w) = (gv, gw) (we omit the ρ here, as is convention), and the representation V ⊗ W
where g(v ⊗ w) = gv ⊗ gw.
We can also define the dual representation, though this takes more care. Every g is
a map V → V . We want an action V ∗ → V ∗ that has a group structure; in particular,
h ◦ g must be hg. Thus we define
g(v ∗ ) = t (g −1 )(v ∗ ).
Note that we have a natural action V ⊗ V ∗ → C with v ⊗ l → l(v).
Theorem 28.5. Let V be any representation of a finite group G and suppose W ⊂ V
is an invariant subspace. Then there exists another invariant subspace U ⊂ V so that
U ⊕W =V.
Corollary 28.6. If V is finite-dimensional, V is a direct sum of irreducible representations.
We present two proofs of the theorem.
Proof.
Lemma 28.7. If V is a representation of G, there exists a positive-definite Hermitian
inner product on V preserved by G. In other words, H(gv, gw) = H(v, w) and all the g
are unitary.
Proof. Choose any Hermitian inner product H0 on V and use an “averaging trick” to set
X
H0 (gv, gw).
H(v, w) =
g∈G
We easily then note that H(gv, gw) = H(v, w).
Note that H’s inner-product-ness depends on the positive definite H0 . Given this
lemma, W ⊥ = {v ∈ V : H(v, w) = 0∀w ∈ W } is a complementary G-invariant subspace.
Proof. Start by choosing any complementary subspace U0 ⊂ V . Let π0 : V → W be the
projection and define a map V → W given by
X
π(v) =
gπ0 (g −1 v).
g∈G
Although π0 is not a homomorphism of representations, it is easy to see that π is; what’s
more, π|W = |G|I where I is the identity. Then π is a surjection and ker π yields the
invariant subspace complementary to W we’re looking for.
77
This fails when our fields have nonzero characteristic, though, which is why modular
representations are more complicated. It also fails when G is not finite: if we don’t have
a finite measure on G, then we can’t do either averaging trick.
Example 28.8. Consider the representation of R on C2 with
1 t
t 7→
.
0 1
This map fixes the x axis and shears the y axis.
We now prove Schur’s Lemma, which is easy to prove but should probably be a theorem,
and tells us about maps between irreducible representations.
Lemma 28.9 (Schur’s). If V, W are irreducible representations of G and ϕ : V → W
any homomorphism of representations, then either ϕ is zero or a multiple of the identity.
Proof. Observe that ker ϕ is an invariant subspace of V . But V is irreducible, so either
ker ϕ = 0 or ker ϕ = V . In the former case, ϕ is injective; in the latter ϕ = 0. We can
apply the same analysis to the image: if the image is 0, ϕ = 0, or the image is all of W
and ϕ is surjective. Thus, either ϕ = 0 or ϕ is both injective and surjective, hence an
isomorphism.
That ϕ = λI for some I relies on the algebraic closure of the complex numbers. Given
ϕ : V → V , it has an eigenvalue λ. (ϕ − λI)(v) = 0, so ϕ − λI = 0 by the above; thus
ϕ = λI.
Now we lay out the goals of our representation theory section. For a given G, we wish
to find all irreducible representations of G, U1 , . . . , Um , as well as how to write every
representation as a sum of these. Moreover, if V, W are representations of G, we would
like to describe representations like V ⊗ W , Symk V , etc. in terms of V, w and their
decomposition.
Example 28.10. Given V a representation of G, we wonder when g : V → V is itself
G-linear. If G is abelian, gh = hg∀g, so the maps g : V → V are homomorphisms of
representations. By Schur, then, each g is just a multiple of the identity. So if V is
irreducible, it must be one-dimensional and our map G → GL(V ) is a map G → C∗ : a
character! So every irreducible representation corresponds to a character χ of G.
Example 28.11. Let’s look at the smallest nonabelian group, S3 . We have three obvious
representations: the trivial representation on U ∼
= C; the alternating representation given
by the sign homomorphism on U 0 ∼
= C; and the permutation representation on V ∼
= C3 .
However, the permutation representation has an invariant subspace, namely the one
spanned by e1 + e2 + e3 . We can easily find a complementary subspace {(z1 , z2 , z3 )|z1 +
z2 + z3 = 0}. This representation is indeed irreducible. It turns out that these are the
only irreducible representations of S3 , and we need to show this.
78
29 11/12/18
Recall what we found last time (or one more class ago): if G is an abelian group, then every
irreducible representation of G is 1-dimensional and corresponds to a unique character
χ : G → C∗ .
The proof is a concatenation of homework problems: i) every g ∈ G is diagonalizable in
any given representation (i.e. ρ(g) ∈ GL(V ) is); ii) commuting diagonalizable operators
are simultaneously diagonalizable. NB: the reason every ρ(g) is diagonalizable is that
each has finite order (because G is finite) and we’re working over C.
Now let’s consider the first non-abelian case: G = S3 . Last time, we constructed three
irreducible representations of S3 :
1) the trivial 1-dimensional representation U ,
ii) the alternating/sign 1-dimensional representation on which G acts by the character
sign : G → Z/Z ,→ C∗ , to be denoted by U 0 ,
iii) The 2-dimensional irreducible subspace of the obvious 3-dimensional representation,
to be denoted V (the obvious 3-dimensional which has basis v1 , v2 , v3 on which S3 acts
by permutation); the complementary subspace to V in this 3-dimensional representation
is a copy of the trivial representation, the span of v1 + v2 + v3 .
An advanced aside: in general, Sn acts on Cn by permuting coordinates, and this
representation has an invariant subspace U = {(z1 , z2 , . . . , zn ) : z1 = z2 = . . . = zn }. The
complementary subspace is V = {(z1 , z2 , . . . , zn ) : z1 + z2 + . . . + zn }. With some work,
one can prove that V is irreducible; it is called the standard representation of Sn .
Back to S3 .
Proposition. The three representations of S3 listed above are the only irreducible representations of S3 .
Corollary 29.1. Any representation W of S3 is isomorphic to a direct sum W = V ⊕n1 ⊕
U ⊕n2 ⊕ U 0⊕n3 for unique ni ∈ N0 .
Proof. Let W . be any representation of S3 .
Recall that we’ve completely classified representations of abelian groups. The idea of
our approach to the representation theory of S3 is to restrict representations to A3 '
Z/3 ⊂ S3 .
Some notation: let τ ∈ S3 be any 3-cycle and σ ∈ S3 be any transposition. We have
that
τ 3 = σ 2 = id, στ σ = τ 2 .
Restrict the representation to the subgroup (isomorphic to Z/3) generated by τ . By
the abelian discussion, there exists a basis of W of eigenvectors of τ , i.e. W = ⊕Vi ,
2
Vi = hvi i and τ (vi ) = wαi vi , where w = e 3 πi is a root of unity.
Very well, we’ve played the abelian card the best we could; now let’s see how σ acts.
Let v ∈ W be an eigenvector for τ , i.e. τ v = αv.
79
Observe that τ (σv) = σ(τ 2 v) = α2 σ(v). In other words, σ sends α-eigenvectors to
α2 -eigenvectors.
This discussion applied to any representation; now let’s specialize to W irreducible.
Choose any eigenvector v ∈ W of τ ; it can have eigenvalue w, w2 or 1.
Case i): τ v = v. By the above, σv = v. There are two possibilities: σ(v) = λv (where
λ = ±1) or σ(v) is linearly independent of v.
In the first case, since W is irreducible we must have W = hvi, and hence W is either
U , the trivial representation, or U 0 , the sign representation (corresponding to λ = ±1).
Let’s see that the second case cannot happen. Indeed, consider hv + σvi. This is a
1-dimensional invariant subspace of W (check!), which implies that it equals the whole
W , which contradicts the linear independence of v and σv.
Case ii): τ v = αv for α = w or w2 . In this case, σ(v) is an eigenvector of τ with
eigenvalue α2 ; in particular, σ(v) and v are linearly independent. Consider hv, σ(v)i, i.e.
the span of v and σ(v). This is a subrepresentation, hence (W is irreducible!) equals W .
Which representation is W ? It is the standard (2-dimensional) representation! Indeed,
it is straightforward to see that they coincide.
Let’s give some applications of this. Recall that if W is any representation of S3 , then
W ' U ⊕a ⊕ U 0⊕b ⊕ V ⊕c .
How do we determine a, b and c? We look at the eigenvalues of τ . Indeed, given a
decomposition of W as above, there are a+b different eigensvectors (linearly independent,
of course) of τ with eigenvalue 1 - they all come from U and U 0 . Similarly, the number
of eigenvectors with eigenvalue w of τ equals c - they all come from V .
We can make similar arguments for σ: there are a + c eigenvectors with eigenvalue 1
and b + c eigenvectors with eigenvalue −1.
The point is that knowing the eigenvalues of σ and τ allows us to determine a, b and c!
Let’s create an application of these ideas: let V be the standard (2-dimensional) representation of S3 . Recall that we can tensor representations together; in particular, we
may consider V ⊗3 . Now, since all representations are completely reducible, we have that
V ⊗3 ' U ⊕a ⊕ U 0⊕b ⊕ V ⊕c .
Let’s try to determine a, b and c. We know that V has a basis e1 , e2 with τ e1 = we1 ,
τ e2 = w2 e2 .
This implies that V ⊗ V has a basis e1 ⊗ e1 , e1 ⊗ e2 , e2 ⊗ e1 , e2 ⊗ e2 . Moreover, we know
how τ acts on these:
τ (e1 ⊗ e1 ) = w2 e1 ⊗ e1 , τ (e1 ⊗ e2 ) = w3 e1 ⊗ e2 = e1 ⊗ e2 , τ (e2 ⊗ e2 ) = we2 ⊗ e2 .
80
Thus, τ has eigenvalues 1, 1, w, w2 . Each 1 corresponds to a 1-dimensional subspace,
and the w, w2 correspond to a standard 2-dimensional representation. If we also consider
the eigenvalues of τ , we find that
V ⊗ V ' U ⊕ U 0 ⊕ V.
We can play a similar game for Sym2 V : Sym2 V has basis e21 , e1 e2 , e22 , and again, we
know how τ acts on these:
τ (e21 ) = w2 e21 , τ (e1 e2 ) = e1 e2 , τ (e22 ) = we22 .
A similar analysis can be carried out for σ, which will allow us to determine whether
Sym2 V ' U ⊕ V or U 0 ⊕ V .
The technique we used relied on S3 having a big abelian subgroup (of index 2, in fact!).
This works also for dihedral groups. For Sn , however, this approach won’t be too helpful.
However, it is still quite useful - in fact, picking maximal abelian subgroups/subalgebras
is central in the theory of Lie groups and algebras.
Instead, we will develop a new technique - that of characters.
Some preliminary discussion first. A funny observation that an efficient way to store
the information of n numbers, unordered and possibly with repetition, is to specify the
coefficients of the polynomial of which they are the roots (i.e. the coefficients of (x − λ1 ) ·
. . .·(x−λn )). These coefficients are expressible in the λi using the symmetric polynomials,
which is why we’ll discuss them first.
Consider the polynomial ring C[z1 , . . . , zn ]. Sn acts on it by permuting the indeterminates.
Definition 29.2. A polynomial f ∈ C[z1 , . . . , zn ] that is a fixed point of Sn under this
action is called a symmetric polynomial.
Remark. A polynomial is an abstract entity, not a function - f1 = f2 means that they’re
equal as elements of C[z1 , . . . , zn ], i.e. have the same degree and same coefficients, and
not that f1 (x1 , . . . , xn ) = f2 (x1 , . . . , xn ) for all (x1 , . . . , xn ) ∈ Cn . Of course, over C this
”doesn’t matter”, but it is an important conceptual point, and over Fp these notions
differ - indeed, xp − x and 0 both evaluate to 0 at all elements of Fp , yet are different
polynomials.
Let’s introduce some symmetric polynomials:
σ1 (z1 , . . . , zn ) =
n
X
zi ,
i=1
σ2 (z1 , . . . , zn ) =
X
1≤i≤j≤n
and so on, until we reach
81
zi zj
σn (z1 , . . . , zn ) =
n
Y
zi .
i=1
The Fundamental theorem of Algebra gives us a bijection between unordered n-tuples of
complex numbers (with possible repetitions) and ordered n-tuples (σ1 , . . . , σn ) (here, the
σi are numbers, not functions - sorry for this notation). Under
Q this bijection, unordered
lets {λi ∈ C} correspond to to coefficients of the polynomial i (x − λi ).
Theorem 29.3. The subring of symmetric polynomials in C[z1 , . . . , zn ], i.e. C[z1 , . . . , zn ]Sn ,
is isomorphic to the polynomial algebra in n variables, with generating set σ1 , . . . , σn :
C[z1 , . . . , zn ]Sn ' C[σ1 , . . . , σn ].
Concretely, this tells us that every symmetric polynomial is uniquely expressible as a
polynomial in σ1 , . . . , σn .
Let’s see why this is believable on an example: n = 2 and f = z12 + z22 . This is clearly
a symmetric polynomial, and indeed, f = (z1 + z2 )2 − 2z1 z2 = σ12 − 2σ2 .
Much more relevant for us, however, is a different generating set of C[z1 , . . . , zn ]Sn the power sums. As suggested by the name, these are:
τ1 (z1 , . . . , zn ) =
n
X
zi ,
i=1
τ2 (z1 , . . . , zn ) =
n
X
zi2 ,
i=1
et cetera, until we reach
τn (z1 , . . . , zn ) =
n
X
zin .
i=1
Theorem 29.4. These functions serve to establish the same isomorphism as the σi :
C[z1 , . . . , zn ]Sn ' C[τ1 , . . . , τn ].
A vaguely philosophical reason this holds is (similarly
an
Pto before)
P 2 thatPspecifying
n
unordered n-tuple {z1 , . . . , zn } is the same as to specify i zi , i zi , . . . , i zi .
30 11/14/18
Last time, we operated on the idea that to understand a representation V of G, we could
look at the collections of eigenvalues of g : V → V for all g ∈ G. But this is rather
inefficient for general representations. Last time, we also saw,Pthough, that to specify a
collection of α ∈ C, it’s enough to specify the power sums of
αik for all k.
82
In fact, it’s enough to specify the sum of the eigenvalues
P fo2 rall g ∈ G.2 Given all g ∈ G
with eigenvalues
αi ? Well, g has eigenvalues
i }, their sum is tr g. What about
P {α
2
k
k
{αi }. Thus,
αi = tr(g ). So if we know the trace of every g ∈ G, we know the
eigenvalues.
Definition 30.1. The character χV of a given representation is the C-valued function
χV (g) = tr g.
Note. When we had an abelian group, we defined characters differently, as homomorphisms G → C∗ . These characters are not homomorphisms. But when G is abelian and
V is irreducible, our definitions of characters coincide.
Notice that χV (g) depends only on the conjugacy class of g by properties of the trace.
Definition 30.2. A function f : G → C invariant under conjugation is called a class
function.
What are the characters of some representations like? Given V, W representations, we
have
i) χV ⊕W (g) = χV (g) + χW (g)
ii) χV ⊗W (g) = χV (g)χW (g)
iii) χV ∗ (g) = χV (g) since each eigenvalue must be a root of unity, so inversion is the
same as complex conjugation.
P
iv) χ∧2 V = 1≤i<j≤n αi αj = 21 (χ(g)2 − χ(g 2 )).
If G acts on a finite set S, we have an associated permutation representation, χV (g) is
just the number of elements fixed by g, since the diagonal will have a 1 when gs = s and
0 otherwise.
We now introduce the notion of character table: we just write out the value of the chare (12) (123)
1
1
acter map on each conjugacy class for each irreducible representation: U0 1
U 1 −1
1
V 2
0
−1
We obtained the last because we know that the permutation representation is U ⊕ V ; the
character of the permutation representation is (3, 1, 0) by the above, so the character of
V must be (2, 0, −1).
What does V ⊗ V look like? We know that χV ⊗V (g) = χV (g)2 , so its character is
(4, 0, 1). We see that this is the sum of the characters of U, U 0 , V . We can also verify that
U, U 0 , V have linearly independent characters. Thus V ⊗ V ∼
= U ⊕ U 0 ⊕ V (this is also
what we found last time with our ad hoc approach).
Definition 30.3. Let V, W be representations of G. Then Hom(V, W ) is all linear maps
V → W and HomG (V, W ) the subspace of G-linear maps. We denote by V G the set
{v ∈ V : gv = v∀g ∈ G}; in other words, every element in G fixes these elements.
83
On your problem set, you showed that Hom(V, W )G = HomG (V, W ).
RecallPthat if V is a representation, g : V → V is not necessarily G-linear. However,
1
φ = |G|
g∈G g is g-linear.
Proposition. ϕ is just projection onto V G ⊂ V . In other words, φ|V G = I and im(ϕ) =
V G.
P
1
χV (g). Now say V, W are
Notice that dim V G = tr ϕ, but by definition tr ϕ = |G|
irreducible representations.
1 X
dim(HomG (V, W )) = dim Hom(V, W )G =
χV (g)χW (g).
|G| g∈G
However, we also know that HomG (V, W ) = 1 if V ∼
= W and 0 otherwise. Also, the
expression we wrote above looks like a Hermitian inner product for class functions. Define
1 X
α(g)β(g);
H(α, β) =
|G|
our observation about irreducibles suggests that the characters of the irreducibles are
orthonormal, and in fact might even be a basis (!!).
Even if we don’t know we have a basis, we now know that there are only finitely many
irreducible representations, and that their number is bounded above by the number of
conjugacy classes.
that if V1 , . . . , Vk are all the irreducible representations of G, then for W =
LObserve
Viai and ai = H(χW , χVi ), so a representation is determined entirely by its character.
31 11/16/18
Recall that if V is a representation of G and
1 X
ϕ=
g:V →V
|G|
P
1
is a projection onto V G , then dim V G = tr ϕ = |G|
χV (g). Notice this tells us that the
average value of the characters is always an integer, even if the individual characters can
take arbitrary complex values. We will now apply this to the representation Hom(V, W ) =
V ∗ ⊗ W with V, W irreducible. We saw in the problem set that
dim HomG (V, W ) = dim Hom(V, W )G
and from our previous work, we know
1 X
1 X
dim Hom(V, W )G =
χV ∗ ⊗W (g) =
χV (g)χW (g).
|G| g∈G
|G| g∈G
Moreover, HomG (V, W ) = 0, 1 depending on whether or not V ∼
= W.
Thus, if we let CF be the space of class functions on G, we’ve defined an inner product
for them and the characters of the irreducibles are orthonormal.
84
Corollary 31.1. If V1 , . . . , Vk are irreducible representations of G, then χVi are linearly
independent and there are at most k ≤ c irreducible representations, where c is the number
of conjugacy classes. We will eventually see that k = c. Moreover, every representation
is determined completely by its character because the multiplicity of each irreducible Vi in
a representation W is just H(χW , χVi ).
L
Using this, we know that for W = i Vi⊕ai ,
X
H(χW , χW ) =
a2i
i
and W is irreducible if and only if this inner product is 1.
We will now apply this to the regular representation R of G; R is the vector space with
basis {eg }g∈G and G acts by left multiplication (having it act on the right is equivalent,
L ⊕ai
but we’ll use left). If V1 , . . . , Vk are the irreducible representations of G and R =
Vi ,
what are the ai ? Since R is a permutation representation, the character of each g is the
number of fixed points. But unless g = e, there are no fixed points. So
H(χR , χVi ) =
1
1
χR (g)χVi (g) =
· |G| · χVi (e) = dim Vi .
|G|
|G|
P
Thus every Vi appears dim Vi times in the regular representation.Thus dim R = (dim Vi )2 ,
but dim R = |G| by definition.
Let’s do some examples. Consider S4 . The conjugacy classes are e (1), the transpositions (6), 3-cycles (8), 4-cycles (6), and pairs of transpositions (3). We know that we
have the trivial, alternating, and standard representations.
U
U0
V
e
(12)
(123)
(1234)
(12)(34)
1
1
3
1
−1
1
1
1
0
1
−1
−1
1
1
−1
Quick reality check: U, U 0 , V have orthogonal characters. These can’t be all the irreducibles, though, since 12 +12 +32 6= 24. There have to be exactly 2 more, in fact, since 13
is most definitely not a square. There’s exactly one way to write 13 as a sum of 2 squares,
namely 22 + 32 . Usually, we can find these by applying some linear algebraic operations
to ones we already know. For example, we could take tensor products of representations;
if we tensor an irreducible representation with a one-dimensional representation, we get
another irreducible (no invariant subspaces still), for example. Thus U 0 ⊗ V = V 0 will be
irreducible and has character (3, −1, 0, 1, −1), so it’s different from V 0 .
We have one more representation to find. We know that if we tensor with U 0 , we get
another 2-dimensional irreducible representation, so we now that W takes the value 0 on
the conjugacy classes of (12) and (1234). We can also find W using the orthogonality
relations. Through some linear algebra, we can determine W up to scalars, and we know
W has dimension 2, so we end up knowing its entire character.
85
U
U0
V
V0
W
e
(12)
(123)
(1234)
(12)(34)
1
1
3
3
2
1
−1
1
−1
0
1
1
0
0
−1
1
−1
−1
−1
0
1
1
−1
1
2
This 2 for (12)(34) acting on W is a bit weird. After all, the eigenvalues must be
roots of unity; thus the entire conjugacy class has eigenvalues 1 on W and must act as
the identity. Thus (12)(34) ∈ ker ρW : S4 → GL2 . This tells us that the ideneity and
the conjugacy class of (12)(34) forms a normal subgroup (the Klein four group) and the
quotient is just S3 . This tells us that this representation can be restricted to S3 , too.
What is V ⊗ V ? Well, H(χV ⊗V , χU ) = 1, H(χV ⊗V , χU 0 ) = 0, H(χV ⊗V , χV ) = 1,
H(χV ⊗V , χV 0 ) = 1, and we’ve only accounted for 7 dimensions so we must also have a
copy of W . Therefore, V ⊗ V = U ⊕ V ⊕ V 0 ⊕ W .
Let’s do one more example, the alternating group on 4 letters; this has conjugacy
classes e, (123), (124), and (12)(34). Our standard representation has character 1 on
each conjugacy class. There are going to be 4 irreducible representations; and it turns
out that the only way we can write 11 as a sum of three squares is 12 +12 +32 . Notice that
when we quotient A4 by the Klein four group generated by the double transpositions, we
get Z/3; we therefore have 2 more representations, each of dimension 1, that restrict to
representations of Z/3. We can then compute the last representation of dimension 3 by
appealing to orthogonality, or notice that when we restrict the standard representation
of S4 to A4 we get an irreducible representation.
U
U0
U 00
V
e
(123)
(132)
(12)(34)
1
1
1
3
1
ω
ω2
0
1
ω2
ω
0
1
1
1
−1
32 11/19/18
32.1 Finishing up characters
P
1
Last time, we observed that if V is any representation of G, the operator ϕ = |G|
g∈G g
G
is in fact a projection
V . By applying this to Hom(V, W ), we deduced that we
L ⊕aonto
i
could write V =
Vi .
Today, we will think about what other linear combinations of g ∈ G (when viewed as
elements of (V )) are G-linear. Suppose α : G → C is any function. For any representation
V of G, set
X
ϕα,V =
α(g)g
g∈G
Proposition. ϕα,V is G-linear for all all representations V of G if and only if α is a
class function.
86
For example, the previous ϕ had α = 1, which is clearly a class function.
Proof. We’ll only prove half of this. For ϕα to be G-linear, we want ϕα (hv) = h(ϕα v).
Writing out terms, the left hand side is the same as
X
X
ϕα (hv) =
α(g)gh(v) =
α(hgh−1 )(hgh−1 )h(v).
g∈G
g∈G
If α is a class function, then α(hgh−1 ) = α(g). So our sum is
!
X
α(g)hgh1 h(v) =
g∈G
X
α(g)hgv = h
g∈G
X
α(g)gv .
g∈G
A consequence of this is that the number of irreducible representations is the same
as the number of conjugacy classes; combined with our previous work, this tells us that
the characters form an orthonormal basis for the space of class functions on G. In other
words, if α is a class function and H(α, χVi ) = 0 for all i, then α = 0.
To start, suppose V is irreducible. If α is a class function, then ϕα,V is G-linear and V
is irreducible of dimension n. By Schur’s lemma, then, ϕα,V is a multiple of the identity
λI, and λ = n1 tr(ϕ). But then
1X
1
λ=
α(g)χ(g) = H(α, χV ).
n g∈G
n
So if H(α, χVi ) = 0 for every irreducible Vi , then ϕα,V is 0 for all representations V of
G. In particular, ϕα,R = 0 with R the regular representation.
But if R is the regular
P
representation, each of the g ∈ (R) are independent, so
β(g)g = 0 =⇒ β(g) = 0.
L ⊕ai
with V1 the trivial
Proposition. If V is any P
representation of G and V =
Vi
representation, then ϕχVi = g∈G χVi (g)g is projection onto the summand Vi⊕ai .
32.2 Representation Ring
Fix a group G and consider the set of representations of G up to isomorphism. We have
two laws of composition: (V, W ) 7→ V ⊕ W and (V, W ) 7→ V ⊗ W . These happen to
satisfy the distributive law since U ⊗ (V ⊕ W ) = (U ⊗ V ) ⊕ (U ⊗ W ). This is starting
to look like a ring, except we lack additive inverses. So we’re going to just add them to
our set by considering a huge set and modding out. Let R̂ be the space of all integral,
finite, formal linear combinations of representations of G. Now consider the subgroup
generated by [V ] + [W ] − [V ⊕ W ]. When we mod out by this subgroup, we get what we
call R(G), the representation ring of G.
P
Observe that as a set, R(G) is just the set of formal linear combinations ki=1 ai Vi with
V1 , . . . , Vk the irreducible representations of G and ai ∈ Z. We call the elements of R(G)
the virtual representations. In this lattice, the actual representations form a cone.
Next, observe that the character defines a ring homomorphism from R(G) to the class
functions of G (elements of the image are virtual characters), and this induces an isomorphism R(G) ⊗Z C ∼
= Cclass (G). There are two theorems that describe the lattice of
virtual characters due to Brauer and Artin.
87
32.3 Examples
33 11/26/18
On Friday, we will discuss what classes you might be interested in taking in the future,
e.g. higher-level math classes and physics classes.
Today, we will discuss the representation theory of S5 and A5 .
Below you will find the character table for S5 . U and U 0 are standard for all finite
groups, V is the standard and V 0 is just V ⊗ U 0 . Then we could consider expressing
86 = |G|2 − dim U 2 − . . . = 84 as a sum of three squares (because we know there are
three irreducibles left). This is a viable strategy for S5 (though fairly inapplicable for
large groups); instead, we will employ multi-linear algebraic operations (such as tensor
products and symmetric/exterior powers) on representations we already have to extract
new ones.
A key formula is that for any representation V ,
1
χΛ2 V (g) = (χV (g)2 − χV (g 2 )).
2
This allows us to add Λ2 V to the table and check that it’s irreducible (by taking its
inner product with itself): 36+24+60
= 1.
120
A similar formula for the symmetric square is
1
χSym2 V (g) = (χV (g)2 + χV (g 2 )).
2
Let us consider Sym2 V .
This allows us to add Sym2 V to the character table. We calculate that
100 + 160 + 20 + 60 + 20
= 3.
120
Thus, Sym2 V is the direct sum of three irreducibles. Those irreducibles might already
be in the table. Well, we know how to check - by taking inner products!
(χSym2 V , χSym2 V ) =
(χSym2 V , χU ) =
10 + 40 + 20 + 30 + 20
= 1,
120
which implies that U is a direct summand of Sym2 V with multiplicity 1. A similar
computation proves that V is also a direct summand of Sym2 V , also with multiplicity 1;
we also see that U 0 and V 0 are not direct summands of Sym2 V .
Thus, Sym2 V = U ⊕V ⊕W , where W is some 5-dimensional irreducible representation!
In fact, we can easily write down the character of W : we just subtract the characters of
U and V off of Sym2 V .
Finally, the last irreducible is W 0 := W ⊗ U 0 .
88
U
U0
V
V ⊗ U0 = V 0
Λ2 V
Sym2 V
W
W0
e
(12)
(123)
(1234)
(12345)
(12)(34)
(12)(345)
1
1
4
4
6
10
5
5
1
−1
2
−2
0
4
1
−1
1
1
1
1
0
1
−1
−1
1
1
0
0
0
0
−1
1
1
−1
−1
−1
1
0
0
0
1
1
0
0
−2
2
1
1
1
−1
−1
1
0
1
1
−1
Let us move on to A5 . A natural operation is to restrict irreducible representations of
S5 to A5 and see how they decompose there. Different irreps of S5 might become equal
when restricted to A5 - for example, U and U 0 are both trivial on A5 .
What happens when we restrict V , the standard representation of S5 , to A5 ? Well, we
can compute its character by hand and then take the inner product with itself to check
that it does remain irreducible.
Similarly, W remains irreducible when restricted to A5 . There are two irreducible
representations of A5 left to find; by the dimensions-square formula, their dimensions
squared sum to 18, which implies that each has dimension 3. To find them, we will
consider Λ2 V . By taking its inner product with itself, we find that it decomposes as the
direct sum of two irreducible representations. Again, by taking inner products we can
check that neither U nor W are direct summands, hence the two are the two irreducibles
we have not found yet - let us call them Y and Z. Finally, to complete the character table
we use orthogonality and the fact that the characters of Y and Z sum to the character
of Λ2 V .
U
V
W
Λ2 V
Y
Z
e
(123)
(12)(34)
(12345)
(12354)
1
4
5
6
3
3
1
1
−1
0
0
0
1
0
1
−2
−1
−1
1
−1
0
1√
1
−1
0
1√
1+ 5
2√
1− 5
2
1− 5
2√
1+ 5
.
2
Recall that A5 is the group of isometries of an icosahedron in R3 . A fun consequence
of the irrationality of some of the entries in the character table of A5 is that there does
not exist an icosahedron in R3 with all vertices having rational coordinates: indeed, that
would give a rational representation of A5 (i.e.√ a homomorphism A5 → GL(3, Q) ,→
GL(3, C), which contradicts the irrationality of 5.
Important disclaimer: all of the fun analysis above was almost completely ad hoc! We
will discuss a more systematic approach in what follows.
Let G be a finite group and H ,→ G be a subgroup. We have a fairly tautological
restriction map
89
ResG
H
representations of G −−−→
representations of H,
or, more categorically, a restriction functor
Rep(G) → Rep(H).
A crucial question to consider is how to move in the opposite direction. To motivate
the actual construction, suppose that V is a representation of G and W ⊂ V is a subspace
invariant under H.
A straightforward observation is that for any g ∈ G, the subspace gW depends only
on the coset gH. The simplest situation we could have is
V =
⊕ σW.
σ∈G/H
If this is true, we will say that V is induced from W .
Theorem 33.1. Given a subgroup H ,→ G and a representation W of H, there exists a
unique induced representation of G, to be denoted IndG
H (W ).
Proof. To be given on Wednesday.
This construction paves a map in the opposite direction:
IndG
H
representations of H −−−→
representations of G.
34 11/30/2018
34.1 Induced Representations
The basic situation is that we have a group G and a subgroup H ⊂ G. We have a maps
{rep of G}
Res
Ind
{rep of H}
This can also be seen as a map of rings of representations:
R(G)
Res
Ind
R(H)
where Res(V ⊗ W ) = Res(V ) ⊗ Res(W ). Ind is not a ring homomorphism:
Ind(V ⊗ W ) 6= Ind(V ) ⊗ Ind(W )
But it is a module homomorphism. If U is a representation of G and W is a representation
of H, then U ⊗ Ind(W ) = Ind(ResU ⊗ W ).
A key fact about induced representations:
90
Proposition. Say U is a representation of G and W a representation of H. Let V =
Ind(W ). Then every H-linear map W → Res(U ) extends uniquely to a G-linear map
V = Ind(W ) → U .
In other words
HomH (W, Res(U )) = HomG (Ind(W ), U )
This is often called Frobenius reciprocity.
L
Proof. V = Ind(W ) = σ∈G/H σW . Given ϕ : W → Res(U ) that is H-linear, to extend
to ϕ̃ : V → U that is G-linear
ϕ̃
σW
U
gσ −1
gσ
ϕ
V
U
commutes. This leaves no choices, so the extension has to be unique. It remains to show
that ϕ̃ is G-linear.
Corollary 34.1.
(χIndW · χU )G = (χW · χResU )H
Proof. It suffices to do this for U, W irreducible. In this case the LHS is the number
of copies of U in Ind(W ). This is the same as dim HomG (IndW, U ). The RHS is the
number of copies of W in Res(U ), which is the same as dim HomH (W, ResU ). It follows
by Frobenius reciprocity that these two are equal.
The bottom line to remember is: The number of times U appears in Ind(W ) is equal
to the number of times W appears in Res(U ) for irreducible U, W .
Example 34.2. Suppose G = S4 and H = S3 .
U4
Res
U3
U40
Res
U30
V4
Res
V3 ⊕ U3
V40
Res
V3 ⊕ U30
W
Res
V3
e.g. if we ask what is Ind(V3 ) we would get V4 ⊕V40 ⊕W - i.e. the irreducible representations
in which V3 appears.
91
35 12/3/2018
S4
e
(12)
(123)
(1234)
(12)(34)
U
U’
V
V’
W
1
1
3
3
2
1
-1
1
-1
0
1
1
0
0
-1
1
-1
-1
1
0
1
1
-1
-1
2
35.1 Frobenius Reciprocity
We have a group G and a subgroup H. W is an irrep of H and U is an irrep of G. The
number of times U appears in Ind(W ) is equal to the number of times that W appears
in Res(U ).
Example 35.1. Let G = S4 and H = h(1234)i ∼
= Z/4 ⊂ S4 . We know that the
irreducible representations of H are 1-dimensional.
• U = trivial rep
• U1 : C, with (1234) acting by i
• U2 : C ” ” ” i2
• ””””” i3
What are the induced representations Ind(Ui )?
Look first at the restriction map using the table above:
• U →U
• U 0 → U2
• V → U1 ⊕ U2 ⊕ U3
• V 0 → U ⊕ U1 ⊕ U3
• W → U ⊕ U2
Now, Frobenius reciprocity says we can just read off the induced representations from
the restrictions:
• U →U ⊕V0⊕W
• U1 → V ⊕ V 0
• U2 → U 0 ⊕ V ⊕ W
92
• U3 → V ⊕ V 0
you should use dimensionality as a coherence check on this.
Now we will introduce some theorems (which you can find in Serre’s book) that provide
much of the motivation for studying representations.
35.2 Some Theorems
Let G be an arbitrary finite group. Note that the set of all representations of G has the
structure of a monoid Nc where c is the number of irreducible representations of G. Let
Cclass (G) ∼
= Cc be the vector space of class functions. We can also form the ring of virtual
representations R(G) ∼
= Zc . We can also form R(G) ⊗ Q ∼
= Qc , completing a containment
series:
{reps of G} ⊂ R(G) ⊂ R(G) ⊗ Q ⊂ Cclass (G)
(54)
You can think of these steps of containment as levels of information, most on the left and
descending to the right.
Theorem 35.2. Artin: Leg G be a finite group. R(G)⊗Q is generated by representations
induced from cyclic subgroups of G.
Definition 35.3. Let G be a finite group and p a prime number. A p-elementary
subgroup H ⊂ G is a subgroup such that H ∼
= A × B where A is a cyclic group of order
not divisible by p, and B is a p-group (order some power of p).
Theorem 35.4. Brauer: Let G be a finite group. Then R(G) is generated by representations induced from p-elementary subgroups of G.
And now for something completely different.
35.3 Real Representations
The goal is to classify, given G, actions of G on real vector spaces up to isomorphism.
This is the same thing as maps G → GLn R up to conjugation. We have a map from
{real reps V0 } → {complex reps V } by V0 7→ V0 ⊗R C. What’s the image of this map?
That is, what complex representations are expressible like this?
Definition 35.5. A representation V of G over C is called real if there exists a representation V0 of G over R such that V = V0 ⊗R C.
So, given a representation V of G over C, when is it real in this sense? A necessary
but insufficient condition is that the character χV is real.
Suppose V is any complex irreducible representation of G such that χV is real. We
observe that χV being real is equivalent to V ∼
= V ? as representations. We then have a
homomorphism of representations V → V ? that commutes with the action of G. Shor’s
lemma says that this is unique up to scalars.
93
Note that if V is in fact real (V = V0 ⊗R C) then we can introduce a positive-definite
bilinear form B on V0 preserved by the action of G, and extend it to a non-degenerate
symmetric bilinear form on V that is preserved by the action of G. B naturally gives an
isomorphism V ∼
= V ? . We conclude that any real representation is isomorphic to its dual
representation.
Where’s the gap? Knowing χV is real, why does this not gives us a non-degenerate
symmetric bilinear form (i.e. why is this not sufficient)? Recall Hom(V, V ? ) = V ? ⊗ V =
Sym2 V ⊕ ∧2 V is not irreducible. So the bilinear form could also be skew-symmetric.
So χV being real implies that there exists a B̃ : V ∼
= V ? that is G linear. Either
B̃ ∈ Sym2 V and V is real, or B̃ ∈ ∧2 V and V is called quaternionic.
36 12/5/2018 Last Class!
I’m so proud of all of you! You’ve accomplished a lot this semester, and I’m looking
forward to watching all of you grow as mathematicians next year! - Brian < 3
Real Representations
Suppose that you’ve never heard of the complex numbers, only R. Then a representation
of G would be a homomorphism G → GLn (R) up to conjugation.
We still have complete reducibility in real representations. Given an action of G on V0
over R, we can introduce an invariant positive-definite symmetric bilinear forms. Complete reducibility follows.
We don’t have Shor’s lemmafor real representations!
As an example, let Z/n act on R2
− sin 2π
cos 2π
n
n
. This is irreducible as a representation
by taking the generator ζ 7→
2π
cos 2π
sin n
n
on R2 .
Consider an action of G on a real vector space V0 . I can associate to it an action of
G on the complex vector space V0 ⊗R C. The first question to ask is what is the image?
That is, which of the complex representations of a group come from real ones?
Definition 36.1. Say a representation of G on a complex vector space V is real if there
exists an action of G on a real vector space V0 such that V = V0 ⊗R C.
Theorem 36.2. If V is an irreducible representation of G, then exactly one of the following three possibilities occurs:
i) χV is complex. This is equivalent to saying V V ? . We say that V is complex
in this case.
ii) χV is real. Equivalently V ∼
= V ? . There exists a non-zero φ ∈ HomG (V, V ? ) ⊂
Hom(V, V ? ) = V ? ⊗ V ? = Sym2 V ? ⊕ ∧2 V ? . By Shor’s lemma φ is either symmetric
or skew-symmetric. In the case that φ is symmetric, we call V real.
iii) Same as ii, but φ is skew-symmetric and we call V quaternionic.
94
Definition 36.3. A skew-field or division ring is a set with operations +, × satisfying
the axioms of a field except for the commutativity of ×.
An example of a skew-field is the quaternions H = R ⊕ Ri ⊕ Rj ⊕ Rk where i2 = j 2 =
k 2 = −1, ij = k = −ji, jk = i = −kj, and ki = j = −ik.
When V is quaternionic, it is essentially a vector space over H that we think of as a
vector space over C.
When V0 is an irreducible real representation of G, then either V0 ⊗R C is irreducible of
type ii in the theorem above, or V0 ⊗R C ∼
= W ⊕ W ? for some pair of conjugate complex
irreducible representations W, W ? .
36.1 Group Algebras
Definition 36.4. Given a finite group G, we define the group algebra
X
CG ≡ {
ag eg : ag ∈ C}
g∈G
where eg eh = egh .
So
X
X
X
(
bg e g ) =
ag eg )(
(
g∈G
g
g∈G
X
ah bh0 )eg
h,h0 : hh0 =g
This is a ring/algebra, and is commutative if and only if G is abelian.
Proposition. A representation of G is equivalent to a left module over CG.
Say G is a finite group, and W1 , · · · , Wc the irreducible representations of G. i.e. We
have ρiL
hom G → GL(Wi ) and CG → End(Wi ). Take the direct sum over i and we get
CG → i End(Wi ).
L
Proposition. The map CG → i End(Wi ) is an isomorphism.
Taking dimensions, we have as a corollary:
X
|G| =
dim(Wi )2
i
36.2 Lie Groups
We want to look at continuous homomorphisms from things like R or S 1 to GLn C. Some
things we’ve seen don’t work in this context.
Example 36.5. Complete reducibility fails for representations of R.
But complete reducibility doesn’t always fail!
Proposition. Complete reducibility holds for continuous representations of S 1 .
Proof. S 1 → GL(V ). Choose any Hermitian inner Rproduct H0 : V ×V → C and define an
invariant Hermitian inner product by H(v, w) = S 1 H0 (gv, gw). The reason this works
is that S 1 is “compact”, which we will understand more in 55B.
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