Study Guide
Harvard Mathematics Qualification Exam
Atanas Atanasov
Charmaine Sia
E-mail address: nasko@math.harvard.edu
E-mail address: sia@math.harvard.edu
Contents
Chapter 1. Notes
1. Algebra
1.1. Dummut & Foote, Preliminaries
1.2. Dummut & Foote, Chapter 1: Introduction to Groups
1.3. Dummut & Foote, Chapter 2: Subgroups
1.4. Dummut & Foote, Chapter 3: Quotient Groups and Homomorphisms
1.5. Dummut & Foote, Chapter 4: Group Actions
1.6. Dummut & Foote, Chapter 5: Direct and Semidirect Products of Abelian Groups
1.7. Dummut & Foote, Chapter 6: Further Topics in Group Theory
1.8. Dummut & Foote, Chapter 7: Introduction to Rings
1.9. Dummut & Foote, Chapter 8: Euclidean Domains, Principal Ideal Domains and Unique
Factorization Domains
1.10. Dummut & Foote, Chapter 9: Polynomial Rings
1.11. Dummut & Foote, Chapter 10: Introduction to Module Theory
1.12. Dummut & Foote, Chapter 11: Vector Spaces
1.13. Dummut & Foote, Chapter 12: Modules over Principal Ideal Domains
1.14. Dummut & Foote, Chapter 13: Field Theory
1.15. Dummut & Foote, Chapter 14: Galois Theory
1.16. Dummut & Foote, Chapter 18: Representation Theory and Character Theory
1.17. Dummut & Foote, Chapter 19: Examples and Applications of Character Theory
2. Algebraic Geometry
2.1. Harris, Lecture 1: Affine and Projective Varieties
2.2. Harris, Lecture 2: Regular Functions and Maps
2.3. Harris, Lecture 3: Cones, Projections, and More About Products
2.4. Harris, Lecture 4: Families and Parameter Spaces
2.5. Harris, Lecture 5: Ideals of Varieties, Irreducible Decomposition, and the Nullstellensatz
2.6. Harris, Lecture 6: Grassmannians and Related Varieties
2.7. Harris, Lecture 7: Rational Functions and Rational Maps
2.8. Harris, Lecture 11: Definitions of Dimension and Elementary Examples
2.9. Harris, Lecture 13: Hilbert Polynomials
2.10. Harris, Lecture 14: Smoothness and Tangent Spaces
2.11. Harris, Lecture 18: Degree
3. Complex Analysis
3.1. Ahlfors, Chapter 1: Complex Numbers
3.2. Ahlfors, Chapter 2: Complex Functions
3.3. Ahlfors, Chapter 3: Holomorphic Functions as Mappings
3.4. Ahlfors, Chapter 4: Complex Integration
3.5. Ahlfors, Chapter 5: Series and Product Developments
3.6. Ahlfors, Chapter 6: Conformal Mapping. Dirichlet’s Problem
4. Algebraic Topology
4.1. Hatcher, Chapter 0: Some Underlying Geometric Notions
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CONTENTS
4.2. Hatcher, Chapter 1: The Fundamental Group
4.3. Hatcher, Chapter 2: Homology
4.4. Hatcher, Chapter 3: Cohomology
5. Differential Geometry
6. Real Analysis
6.1. Rudin, Prologue
6.2. Rudin, Chapter 1: Abstract Integration
6.3. Rudin, Chapter 2: Positive Borel Measures
6.4. Rudin, Chapter 3: Lp -spaces
6.5. Rudin, Chapter 4: Elementary Hilbert Space Theory
6.6. Rudin, Chapter 5: Examples of Banach Space Techniques
6.7. Rudin, Chapter 6: Complex Measures
6.8. Rudin, Chapter 7: Differentiation
6.9. Rudin, Chapter 8: Integration on Product Spaces
6.10. Rudin, Chapter 9: Fourier Transforms
Chapter 2. Past Exams
1. Spring 2010
1.1. Day 1
1.2. Day 2
1.3. Day 3
2. Fall 2009
2.1. Day 1
2.2. Day 2
2.3. Day 3
3. Spring 2009
3.1. Day 1
3.2. Day 2
3.3. Day 3
4. Fall 2008
4.1. Day 1
4.2. Day 2
4.3. Day 3
5. Spring 2008
5.1. Day 1
5.2. Day 2
5.3. Day 3
6. Fall 2007
6.1. Day 1
6.2. Day 2
6.3. Day 3
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CONTENTS
5
Below is a set of guidelines which were used in the compilation of this document. They do not reflect any
absolute practice in typesetting, but merely the combination of my own customs and some rational decisions
which I took to standardize various parts of the document.
• When taking notes from a book, try to copy all definitions, propositions, theorems, and important
remarks. Do not hesitate to add your own relevant observations as these could facilitate the learning
process.
• If there is a change of sentence structure which would simplify the statement of results, then apply
that.
• When the book in question is slightly old (e.g., Rudin, Real and Complex Analysis), we should
change mathematical notation in order to update the text.
• Use as many of the predefined customizations as possible.
• Avoid the use of “one-to-one” and “onto”. Instead replace these with “injective” and “surjective”
respectively.
• Attempt to utilize any standard modern notation, e.g. S 1 , Lie groups, etc.
• If it is customary to use employ certain symbols for an object, then attempt to do so.
• Use \ for the difference of sets instead of −.
• Use \cn (without the customizations \colon) for colons in functions, that is, visually f : X → Y
looks better than f : X → Y (note the difference in spacing before the colon).
• Be consistent with wording and spelling. For example, use holomorphic instead of analytic throughout. On a similar note, hyphenate “non” constructions such as “non-constant” and “non-vanishing”.
TODO:
• Add “subsubsections” which will not display in the table of contents. This is done to logically
separate the various results in a chapter.
• Check all files and convert adjectives “nonX” to “non-X”.
CHAPTER 1
Notes
1. Algebra
Syllabus
Undergraduate: Dummit & Foote, Abstract Algebra, except chapters 15, 16 and 17. (math 122,
123)
1.1. Dummut & Foote, Preliminaries.
• Basics
– Sets
– Functions/maps and related properties – bijective, surjective, left/right inverse, image/preimage,
permutation, restriction/extension
– Relations – reflexive, symmetric, transitive, equivalence relations, partitions, the equivalent of
the last two
• Integers
– Ring structure
– Well ordering
– GCD & LCM
– Division algorithm
– Euclidean algirithm
• Integers modulo n > 1
– Quotient of Z
– Ring structure
1.2. Dummut & Foote, Chapter 1: Introduction to Groups.
• Definition of a group, properties, and examples – associativity, commutativity, inverses, order of
an element (denoted by | − |)
• Dihedral groups
– Notation D2n , and |D2n | = 2n
– D2n = hr, s | rn = s2 = 1, rs = sr−1 i
• Symmetric group – presented as bijections, cycles, order as LCM, conjugacy classes and partitions
• Matrix groups
If |F | = q < ∞, then |GL(n, F )| = (q n − 1)(q n − q) · · · (q n − q n−1 ).
• The Quaternion group
Q8 = {±1, ±i, ±j, ±k}
i2 = j 2 = k 2 = −1
i · j = k, j · k = i, k · i = j, and exchange introduces a sign
• Homomorphisms and isomorphisms – definition
• Group actions – definition
1.3. Dummut & Foote, Chapter 2: Subgroups.
• Subgroup criterion – H ⊂ G is a subgroup if H 6= ∅ and for all x, y ∈ H, we have xy −1 ∈ H
• Centralizer – for any A ⊂ G, its centralizer is CG (A) = {g ∈ G | ga = ag for all a ∈ A}
7
8
1. NOTES
• Center – Z(G) = CG (G)
• Normalizer – for any A ⊂ G, its normalizer is NG (A) = {g ∈ G | gAg −1 = A}
– NG (A) is always a normal subgroup
– If A is a subgroup, then NG (A) is the largest subgroup such that A E NG (A).
– CG (A) E NG (A)
• Cyclic groups and cyclic subgroups
– Any subgroup of a cyclic subgroup is cyclic.
– Consider a cyclic group of size n. For each m|n, there is a unique subgroup of size m.
– The number of generators of a cyclic group of size n is ϕ(n).
• Subgroups generated by subsets of a group – definition, well-defined
• The lattice of a subgroups of a group
1.4.
•
•
•
•
•
•
•
•
•
•
•
Dummut & Foote, Chapter 3: Quotient Groups and Homomorphisms.
Lagrange’s Theorem
First Isomorphism Theorem
Kernels are always normal, and each normal subgroup occurs as the kernel of a homomorphism
(e.g. the quotient one).
Normality is not transitive.
The only group of prime order p is Z/pZ.
Every subgroup of index 2 is normal.
If H, K ≤ G are finite, then |HK| = |H| |K|/|H ∩ K|.
The set HK is a group if and only if HK = KH.
If H ≤ NG (K), then HK is a subgroup of G. In particular, if K E G, then HK is a group for all
H ≤ G.
|G/H| = |G|/|H|
Second (Diamond) Isomorphism Theorem: Let A, B ≤ G and A ≤ NG (B). Then A ∩ B E A and
AB/B ∼
= A/(A ∩ B).
In particular, the conclusion holds true if B E G.
• Third Isomorphism Theorem: If H, K E G and H ≤ K, then K/H E G/H and
∼ G/K.
(G/H)/(K/H) =
• Fourth Isomorphism Theorem: for N E G there is a correspondence between the subgroups of G/N
and those of G containing N given by taking preimages under the quotient map G → G/N ; the
correspondence presernes normality and inclusions, has numerous other expected properties.
• A group G is called simple if it is nontrivial, and its only normal subgroups are 1 and G.
• A sequence of groups
1 = N0 ≤ N1 ≤ · · · ≤ Nk−1 ≤ Nk = G
is called a composition series for G if Ni−1 E Ni and Ni /Ni−1 is simple for all 1 ≤ i ≤ k.
• Jordan-Hölder: Every finite group has a composition series, and its factors are unique up to permutation.
• Hölder program: (1) classify finite simple groups, and (2) the ways of “putting them together”
• Examples of simple groups: Z/pZ, SL(n, F )/Z(SL(n, F )) for any finite field F and n ≥ 2 (except
SL(2, F2 ) and SL(2, F3 )).
• Every element of Sn may be written as a product of transpositions.
• The alternating group An as the kernel of the homomorphism sign : Sn → {±1}.
1.5. Dummut & Foote, Chapter 4: Group Actions.
• |G| = |Gx | |Gx|
• Let H be a subgroup of the finite group G, and G acts on the set of left cosets A = G/H. Then:
(a) G acts transitively on A;
2. ALGEBRAIC GEOMETRY
•
•
•
•
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(b) the stabilizer of the point 1H ∈ A isTH;
(c) the kernel of this representation is x∈G xHx−1 which is the largest normal subgroup of G
contained in H.
Cayley’s Theorem: Every group G of order n is isomorphic to a subgroup of Sn .
If G is a finite subgroup of order n and p is the smallest prime dividing |G|, then any subgroup of
index p is normal.
The number of conjugates of a subset S ⊂ G is [G : NG (S)]. In particular, for s ∈ G we have
NG (s) = CG (s), so the number of conjugates of s is [G : NG (s)].
The Class Equation: Let G be a finite group and g1 , . . . , gr be representatives of the distinct
conjugacy classes of G not contained in the center Z(G). Then
|G| = |Z(G)| +
r
X
[G : CG (gi )].
i=1
• Any group of prime power order has a nontrivial center.
• If G/Z(G) is cyclic, then G is abelian.
• Any group of order p2 is abelian, hence isomorphic to either Z/p2 Z or (Z/pZ)2 .
•
TODO
1.6. Dummut & Foote, Chapter 5: Direct and Semidirect Products of Abelian Groups.
TODO
1.7. Dummut & Foote, Chapter 6: Further Topics in Group Theory. TODO
1.8. Dummut & Foote, Chapter 7: Introduction to Rings. TODO
1.9. Dummut & Foote, Chapter 8: Euclidean Domains, Principal Ideal Domains and
Unique Factorization Domains. TODO
1.10. Dummut & Foote, Chapter 9: Polynomial Rings. TODO
1.11. Dummut & Foote, Chapter 10: Introduction to Module Theory. TODO
1.12. Dummut & Foote, Chapter 11: Vector Spaces. TODO
1.13. Dummut & Foote, Chapter 12: Modules over Principal Ideal Domains. TODO
1.14. Dummut & Foote, Chapter 13: Field Theory. TODO
1.15. Dummut & Foote, Chapter 14: Galois Theory. TODO
1.16. Dummut & Foote, Chapter 18: Representation Theory and Character Theory. TODO
1.17. Dummut & Foote, Chapter 19: Examples and Applications of Character Theory.
TODO
2. Algebraic Geometry
Syllabus
Graduate: Harris, Algebraic geometry, a first course, lectures 1-7, 11, 13, 14, 18. (math 137 and
math 232a)
10
1. NOTES
2.1. Harris, Lecture 1: Affine and Projective Varieties.
• An inclusion-exclusion type formula holds for dimensions of linear spaces in projective space.
Namely, if Λ and Λ0 are linear spaces in a projective space Pn and Λ, Λ0 is their span, then
dim(Λ, Λ0 ) = dim(Λ) + dim(Λ0 ) − dim(Λ ∩ Λ0 ).
Note that we are taking the dimension of the empty set as a linear space to be −1. A similar
formula holds for linear (not affine) subspaces of affine space.
• A set Γ ⊂ Pn of d points may be described as the vanishing set of polynomials of degree d and less.
We say Γ has degree d. If Γ is not contained in a line, then we may describe it by polynomials of
degree d − 1 and less.
• We say that a set of points pi = [vi ] ∈ Pn are independent if so are the corresponding vectors vi .
The space Pn can contain at most n + 1 independent points.
• We say that a set of points Γ ⊂ Pn are in general position if no n + 1 or fewer of them are linearly
dependent.
Theorem 2.1.1. For k ≥ 2, any collection Γ ⊂ Pn of d ≤ kn points in general position may be
described by polynomials of degree k or less. This bound is sharp.
Theorem 2.1.2. Any two ordered sets of n + 2 points in general position in Pn are projectively
equivalent.
• A hypersurface X is a subvariety of Pn described as the zero locus of a single homogeneous polynomial F . We may always choose F without repeated prime factors, and in this case, F is unique
up to multiplication by scalars (this requires the Nullstellensatz and holds only over algebraically
closed fields). When this is done the degree of F is called the degree of the hypersurface X.
• A complex analytic variety is one which is the locus of (homogeneous) holomorphic functions on
AnC or PnC . All varieties in AnC and PnC are complex analytic since polynomials are holomorphic
functions. The converse holds in projective but not in affine complex space.
Theorem 2.1.3 (Chow’s Theorem). Any complex analytic subvariety X ⊂ PnC is an algebraic
variety.
2.2. Harris, Lecture 2: Regular Functions and Maps.
• For an affine variety X ⊂ An , we define the ideal I(X) ⊂ K[z1 , . . . , zn ] of polynomials which vanish
at X. The coordinate ring of X is the quotient A(X) = K[z1 , . . . , zn ]/I(X).
• Consider an open U ⊂ X of an affine variety, and p ∈ U a point. We say a function f on U is regular
at p if in some neighbourhood V of p it is expressible as a quotient g/h where g, h ∈ K[z1 , . . . , zn ]
are polynomials and h(p) 6= 0. We say f is regular on U if it is regular at all p ∈ U .
Lemma 2.2.1. The ring of regular functions regular at every point of an affine variety X is the
coordinate ring A(X). More generally, if U = Uf is a distinguished open, then the ring of regular
function on U is the localization A(X)f = A(X)[1/f ].
• Similarly, for a projective variety X ⊂ Pn , one can define an ideal I(X) generated by all homogeneous polynomials vanishing at X. The homogeneous coordinate ring of X is defined by
S(X) = K[z0 , . . . , zn ]/I(X). Note that (the isomorphism class of) S(X) is invariant under projective equivalence but not under general isomorphism of projective varieties.
• If G is a homogeneous polynomial, then the ring of regular functions on complement of the zero
locus of G is given by the 0-th graded piece of the localization S(X)G = S(X)[1/G].
• For any two positive integers n and d, we define the Veronese map (embedding) of degree d as
νd : Pn → PN
[X0 , . . . , Xn ] 7→ [. . . , X I , . . . ]
2. ALGEBRAIC GEOMETRY
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where X I ranges over all monomials of degree d in X0 , . . . , Xn . It possesses the property that all
hypersurfaces of degree d in Pn are exactly the hyperplane sections of νd (Pn ) ⊂ PN . The Veronese
map carries any subvariety X ⊂ Pn isomorphically
to its image νd (X). The dimension of the
n+d
codomain can be deduced to be N = d − 1. The image of a Veronese map is called a Veronese
variety.
• For any positive integers m and n, we define the associated Segre map
σ : Pn × Pm → P(n+1)(m+1)−1
([X0 , . . . , Xm ], [Y0 , . . . , Yn ]) 7→ [. . . , Xi Yj , . . . ],
where the coordinates in the target space range over all pairwise products of coordinates Xi and
Yj . The image of σ is called a Segre variety and is denoted Σm,n .
2.3. Harris, Lecture 3: Cones, Projections, and More About Products.
• Every quadric hypersurface Q ⊂ Pn can be put in the form X02 + · · · + Xk2 = 0 for some integer
k ≥ 0. We call k + 1 the rank of Q.
• Consider two polynomials f and g of degrees m and n respectively with coefficients in a field K.
We would like to investigate when f and g have a common factor. This happens if and only if
there exists a polynomial h of degree m + n − 1 divisible by both. Equivalently, this means that
the polynomials f , z · f , . . . , z n−1 · f , g, z · g, . . . , z m−1 · g fail to be linearly independent. In turn
this is equivalent to the statement that the determinant
a0 a1 · · am 0
0 · · · 0 0 a0 a1 ·
·
am 0 · · · 0 ..
.
0
·
·
0
a
a
·
·
·
·
a
0
1
m R(f, g) = b
b
·
·
·
b
0
·
·
·
0
1
n
0
0 b0 b1 ·
·
·
b
0
·
·
0
n
.
..
0
·
· 0 b0 b1
· · · · bn is zero. This determinant R(f, g) is often called the resultant of f and g. In fact, the definition
above makes sense for polynomials with coefficients in any ring.
Theorem 2.3.1. Two polynomials f and g in one variable over a field K will have a common factor
if and only if R(f, g) = 0.
The resultant is useful in the proof of the following result.
Theorem 2.3.2. The projection X of X ⊂ Pn from p ∈
/ X to Pn−1 is a projective variety.
• The image of a closed set is not always closed. In special cases, this could however be derived.
Theorem 2.3.3. Let Y be any variety and π : Y × Pn → Y be the projection on the first factor.
Then the image π(X) of any closed subset X ⊂ Y × Pn is a closed subset of Y .
This leads to the following.
Theorem 2.3.4. If X ⊂ Pn is any projective variety and ϕ : X → Pm any regular map, then the
image of ϕ is a projective subvariety of Pm .
Corollary 2.3.5. If X ⊂ Pn is any connected variety and f any regular function on X, then f is
constant.
Corollary 2.3.6. If X ⊂ Pn is any connected variety other than a point and Y ⊂ Pn is any
hypersurface then X ∩ Y = ∅.
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1. NOTES
• We call a set constructible if it is expressible as the union of locally closed subsets. Equivalently,
a subset Z ⊂ Pn is constructible if there exists a nested sequence X1 ⊃ X2 ⊃ · · · ⊃ Xn of closed
subsets of Pn such that
Z = X1 \ (X2 \ (X3 \ · · · \ Xn )).
Theorem 2.3.7 (Chevalley). Let X ⊂ Pm be a quasi-projective variety, f : X → Pn a regular map,
and U ⊂ X any constructible set. Then f (U ) is a constructible subset of Pn .
2.4. Harris, Lecture 4: Families and Parameter Spaces.
• A family of projective varieties in Pn with base B is simply a closed subvariety V ⊂ B × P n . The
fibers Vb = (π1 )−1 (b) of V over points of b are referred to as members of the family.
• Let X ⊂ Pn be any projective variety and {Vb } any family of projective varieties in Pn with base
B. Then the set
{b ∈ B | X ∩ Vb 6= ∅}
is closed in B. More generally, if {Wb } is another family of projective varieties in Pn with base B,
then
{b ∈ B | Vb ∩ Wb 6= ∅}
is a closed subvariety of B.
• Even more generally, if {Wc } is a family of projective varieties in Pn with a possibly different base
C, then the set
{(b, c) ∈ B × C | Vb ∩ Wc 6= ∅}
is a closed subvariety of B × C.
• In a similar vein, for any X ⊂ Pn and family {Vb } the set
{b ∈ B | Vb ⊂ X}
is constructible and the set
{b ∈ B | X ⊂ Vb }
is closed in B.
• Recall that points in Pn ∗ can be thought of as hyperplanes in Pn . Under this identification, the set
Γ = {(H, p) ∈ Pn ∗ × Pn | p ∈ H}
is called the universal hyperplane. The fiber over H ∈ Pn ∗ is precisely the hyperplane H ⊂ Pn . We
call it universal since for any flat family of hyperplanes V ⊂ B × Pn there exists a unique regular
map B → Pn ∗ such that the diagram
V
/Γ
B
/ Pn ∗
is cartesian.
• For any X ⊂ Pn , the universal hyperplane section is
ΩX = {(H, p) ∈ Pn ∗ × Pn | p ∈ H ∩ X} = (π2 )−1 (X),
where π2 : Γ → Pn is the projection on the second factor of t he universal hyperplane Γ.
• A section of a family of projective varieties V ⊂ B × Pn is a map σ : B → V such that π1 ◦ σ = idB .
A rational section is a section defined on some non-empty open subset U ⊂ B.
2. ALGEBRAIC GEOMETRY
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2.5. Harris, Lecture 5: Ideals of Varieties, Irreducible Decomposition, and the Nullstellensatz.
• The following result enables us to establish a correspondence between geometry and algebra. Furthermore, it implies several very useful corollaries.
Theorem 2.5.1 (Nullstellensatz). For any ideal I ⊂ K[z1 , . . . , zn ], the ideal of functions vanishing
on the common zero locus of I is the radical of I, i.e.,
I(V (I)) = rad(I).
Thus, there is a bijective correspondence between subvarieties X ⊂ An and radical ideals I ⊂
K[z1 , . . . , zn ].
Theorem 2.5.2 (Weak Nullstellensatz). Any ideal I ⊂ K[x1 , . . . , xn ] with no common zeros is the
unit ideal.
Theorem 2.5.3. Every prime ideal in K[x1 , . . . , xn ] is the intersection of the ideals of the form
(x1 − a1 , . . . , xn − an ) containing it.
• We say an ideal I cuts out X ⊂ An set-theoretically if V (I) = X; we say I cuts out X schemetheoretically if I = I(X).
• Consider the case of projective varieties. There is almost a bijection between closed varieties X ⊂ Pn
and radical homogeneous ideals I ⊂ K[Z0 , . . . , Zn ]. Define the saturation of I to be
I = {F ∈ K[Z0 , . . . , Zn ] | (Z0 , . . . , Zn )k · F ⊂ I fo rsome k}.
Proposition 2.5.4. The following conditions for a pair of homogeneous ideals I, J ⊂ K[Z0 , . . . , Zn ]
are equivalent:
(i) I and J have the same saturation;
(ii) Im = Jm for all m 0;
(iii) I and J agree locally, that is, they generate the same ideal in each localization K[Z0 , . . . , Zn , Zi−1 ]
of K[Z0 , . . . , Zn ].
•
•
•
•
•
•
We say that an ideal I cuts out X ⊂ Pn scheme-theoretically if I = I(X).
A variety is called irreducible if for any pair of closed subvarieties Y, Z ⊂ X such that Y ∪ Z = X,
either Y = X or Z = X.
An affine variety X ⊂ An is irreducible if and only if I(X) is prime.
If a projective variety X ⊂ Pn is irreducible, then so is every non-empty open affine U = X ∩ An .
The converse is also true if we consider all hyperplane compliments of X, but not if we restrict our
attention only to the standard opens.
A variety is irreducible if and only if every Zariski open subset is dense, i.e., every two Zariski
non-empty open subsets intersect.
The image of an irreducible variety under a regular map is irreducible.
The following are two potentially useful criteria for irreducibility.
Theorem 2.5.5. Let X ⊂ Pn be an irreducible variety, and let ΩX ⊂ Pn ∗ × X be its universal
hyperplane section. Then ΩX is irreducible.
Theorem 2.5.6. Let f : Z → Y is a regular map. Assume that
(i) f is open;
(ii) Y is irreducible;
(iii) for a dense open set of points p ∈ Y , the fiber f −1 (p) is irreducible.
Then Z is irreducible.
• The following is useful in discussing fibers.
14
1. NOTES
Proposition 2.5.7. Let π : X → Y be any regular map with Y irreducible, and let Z ⊂ X be any
locally closed subset. Then for a general point p ∈ Y the closure of the fiber Zp = Z ∩ π −1 (p) is
the intersection of the closure Z of Z with the fiber Xp = π−1 (p).
2.6. Harris, Lecture 6: Grassmannians and Related Varieties.
• TODO
TODO
2.7. Harris, Lecture 7: Rational Functions and Rational Maps. TODO
2.8. Harris, Lecture 11: Definitions of Dimension and Elementary Examples. TODO
2.9. Harris, Lecture 13: Hilbert Polynomials. TODO
2.10. Harris, Lecture 14: Smoothness and Tangent Spaces. TODO
2.11. Harris, Lecture 18: Degree. TODO
3. Complex Analysis
Syllabus
Undergraduate: Ahlfors, Complex Analysis (2nd ed), chapters 1-4 and section 5.1. (math 113)
Graduate: Ahlfors, Complex Analysis (2nd ed), chapter 5, section 6.1, and 6.2. (math 213a)
3.1. Ahlfors, Chapter 1: Complex Numbers.
Proposition 3.1.1 (Lagrange’s identity). The equality
n
2
n
n
X
X
X
|ai |2
|bi |2 −
ai bi =
i=1
i=1
i=1
X
ai bj − aj bi 2 .
1≤i<j≤n
holds for all a1 , b1 , . . . , an , bn ∈ C.
The following inequalities hold true for all complex values.
−|a| ≤ Re a ≤ |a|
−|a| ≤ Im a ≤ |a|
|a + b| ≤ |a| + |b|
|a − b| ≥ ||a| − |b||
2
|a1 b1 + · · · + an bn | ≤ (|a1 |2 + · · · + |an |2 )(|b1 |2 + · · · + |bn |2 )
3.2. Ahlfors, Chapter 2: Complex Functions.
3.2.1. Introduction to the Concept of Holomorphic Function.
Definition 3.2.1. A function u(x, y) is called harmonic if it satisfies the differential equation
∆u =
∂2u ∂2u
+ 2 = 0.
∂x2
∂y
The real and imaginary parts of a holomorphic function are harmonic.
Proposition 3.2.2. If u(x, y) and v(x, y) have continuous first-order partial derivatives which satisfy the
Cauchy-Riemann differential equations
∂u
∂v
=
,
∂x
∂y
∂u
∂v
=−
∂y
∂x
3. COMPLEX ANALYSIS
15
then f (z) = u(z) + iv(z) is holomorphic with homomorphic derivative f 0 (z), and conversely. In complex
form, these two equations become
∂f
∂f
= −i .
∂x
∂y
We will use the informal notation
∂f
1 ∂f
∂f
=
−i
,
∂z
2 ∂x
∂y
∂f
1
=
∂z
2
∂f
∂f
+i
∂x
∂y
.
Theorem 3.2.3 (Gauss-Lucas). For any polynomial P (z) ∈ C[z], the roots of P 0 (z) lie in the convex hull of
the roots of P (z).
Definition 3.2.4. The order of a rational function R(z) = P (z)/Q(z) for P, Q ∈ C[z] is max{deg P, deg Q}.
Proposition 3.2.5. A rational function R of order p has p zeros and p poles, and the equation R(z) = a
has exactly p roots.
When counting poles, zeros, and solutions in the result above, we considered R as a function on the Riemann
sphere CP1 = C ∪ {∞}.
Proposition 3.2.6. If Q is a polynomial with distinct roots α1 , . . . , αn , and if P is a polynomial of degree
< n, then
n
P (αk )
P (z) X
=
.
Q(z)
Q0 (αk )(z − αk )
k=1
3.2.2. Elementary Theory of Power Series.
Corollary 3.2.7 (Lagrange’s interpolation problem). There exists a unique polynomial of degree < n with
given values ck at the points αk .
The inequalities
lim inf αn + lim inf βn ≤ lim inf(αn + βn ) ≤ lim inf αn + lim sup βn ,
lim inf αn + lim sup βn ≤ lim sup(αn + βn ) ≤ lim sup αn + lim sup βn
hold for any two real sequences {αn } and {βn }.
Proposition 3.2.8. The limit of a sequence of a uniformly converging continuous functions is continuous.
Definition 3.2.9. Let {fn } be a sequence of functions. We say that a positive sequence {an } is majorant
for {fn } if there exists a constant M > 0 such that |fn | ≤ M an for all sufficiently large n. Conversely {fn }
is called the minorant of {an }.
P
P
Proposition 3.2.10 (Weierstrass M -test). If
an converges, then
fn converges uniformly.
P
Theorem 3.2.11. For each power series n an z n there exists a number
R=
1
lim sup
p
∈ [0, ∞],
n
|an |
called the radius of convergence, with the following properties:
(i) the series converges absolutely for all |z| < R;
(ii) if 0 ≤ ρ < R, then the convergence is uniform for |z| ≤ ρ;
(iii) if |z| > R, then the terms of the series are unbounded, hence the series does not converge;
(iv) in |z| < R the sum of the series is a holomorphic function; its derivative can be obtained via termwise
derivation, and has identical radius of convergence.
P
P
P
Proposition 3.2.12. If an z n and bn z n have radii of convergence R1 and R2 , then an bn z n has radius
of convergence at least R1 R2 .
16
1. NOTES
Proposition 3.2.13. If limn |an |/|an+1 | = R, then
P
an z n has radius of convergence R.
The following result can be used to analyze when a power series converges at a point of its circle of convergence. Without loss of generality, we assume R = 1 and z = 1.
P
P
Theorem 3.2.14 (Abel’s Limit Theorem). If
an converges, then f (z) =
an z n tends to f (1) as z
approaches 1 in such a way that |1 − z|/(1 − |z|) remains bounded. Geometrically this means that z stays
in an angle < 180◦ with vertex 1, symmetrically to the part (−∞, 1) of the real axis (a Stolz angle).
3.3. Ahlfors, Chapter 3: Holomorphic Functions as Mappings.
3.3.1. Elementary Point Set Topology.
Definition 3.3.1. A metric space is called totally bounded if, for every ε > 0, X can be covered by finitely
map balls of radius ε.
Theorem 3.3.2. A metric space is compact if and only if it is complete and totally bounded.
Corollary 3.3.3 (Heine-Borel). A subset of Rn is compact if and only if it is closed and bounded.
Theorem 3.3.4. A metric space is compact if and only if every infinite sequence has a limit point.
Corollary 3.3.5 (Bolzano-Weierstrass). A metric space is compact if every bounded sequence has a convergent subsequence.
Proposition
T 3.3.6 (Cantor’s Lemma). If E1 ⊃ E2 ⊃ · · · is a decreasing sequence of non-empty compact
sets, then n En is not empty.
Proposition 3.3.7. The following properties hold for any continuous map:
(i) the image of any compact sets is compact;
(ii) the image of any connected set is connected.
Corollary 3.3.8. A continuous bijective map with compact domain is a homeomorphism.
Proposition 3.3.9. On a compact metric space, every continuous function is uniformly continuous.
Definition 3.3.10. A connected open set Ω ⊂ C is called a region.
Theorem 3.3.11. A holomorphic function in a region Ω whose derivative vanishes identically is constant.
The same is true if either the real part, the imaginary part, the modulus, or the argument is constant.
3.3.2. Conformality.
Definition 3.3.12. A holomorphic function f : U → C is called conformal if f 0 does not vanish in U .
Conformal maps are interesting for the following two properties.
Proposition 3.3.13. Let f be a conformal map.
(i) The angle between the image of a tangent vector at a point z and the original vector is always arg f 0 (z).
(ii) Infinitesimal segments near a point z are scaled by |f 0 (z)| independently of their direction.
Corollary 3.3.14. Conformal maps preserve angles between curves.
Consider a C 1 path γ and a region E. It is customary to write
Z
Z
0
L(γ) =
|γ (t)|dt =
|dz|,
γ
γ
ZZ
A(E) =
dxdy
E
for the length of γ and the area of E. The image path f ◦ γ and image region f (E) under a conformal map
f satisfy
Z
Z
ZZ
L(f ◦ γ) =
|f 0 (γ(t))||γ 0 (t)|dt =
|f 0 (z)||dz|,
A(f (E)) =
|f 0 (z)|2 dxdy.
γ
γ
E
3. COMPLEX ANALYSIS
17
In the case of a region, the image area is computed with multiplicity.
3.3.3. Linear Transformations.
Definition 3.3.15. Any a, b, c, d ∈ C satisfying ad − bc 6= 0 determine a biholomorphic map CP1 → CP1
given by
az + b
.
z 7→
cz + d
Such maps are called linear transformations. The group of linear transformations is isomorphic to P GL(2, C)
via
az + b
a b
7
→
z 7→
.
c d
cz + d
A linear transformation is called normalized if some corresponding matrix has determinant 1. The group of
normalized linear transformations is isomorphic to SL(2, C)/{±I}.
Proposition 3.3.16. For any distinct z2 , z3 , z4 ∈ CP1 , there exists a unique linear transformation T satisfying T z2 = 1, T z3 = 0, T z4 = ∞.
Definition 3.3.17. The cross ratio (z1 , z2 , z3 , z4 ) is the image of z1 under the linear transformation which
carries z2 , z3 , z4 into 1, 0, ∞.
Proposition 3.3.18.
(i) Any distinct z1 , z2 , z3 , z4 ∈ CP1 and any linear transformation T satisfy
(T z1 , T z2 , T z3 , T z4 ) = (z1 , z2 , z3 , z4 ).
(ii) A cross ratio (z1 , z2 , z3 , z4 ) is real if and only if the four points lie on a circle.
(iii) A linear transformation carries circles into circles.
Note that in (ii) and (iii) above, a circle in CP1 refers to either a circle in C or a straight line in C together
with ∞.
Proposition 3.3.19 (Ptolemy’s Theorem). If the consecutive vertices z1 , z2 , z3 , z4 of a quadrilateral lie on
a circle, then
|z1 − z3 | · |z2 − z4 | = |z1 − z2 | · |z3 − z4 | + |z1 − z4 | · |z2 − z3 |.
Definition 3.3.20. Consider a linear transformation T which carries the real axis to a circle C. For any w,
we say the points z = T w and z ∗ = T w are symmetric with respect to C. Equivalently, if C passes through
z1 , z2 , z3 then any pair of symmetric points z, z ∗ satisfy (z ∗ , z1 , z2 , z3 ) = (z, z1 , z2 , z3 ).
Note that the definition of symmetry is independent of the chosen linear transformation T or the three points
z1 , z2 , z3 on C.
Proposition 3.3.21.
(i) If C is a line, then the symmetry operation is reflection in C.
(ii) If C is a circle, then the symmetry operation is what we call inversion in plane geometry.
Proposition 3.3.22 (Symmetry principle). Let T be a linear transformation which carries the circle C1 into
the circle C2 . If z and z ∗ are symmetric about C1 , then T z and T z ∗ are symmetric about C2 .
Definition 3.3.23. An orientation of a circle C is determined by an oriented triple of distinct points
z1 , z2 , z3 ∈ C. We say that a point z ∈
/ C is on the right of C if Im(z, z1 , z2 , z3 ) > 0, and on the left of C if
Im(z, z1 , z2 , z3 ) < 0.
Each circle C in the complex plane C can be endowed with the counterclockwise orientation. With respect
to it, the interior of C is on its left, and it exterior on its right.
18
1. NOTES
Definition 3.3.24. Consider a linear transformation
z−a
.
z−b
T (z) = k ·
The preimages under T of straight lines through the origin are circles through a and b (denoted by C1 ). The
preimages of concentric circles about the origin are circles with the equation
z − a
z − b = ρ/|k|,
where ρ is the radius of the original circle. These are called circles of Apollonius with limit points a and b
(denoted by C2 ). The configuration formed by all the circles C1 and C2 is referred to as the circular net or
the Steiner circles determined by a and b.
Proposition 3.3.25.
(i) There is exactly one C1 and one C2 through each point in the plane with the exception of a and b.
(ii) Every C1 meets every C2 at right angles.
(iii) Reflection in a C1 transforms every C2 into itself and every C1 into another C1 . Reflection in a C2
transforms every C1 into itself and every C2 into another C2 .
(iv) The limits points are symmetric with respect to each C2 , but not with respect to any other circle.
The following two maps are often useful in constructing conformal isomorphisms.
Proposition 3.3.26. Let H = {z ∈ C | Im z > 0} be the upper half-plane and ∆ = {z ∈ C | |z| < 1} the
unit disk. The maps F : H → ∆ and G : ∆ → H given by
F (z) =
i−z
,
i+z
G(w) = i
1−w
1+w
are conformal isomorphisms and inverses to each other.
Theorem 3.3.27. If f : ∆ → ∆ is a conformal automorphism of the unit disk ∆, then there exist θ ∈ R and
α ∈ ∆ such that
α−z
.
f (z) = eiθ
1 − αz
Corollary 3.3.28. Each conformal automorphism of the upper half-plane H is given by the action of a
matrix in SL(2, R)
3.4. Ahlfors, Chapter 4: Complex Integration.
3.4.1. Fundamental Theorems. A definite integral of any complex-valued function f satisfies
Z
Z
b
b
f (t)dt ≤
|f (t)|dt.
a
a
The line (path) integral of f along a path γ : [a, b] → C is defined as
Z
Z b
f dz =
f (γ(t))γ 0 (t)dt,
γ
a
and it is independent of the parametrization of γ. This has the properties
Z
Z
Z
Z
Z
f dz = − f dz,
f dz =
f dz + · · · +
−γ
We can further define
Z
Z
f dz =
f dz,
γ
γ
γ
γ1 +···+γn
Z
f dx =
γ
1
2
Z
Z
f dz +
γ
γ
f dz ,
γ1
Z
f dy =
γ
f dz.
γn
1
2i
Z
Z
f dz −
γ
γ
f dz .
3. COMPLEX ANALYSIS
19
The integral with respect to arc length is defined as
Z
Z
Z
f ds =
f |dz| =
f (γ(t))|γ 0 (t)|dt,
γ
γ
γ
and it is again independent of parametrization. Among others, it satisfies the following two properties:
Z
Z
Z
Z
|f | · |dz|.
f |dz| = − f |dz|,
f dz ≤
−γ
γ
γ
γ
The length of a path γ is defined as
Z
|dz|.
`(γ) =
γ
R
Theorem 3.4.1. The line integral γ pdx + qdy, defined in U ⊂ C, depends only on the endpoints of γ if
and only if there exists a function F (x, y) on U with partial derivatives ∂F/∂x = p, ∂F/∂y = q.
R
Corollary 3.4.2. The integral γ f dz, with continuous f , depends only on the endpoints of γ if and only
if f is the derivative of a holomorphic function on U .
3.4.2. Cauchy’s Integral Formula.
Definition 3.4.3. Let γ be a piecewise differentiable closed curve in a region Ω, and a a point in Ω \ γ. The
index (also called winding number ) of a point a with respect to a curve γ is given by the equation
Z
dz
1
.
n(γ, a) =
2πi γ z − a
Proposition 3.4.4.
(i)
(ii)
(iii)
(iv)
The index n(γ, a) is an integer.
n(−γ, a) = −n(γ, a)
If γ lies inside a circle, then n(γ, a) = 0 for all points outside of the same circle.
As a function of a the index n(γ, a) is constant in each of the regions determined by γ, and zero in the
unbounded region.
Theorem 3.4.5. Suppose f is holomorphic in an open disk ∆, and let γ be a closed path in ∆. For any
point z ∈ ∆ \ γ,
Z
1
f (ζ)
n(γ, z) · f (z) =
dζ.
2πi γ ζ − z
When γ is a circle centered at a, then n(γ, a) = 1. The resulting equality
Z
1
f (ζ)
dζ
f (z) =
2πi γ ζ − z
is called Cauchy’s integral formula. The following is a generalization which allows us to compute higher
derivatives in this fashion.
Proposition 3.4.6. For any n ≥ 0,
f (n) (z) =
n!
2πi
Z
γ
f (ζ) dζ
.
(ζ − z)n+1
The following two results follow.
Theorem 3.4.7 (Morera’s Theorem). If f is defined and continuous in a region Ω, and if
closed curves γ in Ω, then f is holomorphic in Ω.
Theorem 3.4.8 (Liouville’s Theorem). A bounded entire function is constant.
R
γ
f dz = 0 for all
20
1. NOTES
3.4.3. Local Properties of Holomorphic Functions.
Theorem 3.4.9. Suppose that f is holomorphic in Ω0 = Ω\{a} where Ω is region. A necessary and sufficient
condition that there exist a holomorphic function in Ω which coincides with f in Ω0 is that limz→a (z−a)f (z) =
0. The extended function is uniquely determined.
A recursive application of the previous result yields the following.
Theorem 3.4.10. If f is holomorphic in a region Ω, containing a, it is possible to write
f (z) = f (a) +
f 0 (a)
f 00 (a)
f (n−1) (a)
(z − a) +
(z − a)2 + · · · +
(z − a)n−1 + fn (z)(z − a)n
1!
2!
(n − 1)!
where fn is holomorphic in Ω. Furthermore, inside a circle γ the function fn can be computed as
Z
1
f (ζ)dζ
fn (z) =
.
2πi γ (ζ − a)n (ζ − z)
Proposition 3.4.11. Let f be a holomorphic function on a region Ω and a ∈ Ω. If f (n) (a) = 0 for all n ≥ 0,
then f vanishes identically on Ω.
It follows that if a holomorphic function f is not identically zero, then at least one of its derivatives does
not vanish. Let n ≥ 0 be the smallest integer such that f (n) (a) 6= 0. It follows that we can write f (z) =
(z − a)n fn (z) where fn is holomorphic and fn (a) 6= 0. We say that f has a zero of order n at a. An
immediate consequence is that zeros are isolated points. This can be reformulated as follows.
Proposition 3.4.12 (Analytic continuation). Let f and g be two holomorphic functions in a region Ω. If f
and g agree on a set with an accumulation point, then f = g on all of Ω.
Corollary 3.4.13. A holomorphic function is uniquely determined by its values on any set with an accumulation point in the region of holomorphicity.
Definition 3.4.14. If a function f is holomorphic in a neighbourhood of a point a, except perhaps at a,
then we say f has an isolated singularity at a. If limz→a f (z) = ∞, then we say f has a pole at a.
If f has a pole at a, then the function g = 1/f is holomorphic in a neighbourhood of a. The order of the
pole of f at a is defined as the order of the zero of g at a. In other words, the order of the pole is the unique
integer such that h(z) = (z − a)n f (z) is holomorphic at a and h(a) 6= 0. Even even more detail we could
consider the conditions
(1) limz→a |z − a|α |f (z)| = 0,
(2) limz→a |z − a|α |f (z)| = ∞
for real values α. If (1) holds for some α, then it holds for all α0 ≥ α. Similarly, if (2) holds for some α, then
it holds for all α0 ≤ α. There are three possibilities:
(i) condition (1) holds for all α, hence f vanishes identically;
(ii) there exists some α0 such that (1) holds for α > α0 and (2) for α < α0 ; then f has either a zero or a
pole at a;
(iii) neither (1) nor (2) holds; we say f has an essential singularity at a.
Theorem 3.4.15. A holomorphic function comes arbitrarily close to any complex value in every neighbourhood of an essential singularity.
Definition 3.4.16. A function which is holomorphic in a region Ω with the exception of finitely many poles
is called meromorphic.
Proposition 3.4.17. Let zj be the zeros of a function f which is holomorphic in a disk ∆ and does not
vanish identically, each zero being counted as many times as its order indicates. For every closed curve γ in
3. COMPLEX ANALYSIS
21
∆ which does not pass through a zero
X
j
1
n(γ, zj ) =
2πi
Z
γ
f0
dz,
f
where the sum has only finitely many terms 6= 0.
Theorem 3.4.18. Suppose that f is holomorphic at z0 , f (z0 ) = w0 , and that f (z) − w0 has a zero of order
n at z0 . If ε > 0 is sufficiently small, there exists a corresponding δ > 0 such that for all a with |a − w0 | < δ
the equation f (z) = a has exactly n roots in the disk |z − z0 | < ε.
Corollary 3.4.19. A non-constant holomorphic map is open.
Corollary 3.4.20. If f is holomorphic at z0 with f 0 (z0 ) 6= 0, then it maps a neighbourhood of z0 conformally
and homeomorphically onto a connected open.
Theorem 3.4.21 (The maximum principle). If f is holomorphic and non-constant in a region Ω, then |f |
has no maximum in Ω.
Alternatively, one can restate this as follows.
Corollary 3.4.22. If |f | is defined and continuous on a closed bounded set E and holomorphic on the
interior of E, then the maximum of |f | on E is attained somewhere in ∂E.
Theorem 3.4.23 (Schwartz’s Lemma). If f (z) is holomorphic for |z| < 1 and satisfies the conditions |f (z)| ≤
1, f (0) = 0, then |f (z)| ≤ |z| and |f 0 (0)| ≤ 1. If |f (z)| = |z| for some z 6= 0, or if |f 0 (0)| = 1, then f (z) = cz
where c is a constant satisfying |c| = 1.
Corollary 3.4.24. Every bijective conformal mapping of a disk onto another (or a half plane) is given by
a linear transformation.
3.4.4. The General Form of Cauchy’s Theorem.
Theorem 3.4.25 (Cauchy’s Theorem). If f is holomorphic in a region Ω, then
Z
f dz = 0
γ
for every null-homologous cycle γ in Ω.
Corollary 3.4.26. If Ω is simply-connected, then
R
γ
f dz = 0 for all cycles γ.
Corollary 3.4.27. In a simply-connected region, every holomorphic function has an antiderivative.
Corollary 3.4.28. If f is holomorphic and non-vanishing inp
a simply-connected region, then it is possible
to define single-valued holomorphic branches of log f (z) and n f (z) in that region.
3.4.5. The Calculus of Residues.
Definition 3.4.29. The residue of f at an isolated singularity a is the unique complex number R which
makes f (z) − R/(z − a) the derivative of a single-valued holomorphic function in an annulus 0 < |z − a| < δ.
We denote it by R = resz=a f (z) = resa f .
Theorem 3.4.30. If f has a pole of order n at z0 , then
n−1
1
d
resz0 f = lim
(z − z0 )n f (z).
z→z0 (n − 1)!
dz
Theorem 3.4.31 (Cauchy Residue Theorem). Let f be holomorphic except for isolated singularities aj in a
region Ω. Then
Z
X
1
f dz =
n(γ, aj ) resaj f
2πi γ
j
22
1. NOTES
for any null-homologous cycle γ in Ω which does not pass through any of the points aj .
Corollary 3.4.32 (Argument principle). If f is meromorphic in a region Ω with zeros aj and poles bk , then
Z 0
X
X
1
f
dz =
n(γ, aj ) −
n(γ, bk )
2πi γ f
j
k
for every null-homologous cycle γ in Ω which does not pass through any of the zeros or poles.
Corollary 3.4.33 (Rouché’s Theorem). Let γ be a null-homologous cycle in a region Ω such that n(γ, z)
is either 0 or 1 for any point z ∈ U \ γ. Suppose that f and g are holomorphic in Ω and satisfy |g| < |f | on
γ. Then f and f + g have the same number of zeros enclosed by γ.
The following few paragraph summarize a few applications of Cauchy Residue Theorem to the evaluation of
definite integrals.
1. Integrals of the form
Z
I=
2π
R(cos θ, sin θ)dθ
0
where R is a rational function may be evaluated by
Z
1
z+
I = −i
R
2
|z|=1
the substitution z = eiθ . Then
1
1
1
dz
,
z−
.
z
2i
z
z
Provided there are no poles of R on the unit circle, an application of Cauchy Residue Theorem finishes
the job.
2. Integrals of the form
Z ∞
I=
R(x)dx
−∞
for rational R converge if and only if in the rational function R the degree of the denominator is at least
two units greater than the degree of the numerator, and if no poles lie on the real axis. Integrating along
a semicircular contour whose diameter is along the real axis yields
X
I = 2πi
resz R.
Im z>0
3. The integrals
Z
∞
Z
∞
R(x) cos xdx,
R(x) sin xdx
−∞
−∞
can be evaluated as the real and imaginary parts of
Z ∞
I=
R(x)eix dx.
−∞
Provided R has a zero of at least order two at infinity, an analogous semicircle computation aided by the
Cauchy Residue Theorem yields
X
I = 2πi
resz R(z)eiz .
Im z>0
The statement holds even under the weaker hypothesis that R(∞) = 0, not necessarily of order two or
higher (a rectangular contour is better suited for this case).
3. COMPLEX ANALYSIS
23
3.4.6. Harmonic Functions.
Lemma 3.4.34. All linear functions ax + by are harmonic.
Laplace’s equation
∆u =
∂2u ∂2u
+ 2 =0
∂x2
∂y
in polar coordinates becomes
∂
r
∂r
∂u
r
∂r
+
∂2u
= 0.
∂θ2
Corollary 3.4.35. Any harmonic function which depends only on r must be of the form a log r + b.
Lemma 3.4.36. If u is a harmonic function on a region Ω, then
f (z) =
∂u
∂u
−i
∂x
∂y
is holomorphic on Ω.
Definition 3.4.37. The conjugate differential to
du =
∂u
∂u
dx +
dy
∂x
∂y
is
∗
du = −
∂u
∂u
dx +
dy.
∂y
∂x
In general however, there is no single-valued function v such that dv = ∗ du. It is easy to see that
f dz = du + ∗ du.
Since by Cauchy’s Theorem the integral of f dz vanishes along any null-homologous cycle, and the exact
differential du vanishes along any cycle, it follows that
Z
Z
∂u
∂u
∗
dy
du =
− dx +
∂y
∂x
γ
γ
is zero
R for all null-homologous cycles γ. If the domain over which we are working in is simply-connected,
then γ ∗ du = 0 for all cycles γ, hence there is a well-defined (up to additive constant) single-valued function
v such that dv = ∗ du. The following is an important generalization.
Theorem 3.4.38. If u1 and u2 are harmonic in a region Ω, then
Z
u1 ∗ du2 − u2 ∗ du1 = 0
γ
for all null-homologous cycles γ.
Applying the previous result to u1 = log r, u2 = u, and a cycle γ made up of two counterclockwise oriented
circles, we obtain the following.
Theorem 3.4.39. The arithmetic mean of a harmonic function over concentric circles |z| = r is a linear
function of log r,
Z
1
udθ = α log r + β,
2π |z|=r
and if u is harmonic in a disk then α = 0 and the arithmetic mean is constant.
The following is a useful consequence.
24
1. NOTES
Theorem 3.4.40 (Maximum principle for harmonic functions). A non-constant harmonic function has neither a maximum nor a minimum in its region of definition. Consequently, the maximum and minimum on
a closed bounded set E are taken on the boundary of E.
Note that if f is non-vanishing and holomorphic, then log |f | is a well-defined harmonic function. The
maximum principle for holomorphic functions can be derived as a consequence of the previous result by
applying it to log |f |.
Corollary 3.4.41. A function u continuous on a closed bounded set E and harmonic on its interior is
uniquely determined by its values on ∂E.
Theorem 3.4.42 (Poisson’s formula). Suppose that u is harmonic for |z| < R, continuous for |z| ≤ R. Then
Z
R2 − |a|2
1
u(z)dθ
u(a) =
2π |z|=R |z − a|2
for all |a| < R.
An interesting consequence is that we can express u(z) as the real part of
Z
ζ +z
dζ
1
u(ζ) + iC.
f (z) =
2πi |ζ|=R ζ − z
ζ
Taking the imaginary part yields an explicit formula for the conjugate harmonic function.
Definition 3.4.43. For any piecewise continuous function U (θ) in 0 ≤ θ ≤ 2π, define the Poisson integral
of U as
Z 2π
1
eiθ + z
PU (z) =
U (θ)dθ,
Re iθ
2π 0
e −z
a function of |z| ≤ 1.
It is not hard to see that
PU +V = PU + PV ,
PcU = cPU ,
Pc = c,
and U ≥ 0 implies PU ≥ 0. These properties can be summed by stating P is a positive linear functional.
Furthermore, a ≤ U ≤ b implies a ≤ PU ≤ b.
Theorem 3.4.44 (Schwarz’s Theorem). The function PU (z) is harmonic for |z| < 1, and
lim PU (z) = U (θ0 )
z→eiθ0
provided that U is continuous at θ0 .
The following is a restatement of Poisson’s integral and Schwarz’s Theorem for the half-plane.
Theorem 3.4.45. If U (ξ) is piecewise continuous and bounded for all real ξ, then
Z
y
1 ∞
U (ξ)dξ
PU (z) =
π −∞ (x − ξ)2 + y 2
represents a harmonic function in the upper half-plane with boundary values U (ξ) at points of continuity.
Theorem 3.4.46. Let Ω+ be the part in the upper half-plane of a symmetric region Ω (that is, satisfying
Ω = Ω), and let σ be the real axis in Ω. Suppose v is continuous in Ω+ ∪ σ, harmonic in Ω+ , and zero on
σ. Then v has a harmonic extension to Ω which satisfies the symmetry relation v(z) = −v(z). In the same
situation, if v is the imaginary part of a holomorphic function f in Ω+ (that is, the function f is real on the
real axis), then f has a holomorphic extension which satisfies f (z) = f (z).
3.5. Ahlfors, Chapter 5: Series and Product Developments.
3. COMPLEX ANALYSIS
25
3.5.1. Power Series Expansions.
Theorem 3.5.1. Suppose that fn is holomorphic in the region Ωn , and that the sequence {fn } converges
to a limit function f in a region Ω, uniformly on every compact subset of Ω. Then f is holomorphic in Ω.
Moreover, fn0 converges uniformly to f 0 on every compact subset of Ω.
P
Corollary 3.5.2. If a series of holomorphic terms f =
fn converges uniformly on every compact subset
of a region Ω, then the sum f is holomorphic in Ω, and the series can be differentiated term by term.
Proving uniform convergence can be facilitated by the maximum principle. For example, if fn are holomorphic
in the disk |z| < 1, and if it can be shown that the sequence converges uniformly on each circle |z| = rm for
limm→∞ rm = 1, then the uniform convergence hypothesis follows.
Theorem 3.5.3. If the functions fn are holomorphic and non-vanishing in a region Ω, and if fn converges
to f , uniformly on every compact subset of Ω, then f is either identically zero or non-vanishing in Ω.
Remark. The previous result holds if we replace “non-vanishing” with “has at most m zeros” where m is
a non-negative integer.
Theorem 3.5.4. If f is holomorphic in the region Ω, containing z0 , then the representation
∞
X
f (n) (z0 )
(z − z0 )n
f (z) =
n!
n=0
is valid in the largest open disk of center z0 contained in Ω.
3.5.2. Partial Fractions and Factorization.
Theorem 3.5.5. Let {br } be a sequence of complex numbers with limr→∞ br = ∞, and let Pr (ζ) be polynomials without constant term. Then there are functions which are meromorphic in the whole plane with
poles at the points br , and the corresponding singular parts are Pr (1/(z − br )). Moreover, the most general
meromorphic function of this kind can be written in the form
X 1 f (z) =
Pr
− pr (z) + g(z),
z − br
r
where the pr (z) are suitably chosen polynomials and g(z) is holomorphic in the whole plane.
Using the previous theorem, the following series can be derived.
∞
X
π2
1
=
2
sin πz n=−∞ (z − n)2
∞
1 X
1
1
1 X 2z
π cot πz = +
+
= +
z
z−n n
z n=1 z 2 − n2
n6=0
m
X
π
(−1)n
= lim
sin πz m→∞ n=−m z − n
Q∞
Definition 3.5.6. An infinite product n=1 pn is said to converge if only at most a finite number of the
factors are zero, and if the partial products formed by the non-vanishing factors tend to a finite limit which
is different from zero.
Q
If pn converges, then it is clear that limn→∞ pn = 1. Therefore, it is customary to write an infinite product
as
∞
Y
(1 + an ).
n=1
A necessary condition for convergence is then limn→∞ an = 0.
26
Theorem 3.5.7. The infinite product
1. NOTES
Q
(1 + an ) with 1 + an 6= 0 converges if and only if so does the series
∞
X
log(1 + an ).
n=1
whose terms represent the values of the principal branch of the logarithm.
Q
P
Definition 3.5.8. We say the product (1 + an ) converges absolutely if so does the series
log(1 + an ).
Q
Theorem 3.5.9. A necessary
and sufficient condition for the absolute convergence of (1 + an ) is the
P
absolute convergence of
|an |.
The previous result has a counterpart for uniform convergence if we allow to replace an by functions fn .
This requires a slightly more technical statement if we allow zeros of these functions. More precisely, we
consider only sets on which only finitely many of the factors can vanish. If these factors are omitted, then
it is sufficient to study the uniform convergence of the remaining product.
Proposition 3.5.10. The value of an absolutely convergent product does not change if the factors are
reordered.
Definition 3.5.11. A function which is holomorphic in the entire complex plane is called entire.
Proposition 3.5.12. Every entire function can be expressed as the derivative of some entire function.
Proposition 3.5.13. Every non-vanishing entire function is of the form ef where f is an entire function.
It follows that every entire function f with finitely many zeros can be expressed in the form
N Y
z
.
f (z) = z m eg(z)
1−
an
n=1
If there are infinitely many zeros, then the obvious generalization is
∞ Y
z
f (z) = z m eg(z)
.
1−
an
n=1
This is a valid expression only if the infinite product converges uniformly on every compact set. If this is so,
then the product represents an entire function with zeros at the same points, and with the same multiplicities
as f (z).
Q
P
Proposition 3.5.14. The product (1 − z/an ) converges absolutely if and only if
1/|an | is convergent,
and in this case the convergence is uniform in every compact set.
In general, convergence may not be attained. To meet the needs, convergence-producing factors must be
introduced.
Theorem 3.5.15. There exists an entire function with arbitrarily prescribed zeros an provided that, in the
case of infinitely many zeros, an → ∞. Every entire function with these and no other zeros can be written
in the form
2
mn !
∞ Y
z
z
1 z
1
z
m g(z)
f (z) = z e
1−
exp
+
+ ··· +
a
a
2
a
m
a
n
n
n
n
n
n=1
where the product is taken over all an 6= 0, then mn are certain integers, and g(z) is an entire function.
Furthermore,
we can choose all mn to be equal to some fixed integer h. Pick h as the smallest integer for
P
which
1/|an |h+1 converges. Then the infinite product above is called the canonical product associated
with the sequence {an }, and the integer h is called the genus of the canonical product.
Whenever possible, we attempt to work with canonical products. If in this presentation g is a polynomial,
then f is said to have finite genus max{deg g, h}.
3. COMPLEX ANALYSIS
27
Corollary 3.5.16. Every function which is meromorphic in the whole plane is the fraction of two entire
functions.
Using the previous theorem, the following infinite product can be derived.
∞ Y
Y
z z/n
z2
1−
sin πz = πz
e
= πz
1− 2
n
n
n=1
n6=0
Consider the function
∞ Y
G(z) =
1+
n=1
z −z/n
e
n
which has zeros at all negative integers. It is evident G(−z) has the positive integers for zeros. Combining
these facts with the infinite product expansion of sin πz given above, we obtain
zG(z)G(−z) =
sin πz
.
π
By comparing G(z) and G(z − 1), it can be shown that
G(z − 1) = zeγ G(z)
where
γ = lim
n→∞
1
1 1
1 + + + · · · + − log n
2 3
n
is called Euler’s constant. The function
Γ(z) =
e−γz
zG(z)
is called Euler’s gamma function and satisfies
Γ(z + 1) = zΓ(z).
A more explicit representation is
Γ(z) =
∞
e−γz Y z −1 z/n
e ,
1+
z n=1
n
and one of the previous formulas takes the form
π
.
sin πz
Proposition 3.5.17. The function Γ is meromorphic with simple poles at 0, −1, −2, . . . and no zeros. For
any n ≥ 0,
(−1)n
res−n Γ =
.
n!
Proposition 3.5.18. For all z with Re z > 0,
Z ∞
Γ(z) =
tz−1 e−t dt.
Γ(z)Γ(1 − z) =
0
A few notable values are
Γ(1) = Γ(2) = 1,
Γ(1/2) =
√
π.
Combining the first equality with the functional equation we deduce the following.
Proposition 3.5.19. For all positive integers n,
Γ(n) = (n − 1)!.
28
1. NOTES
Theorem 3.5.20 (Legendre’s duplication formula). The equality
√
1
π Γ(2z) = 22z−1 Γ(z)Γ z +
2
holds for all z ∈ C.
Theorem 3.5.21 (Stirling’s formula). There exists an entire function J which tends to 0 as z → ∞ in a half
plane Re z ≥ c > 0 satisfying
r
2π z z J(z)
Γ(z) =
e
.
z e
Therefore
n!
n = 1.
lim √
n→∞
2πn ne
3.5.3. Entire Functions.
Theorem 3.5.22 (Jensen’s formula). Let f be an entire function not vanishing at the origin. Pick some
ρ > 0 and let a1 , . . . , an be the zeros of f in the closed disk |z| ≤ ρ, multiple zeros being repeated. Then
Z 2π
n
X
ρ
1
log |f (0)| = −
log
+
log |f (ρeiθ )|dθ.
|a
|
2π
i
0
i=1
Definition 3.5.23. Let f be an entire function with zeros {an } satisfying f (0) 6= 0. Let M (r) be the
maximum of |f (z)| on |z| = r. The order of the entire function f is defined as
λ = lim sup
r→∞
log log M (r)
.
log r
According to the definition λ is the smallest number such that
M (r) ≤ er
λ+ε
for any given ε > 0 as soon as r is sufficiently large.
Theorem 3.5.24 (Hadamard’s Theorem). The genus and the order of an entire function satisfy the inequality
h ≤ λ ≤ h + 1.
Corollary 3.5.25. An entire function of non-integer order assumes every finite value infinitely many times.
P −σ
3.5.4. The Riemann Zeta Function. Since the series
n
converges uniformly for all real σ ≥ σ0 for
any fixed σ0 > 0, it follows that the Riemann zeta function
∞
X
1
ζ(s) =
s
n
n=1
is holomorphic in the half-plane Re s > 1.
Theorem 3.5.26. If Re s > 1, then
ζ(s) =
Y
p
1
.
1 − p−s
The following allows us to extend ζ to the entire complex plane.
Theorem 3.5.27. If Re s > 1, then
Γ(1 − s)
ζ(s) = −
2πi
s−1
Z
C
(−z)s−1
dz
ez − 1
where (−z)
is defined on the complement of the positive real axis as e(s−1) log(−z) with −π < Im log(−z) <
π, and C is a path traversing above and below the positive real axis.
3. COMPLEX ANALYSIS
29
Corollary 3.5.28. The function ζ can be extended to a meromorphic function in the whole complex plane
whose only pole is a simple at s = 1 with residue 1.
Theorem 3.5.29. If n is a non-negative integer, then
Z −n−1
(−1)n n!
z
dz.
z
2πi
C e −1
In particular, ζ vanishes at all even negative integers, at the odd ones it is given by
(−1)m Bm
,
ζ(−2m + 1) =
2m
and
1
ζ(0) = − .
2
Theorem 3.5.30 (Functional equation for the Riemann zeta function).
πs
ζ(s) = 2s π s−1 sin Γ(1 − s)ζ(1 − s).
2
Corollary 3.5.31. The function
1
ξ(s) = s(1 − s)π −s/2 Γ(s/2)ζ(s)
2
is entire and satisfies ξ(s) = ξ(1 − s).
ζ(−n) =
Proposition 3.5.32. The order of ξ is 1.
3.5.5. Normal Families. In what follows, a family of functions F would refer to a set of functions defined
in a fixed region Ω ⊂ C, and with values in a metric space (S, d).
Definition 3.5.33. The functions in a family F are said to be equicontinuous on a set E ⊂ Ω if for each
ε > 0, there exists δ > 0, such that for all f ∈ F, |z − z0 | < δ implies d(f (z), f (z0 )) < ε for all z0 , z ∈ E.
Note that each function f in an equicontinuous family F on E is itself uniformly continuous on E.
Definition 3.5.34. A family F is said to be normal in Ω if every sequence of functions {fn } ⊂ F contains
a subsequence which converges uniformly on every compact subset of Ω.
Note that the above definition does not require the limit functions of convergent subsequences to lie in F.
Definition
S 3.5.35. An exhaustion of a region Ω is an increasing sequence of compact subspaces Ek ⊂ Ω
such that k Ek = Ω satisfying the following condition: for every compact subset E ⊂ Ω there exists some
k such that E ⊂ Ek .
It can be shown that exhaustions always exists for regions in the complex plane. Our aim is to define a
metric on the space of functions Ω → S such that convergence in this metric would be unanimous with
uniform convergence on all compact subsets E ⊂ Ω. We start by defining a new metric δ on S given by
d(a, b)
.
δ(a, b) =
1 + d(a, b)
Then for each two functions f, g : Ω → S we define
δk (f, g) = sup δ(f (x), g(x))
x∈Ek
which may be regarded as the distance between f and g on Ek . Finally, we set
∞
X
ρ(f, g) =
δk (f, g)2−k .
k=1
Proposition 3.5.36. The so defined function ρ is a metric on the space of functions Ω → S. Furthermore,
if S is a complete metric space, then so is the space of functions endowed with the metric ρ.
30
1. NOTES
Proposition 3.5.37. A sequence of functions {fn } converges with respect to ρ if and only if it converges
uniformly on every compact subset E ⊂ Ω.
An application of the Bolzano-Weierstrass Theorem to the metric space F endowed with the restriction of ρ
yields the following.
Theorem 3.5.38. A family F is normal if and only if its closure F with respect to ρ is compact.
If F is compact, it is customary to say F is relatively compact.
Theorem 3.5.39. If S is complete, then F is normal if and only if it is totally bounded.
The following is a restatement using the original metric d.
Theorem 3.5.40. A family F is totally bounded if and only if to every compact set E ⊂ Ω and every ε > 0
it is possible to find f1 , . . . , fn ∈ F such that every f ∈ F satisfies d(f, fi ) < ε on E for some fi .
Theorem 3.5.41 (Arzela’s Theorem). A family F of continuous functions with values in a metric space S
is normal in the region Ω ⊂ C if and only if:
(i) F is equicontinuous on every compact set E ⊂ Ω;
(ii) for any z ∈ Ω the set {f (z) | f ∈ F} lies in a compact subset of S.
Theorem 3.5.42. A family of holomorphic functions is normal with respect to C if and only if the functions
in F are uniformly bounded on every compact set.
If a family of holomorphic functions satisfies the conditions of the previous result, we say it is locally bounded.
Theorem 3.5.43. A locally bounded family of holomorphic functions has locally bounded derivatives.
Lemma 3.5.44. If a sequence of meromorphic functions converges in the sense of CP1 , uniformly on every
compact set, then the limit functions is meromorphic or identically equal to ∞. If a sequence of holomorphic
functions converges in the same sense, then the limit function is either holomorphic or identically equal to
∞.
Theorem 3.5.45. A family of holomorphic or meromorphic functions {f } is normal in the spherical sense
if and only if the expressions
ρ(f ) =
2|f 0 (z)|
1 + |f (z)|2
are locally bounded.
3.6. Ahlfors, Chapter 6: Conformal Mapping. Dirichlet’s Problem.
3.6.1. The Riemann Mapping Theorem.
Theorem 3.6.1 (Riemann Mapping Theorem). Given any simply-connected region Ω which is not the
whole plane, and a point z0 ∈ Ω, there exists a unique holomorphic function f : Ω → C, normalized by the
conditions f (z0 ) = 0, f 0 (z0 ) > 0, such that f is bijective from Ω onto the unit disk |w| < 1.
Theorem 3.6.2. Let f : Ω → Ω0 be a homeomorphism between two regions. If {zn } or z(t) tends to the
boundary of Ω, then {f (zn )} or f ◦ z tends to the boundary of Ω0 .
Theorem 3.6.3. Suppose that the boundary of a simply-connected region Ω contains an analytic arc γ as
a one-sided free boundary arc. Then the function f which maps Ω onto the unit disk can be extended to a
function which is holomorphic and bijective onto Ω ∪ γ. The image of γ is an arc γ 0 on the unit circle.
3.6.2. Conformal Mappings of Polygons.
4. ALGEBRAIC TOPOLOGY
31
Theorem 3.6.4. The functions F which map |w| < 1 conformally onto polygons with angles αk π, k =
1, . . . , n, are of the form
Z wY
n
F (w) = C
(w − wk )−βk dw + C 0
0
k=1
where βk = 1 − αk , the wk are points on the unit circle, and C,C 0 are complex constants.
TODO: Read about infinite products in Stein & Shakarchi
4. Algebraic Topology
Syllabus
Undergraduate: Hatcher, Algebraic Topology, chapter 1 (but not the additional topics). (math 131)
Graduate: Hatcher, Algebraic Topology, chapter 2 (including additional topics) and chapter 3 (without additional topics). (math 231a)
4.1. Hatcher, Chapter 0: Some Underlying Geometric Notions.
Definition 4.1.1. The join of two topological spaces X and Y , denoted X ∗ Y , is the quotient of X × Y × I
under the identifications (x, y1 , 0) ∼ (x, y2 , 0) and (x1 , y, 1) ∼ (x2 , y, 1). In other words, this amount to
collapsing X × Y × {0} to X and X × Y × {1} to Y . On can also think of the join as the set of formal linear
combinations t1 x + t2 y for x ∈ X, y ∈ Y , t1 , t2 ∈ R satisfying t1 + t2 = 1. Alternatively, one may think of it
as the collection of all line segments joining points in X with points in Y .
The join is an associative operation. Two useful examples are:
(i) the n-fold join of the one point space is the (n − 1)-simplex ∆n−1 ;
(ii) the n-fold join of the two point space S 0 is the (n − 1)-sphere S n−1 .
Proposition 4.1.2. If (X, A) is a CW pair and A is a contractible subcomplex, then the quotient map
X → X/A is a homotopy equivalence.
Compare this with the following.
Proposition 4.1.3. If (X, A) is a CW pair and A is a contractible in X, that is, the inclusion A ,→ X is
homotopic to the constant map, then X/A ' X ∨ SA.
Proposition 4.1.4. If (X1 , A) is a CW pair and the two attaching maps f, g : A → X0 are homotopic, then
X0 tf X1 ' X0 tg X1 .
A pair of topological spaces (X, A) is said to have the homotopy extension property if given a map f0 : X → Y
and a homotopy gt : A → Y for 0 ≤ t ≤ 1 satisfying f0 |A = g0 , then there exists a homotopy ft : X → A for
0 ≤ t ≤ 1 satisfying ft |A = gt for all t.
Proposition 4.1.5. A pair (X, A) has the homotopy extension property if and only if X × {0} ∪ A × I is a
deformation retract of X × I.
Proposition 4.1.6. All CW pairs (X, A) have the homotopy extension property.
Proposition 4.1.7. Suppose the pairs (X, A) and (Y, A) satisfy the homotopy extension property, and
f : X → Y is a homotopy equivalence with f |A = idA . Then f is a homotopy equivalence rel A.
Corollary 4.1.8. If (X, A) satisfies the homotopy extension property and the inclusion A ,→ X is a
homotopy equivalence, then A is a deformation retract of X.
Corollary 4.1.9. A map f : X → Y is a homotopy equivalence if and only if X is a deformation retract
of the mapping cylinder Mf . Hence, two spaces X and Y are homotopy equivalent if and only if there is a
third space containing both X and Y as deformation retracts.
4.2. Hatcher, Chapter 1: The Fundamental Group.
32
1. NOTES
4.2.1. Basic Constructions.
Proposition 4.2.1. Let X be a topological space, and h a path from x0 to x1 in X. Then the map
βh : π1 (X, x1 ) → π1 (X, x0 ) given by βh [f ] = [h · f · h] is an isomorphism.
Theorem 4.2.2 (Brouwer Fixed Point Theorem). Every continuous map h : D2 → D2 has a fixed point.
Theorem 4.2.3 (Borsuk-Ulam). For every continuous map f : S 2 → R2 there exists a pair of antipodal
points x and −x with f (x) = f (−x).
Corollary 4.2.4. There is no injective continuous map S 2 → R2 , and hence S 2 is not homeomorphic to
any subset of R2 .
Corollary 4.2.5. Whenever S 2 is expressed as the union of three closed sets A1 , A2 , and A3 , then at least
one of these sets must contain a pair of antipodal points {x, −x}.
Proposition 4.2.6. π1 (X × Y, (x0 , y0 )) ∼
= π1 (X, x0 ) × π1 (Y, y0 )
Proposition 4.2.7. If n ≥ 2, then π1 (S n ) = 0.
Proposition 4.2.8. If a space X retracts onto a space A, then the homomorphism i∗ : π1 (A, x0 ) → π1 (X, x0 )
induced by the inclusion i : A → X is injective. If A is a deformation retract of X, then i∗ is an isomorphism.
Proposition 4.2.9. If ϕ : X → Y is a homotopy equivalence, then the induced homomorphism ϕ∗ : π1 (X, x0 ) →
π1 (Y, ϕ(x0 )) is an isomorphism for all x0 ∈ X.
4.2.2. Van Kampen’s Theorem.
Theorem 4.2.10 (Seifert–van Kampen). If X is the union of path-connected open sets Aα each containing `
the basepoint x0 ∈ X and if each intersection Aα ∩ Aβ is path-connected, then the homomorphism
Φ : α π1 (Aα ) → π1 (X) is surjective. If in addition each intersection Aα ∩ Aβ ∩ Aγ is path-connected,
then the kernel of Φ is the normal subgroup N generated by all`
elements of the form iαβ (ω)iβα (ω)−1 for
∼
ω ∈ π1 (Aα ∩ Aβ ), and hence Φ induces an isomorphism π1 (X) = α π1 (Aα )/N .
Proposition 4.2.11. Suppose we attach a collection of 2-cells e2α to a path-connected space X via maps
ϕα : S 1 → X, producing a space Y . Fix basepoints s0 ∈ S 1 and x0 ∈ X. Choose a path γα from x0 to
ϕ(s0 ) for each α so that γα ϕα γα is a loop in X based at x0 . Then the inclusion X ,→ Y induces a surjection
π1 (X, x0 ) → π1 (Y, x0 ) whose kernel is generated by the loops γα ϕα γα .
Corollary 4.2.12. For every group G there is a 2-dimensional CW complex XG with π1 (XG ) ∼
= G.
For Σg = (T 2 )#g and Ng = (RP2 )#g , we have
−1
−1 −1
π1 (Σg ) = ha1 , b1 , . . . , ag , bg | a1 b1 a−1
1 b1 · · · ag bg ag bg i,
π1 (Ng ) = ha1 , . . . , ag | a21 · · · a2g i.
Corollary 4.2.13. The surfaces Σg and Σh are homotopy equivalent if and only if g = h.
4.2.3. Covering Spaces.
e → X, a homotopy ft : Y →
Proposition 4.2.14 (Homotopy lifting property). Given a covering space p : X
e
e
e
e of fe0 that lifts ft .
X, and a map f0 : Y → X lifting f0 , then there exists a unique homotopy ft : Y → X
When Y is a point, the previous results is also known as the path lifting property. Concretely, for each path
e lifting f and
f : I → X and each lift x
f0 of the starting point x0 = f (0), there is a unique path fe: I → X
starting at x
f0 .
e x
e x
Proposition 4.2.15. The map p∗ : π1 (X,
f0 ) → π1 (X, x0 ) induced by a covering p : (X,
f0 ) → (X, x0 ) is
e x
injective. The image subgroup p∗ (π1 (X,
f0 )) in π1 (X, x0 ) consists of the homotopy classes of loops in X
e starting at x
based at x0 whose lifts to X
f0 are loops.
4. ALGEBRAIC TOPOLOGY
33
e x
e pathProposition 4.2.16. The number of sheets of a covering space p : (X,
f0 ) → (X, x0 ) with X and X
e
connected equals the index of p∗ (π1 (X, x
f0 )) in π1 (X, x0 ).
e x
Proposition 4.2.17 (Lifting criterion). Let p : (X,
f0 ) → (X, x0 ) be a covering space and f : (Y, y0 ) →
e x
(X, x0 ) a map with Y path-connected and locally path-connected. Then a lift fe: (Y, y0 ) → (X,
f0 ) of f
e x
exists if and only if f∗ (π1 (Y, y0 )) ⊂ p∗ (π1 (X,
f0 )).
e → X and a map f : Y → X, if
Proposition 4.2.18 (Unique lifting property). Given a covering space p : X
e of f agree at one point of Y and Y is connected, then fe1 = fe2 .
two lifts fe1 , fe2 : Y → X
Proposition 4.2.19. Suppose X is a path-connected, locally path-connected, and semilocally simplyconnected. Then for every subgroup H ⊂ π1 (X, x0 ) there is a covering space p : XH → X such that
p∗ (π1 (XH , x
f0 )) = H for a suitably chosen basepoint x
f0 ∈ XH .
Corollary 4.2.20. Any topological space which is path-connected, locally-path connected, and semilocally
simply-connected admits a universal cover.
Proposition 4.2.21. If X is path-connected and locally path-connected, then two path-connected covering
f1 → X and p2 : X
f2 → X are isomorphic via an isomorphism f : X
f1 → X
f2 taking a basepoint
spaces p1 : X
−1
−1
f1 , x
f2 , x
x
f1 ∈ p1 (x0 ) to a basepoint x
f2 ∈ p2 (x0 ) if and only if (p1 )∗ (π1 (X
f1 )) = (p2 )∗ (π1 (X
f2 )).
Theorem 4.2.22. Let X be path-connected, locally path-connected, and semilocally path-connected. Then
there is a bijection between the set of basepoint-preserving isomorphism classes of path-connected covering
e x
spaces p : (X,
f0 ) → (X, x0 ) and the set of subgroups of π1 (X, x0 ), obtained by associating the subgroup
e
e x
p∗ (π1 (X, x
f0 )) to the covering space (X,
f0 ). If basepoints are ignored, this correspondence gives a bijection
e → X and conjugacy classes of
between the isomorphism classes of path-connected covering spaces p : X
subgroups of π1 (X, x0 ).
Theorem 4.2.23. Let X be a topological space satisfying the hypothesis of the previous result. The set
of n-sheeted covering spaces (not necessarily connected) of X endowed with basepoints is in bijection with
the set of homomorphisms π1 (X, x0 ) → Sn , where Sn stands for the symmetric group on n symbols. If we
drop basepoints then there is a bijection with the set of homomorphisms up to conjugation, that is, two
homomorphisms ϕ1 , ϕ2 : π1 (X, x0 ) → Sn correspond to equivalent covers if and only if ϕ1 ◦ ϕ−1
2 is an inner
automorphism of Sn .
e → X, an automorphism of covers X
e →X
e is called a deck
Definition 4.2.24. For a covering space p : X
e
transformation. These form a group denoted Aut(X/X).
e → X is called normal if for each x1 , x2 ∈ X
e satisfying p(x1 ) =
Definition 4.2.25. A covering space p : X
e
p(x2 ) there exists a deck transformation ϕ ∈ Aut(X/X) such that ϕ(x1 ) = x2 .
e x
Proposition 4.2.26. Let p : (X,
f0 ) → (X, x0 ) be a path-connected covering space of a path-connected,
e x
locally path-connected space X, and let H be the subgroup p∗ (π1 (X,
f0 )) ⊂ π1 (X, x0 ). Then:
(a) this covering space is normal if and only if H is a normal subgroup of π1 (X, x0 );
e
(b) Aut(X/X)
is isomorphic to the quotient N (H)/H where N (H) is the normalizer of H in π1 (X, x0 ).
e
e is a normal covering. Hence for the universal
In particular, Aut(X/X)
is isomorphic to π1 (X, x0 )/H if X
∼ π1 (X).
e → X we have Aut(X/X)
e
cover X
=
Definition 4.2.27. A group action of a group G on a topological space Y is called a covering space action if
the following condition holds: each y ∈ Y has a neighbourhood U such that all the images g(U ) for varying
g ∈ G are disjoint. In other words, g1 (U ) ∩ g2 (U ) 6= ∅ implies g1 = g2 .
e → X the action of Aut(X/X)
e
e is a covering space action.
Note that for each covering X
on X
Proposition 4.2.28. Each covering action of a group G on a space Y satisfies the following:
34
1. NOTES
(a) the quotient map p : Y → Y /G, p(y) = Gy, is a normal covering space;
(b) if Y is path-connected, then G is the group of deck transformations of this covering space Y → Y /G;
(c) if Y is path-connected and locally path-connected, then G is isomorphic to π1 (Y /G)/p∗ (π1 (Y )).
For each m, n ≥ 1 there is a covering space Σmn+1 → Σm+1 . Conversely, if there is a covering Σg → Σh
then g = mn + 1 and h = m + 1 for some m, n ≥ 1.
Proposition 4.2.29. Consider maps X → Y → Z such that both Y → Z and the composition X → Z are
covering spaces. If Z is locally path-connected, then X → Y is a covering space. Furthermore, if X → Z is
normal, then so is X → Y .
Proposition 4.2.30. Consider a covering action of a group G on a path-connected, locally path-connected
space X. Then any subgroup H ⊂ G determines a composition of covering spaces X → X/H → X/G.
Furthermore, the following properties hold.
(a) Every path-connected covering space between X and X/G is isomorphic to X/H for some subgroup
H ⊂ G.
(b) Two such covering spaces X/H1 and X/H2 of X/G are isomorphic if and only if H1 and H2 are conjugate
subgroups of G.
(c) The covering space X/H → X/G is normal if and only if H is a normal subgroup of G, in which case
the group of deck transformations of this cover is G/H.
4.3. Hatcher, Chapter 2: Homology.
4.3.1. Simplicial and Singular Homology.
Proposition
4.3.1. Consider
the decomposition of a topological space X into its path-components X =
F
∼ L H• (Xα ).
α Xα . Then H• (X) =
α
Theorem 4.3.2 (Exact sequence of a pair). For every pair (X, A), we have a long exact sequence
···
/ Hn (X)
/ Hn (A)
/ Hn (X, A)
/ Hn−1 (A)
∂
/ ··· .
The connecting homomorphism ∂ : Hn (X, A) → Hn−1 (A) has a simple description: if a class [α] ∈ Hn (X, A)
is represented by a relative cycle α, then ∂[α] = [∂α] ∈ Hn−1 (A).
The following is a mild generalization.
Theorem 4.3.3 (Exact sequence of a triple). For every triple (X, A, B), we have a long exact sequence
/ Hn (A, B)
···
/ Hn (X, B)
/ Hn (X, A)
∂
/ Hn−1 (A, B)
/ ··· .
Theorem 4.3.4 (Excision). Given subspaces Z ⊂ A ⊂ X such that the closure of Z is contained in the
interior of A, then the inclusion (X \ Z, A \ Z) ,→ (X, A) induces isomorphisms Hn (X \ Z, A \ Z) → Hn (X, A)
for all n. Equivalently, for subspaces A, B ⊂ X whose interiors cover X, the inclusion (B, A ∩ B) ,→ (X, A)
induces isomorphisms Hn (B, A ∩ B) → Hn (X, A) for all n.
We call a pair (X, A) good if A is a nonempty closed subspace and it is the deformation retract of a
neighbourhood.
Proposition 4.3.5. For good pairs (X, A), the quotient map q : (X, A) → (X/A, A/A) induces isomorphisms
e n (X/A) for all n.
q∗ : Hn (X, A) → Hn (X/A, A/A) ∼
=H
Theorem 4.3.6. If (X, A) is a good pair, then there is an exact sequence
···
e n (A)
/H
e n (X)
/H
e n (X/A)
/H
∂
e n−1 (A)
/H
/ ··· .
Corollary 4.3.7. If the CW complex X is the union of subcomplexes A and B, then the inclusion (B, A ∩
B) ,→ (X, A) induces isomorphisms Hn (B, A ∩ B) → Hn (X, A) for all n.
4. ALGEBRAIC TOPOLOGY
W
Corollary 4.3.8. For a wedge sum
α
35
Xα , the inclusions iα : Xα →
W
α
Xα induce an isomorphism
!
M
(iα )∗ :
α
M
e • (Xα ) → H
e•
H
α
_
Xα
,
α
provided that the wedge sum is formed at basepoints xα ∈ Xα such that the pairs (Xα , xα ) are good.
Theorem 4.3.9. If nonempty subsets U ⊂ Rm and V ⊂ Rn are homeomorphic, then m = n.
Proposition 4.3.10. If A is a retract of X, then the maps Hn (A) → Hn (X) induced by the inclusion
A ,→ X are injective.
Corollary 4.3.11. There exists no retraction Dn → S n .
Corollary 4.3.12 (Brouwer Fixed Point Theorem). Every continuous map h : Dn → Dn has a fixed point.
e n (X) ∼
e n+1 (SX).
Proposition 4.3.13. For all n, there are isomorphisms H
=H
Proposition 4.3.14. Let X be a finite-dimensional CW complex.
(a) If X has dimension n, then Hi (X) = 0 for i > n and Hn (X) is free.
(b) If there are no cells of dimension n − 1 or n + 1, then Hn (X) is free with basis in bijective correspondence
with the n-cells.
(c) If X has k n-cells, then Hn (X) is generated by at most k elements.
4.3.2. Computations and Applications.
Definition 4.3.15. Each map f : S n → S n induces a homomorphism f∗ : Hn (S n ) → Hn (S n ) which is
multiplication by an integer d called the degree of f denoted deg f .
Proposition 4.3.16.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
deg idS n = 1.
If f is not surjective, then deg f = 0.
If f ' g, then deg f = deg g.
deg(f ◦ g) = deg f deg g.
If f is the reflection in S n−1 of S n , then deg f = −1.
The antipodal map has degree (−1)n+1 .
If f : S n → S n has no fixed points, then deg f = (−1)n+1 .
If Sf : S n+1 → S n+1 denotes the suspension map of f : S n → S n , then deg Sf = deg f .
Theorem 4.3.17. A continuous nonvanishing vector field on S n exists if and only if n is odd.
Proposition 4.3.18. If n is even, then Z/2 is the only nontrivial group that can act freely on S n .
Proposition 4.3.19. Let f : S n → S n be a continuous map, and y ∈ S n be a point whose preimage is
finite, say f −1 (y) = {x1 , . . . , xm }. Let U1 , . . . , Um be disjoint neighbourhoods of the xi mapped homeomorphically to a neighbourhood V of y. The local degree of f at xi , denoted deg P
f |xi , is an integer d such that
f∗ : Hn (Ui , Ui \ {xi }) → Hn (V, V \ {y}) is multiplication by d. Then deg f = i deg f |xi .
Proposition 4.3.20. Let X be a CW complex.
(a) H• (X n , X n−1 ) ∼
= Z`(n) where ` is the number of n-cells of X (potentially infinite).
n
(b) Hk (X ) = 0 if k > n. In particular, if X is finite dimensional, then Hk (X) = 0 for k > dim X.
(c) The inclusion i : X n ,→ X induces isomorphisms i∗ : Hk (X n ) → Hk (X) for k < n.
There is an alternative formulation of homology which makes it easy to compute. For any CW complex X
there is a chain complex
···
/ Hn+1 (X n+1 , X n )
dn+1
/ Hn (X n , X n−1 )
dn
/ Hn−1 (X n−1 , X n−2 )
/ ··· .
36
1. NOTES
The homology of this complex, denoted HnCW (X), is called cellular homology. Note that Hn (X n , X n−1 )
is
abelian group generated by the n-cells of X. The differentials dn can be computed by dn (enα ) =
Pa free n−1
where dαβ is the degree of the map Sαn−1 → X n−1 → Sβn−1 that is the composition of the
β dαβ eβ
attaching map of enα with the quotient map collapsing X n−1 \ en−1
to a point.
β
The following computations follow.
2g
H• (Σg ) ∼
= Z(0) ⊕ Z(1) ⊕ Z
H• (Ng ) ∼
= Z(0) ⊕ (Zg−1 ⊕ Z/2)(1)


if k = 0 and if k = n is odd,
Z
Hk (RPn ) ∼
= Z/2 if k is odd and 0 < k < n,


0
otherwise.


if k = 0 or 2n − 1,
Z
∼
Hk (Lm (`1 , . . . , `n )) = Z/m if k is odd and 0 < k < 2n − 1,


0
otherwise.
Theorem 4.3.21. H•CW (X) ∼
= H• (X).
e • (X) ∼
For an abelian group G and an integer n ≥ 1, we can construct a CW complex X satisfying H
= G(n) .
Furthermore, it can be shown that the homotopy type of X is uniquely determined by the previous condition
(provided that for n > 1, we require X to be simply-connected), hence we will refer to X as a Moore space
and denote it by M (G, n).
Theorem 4.3.22. For finite CW complexes X, the Euler characteristic is
X
X
χ(X) =
(−1)n rank Hn (X n , X n−1 ) =
(−1)n rank Hn (X).
n
n
For example,
χ(Σg ) = 2 − 2g,
χ(Ng ) = 2 − g.
Suppose r : X → A is a retraction and i : A → X the associated inclusion.
i∗ : H• (A) → H• (X) is injective, hence the long exact sequence of the pair
sequences
/ Hn (X)
/ Hn (X, A)
/ Hn (A)
0
We have already shown that
(X, A) splits into short exact
/ 0.
Furthermore, the relation r∗ i∗ = idH• (A) implies the above short exact sequences are split.
Proposition 4.3.23 (Split short exact sequences). For a short exact sequence
0
/A
/B
i
/C
j
/0
of abelian groups the following statements are equivalent:
(a) there is a homomorphism p : B → A such that p ◦ i = idA ;
(b) there is a homomorphism s : C → B such that j ◦ s = idC ;
(c) there is an isomorphism B ∼
= A ⊕ C making the diagram
6B
i
0
/A
(
∼
=
A⊕C
j
(
6C
/0
commute, where the maps in the lower row are the obvious projections.
4. ALGEBRAIC TOPOLOGY
37
Theorem 4.3.24 (Mayer-Vietoris sequence). Let X be a topological space and A, B ⊂ X two subspaces
such that their interiors cover X. Then there is a long exact sequence as follows.
/ Hn (A ∩ B)
···
/ Hn (A) ⊕ Hn (B)
/ Hn (X)
∂
/ Hn−1 (A ∩ B)
/ ···
The connecting homomorphism ∂ : Hn (X) → Hn−1 (A ∩ B) has the following explicit form. Consider a class
α ∈ Hn (X) represented by a cycle z. By further subdivision we can ensure that z = x + y such that x lies
in A and y in B. Note that x and y need not be cycles themselves, but ∂x = −∂y. Then ∂α ∈ Hn−1 (A ∩ B)
is represented by a the cycle ∂x = −∂y.
There is also a similar sequence in reduced homology. Even further, we can take A and B to be deformation
retracts of neighbourhoods U and V respectively satisfying X = U ∩ V . This is particularly useful when X
is a CW complex and A and B subcomplexes since U and V as described always exist. The following can
often be viewed as a generalization of the Mayer-Vietoris sequence.
Theorem 4.3.25. Consider two maps f, g : X → Y and form the space Z = (X × I t Y )/∼ by identifying
(x, 0) ∼ f (x) and (x, 1) ∼ g(x) for all x ∈ X. Less formally, we can describe Z as X × I glued to Y at
one end via f and at the other via g. Let i : Y ,→ Z be the evident inclusion. Then there is a long exact
sequence as follows.
/ Hn (X)
···
f∗ −g∗
/ Hn (Y )
/ Hn (Z)
i∗
∂
/ Hn−1 (X)
/ ···
Theorem 4.3.26 (Relative Mayer-Vietoris sequence). Let (X, Y ) = (A ∪ B, C ∪ D) such that: (1) C ⊂ A,
(2) D ⊂ B, (3) X is the union of the interiors of A and B, and (4) Y the union of the interiors of C and D.
Then there is a long exact sequence in relative homology as follows.
···
/ Hn (A ∩ B, C ∩ D)
/ Hn (A, C) ⊕ Hn (B, D)
/ Hn (X, Y )
/ Hn−1 (A ∩ B, C ∩ D)
/ ···
All variants of the Mayer-Vietoris sequence hold for reduced homology too. Furthermore, all preceding
results in this section generalize to homology with coefficients.
4.3.3. The Formal Viewpoint.
Definition 4.3.27. A (reduced) homology theory is a sequence is covariant functors e
hn from the category of
CW complexes to the category of abelian groups which satisfy the following axioms.
(1) If f ' g, then f∗ = g∗ : e
hn (X) → e
hn (Y ).
(2) There are boundary homomorphisms ∂ : e
hn (X/A) → e
hn−1 (A) defined for each CW pair (X, A), fitting
into an exact sequence
···
∂
/e
hn (A)
i∗
/e
hn (X)
q∗
/e
hn (X/A)
∂
/e
hn−1 (A)
i∗
/ ··· ,
where i : A → X and q : X → X/A are respectively the evident inclusion and quotient maps. Furthermore, the boundary mas are natural: for f : (X, A) → (Y, B) inducing a quotient map f : X/A → Y /B,
the diagrams
∂ / e
e
hn (X/A)
hn−1 (A)
f∗
e
hn (Y /B)
∂
f∗
/e
hn−1 (B)
commute.
W
(3) For a wedge sum X = α Xα with inclusions iα : Xα ,→ X, the direct sum map
M
M
e
(iα )∗ :
hn (Xα ) → e
hn (X)
α
is an isomorphism for all n.
α
38
1. NOTES
4.3.4. Homology and Fundamental Group.
Theorem 4.3.28 (Hurewicz Theorem). By regarding loops as singular 1-cycles, we obtain a homomorphism
h : π1 (X, x0 ) → H1 (X). If X is path-connected, then h is surjective and has kernel the commutator subgroup
of π1 (X), so h induces an isomorphism between the abelianization of π1 (X) onto H1 (X). If γ : S 1 → X is a
loop, then the map h may alternatively be described as h[γ] = γ∗ (α) where α is the (oriented) generator of
H1 (S 1 ).
4.3.5. Classical Applications.
Proposition 4.3.29.
e • (S n \ h(Dk )) = 0.
(a) For an embedding h : Dk → S n , we have H
k
n
e • (S n \ h(S k )) ∼
(b) For an embedding h : S → S with k < n, we have H
= Z(n−k−1) .
Theorem 4.3.30 (Invariance of Domain). If U is an open set in Rn , then for any embedding h : U → Rn
the image h(U ) must be an open set in Rn . The statement holds if we replace Rn with S n throughout.
Corollary 4.3.31. If M is a compact n-manifold and N a connected n-manifold, then an embedding
h : M → N must be surjective, hence a homeomorphism.
Theorem 4.3.32 (Hopf). The only finite-dimensional division algebras over R which are commutative and
have an identity are R and C.
Proposition 4.3.33. An odd map f : S n → S n , satisfying f (−x) = −f (x) for all x ∈ S n , must have odd
degree. The claim also holds if we replace odd with even throughout.
e → X is a two-sheeted cover, then there is a long exact sequence as follows.
Proposition 4.3.34. If p : X
···
/ Hn (X; Z/2)
τ∗
e Z/2)
/ Hn (X;
p∗
/ Hn (X; Z/2)
/ Hn−1 (X; Z/2)
/ ···
The map τ∗ , called the transfer homomorphism, is given by summing up the two lifts of a given chain.
Corollary 4.3.35 (Borsuk-Ulam). For every map g : S n → Rn there exists a point x ∈ S n such that
g(x) = g(−x).
4.3.6. Simplicial Approximation.
Theorem 4.3.36 (Simplicial approximation). If K is a finite simplicial complex and L an arbitrary simplicial
complex, then any map f : K → L is homotopic to a map that is simplicial with respect to some iterated
barycentric subdivision of K.
For a map ϕ : Zn → Zn we define its trace tr ϕ in the usual sense. More generally, consider a finitely generated
abelian group A with torsion part AT . For any ϕ : A → A, we define tr ϕ = tr ϕ where ϕ : A/AT → A/AT is
the induced map mod torsion.
Definition 4.3.37. Let X be a finite CW complex and f : X → X a continuous map. The Lefschetz number
of f is
X
τ (f ) =
(−1)n tr(f∗ : Hn (X) → Hn (X)).
i
Note that τ (idX ) = χ(X).
Theorem 4.3.38 (Lefschetz Fixed Point Theorem). If X is a finite simplicial complex, or more generally a
retract of a finite simplicial complex, and f : X → X a map with τ (f ) 6= 0, then f has a fixed point.
It is the case that every compact, locally contractible space that can be embedded in Rn for some n is a retract
of a finite simplicial complex. In particular, this includes compact manifolds and finite CW complexes.
4. ALGEBRAIC TOPOLOGY
39
Theorem 4.3.39 (Simplicial Approximation to CW Complexes). Every CW complex X is homotopy equivalent to a simplicial complex, which can be chosen to be of the same dimension as X, finite if X is finite,
and countable if X is countable.
4.4. Hatcher, Chapter 3: Cohomology.
4.4.1. Cohomology Groups. The homology of a space X is customarily constructed in two steps: (1)
form a chain complex C• (X) (simplicial, singular, cellular, etc.), and then (2) take its homology. To make
the transition to cohomology, one needs to dualize after step (1). In other words, assuming we are going
to work with coefficients over an abelian group G, we first form C • (X; G) = Hom(C• ; G). The homology
of this complex is then denoted H • (X; G). We proceed to investigate the relation between H • (X; G) and
Hom(H• (X), G).
Theorem 4.4.1. If a chain complex C• of free abelian groups has homology groups H• (C), then the cohomology groups H • (C; G) of the cochain complex Hom(C• , G) are determined by the split exact sequences
0
/ Ext(Hn−1 (C), G)
/ H n (C; G)
h
/ Hom(Hn (C), G)
/ 0.
The Ext(H, G) groups are defined in the following fashion. Every abelian group has a free resolution 0 →
F1 → F0 → H → 0. We apply the functor Hom(−, G) to it and take homology. It turns out that the
cohomology is nontrivial only in degree 1 (this is specific to the category of abelian groups Z-mod), and we
define Ext(H, G) = H 1 (H; G) = H1 (Hom(F• , G)). For computational purposes, the following properties are
useful.
(a) Ext(H ⊕ H 0 , G) ∼
= Ext(H, G) ⊕ Ext(H 0 , G).
(b) Ext(H, G) = 0 if H is free.
(c) Ext(Z/n, G) ∼
= G/nG.
The following result summarizes the above facts when G = Z.
Corollary 4.4.2. If the homology groups Hn and Hn−1 of a chain complex C of free abelian groups are
finitely generated, with torsion subgroups Tn ⊂ Hn and Tn−1 ⊂ Hn , then
H n (C; Z) ∼
= (Hn /Tn ) ⊕ Tn−1 .
Corollary 4.4.3. If a chain map between chain complexes of free abelian groups induces an isomorphism
on homology groups, then it induces an isomorphism on cohomology groups with any coefficient group G.
A large number of the results from the previous chapter hold for cohomology – one only needs to reverse the
direction of sequences.
4.4.2. Cup Product. Unlike the general theories of homology and cohomology in which coefficients could
be taken in an arbitrary abelian group, we are required to work over a commutative ring R in order to define
cup products. For cochains ϕ ∈ C k (X; R) and ψ ∈ C ` (X; R), the cup product ϕ ` ψ ∈ C k+` (X; R) is the
cochain whose value on a singular simplex σ : ∆k+` → X is given by the formula
(ϕ ` ψ)(σ) = ϕ(σ|[v0 ,...,vk ] ) · ψ(σ|[vk ,...,vk+` ] ).
Proposition 4.4.4. For ϕ ∈ C k (X; R) and ψ ∈ C ` (X; R), we have
δ(ϕ ` ψ) = δϕ ` ψ + (−1)k ϕ ` δψ.
The previous result implies that the cup product of two cocycles is again a cocycle, and the cup product of
a cocycle and a coboundary (in either order) is again a coboundary. Therefore, the cup product on cochains
induces a cup product on cohomology, namely
H k (X; R) × H ` (X; R)
`
/ H k+` (X; R).
Associativity and distributivity on the level of cochains implies these properties for the product on cohomology too. If R is unital, then there is an identity for the cup product – the class 1 ∈ H 0 (X; R) defined by the
40
1. NOTES
0-cocycle taking the value 1 on each singular 0-simplex. There is a relative version of the cup product which
takes the form
`
/ H k+` (X, A ∪ B; R),
H k (X, A; R) × H ` (X, B; R)
and this specializes to
H k (X, A; R) × H ` (X; R)
`
/ H k+` (X, A; R),
H k (X; R) × H ` (X, A; R)
`
/ H k+` (X, A; R),
H k (X, A; R) × H ` (X, A; R)
`
/ H k+` (X, A; R).
Proposition 4.4.5. For a map f : X → Y , the induced maps f ∗ : H n (Y ; R) → H n (X; R) satisfy f ∗ (α `
β) = f ∗ (α) ` f ∗ (β), and similarly in the relative case.
This prompts us to note that we can regard H • (−; R) as a functor from the category of topological spaces
to the category of graded R-algebras. We proceed summarize a few common cohomology rings.
H • (RPn ; Z/2) ∼
= (Z/2)[α]/(αn+1 ), |α| = 1
H • (RP∞ ; Z/2) ∼
= (Z/2)[α],
|α| = 1
H • (CPn ; Z) ∼
= Z[α]/(αn+1 ), |α| = 2
H • (CP∞ ; Z) ∼
= Z[α], |α| = 2
H • (HPn ; Z) ∼
= Z[α]/(αn+1 ), |α| = 4
H • (HP∞ ; Z) ∼
= Z[α], |α| = 4
The cohomology of real projective spaces over Z are slightly more delicate.
H • (RP2n ; Z) ∼
= Z[α]/(2α, αn+1 ), |α| = 2
H • (RP2n+1 ; Z) ∼
= Z[α, β]/(2α, αn+1 , β 2 , αβ),
H (RP ; Z) ∼
= Z[α]/(2α),
•
∞
|α| = 2, |β| = 2n + 1
|α| = 2
Proposition 4.4.6. The isomorphisms
/ Q H • (Xα ; R)
α
e • (Xα ; R)
e • (W Xα ; R)
/Q H
H
α
α
F
W
whose coordinates are induced by the inclusions iα : Xα ,→ α Xα and iα : Xα ,→ α Xα respectively are
ring isomorphisms.
H •(
F
α
Xα ; R)
The second isomorphism above provides us with a tool to reject spaces as being homotopy equivalent to
wedge products.
Theorem 4.4.7. When R is commutative, the rings H • (X, A; R) are graded commutative, that is, the
identity
α ` β = (−1)k` β ` α
k
`
holds for all α ∈ H (X, A; R) and β ∈ H (X, A; R).
The cross product, also known as the external cup product, is a map
H k (X; R) × H ` (Y ; R)
×
/ H k+` (X × Y ; R)
given by
a × b = p∗1 (a) ` p∗2 (b),
where p1 : X × Y → X and p2 : X × Y → Y are the evident projections. It is not hard to see this map is
bilinear, hence induces a linear map H k (X; R) ⊗ H ` (Y ; R) → H k+1 (X × Y ; R).
4. ALGEBRAIC TOPOLOGY
41
Theorem 4.4.8 (Künneth formula). The cross product H • (X; R) ⊗R H • (Y ; R) → H • (X × Y ; R) is an
isomorphism of rings if X and Y are CW complexes and H k (Y ; R) is a finitely generated free R-module for
all k.
It turns out the hypothesis X and Y are CW complexes is unnecessary. The result also hold in a relative
setting.
Theorem 4.4.9 (Relative Künneth formula). For CW pairs (X, A) and (Y, B) the cross product homomorphism H • (X, A; R)⊗R H • (Y, B; R) → H • (X ×Y, A×Y ∪X ×B; R) is an isomorphism of rings if H k (Y, B; R)
is a finitely generated free R-module for each k.
The relative version yields a reduced one which involves the smash product X ∧ Y .
Theorem 4.4.10. If Rn has the structure of a division algebra over R, then n must be a power of 2.
4.4.3. Poincaré Duality. Fix a coefficient ring R. For any space X and any subset A ⊂ X, the local
homology of X at A is H• (X|A; R) = H• (X, X \ A; R) and similarly for cohomology. As usual, omitting R
means we are working over Z.
Definition 4.4.11. A compact manifold without boundary is called closed.
Theorem 4.4.12. The homology groups of a closed manifold are finitely generated.
∼ R(n) for all x ∈ M . An RFrom now on, consider a manifold M . It is easy to see that H• (M |x; R) =
orientation of M at x is a choice of a generator µx ∈ Hn (M |x; R). An R-orientation at a point x determines
R-orientations for all y in a neighbourhood B (open ball of finite radius) of x via the canonical isomorphisms
∼ Hn (M |B; R) =
∼ Hn (M |x; R).
Hn (M |y; R) =
An R-orientation of M is a consistent choice of local R-orientations at all x ∈ M . This is better handled
via the space
MR = {µx ∈ Hn (M |x; R) | x ∈ M }
topologized in an appropriate manner. There is a canonical projection MR → M which turns this into a
covering space. Since Hn (M |x; R) ∼
= Hn (M |x) ⊗ R, each r ∈ R determines a subcovering space
Mr = {±µx ⊗ r ∈ Hn (M |x; R) | x ∈ M, µx is a generator of Hn (M |x)} ⊂ MR .
The space
f = {µx ∈ Hn (M |x) | µx is a generator of Hn (M |x)}
M
F
f for all k ≥ 1. Similarly,
is also of great interest. For example MZ = k≥0 Mk where M0 ∼
= M and Mk ∼
=M
×
f. Note that M
f is a two-sheeted cover of M
if r ∈ R has order 2, then Mr ∼
= M , and otherwise Mr ∼
=M
f of this cover.
(potentially not connected), and a (Z-)orientation of M is nothing but a section M → M
f has two components.
Proposition 4.4.13. If M is connected, then M is orientable if and only if M
Corollary 4.4.14. If M is simply-connected, or more generally if π1 (M ) has no subgroup of index two,
then M is orientable.
An orientable manifold is R-orientable for all R, while a non-orientable manifold is R-orientable if and only
if R contains a unit of (additive) order 2, which is equivalent to having 2 = 0 in R. It follows that every
manifold is Z/2-orientable. In practice, the important cases are R = Z and R = Z/2.
Theorem 4.4.15. Let M be a closed connected n-manifold.
(a) If M is R-orientable, then the natural map Hn (M ; R) → Hn (M |x; R) ∼
= R is an isomorphism for all
x ∈ M.
(b) If M is not R-orientable, then the natural map Hn (M ; R) → Hn (M |x; R) ∼
= R is injective with image
{r ∈ R | 2r = 0} for all x ∈ M .
(c) Hi (M ; R) = 0 for i > n.
42
1. NOTES
Corollary 4.4.16. Let M be an n-manifold. If M is orientable, then Hn (M ) ∼
= Z, and if not, then
Hn (M ) = 0. In either case Hn (M ; Z/2) ∼
Z/2.
=
Definition 4.4.17. A fundamental (orientation) class for M with coefficients in R is an element of Hn (M ; R)
whose image in Hn (M |x; R) is a generator for all x ∈ M .
Corollary 4.4.18. If M is a closed connected n-manifold, the torsion subgroup of Hn−1 (M ) is trivial if M
is orientable and Z/2 if M is non-orientable.
Proposition 4.4.19. If M is a connected non-compact n-manifold, then Hi (M ; R) = 0 for i ≥ n.
For an arbitrary space X and a coefficient ring R, define an R-bilinear cap product
a : Ck (X; R) × C ` (X; R) → Ck−` (X; R)
for k ≥ ` by setting
σ a ϕ = ϕ(σ|[v0 ,...,v` ] )σ|[v` ,...,vk ] .
The formula
∂(σ a ϕ) = (−1)` (∂σ a ϕ − σ a δϕ)
implies this induces a map on homology and cohomology
/ Hk−` (X; R).
a
Hk (X; R) × H ` (X; R)
There are relative forms
/ Hk−` (X, A; R)
a
Hk (X, A; R) × H ` (X; R)
/ Hk−` (X; R),
a
Hk (X, A; R) × H ` (X, A; R)
and more generally
Hk (X, A ∪ B; R) × H ` (X, A; R)
a
/ Hk−` (X, B; R)
defined for open A, B ⊂ X. The naturality of the cap product is expressed via the following diagram.
Hk (X) × H ` (X)
O
f∗
a
/ Hk−` (X)
f∗
Hk (Y ) × H ` (Y )
a
f∗
/ Hk−` (Y )
More precisely, we have
f∗ (α) a ϕ = f∗ (α a f ∗ (ϕ)).
We are now ready to state our main result.
Theorem 4.4.20 (Poincaré Duality). Let M be a closed R-orientable n-manifold with fundamental class
[M ] ∈ Hn (M ; R). Then the map D : H k (M ; R) → Hn−k (M ; R) defined by D(α) = [M ] a α is an isomorphism for all k.
One can define cohomology with compact support, denoted Hc• (X), as the cohomology of the cochain complex
which is formed by all compactly supported cochains (that is, vanishing on chains outside a compact set).
It is clear that H • (X) ∼
= Hc• (X) for all compact spaces X.
Proposition 4.4.21. If a space X is the union of a directed set of subspaces Xα with the property that
each compact set in X is contained in some Xα , then the natural map lim Hi (Xα ; G) → Hi (X; G) is an
−→
isomorphism for all i and G.
For any space X, the compact sets K ⊂ X form a directed system since the union of any two compact
sets is compact. If K ⊂ L is an inclusion if compact sets, then there is a natural map H • (X, X \ K; G) →
H • (X, X \ L; G). It is possible to check that the resulting limit lim H • (X, X \ K; G) equals Hc• (X; G).
−→
4. ALGEBRAIC TOPOLOGY
43
This could be a useful property, for example, one can compute that Hc• (Rn ) ∼
= Z(n) . Poincaré Duality then
generalizes in the following way.
Theorem 4.4.22. The duality map DM : Hck (M ; R) → Hn−k (M ; R) is an isomorphism for all k whenever
M is an R-oriented n-manifold.
Corollary 4.4.23. A closed manifold of odd dimension has Euler characteristic zero.
The cup and cap product are related by the formula
ψ(α a ϕ) = (ϕ ` ψ)(α)
for α ∈ Ck+` (X; R), ϕ ∈ C k (X; R), and ψ ∈ C ` (X; R). For a closed R-orientable n-manifold M , consider
the cup product pairing
H k (M ; R) × H n−k (M ; R) → R,
(ϕ, ψ) 7→ (ϕ a ψ)[M ].
Proposition 4.4.24. The cup product pairing is non-singular for closed R-orientable manifolds when R is
a field, or when R = Z and torsion in H • (X; R) is factored out.
Corollary 4.4.25. If M is a closed connected orientable n-manifold, then for each element α ∈ H k (M )
of infinite (additive) order that is not a proper multiple of another element, there exists an element β ∈
H n−k (M ) such that α ` β is a generator of H n (M ). With coefficients in a field the same conclusion holds
for any α 6= 0.
k
Let Hfree
(M ) denote H k (M ) modulo torsion. If M is closed orientable manifold of dimension 2n, then
n
n
the middle-dimensional cup product pairing Hfree
(M ) × Hfree
(M ) → Z is a nonsingular bilinear form on
n
Hfree (M ). This form is symmetric when n is even, and skew-symmetric when n is odd. In the latter case, it
is always possible to chose a basis so the bilinear form is given by a matrix formed by 2 × 2 blocks 10 −10
along the diagonal and 0 everywhere else. In particular, this implies that if n is odd, then the rank of H n (M )
is even. If n is even, classifying symmetric bilinear forms is an interesting algebraic question. For any given
n, there are finitely many such but their number grows quickly with n.
Theorem 4.4.26 (J.H.C. Whitehead). The homotopy type of a simply-connected closed 4-manifold is
uniquely determined by its cup product structure.
A compact manifold M with boundary is defined to be R-orientable if M \ ∂M is R-orientable as a manifold
without boundary. Orientability in the case of boundary implies there exists a fundamental class [M ] in
Hn (M, ∂M ; R) restricting to a given orientation at each point of M \ ∂M . The following is a generalization
of Poincaré duality.
Theorem 4.4.27. Suppose M is a compact R-orientable n-manifold whose boundary ∂M is decomposed as
the union of two compact (n−1)-dimensional manifolds A and B with common boundary ∂A = ∂B = A∩B.
Then cap product with a fundamental class [M ] ∈ Hn (M, ∂M ; R) gives isomorphisms DM : H k (M, A; R) →
Hn−k (M, B; R) for all k.
The possibility that A, B, or A ∩ B is empty is not excluded. The cases A = ∅ and B = ∅ are sometimes
called Lefschetz duality.
Theorem 4.4.28 (Alexander Duality). If K is a compact, locally contractible, non-empty, proper subspace
e i (S n \ K) ∼
e n−i−1 (K) for all i.
of S n , then H
=H
Corollary 4.4.29. If X \ Rn is compact and locally contractible then Hi (X) is 0 for i ≥ n and torsion-free
for i = n − 1 and n − 2.
Proposition 4.4.30. If K is a compact, locally contractible subspace of an orientable n-manifold M , then
there are isomorphisms Hi (M, M \ K) ∼
= H n−i (K) for all i.
The condition of local contractibility can be removed if one uses Čech instead of singular cohomology.
44
1. NOTES
Definition 4.4.31. Let M and N be connected closed orientable n-manifolds with fundamental classes
[M ] ∈ Hn (M ) and [N ] ∈ Hn (N ) respectively. The degree of a map f : M → N , denoted d = deg f , is such
an integer that f∗ [M ] = d[N ].
Proposition 4.4.32. For any closed orientable n-manifold M there is a degree 1 map M → S n .
Proposition 4.4.33. Let f : M → N be a map between connected closed orientable n-manifolds. Suppose
B ⊂ N is a ball such that f −1 (B)
P is the disjoint union of balls Bi each mapped homeomorphically by f
onto B. Then the degree of f is i εi where εi is +1 or −1 according to whether f |Bi : Bi → B preserves
or reverses local orientations induced from the given fundamental classes [M ] and [N ].
Proposition 4.4.34. A p-sheeted covering M → N of connected closed orientable manifolds has degree ±p.
4.4.4. Universal Coefficients for Homology.
Theorem 4.4.35 (Universal Coefficients for Homology). For each pair of spaces (X, A) there are split exact
sequences
/ Hn (X, A; G)
/ Tor(Hn−1 (X, A), G)
/0
/ Hn (X, A) ⊗ G
0
for all n, and these sequences are natural with respect to maps (X, A) → (Y, B).
The following result enables us to compute the Tor groups.
Proposition 4.4.36.
∼
(a) Tor(A,
L B) = Tor(B,
L A).
(b) Tor( i Ai , B) ∼
= i Tor(Ai , B).
(c) Tor(A, B) = 0 if A or B is free, or more generally torsion-free.
(d) Tor(A, B) ∼
= Tor(T (A), B) where T (A) is the torsion subgroup of A.
n·
(e) Tor(Z/n, A) ∼
= Ker(A −→ A).
(f) For each short exact sequence 0 → B → C → D → 0 there is a natural associated exact sequence
0
/ Tor(A, B)
/ Tor(A, C)
/ Tor(A, D)
/ A⊗B
/ A⊗C
/ A⊗D
/ 0.
Corollary 4.4.37.
(a) Hn (X; Q) ∼
= Hn (X; Z) ⊗ Q, so when Hn (X; Z) is finitely generated, the dimension of Hn (X; Q) as a
Q-vector space equals the rank of of Hn (X; Z).
(b) If Hn (X; Z) and Hn−1 (X; Z) are finitely generated, then for p prime, Hn (X; Z/p) consists of
(i) a Z/p summand for each Z summand of Hn (X; Z),
(ii) a Z/p summand for each Z/pk summand in Hn (X; Z), k ≥ 1,
(iii) a Z/p summand for each Z/pk summand in Hn−1 (X; Z), k ≥ 1.
Corollary 4.4.38.
e • (X; Z) = 0 if and only if H
e • (X; Q) = 0 and H
e • (X; Z/p) = 0 for all primes p.
(a) H
(b) A map f : X → Y induces isomorphisms on homology with Z coefficients if and only if it induces
isomorphisms on homology with Q and Z/p coefficients for all primes p.
4.4.5. The General Künneth Formula.
Theorem 4.4.39 (Künneth formula for PID). If X and Y are CW complexes and R is a principal ideal
domain, then there are split short exact sequences
/ L TorR (Hi (X; R), Hn−i−1 (Y ; R))
/ L Hi (X; R) ⊗R Hn−i (Y ; R)
/ Hn (X × Y ; R)
/0
0
i
i
natural in X and Y .
Corollary 4.4.40. If F is a field and X and Y are CW complexes, then the cross product map
M
h:
Hi (X; F ) ⊗F Hn−i (Y ; F ) → Hn (X × Y ; F )
i
6. REAL ANALYSIS
45
is an isomorphism for all n.
There is a relative version of the Künneth formula which reads
L
/
/ Hn (X × Y, A × Y ∪ X × B; R)
0
i Hi (X, A; R) ⊗R Hn−i (Y, B; R)
L
/
TorR (Hi (X, A; R), Hn−i−1 (Y, B; R))
i
In the relative case, this reduces to
L e
e
/
0
i Hi (X; R) ⊗R Hn−i (Y ; R)
e n (X ∧ Y ; R)
/H
/
L
i
/
/ 0.
e i (X; R), H
e n−i−1 (Y ; R))
TorR (H
/ 0,
where X ∧ Y stands for the smash product of X and Y . It is possible to combine the Künneth formula to
the more concise form
M
Hn (X × Y ; R) ∼
Hi (X; Hn−i (Y ; R)),
=
i
and similarly for relative and reduced homology. There is a version for cohomology which reads
M
H n (X × Y ; R) ∼
H i (X; H n−i (Y ; R)).
=
i
Both of these hold if we replace R with an arbitrary coefficient group G.
5. Differential Geometry
Syllabus
Undergraduate: Boothby, An introduction to differentiable manifolds and Riemannian geometry,
sections VII.1 , VIII.1 and VIII.2. (math 136)
Graduate: Boothby, An introduction to differentiable manifolds and Riemannian geometry, chapters
I-V and VII. (math 132 and 230a)
TODO
6. Real Analysis
Syllabus
Undergraduate: Royden, Real Analysis (3rd ed), chapters 1-10. (math 114).
Graduate: Rudin, Real and Complex Analysis, chapters 1-9. (math 212a).
Additional: Stein and Shakarchi, Real Analysis: Measure Theory, Integration and Hilbert Spaces
may also be a good source for some of this material.
TODO: Royden
6.1. Rudin, Prologue. We define the complex function
∞
X
zn
,
exp(z) =
n!
n=0
and often write ez = exp(z). The given series converges absolutely for every z, and uniformly on every
bounded set in the complex plane. It satisfies exp(a + b) = exp(a) exp(b) for all a, b ∈ C.
Theorem 6.1.1.
(a) For every z ∈ C, we have ez 6= 0.
(b) exp is its own derivative.
(c) The restriction of exp to the real axis is a monotonically increasing positive function, and
lim ex = ∞,
x→∞
lim ex = 0.
x→−∞
(d) There exists a positive number π such that eπi/2 = i, and such that ez = 1 if and only if z/(2πi) is an
integer.
(e) exp is a periodic function with period 2πi.
46
1. NOTES
(f) The mapping t 7→ eit maps the real axis onto the unit circle.
(g) If w is a nonzero complex number, then w = ez for some z ∈ C.
6.2. Rudin, Chapter 1: Abstract Integration.
Definition 6.2.1. A collection T of subsets of a set X is said to be a topology on X if T has the following
three properties:
(i) ∅ ∈ T and X ∈ T ;
(ii) if Vi ∈ T for i = 1, . . . , n, then V1 ∩ · · · ∩ Vn ∈ T ;
S
(iii) if {Vα } is an arbitrary collection of members of T , then α Vα ∈ T .
The elements of T are called open sets in X, and X is called a topological space.
Definition 6.2.2. A map f : X → Y between two topological spaces is called continuous if f −1 (U ) is open
in X for all open U ⊂ Y .
A map f : X → Y between two topological spaces is called continuous at x ∈ X if for every open V ⊂ Y
around f (x), there exists an open U ⊂ X such that f (U ) ⊂ V .
Proposition 6.2.3. A map f : X → Y between two topological spaces is continuous if and only if it is
continuous around all x ∈ X.
Definition 6.2.4. A collection M of subsets of a set X is said to be a σ-algebra on X if M has the following
three properties:
(i) X ∈ M;
c
(ii) if A ∈ M,
S∞ then A = X \ A ∈ M;
(iii) if A = n=1 An for some An ∈ M, then A ∈ M.
We call the elements of M measurable sets in X, and X a measurable space.
Definition 6.2.5. Let X be a measurable space, and Y a topological space. A map f : X → Y is called
measurable if f −1 (U ) is measurable in X for every open U ⊂ Y .
Remark. Rudin could have alternatively defined a measurable map between two measurable spaces. What
is the motivation behind the more restrictive definition above?
Proposition 6.2.6 (Further properties of metric spaces). Let X be a measurable space with σ-algebra M.
(a)
(b)
(c)
(d)
∅ ∈ M.
Sn
If Ai ∈ M for 1 ≤ i ≤ n, then
T∞ i=1 An ∈ M.
If Ai ∈ M for i ≥ 1, then n=1 An ∈ M.
If A, B ∈ M, then A \ B ∈ M.
Theorem 6.2.7. Let g : Y → Z be a continuous function between topological spaces.
(a) If f : X → Y is a continuous function for a topological space X, then h = g ◦ f : X → Z is continuous.
(b) Let f : X → Y is a measurable function for a measurable space X, then h = g ◦ f : X → Z is measurable.
Theorem 6.2.8. Let u and v be real measurable functions on a measurable space X, and Φ : R2 → Y a
continuous map for a topological space Y . Then h : X → Y defined by h(x) = Φ(u(x), v(x)) is measurable.
Corollary 6.2.9. Let X be a measurable space.
(a) If f = u+iv, where u and v are real measurable functions on X, then f is a complex measurable function
on X.
(b) If f = u + iv is a complex measurable function on X, then u, v, and |f | are real measurable functions
on X.
(c) If f and g are complex measurable functions on X, then so are f + g and f g.
6. REAL ANALYSIS
47
(d) If E ⊂ X is a measurable set, then the function χE : X → C given by
(
1 if x ∈ E,
χE (x) =
0 if x ∈
/E
is measurable.
(e) If f is a complex measurable function on X, there is a complex measurable function α on X such that
|α| = 1 and f = α|f |.
Theorem 6.2.10. If F is any collection of subsets of X, then there exists a smallest σ-algebra M in X such
that F ⊂ M. We say M is generated by F.
Definition 6.2.11. Let X be a topological space. The Borel measure B on X is generated by the opens in
X. The elements of B are called Borel sets.
Proposition 6.2.12. Every continuous map is Borel measurable, that is, measurable when the domain is
endowed with the Borel measure.
Theorem 6.2.13. Let X be a measurable space with σ-algebra M, Y a topological space, and f : X → Y
a map.
(a) If Ω is the collection of all sets E ⊂ Y such that f −1 (E) ∈ M, then Ω is a σ-algebra on Y . In short,
pushforwards of σ-algebras are σ-algebras.
(b) If f is measurable and E is a Borel set in Y , then f −1 (E) ∈ M.
(c) If Y = [−∞, ∞] and f −1 ((α, ∞]) ∈ M for all α ∈ R, then f is measurable.
Definition 6.2.14. Let {an } be a sequence in [−∞, ∞], and put
lim sup an = inf{sup{ak , ak+1 , . . . } | k ≥ 1},
n→∞
lim inf an = sup{inf{ak , ak+1 , . . . } | k ≥ 1}.
n→∞
Then
lim inf an = − lim sup(−an ).
n
n
If {an } converges, then
lim an = lim sup an = lim inf an .
n
n
n
Suppose {fn } is a sequence of extended-real function on a set X. Then supn fn and lim supn fn are functions
defined on X by
sup fn (x) = sup(fn (x)),
n
n
lim sup fn (x) = lim sup(fn (x)).
n
n
Similarly for inf n fn and lim inf n fn . If f (x) = limn fn (x) is well-defined for all x ∈ X, then we call f the
pointwise limit of the sequence {fn }.
Theorem 6.2.15. If fn : X → [−∞, ∞] is measurable for n ≥ 1, then so are supn fn and lim supn fn .
Corollary 6.2.16.
(a) The limit of every pointwise convergent sequence of complex measurable functions is measurable.
(b) If f and g are measurable (with range [−∞, ∞]), then so are max{f, g} and min{f, g}. In particular,
this is true of the functions f + = max{f, 0} and f − = − min{f, 0}. These are respectively called the
positive and negative parts of f . We have f = f + − f − and |f | = f + + f − .
Proposition 6.2.17. If f = g − h, g ≥ 0, and f ≥ 0, then f + ≤ g and f − ≤ h.
48
1. NOTES
Definition 6.2.18. A function s on a measurable space X whose range consists of only finitely many points
in [0, ∞) is called simple.
Pn
If α1 , . . . , αn are the distinct values assumed by s, and Ai = f −1 (αi ), then s = i=1 αi χAi . The function s
is measurable if and only if all Ai are measurable.
Theorem 6.2.19. Let f : X → [0, ∞] be a measurable function. There exists simple measurable functions
sn on X such that:
(a) 0 ≤ s1 ≤ s2 ≤ · · · ≤ f ;
(b) sn → f pointwise.
Remark. It can be shown using the construction in the proof of the theorem above, that if f is bounded,
then sn is a uniformly convergent sequence.
Definition 6.2.20.
(a) A (positive) measure is a function µ, defined on a σ-algebra M, whose range is in [0, ∞] and which is
countably additive. This means that is {Ai } is a disjoint countable collection of elements of M, then
µ
∞
[
!
Ai
=
i=1
∞
X
µ(Ai ).
i=1
To avoid trivialities, we shall also assume that µ(A) < ∞ for at least one A ∈ M.
(b) A measure space is a measurable space which has a positive measure defined on the σ-algebra of its
measurable sets.
(c) A complex measure is a complex-valued countable additive function defined on a σ-algebra.
Theorem 6.2.21. For any positive measure µ on a σ-algebra M, the following holds:
(a)
(b)
(c)
(d)
(e)
µ(∅) = 0;
if A1 , . . . , An are pairwise disjoint elements of M, then µ(A1 ∪ · · · ∪ An ) = µ(A1 ) + · · · + µ(An );
if A, B S
∈ M and A ⊂ B, then µ(A) ≤ µ(B);
∞
if A = Tn=1 An for A1 ⊂ A2 ⊂ · · · where all An are elements of M, then µ(An ) → µ(A) as n → ∞;
∞
if A = n=1 An for A1 ⊃ A2 ⊃ · · · where all An are elements of M, and µ(A1 ) is finite, then µ(An ) →
µ(A) as n → ∞.
Example 6.2.22. Let X be any set.
(a) For any E ⊂ X define µ(E) = ∞ if E is infinite, and µ(E) = |E| if E is finite. This µ is called the
counting measure on X.
(b) For x0 ∈ X, define µ(E) = 1 is x0 ∈ E, and µ(E) = 0 otherwise. This µ is called the unit mass
concentrated at x0 ∈ X.
To make sense of arithmetic on [0, ∞], we define additionally a + ∞ = ∞ + a = ∞ if 0 ≤ a ≤ ∞, and
(
a·∞=∞·a=
∞ if 0 < a ≤ ∞,
0 if a = 0.
This ensures the commutative, associative, and distributive laws hold in [0, ∞] without any restriction. It is
important to note that cancellation laws have to be treated with some care: a + b = a + c implies b = c only
when a < ∞, and ac = ac implies b = c only when 0 < a < ∞.
Proposition 6.2.23. If 0 ≤ a1 ≤ a2 ≤ · · · , 0 ≤ b1 ≤ b2 ≤ · · · , an → a, and bn → b, then an bn → ab.
Corollary 6.2.24. Sums and products of measurable functions into [0, ∞] are measurable.
6. REAL ANALYSIS
49
Definition 6.2.25. If s : X → [0, ∞) is a measurable simple function of the form s =
α1 , . . . , αn are distinct values of s, and if E ∈ M, we define
Z
n
X
sdµ =
αi µ(Ai ∩ E).
E
Pn
i=1
αi χAi , where
i=1
If f : X → [0, ∞] is measurable, and E ∈ M, we define
Z
Z
f dµ = sup
sdµ,
E
E
where the supremum is taken over all simple measurable functions s such that 0 ≤ s ≤ f .
This is called the Lebesgue integral of f over E, with respect to the measure µ.
Proposition 6.2.26.
(a)
(b)
(c)
(d)
(e)
(f)
If
If
If
If
If
If
R
R
0 ≤ f ≤ g, then E f dµ ≤
R E gdµ. R
A ⊂ B and f ≥ 0, then A f dµ ≤ B f dµ.R
R
f ≥ 0 and c ∈ [0, ∞) is a constant,
then E cf dµ = c E f dµ.
R
f (x) = 0 for all Rx ∈ E, then E f dµ = 0, even if µ(E) = ∞.
µ(E) = 0, then
f dµR = 0, even if f (x) = ∞ for all x ∈ E.
E
R
f ≥ 0, then E f dµ = X χE f dµ.
Proposition
6.2.27. Let s and t be nonnegative Rmeasurable simple
functions
on X. For E ∈ M, define
R
R
R
ϕ(E) = E sdµ. Then ϕ is a measure on M. Also X (s + t)dµ = X sdµ + E tdµ.
Theorem 6.2.28 (Lebesgue’s Monotone Convergence Theorem). Let {fn } be a sequence of measurable
functions on X, and suppose that
(a) 0 ≤ f1 ≤ f2 ≤ · · · ,
(b) fn → f pointwise on X.
Then f is measurable, and
Z
lim
n
Z
fn dµ =
X
f dµ.
X
Theorem 6.2.29. If fn : X → [0, ∞] is measurable for n ≥ 1, and f =
P∞
n=1
fn , then
R
X
f dµ =
P∞ R
n=1 X
fn dµ.
The following follows by employing the counting measure on Z+ .
Corollary 6.2.30. If aij ≥ 0 for all i, j ≥ 0, then
∞ X
∞
X
i=1 j=1
aij =
∞ X
∞
X
aij .
j=1 i=1
Proposition 6.2.31 (Fatou’s Lemma). If fn : X → [0, ∞] is measurable for all n ≥ 1, then
Z Z
lim inf fn dµ ≤ lim inf
fn dµ.
X
n
n
X
Theorem 6.2.32. Suppose f : X → [0, ∞] is measurable, and ϕ(E) =
measure on M, and
Z
Z
f dϕ =
df dµ.
X
1
R
E
f dµ for E ∈ M. Then ϕ is a
X
RDefinition 6.2.33. Let L (µ) be1 the collection of all complex measurable functions f on X for which
|f |dµ < ∞. The elements of L (µ) are called Lebesgue measurable functions with respect to µ.
X
50
1. NOTES
Definition 6.2.34. If f = u + iv for u and v real measurable functions on X, and if f ∈ L1 (µ), then we
define
Z
Z
Z
Z
Z
f dµ =
u+ dµ −
u− dµ + i
v + dµ − i
v − dµ
E
E
E
E
E
for every E ∈ M. In some circumstances it is Rdesirable to
R extend this
R to the case when f is a measurable
function with range in [−∞, ∞]. Then, we set E f dµ = E f + dµ − E f − dµ, provided that at least one of
the integrals on the right is finite.
Theorem 6.2.35. If f, g ∈ L1 (µ) and α, β ∈ C, then αf + βg ∈ L1 (µ), and
Z
Z
Z
(αf + βg)dµ = α
f dµ + β
f dµ.
X
X
X
R
R
Theorem 6.2.36. If f ∈ L (µ), then X f dµ ≤ X |f |dµ.
1
Theorem 6.2.37 (Lebesgue’s Dominated Convergence Theorem). Suppose {fn } is a sequence of complex
measurable functions on X such that f = limn fn is defined pointwise on all X. If there is a function
g ∈ L1 (µ) such that |fn | ≤ g for all n ≥ 1, then f ∈ L1 (µ),
Z
lim
|fn − f |dµ = 0,
n
and
X
Z
lim
n
Z
fn dµ =
X
f dµ.
X
Definition 6.2.38. Let P be a property which a point x ∈ X may or may not have. If µ is a measure on a
σ-algebra M and if E ∈ M, the statement “P holds almost everywhere on E” (abbreviated “P holds a.e. on
E”) means that there exists an N ∈ M such that µ(N ) = 0, N ⊂ E, and P holds at every point of E \ N .
Note that the concept depends strongly on the used measure µ.
Definition 6.2.39. If f and g are measurable functions and if µ({x | f (x) 6= g(x)}) = 0, we say that f = g
a.e. on X, and write f ∼ g. Note that ∼ is an equivalence relation.
R
R
Proposition 6.2.40. If f ∼ g, then, for every E ∈ M, we have E f dµ = E gdµ.
Theorem 6.2.41. Let (X, M, µ) be a measure space, and M∗ be the collection of all E ⊂ X for which there
exist A, B ∈ M with A ⊂ E ⊂ B and µ(B \ A) = 0, and define µ(E) = µ(A) in this situation. Then M∗ is
a σ-algebra, and µ is a measure on M∗ .
Definition 6.2.42. The extended measure µ is called complete, since all subsets of sets of measure 0 are
now measurable; the σ-algebra M∗ is called the µ-completion of M.
In light of the previous theorem and definition, we may only require functions to be well-defined and measurable on the complement of a zero measure set. This is sometimes a very useful trick.
Theorem 6.2.43. Suppose {fn } is a sequence of complex measurable functions defined a.e. on X such that
∞ Z
X
|fn |dµ < ∞.
n=1
Then the series f (x) =
P∞
n=1
X
fn (x) converges for almost all x, f ∈ L1 (µ), and
Z
∞ Z
X
f dµ =
fn dµ.
X
n=1
X
Theorem 6.2.44.
R
(a) If f : X → [0, ∞] Ris measurable, E ∈ M, and E f dµ = 0, then f = 0 a.e. on E.
(b) If f ∈ L1 (µ) and E f dµ = 0 for every E ∈ M, then f = 0 a.e. on X.
6. REAL ANALYSIS
51
R
R
(c) If f ∈ L1 (µ) and X f dµ = X |f |dµ, then there is a constant α such that αf = |f | a.e. on X.
Theorem 6.2.45. Suppose µ(X) < ∞, f ∈ L1 (µ), S is a closed set in the complex plane, and the averages
Z
1
AE (f ) =
f dµ
µ(E) E
lie in S for all E ∈ M with µ(E) > 0. Then f (x) ∈ S for almost all x ∈ X.
Theorem 6.2.46. Let {Ek } be a sequence of measurable sets in X, such that
almost all x ∈ X lie in at most finitely many of the sets Ek .
P∞
k=1
µ(Ek ) < ∞. Then
6.3. Rudin, Chapter 2: Positive Borel Measures.
Definition 6.3.1. Let X be a topological space.
(a) A set E ⊂ X is called closed if its complement E c is open.
(b) The closure E of E ⊂ X is the smallest closed set in X which contains E.
(c) A topological space is called compact if every open cover has a finite subcover.
(d) A neighbourhood of a point p ∈ X is an open set which contains p.
(e) X is Hausdorff if for every two points can be separated by opens.
(f) X is locally compact if every point of X has a neighbourhood whose closure is compact (alternatively,
for every p ∈ X, there is a neighbourhood U of p, and compact K ⊂ X satisfying U ⊂ K).
Theorem 6.3.2 (Heine-Borel Theorem). A subset of an Euclidean space is compact if and only it is closed
and bounded.
Proposition 6.3.3. Every metric space is Hausdorff.
Theorem 6.3.4. A closed subset of a compact space is also compact.
Corollary 6.3.5. If A ⊂ B ⊂ X and B has a compact closure, then so does A.
Theorem 6.3.6. Suppose X is Hausdorff, K ⊂ X is compact, and p ∈ K c . Then there exist open sets U
and V such that p ∈ U , K ⊂ V , and U ∩ V = ∅.
Corollary 6.3.7.
(a) Compact subsets of Hausdorff spaces are closed.
(b) In a Hausdorff space, the intersections of a closed and a compact subset is compact.
Theorem 6.3.8. If {Kα } is a collection of compact subsets of a Hausdorff space and their intersection is
empty, then there exists a finite subcollection with empty intersection.
Theorem 6.3.9. Suppose X is a locally compact Hausdorff space, U ⊂ X, and K ⊂ X is compact. Then
there exists an open V ⊂ X with compact closure which satisfies
K ⊂ V ⊂ V ⊂ U.
Definition 6.3.10. Let X be a topological space, and consider a function f : X → R.
(i) We call f lower semicontinuous if f −1 ((α, ∞)) is open for all α ∈ R.
(ii) We call f upper semicontinuous if f −1 ((−∞, α)) is open for all α ∈ R.
An alternative definition could be made by replacing R with [−∞, ∞].
Proposition 6.3.11. Let X be a topological space. A function f : X → R is continuous if and only if it is
both upper and lower semicontinuous.
Example 6.3.12.
(a) Characteristic functions of open sets are lower semicontinuous.
(b) Characteristic functions of closed sets are upper semicontinuous.
(c) The supremum of any collection of lower semicontinuous function is lower semicontinuous.
52
1. NOTES
(d) The infimum of any collection of upper semicontinuous function is upper semicontinuous.
Definition 6.3.13. Let X be a topological space, and f : X → C a function. The support of f is
Supp f = f −1 (C \ {0}).
The collection of all continuous complex functions on X with compact support is denoted Cc (X).
Note that Cc (X) is a vector space over C. We could have also replaced “continuous” above with “measurable”
to yield another interesting object of study.
Theorem 6.3.14. The image of a compact set under a continuous map is compact.
Corollary 6.3.15. The image of any f ∈ Cc (X) is compact in C.
Definition 6.3.16. For any compact K ⊂ X, the notation
K≺f
will mean that f ∈ Cc (X), 0 ≤ f ≤ 1, and f |K = 1. Similarly, for any open U ⊂ X, the notation
f ≺U
will mean that f ∈ Cc (X), 0 ≤ f ≤ 1, and Supp f ⊂ U . Then, K ≺ f ≺ U will simultaneously mean K ≺ f
and f ≺ U .
Proposition 6.3.17 (Urysohn’s Lemma). Let X be a locally compact Hausdorff space, U ⊂ X an open,
and K ⊂ X a compact. Then there exists f ∈ Cc (X) such that K ≺ f ≺ U .
Theorem 6.3.18. Compact Hausdorff spaces admit partitions of unity. That is, for
P any compact space K,
and a (finite) open cover U1 , . . . , Un , there exist fi ∈ C(X) satisfying fi ≺ Ui , and i fi = 1. The statement
also holds true for compact subspaces of locally compact Hausdorff spaces.
Theorem 6.3.19 (Riesz Representation Theorem). Let X be a locally compact Hausdorff space, and Λ : Cc (X) →
C be a positive linear functional, that is, Λ is a morphism of vector spaces over C, and for every f ≥ 0, we
have Λ(f ) ≥ 0. Then there exists a σ-algebra M on X containing all Borel sets of X, and a unique positive
measure µ on M which represents Λ in the sense that:
R
(a) Λ(f ) = X f dµ for every f ∈ Cc (X);
(b) µ(K) < ∞ for every compact K ⊂ X;
(c) for every E ∈ M, we have
µ(E) = inf{µ(U ) | E ⊂ U, U is open};
(d) the relation
µ(E) = sup{µ(K) | K ⊂ E, K is compact}
holds for every open E ⊂ X, and for every E ∈ M with µ(E) < ∞;
(e) M is complete, i.e., if E ∈ M, A ⊂ E, and µ(E) = 0, then A ∈ M.
Definition 6.3.20. Let X be a locally compact Hausdorff space X. A measure µ defined on the σ-algebra
of Borel sets is called a Borel measure.
Definition 6.3.21. Assume µ is a positive Borel measure on X. We call it outer (resp. inner) regular if
every Borel set E ⊂ X has property (c) (resp. property (d)) from the Riesz Representation Theorem. If a
measure is both inner and outer regular, then we call it regular.
Definition 6.3.22. A topological space is called σ-compact if it is the union of finitely many compacts, or
alternatively, every open cover has a countable subcover. We call a topological space σ-locally compact if it
is both σ- and locally compact.
A set E in a measure space is said to have σ-finite measure if R is a countable union of measurable sets Ei
with µ(Ei ) < ∞.
6. REAL ANALYSIS
53
Theorem 6.3.23. Let X be a σ-locally compact Hausdorff space. If M and µ as in the Riesz Representation
Theorem, then they have the following properties:
(a) for any E ∈ M and ε > 0, there is a closed F ⊂ X and an open U ⊂ X satisfying F ⊂ E ⊂ U and
µ(U \ F ) < ε;
(b) µ is a regular Borel measure on X;
(c) for any E ∈ M, there are sets A and B such that A is a countable union of closed sets, B is a countable
intersection of open sets, A ⊂ E ⊂ B, and µ(B \ A) = 0.
Theorem 6.3.24. Let X be a locally compact Hausdorff space in which every open is σ-compact. If µ is a
positive Borel measure on X such that µ(K) < ∞ for all compact K ⊂ X, then µ is regular.
Qk
Definition 6.3.25. Let W = i=1 Ii ⊂ Rk where Ii ⊂ R are intervals (open, closed, or any combination
thereof) with endpoints αi and βi . We call such a set a k-cell. The volume of W is defined to be vol(W ) =
Qk
i=1 (βi − αi ).
If a = (a1 , . . . , ak ) ∈ Rk and δ > 0, we call the set
Q(a, δ) =
k
Y
[ai , ai + δ)
i=1
a δ-box with corner at a.
For an integer n ≥ 1, let Ωn denote the set of 2−n -boxes with corners at points of Pn = 2−n Zk .
Proposition 6.3.26.
(a)
(b)
(c)
(d)
For any fixed n ≥ 1, the set Ωn is a partition of Rk .
If Q ∈ Ωn , Q0 ∈ Ωr , and r < n, then either Q ⊂ Q0 or Q ∩ Q0 = ∅.
If Q ∈ Ωr , then vol(Q) = 2−rk ; and if r < n, then |Pn ∩ Q| = 2(n−r)k .S
Every nonempty open in Rk is a countable union of disjoint boxes in n≥1 Ωn .
Theorem 6.3.27. There exists a positive complete measure m defined on a σ-algebra M on Rk with the
following properties:
(a) m(W ) = vol(W ) for every k-cell W ;
(b) M contains all Borel sets of Rk ; more precisely, E ∈ M if and only if there exists A, B ⊂ Rk such that
A ⊂ E ⊂ B, A is a countable union of closed sets, B is a countable union of opens, and m(B \ A) = 0;
also, m is regular;
(c) m is translation-invariant, i.e., for every E ∈ M and x ∈ Rk , we have m(E + x) = m(E);
(d) if µ is any positive translation-invariant Borel measure on Rk such that µ(K) < ∞ for every compact
K ⊂ Rk , then there is a constant c > 0 such that µ(E) = c m(E) for all Borel sets E ⊂ Rk ;
(e) for every linear transformation T : Rk → Rk , we have m(T (E)) = det T m(E) for all E ∈ M; in
particular, m(T (E)) = m(E) when T is a rotation.
Definition 6.3.28. The members of M are called Lebesgue measurable sets in Rk ; m is the Lebesgue measure
on Rk . When clarity requires it, we shall write mk in place of m.
For any E ⊂ Rk , the measure m induces a restricted measure mE on E (and also a σ-algebra ME ). We will
write L1 (E) = L1 (mE ), and, in particular, L1 (Rk ) = L1 (m).
If k = 1, I is an interval (open, closed, or any combination thereof) with endpoints a and b, and f ∈ L1 (I),
it is customary to denote
Z b
Z
f (x)dx = f dm.
a
I
Proposition 6.3.29. For any continuous complex function f on [a, b], the Riemann integral of f and the
Lebesgue integral of f over [a, b] agree.
54
1. NOTES
Remark. A natural question to ask is whether all elements of M are Borel. Another such is whether all
subsets of Rk are measurable. The answer of both of these is negative (for the latter one see the following
result).
Theorem 6.3.30. If A ⊂ R and every subset of A is Lebesgue measurable, then m(A) = 0.
Corollary 6.3.31. Every set of positive measure has a nonmeasurable subset.
Theorem 6.3.32 (Lusin’s Theorem). Suppose f is a complex measurable function on X, µ(A) < ∞, f |X\A =
0, and ε > 0. Then there exists a f ∈ Cc (X) such that
µ({x | f (x) 6= g(x)}) < ε.
Furthermore, we may arrange so that
sup |g(x)| ≤ sup |f (x)|.
x∈X
x∈X
Corollary 6.3.33. Assume the hypotheses of Lusin’s Theorem are satisfied and that |f | ≤ 1. Then there
is a sequence {gn } such that gn ∈ Cc (X), |gn | ≤ 1, and
f (x) = lim gn (x)
n
a.e.
Theorem 6.3.34 (Vitali-Carathéodory Theorem). Suppose f ∈ L1 (µ), f is real-valued, and ε > 0. Then
there exist functions u, v : X → R such that u ≤R f ≤ v, u is upper semicontinuous and bounded above, v is
lower semicontinuous and bounded below, and X (v − u)dµ < ε.
6.4. Rudin, Chapter 3: Lp -spaces.
Definition 6.4.1. Consider an interval (a, b) ⊂ [−∞, ∞]. A function ϕ : (a, b) → R is called convex if the
inequality
ϕ((1 − λ)x + λy) ≤ (1 − λ)ϕ(x) + λϕ(y)
holds for all a < x, y < b, and 0 ≤ λ ≤ 1.
Graphically, the above means for any x, y ∈ (a, b) the graph of ϕ between these two points is below the
straight line connecting (x, ϕ(x)) and (y, ϕ(y)). Alternatively, convexity is equivalent to the inequality
ϕ(t) − ϕ(s)
ϕ(u) − ϕ(t)
≤
t−s
u−t
for all a < s < t < u < b.
The following is a consequence of the mean value theorem for differentiation and the alternative formulation
of convexity above.
Proposition 6.4.2. A differentiable ϕ : (a, b) → R is convex if and only if ϕ0 is monotonically increasing.
Theorem 6.4.3. If ϕ : (a, b) → R is convex, then it is also continuous.
Remark. The statement above would not hold had the domain not been an open interval.
Theorem 6.4.4 (Jensen’s Inequality). Let µ be a positive measure on a σ-algebra M on a set X with
µ(X) = 1. If f ∈ L1 (µ), a < f < b, and ϕ is convex on (a, b), then
Z
Z
ϕ
f dµ ≤
(ϕ ◦ f )dµ.
X
X
Remark. The cases a = −∞ and b = ∞ are not excluded. In either of these, it may occur that ϕ ◦ f is
not an element of L1 (µ). Then its integral exists in the extended sense and equals +∞, in other words, the
equality is vacuous.
Example 6.4.5.
6. REAL ANALYSIS
55
(a) Taking ϕ(x) = ex , we obtain
Z
Z
f dµ ≤
ef dµ.
exp
X
X
(b) Taking X = {p1 , . . . , pn }, µ({pi }) = 1/n for all 1 ≤ i ≤ n, and f (pi ) = xi , we obtain
ex1 + · · · + exn
x1 + · · · + xn
≤
.
n
n
Taking yi = exi , this turns into the AM-GM inequality
y1 + · · · + yn
(y1 · · · yn )1/n ≤
.
n
Pn
(c) If we modify the previous example only by setting µ({pi }) = αi > 0 with i=1 αi = 1, we obtain
exp
y1α1 · · · ynαn ≤ α1 y1 + · · · + αn yn .
Definition 6.4.6. If two positive real numbers p and q satisfy p + q = pq, or alternatively 1/p + 1/q = 1,
we say they are conjugate exponents.
Remark. Note that 1/p + 1/q = 1 implies 1 < p, q < ∞.
Theorem 6.4.7. Let p and q be conjugate exponents, X a measure space with measure µ, and f, g : X →
[0, ∞] measurable functions. Then
Z
1/p Z
1/q
Z
(Hölder’s inequality)
f gdµ ≤
f dµ
gdµ
,
X
X
X
and
Z
(Minkowski’s inequality)
1/p Z
1/p Z
1/p
p
p
(f + g) dµ
≤
f dµ
+
g dµ
.
p
X
X
X
Remark. When p = q = 2, Hölder’s inequality becomes the well-known (Cauchy-)Schwartz inequality.
R
R
Remark. Assuming X f p dµ < ∞ and X g q dµ < ∞, equality holds in Hölder’s inequality if and only if
there exists constants α and β, not both 0, such that αf p = βg q a.e. An analogous statement could be made
regarding Minkowski’s inequality.
Definition 6.4.8. Let X be a measure space with a positive measure µ. For any 0 < p < ∞ and f : X → C
measurable, define its Lp -norm
Z
1/p
kf kp =
|f |p dµ
.
X
Let Lp (µ) consist of all such functions with finite Lp -norm.
As before, we write Lp (Rk ) = Lp (µ) if µ is the Lebesgue measure on Rk , and similarly for measurable subsets
of Rk .
If µ is the counting measure on a set A, then we denote `p (A) = Lp (A). If A is countably infinite, we may
even write `p . In this case, we may imagine the elements of `p as complex sequences x = {xn } satisfying
!1/p
∞
X
p
kxkp =
|xn |
< ∞.
n=1
Definition 6.4.9. Suppose f : X → [0, ∞] is measurable. Let S be the set of all α ∈ R satisfying
µ(f −1 ((α, ∞])) = 0.
If S = ∅, then put β = 0. If S 6= ∅, then put β = inf S. Since
n
[
1
−1
−1
g ((β, ∞]) =
g
β + ,∞ ,
n
n=1
56
1. NOTES
and since the union of a countable collection of measure 0 sets has measure 0, it follows that β ∈ S. We call
β the essential supremum of f .
For any measurable f : X → C, we define its L∞ -norm kf k∞ to be the essential supremum of |f |. Let L∞ (µ)
consist of all such functions with finite L∞ -norm. We call these functions essentially bounded on X.
Remark. It follows from the definition that |f | ≤ λ a.e. if and only if kf k∞ ≤ λ.
Definition 6.4.10. As before, we will write L∞ (Rk ) = L∞ (µ) if µ is the Lebesgue measure on Rk . Similarly,
we write `∞ (A) = L∞ (µ) if µ is the counting measure on a set A. We may write `∞ alone if A is countable.
Theorem 6.4.11. Let p and q be conjugate exponents. If f ∈ Lp (µ), g ∈ Lq (µ), then
kf gk1 ≤ kf kp kgkq ,
hence f g ∈ L1 (µ).
Theorem 6.4.12. Let 1 ≤ p ≤ ∞. If f, g ∈ Lp (µ), then
kf + gkp ≤ kf kp + kgkp ,
hence f + g ∈ Lp (µ).
Remark. If f ∈ Lp (µ), and α ∈ C, then kαf kp = |α| kf kp . This remark combined with the previous result
implies that Lp (µ) is a vector space over C.
Remark. Let d : Lp (µ) × Lp (µ) → [0, ∞) be given by d(f, g) = kf − gkp . It is not hard to check that
d(f, g) = 0 if and only if f ∼ g, that is, f = g a.e. In other words, d descends to a metric on Lp (µ)/∼. It
can also be checked that ∼ respects linearity, hence Lp (µ)/∼ is a vector space over C.
Remark. To avoid cumbersome details, we may sometimes treat Lp (µ) as a metric space and actually refer
to Lp (µ)/∼. In this case the elements of Lp (µ) are not functions, but equivalence classes of functions.
Remark. It is elementary though essential to define convergence and being Cauchy in Lp (µ).
Theorem 6.4.13. For every 1 ≤ p ≤ ∞, and every positive measure µ, the metric space Lp (µ) is complete.
Theorem 6.4.14. If 1 ≤ p ≤ ∞ and {fn } is a Cauchy sequence in Lp (µ) with limit f , then {fn } has a
subsequence which converges pointwise a.e. to f .
Theorem 6.4.15. Let S be the class of all complex, measurable, simple functions on X such that µ(s−1 (C \
{0})) < ∞. If 1 ≤ p < ∞, then S is dense in Lp (µ).
Theorem 6.4.16. For 1 ≤ p < ∞, Cc (X) is dense in Lp (µ).
Remark. Fix k ≥ 1, and consider the spaces Cc (Rk ) ⊂ Lp (Rk ). For every 1 ≤ p ≤ ∞, the Lp -norm induces
a genuine metric on Cc (Rk ) (unlike in Lp (Rk ), we do not need equivalence classes). If 1 ≤ p < ∞, then
Cc (Rk ) is dense in Lp (Rk ) and the latter is complete, hence we may treat Lp (Rk ) as the completion of Cc (Rk )
with respect to the metric induced by the Lp -norm. In the case p = ∞, the completion of Cc (Rk ) is not
Lp (Rk ) but C0 (Rk ), the space of all continuous functions on Rk which “vanish at infinity” (see below). It is
also interesting to note that for f ∈ C c (Rk ), we have
kf k∞ = sup |f (x)|.
x∈Rk
Definition 6.4.17. Let X be a locally compact Hausdorff space. A continuous function f : X → C is said
to vanish at infinity if for every ε > 0 there exists a compact K ⊂ X such that |f |X\K | < ε. The set of all
functions is denoted C0 (X).
Remark. It is clear that Cc (X) ⊂ C0 (X), and equality is attained if X is compact. In that case, we may
write C(X) for either of them.
6. REAL ANALYSIS
57
Theorem 6.4.18. Let X be a locally compact Hausdorff space. The completion of Cc (X) with respect to
the supremum norm
kf k = sup |f (x)|
x∈X
is C0 (X).
6.5. Rudin, Chapter 4: Elementary Hilbert Space Theory.
Definition 6.5.1. An inner product for a complex vector space H is a map (−, −) : H × H → C which
satisfies the following properties:
(a) (y, x) = (x, y) for all x, y ∈ H;
(b) (x + y, z) = (x, z) + (y, z) for all x, y, z ∈ H;
(c) (αx, y) = α(x, y) for all α ∈ C, x, y ∈ H;
(d) (x, x) ≥ 0 for all x ∈ H;
(e) (x, x) = 0 only if x = 0.
A complex vector space endowed with an inner product is called an inner product space.
p
√
Definition 6.5.2. The norm of x ∈ H is kxk = (x, x), where − denotes the non-negative square root.
The following are immediate consequences of the given properties:
(a) (0, x) = 0 for all x ∈ H;
(b) for every y ∈ H, the map H → C given by x 7→ (x, y) is linear;
(c) (x, αy) = α(x, y) for all α ∈ C, x, y ∈ H;
(d) kxk2 = (x, x) for all x ∈ H.
Proposition 6.5.3 ((Cauchy-)Schwartz Inequality). For all x, y ∈ H, we have
|(x, y)| ≤ kxk kyk.
Proposition 6.5.4 (Triangle Inequality). For all x, y ∈ H, we have
kx + yk ≤ kxk + kyk.
Remark. Any inner product space H possesses a natural metric structure in which the distance between
x, y ∈ H is given by kx − yk. This can be verified using the properties and inequalities listed above.
Definition 6.5.5. A Hilbert space is an inner product space which is complete in its natural metric structure.
Let H denote a Hilbert space for the remainder of this chapter.
Example 6.5.6.
Pn
(a) The vector space Cn endowed with (x, y) = i=1 xi yi is a Hilbert space.
(b) If µ is any positive measure on a set X, then L2 (µ) is a Hilbert space with inner product
Z
(f, g) =
f gdµ.
X
Note that in this case the induced norm agrees with the L2 -norm.
(c) The vector space of all continuous functions [0, 1] → C is an inner product space when endowed with
Z 1
(f, g) =
f (t)g(t)dt.
0
This however is not a Hilbert space since completeness fails to hold.
Theorem 6.5.7. For any fixed y ∈ H, the maps H → C, H → C, and H → R respectively given by
x 7→ (x, y),
are all continuous.
x 7→ (y, x),
x 7→ kxk
58
1. NOTES
Definition 6.5.8. A closed subspace of a Hilbert space is a subspace closed in the natural metric structure.
Remark. If M ⊂ H is a subspace, then so is its closure M .
Definition 6.5.9. Let V be a vector space. A subset E ⊂ V is called convex if the straight segment joining
any two points of E is also contained in E. Formally this means that if x, y ∈ E and 0 < t < 1, then
(1 − t)x + ty ∈ E.
Remark. The following properties follow immediately from the definition of convexity:
(a) every subspace of a vector space is convex;
(b) any translate of a convex set is also convex.
Definition 6.5.10. Let V be an inner product space. We call two vectors x, y ∈ V orthogonal, and write
x ⊥ y, if (x, y) = 0. For x ∈ V , let x⊥ denote the set all vectors in V perpendicular to it. Similarly, for any
subspace M ⊂ V , let M ⊥ denote the set of all vectors in V perpendicular to everything in M .
Remark.
(a) The relation ⊥ on V is symmetric.
(b) Both x⊥ and M ⊥ are subspaces.
T
(c) For any x ∈ H, the space x⊥ ⊂ H is closed. Similarly, so is M ⊥ = x∈M x⊥ .
Theorem 6.5.11. Every nonempty, closed, convex set in a Hilbert space contains a unique element of
smallest norm.
Theorem 6.5.12. Let M be a closed subspace of a Hilbert space H.
(a) Every x ∈ H has a unique decomposition x = P (x) + Q(x), where P (x) ∈ M and Q(x) ∈ M ⊥ .
(b) P (x) and Q(x) are the nearest points to x in M and M ⊥ respectively.
(c) The maps P : H → M and Q : H → M ⊥ are both linear.
(d) Every x ∈ H satisfies kxk2 = kP (x)k2 + kQ(x)k2 .
The maps P and Q are called the orthogonal projections of H onto M and M ⊥ .
Corollary 6.5.13. If M 6= H, then there exists y ∈ H \ {0} such that y ⊥ M .
Theorem 6.5.14. If L : H → C is continuous and linear, then there exists a unique y ∈ H such that
L(x) = (x, y) for all x ∈ H.
Definition 6.5.15. Let V be a vector space. For any subset S ⊂ V , let hSi denote the set of finite linear
combinations of elements of S, also called its span.
Definition 6.5.16. A collection of vectors {uα | α ∈ A} in a Hilbert space H is called orthonormal if
(uα , uβ ) = δαβ for all α, β ∈ A. To each x ∈ H, we associate a function x
b : H → C given by α 7→ (x, uα ).
The numbers x
b(α) are called the Fourier coefficients of x relative to {uα }.
Theorem 6.5.17. Suppose {uα | α ∈ A} is an orthonormal set in a Hilbert space H and F ⊂ A is a finite
subset. Let MF be the span of {uα | α ∈ F }.
(a) For any ϕ : A → C satisfying ϕ|A\F = 0, the vector
X
y=
ϕ(α)uα
α∈F
satisfies yb(α) = ϕ(α) for all α ∈ A, and
kyk2 =
X
|ϕ(α)|2 .
α∈F
(b) Define sF : H → H by
sF (x) =
X
α∈F
x
b(α)uα .
6. REAL ANALYSIS
59
Then for all x ∈ H, we have
kx − sF (x)k < kx − sk
for all s MF , except for s = sF (x), and
X
|b
x(α)|2 ≤ kxk2 .
α∈F
P
Before delving into the next
P result, let us clarify the meaning of α∈A ϕ(α) for infinite A. For this, we
assume 0 ≤ ϕ ≤ ∞. Then α∈A ϕ(α) will denote the supremum
of finite sums ϕ(α1 ) + · · · + ϕ(αn ), where
R
α1 , . . . , αn ∈ A are distinct. Note that this agrees with A ϕdµ if µ is the counting measure on A. We can
then refer to `2 (A) which is a Hilbert space with inner product
X
(ϕ, ψ) =
ϕ(α)ψ(α).
α∈A
We know that the functions A → C which vanish at all but finitely many points form a dense subset of
`2 (A). Furthermore, for all ϕ ∈ `2 (A), the set ϕ−1 (C \ {0}) is at most countable.
Lemma 6.5.18. Assume the following:
(a) X and Y are metric spaces, and X is complete;
(b) f : X → Y is continuous;
(c) X has a dense subset X0 on which f is an isometry;
(d) f (X0 ) is dense in Y .
Then f : X → Y is an isometry, and in particular, a bijection.
Theorem 6.5.19 (Riesz-Fischer Theorem). Let {uα | α ∈ A} be an orthonormal set in the Hilbert space H,
and P the span of {uα }. The inequality
X
|b
x(α)|2 ≤ kxk2
α∈A
holds for all x ∈ H, and x 7→ x
b is a continuous map H → `2 (A) whose restriction to P is an isometry.
Theorem 6.5.20. Let {uα | α ∈ A} be an orthonormal set in the Hilbert space H, and P the span of {uα }.
The following are equivalent:
(a) {uα } is a maximal orthonormal set in H;
(b) the span of {uα } is dense in H;
(c) the equality
X
|b
x(α)|2 = kxk2
α∈A
holds for all x ∈ H;
(d) the equality
(Parseval’s identity)
X
x
b(α)b
y (α) = (x, y)
α∈A
holds for all x, y ∈ H.
Definition 6.5.21. Maximal orthonormal sets are called complete orthonormal set or orthonormal base.
Remark. Note that an orthonormal base in a Hilbert space need not be a basis for the underlying vector
space. Actually, it will only be one in the finite-dimensional case.
Corollary 6.5.22. If {uα | α ∈ A} is a maximal orthogonal set in a Hilbert space H, then the map
H → `2 (A) given by x →
7 x
b is an isomorphism of Hilbert spaces.
Theorem 6.5.23 (Hausdorff Maximality). Every nonempty partially ordered set contains a maximal totally
ordered subset.
60
1. NOTES
Theorem 6.5.24. Every orthonormal set B in a Hilbert space H is contained in a maximal orthonormal set
in H.
Corollary 6.5.25. Every Hilbert space admits a maximal orthonormal set.
In what follows, we will identify maps S 1 → X with periodic maps R → X with period 2π. More concretely,
we shall sometimes write f (t) rather than f (eit ), even if we think of f as defined on S 1 .
Consider the class of all Lebesgue measurable, 2π-periodic functions on R for which the norm
1/p
Z π
1
|f (t)|p dt
kf kp =
2π −π
is finite. Under the identification above, the set of these functions is Lp (S 1 ). Note the normalizing factor
1/2π inserted for the sake of convenience. Similarly for L∞ (S 1 ) and C(S 1 ).
Definition 6.5.26. A trigonometric polynomial is a finite sum of the form
f (t) = a0 +
N
X
(an cos nt + bn sin nt)
n=1
for defined for t ∈ R where ai , bj ∈ C. These coincide with finite sums of the form
f (t) =
N
X
cn eint
n=−N
where cn ∈ C which are often more convenient.
int
It is clear that all trigonometric polynomials have
R π period 2π. For n ∈ Z, we set un (t) = e . If we define
2
1
the inner product in L (S ) as (f, g) = 1/2π −π f (t)g(t)dt, then (um , un ) = δmn . The orthonormal set
{un | n ∈ Z} in L2 (S 1 ) is called the trigonometric system.
Theorem 6.5.27. For any f ∈ C(S 1 ) and ε > 0, there exists a trigonometric polynomial P such that
|f − P | < ε.
Combining the above with the fact C(S 1 ) is dense in L2 (S 1 ), we obtain the following.
Corollary 6.5.28. The trigonometric system is a maximal orthonormal set in L2 (S 1 ).
Definition 6.5.29. For any f ∈ L1 (S 1 ), we define the Fourier coefficients of f by the formula
Z π
1
b
f (n) = (f, un ) =
f (t)e−int dt.
2π −π
PN
P∞
The Fourier series of f is n=−∞ fb(n)eint , and its partial sums are sN (t) = n=−N fb(n)eint .
The following are consequences of previous results for Lp spaces.
Corollary 6.5.30.
(a) For any f, g ∈ L2 (S 1 ), we have
∞
X
1
fb(n)b
g (n) =
2π
n=−∞
Z
π
f (t)g(t)dt,
−π
and the series on the left converges absolutely.
(b) For any f ∈ L2 (S 1 ), we have
lim kf − sN k2 = 0,
N →∞
6. REAL ANALYSIS
61
since a special case of (a) implies
kf − sN k22 =
2
X fb(n) .
|n|>N
6.6. Rudin, Chapter 5: Examples of Banach Space Techniques.
Definition 6.6.1. A vector space X is called normed if it is endowed with a norm, that is, a map k−k : X →
[0, ∞) satisfying the following conditions:
(a) kx + yk ≤ kxk + kyk for all x, y ∈ X;
(b) kαxk = |α| kxk for all α ∈ C, x ∈ X;
(c) kxk = 0 implies x = 0 for all x ∈ X.
Remark. Any normed vector space X possesses a natural metric structure in which the distance between
x, y ∈ X is given by kx − yk. This can be verified using the properties listed above.
Definition 6.6.2. A Banach space is a normed vector space which is complete with respect to the metric
induced by its norm.
Remark. Every Hilbert space is a Banach space in a natural way.
Definition 6.6.3. Let X and Y be normed vector spaces. The norm of a linear map Λ : X → Y is given by
kΛk = sup{kΛ(x)k | x ∈ X, kxk ≤ 1}.
Geometrically, Λ maps the closed unit ball in X to the closed ball of radius kΛk around 0 in Y .
Definition 6.6.4. If kΛk < ∞, then Λ is called bounded.
Theorem 6.6.5. Let X and Y be normed vector spaces. For a linear transformation Λ : X → Y , the
following are equivalent:
(a) Λ is bounded;
(b) Λ is continuous;
(c) Λ is continuous at 0 ∈ X.
Theorem 6.6.6 (Baire’s Theorem). In a complete metric space X, the intersection of every countable
collection of dense opene subsets is dense. In particular (except in the trivial case X = ∅), the intersection
is not empty.
Corollary 6.6.7. In a complete metric space, the intersection of any countable collection of dense countable
intersections of opens is again a dense countable intersection of opens.
The following describes why Baire’s Theorem is sometimes called the Category Theorem.
Definition 6.6.8. Let X be a metric space. We call a set E ⊂ X nowhere dense if E contains no nonempty
opens.
Definition 6.6.9 (Baire’s terminology). Any countable union of nowhere dense sets is called to be of the
first category. All other subsets are called to be of the second categoty.
Using this language Baire’s Theorem becomes.
Corollary 6.6.10. No complete metric space is of the first category.
Theorem 6.6.11 (Banach-Steinhauss Theorem). Let X be a Banach space, Y a normed vector space, and
{Λα : X → Y | α ∈ A} a collection of bounded linear transformations. Then there exists an M < ∞ such
that
kΛα k ≤ M
for all α ∈ A, or
sup kΛα (x)k = ∞
α∈A
62
1. NOTES
for all x belonging to some countable intersection of opens in X which is also dense.
Remark. The Banach-Steinhauss Theorem is sometimes referred to as the uniform boundedness principle.
Theorem 6.6.12 (Open Mapping Theorem). Let Y and V be the open units of two Banach spaces X and
Y respectively. To every surjective bounded linear transformation Λ : X → Y there corresponds a δ > 0
satisfying Λ(U ) ⊃ δV .
Theorem 6.6.13. If X and Y are Banach spaces and Λ : X → Y is a bijective bounded linear transformation,
then there is a δ > 0 such that kΛ(x)k ≥ δkxk. In other words, Λ−1 : Y → X is a bounded linear
transformation too.
TODO here
6.7. Rudin, Chapter 6: Complex Measures. TODO
6.8. Rudin, Chapter 7: Differentiation. TODO
6.9. Rudin, Chapter 8: Integration on Product Spaces. TODO
6.10. Rudin, Chapter 9: Fourier Transforms. TODO
CHAPTER 2
Past Exams
S 2010
D1 D2 D3
Algebra
Algebraic Geometry
Complex Analysis
Algebraic Topology
Differential Geometry
Real Analysis
F 2009
D1 D2 D3
S 2009
D1 D2 D3
F 2008
D1 D2 D3
•
• • •
• • •
S 2008
D1 D2 D3
F 2007
D1 D2 D3
•
• • •
◦ • ·
· • ·
• • •
• • ·
· • •
• · •
• ◦ ◦
• • ·
• • •
Legend: • done, ◦ partly done, · to be done
1. Spring 2010
1.1. Day 1.
Problem 1
TODO
Problem 2
TODO
Problem 3
Let λ be a real number greater than 1. Show that the equation zeλ−z = 1 has exactly one solution with
|z| < 1, and that this solution z is real.
Solution. We first show that zeλ−z − 1 has one solution. Consider the entire functions f (z) = zeλ−z
and g(z) = −1, so that we are interested in the roots of f + g in the unit disk ∆. First of all z has one root
in ∆, and eλ−z does not vanish, hence f has exactly one root in ∆. To apply Rouché’s Theorem, it suffices
to observe that
|f | = |z||eλ−z | = eλ−Re z > 1 = |g| ≤ |f |
on the unit circle ∂∆. This completes the first part of the problem. To conclude the one root is real, consider
the restriction of f + g to the line segment [−1, 1], namely
(f + g)(−1) = −eλ+1 < 0,
(f + g)(1) = eλ−1 > 0.
Since (f + g)|R is real-values, the Intermediate Value Theorem shows f + g vanishes somewhere on along
(−1, 1), and since there is a unique root in ∆, this completes our claim.
Problem 4
TODO
Problem 5
63
64
2. PAST EXAMS
TODO
Problem 6
Let X be a topological space. We say that two covering spaces f : Y → X and g : Z → X are isomorphic if
there exists a homeomorphism h : Y → Z such that g ◦ h = f . If X is compact oriented surface of genus g,
how many connected 2-sheeted covering spaces does X have, up to isomorphism?
Solution. It is a classical theorem that covering spaces of a connected (and sufficiently nice, such as
a manifold) space X are in a bijective correspondence with subgroups of π1 (X) up to conjugation. To a
subgroup H ⊂ π1 (X) corresponds to covering space with [π1 (X) : H] sheets. Therefore we are interested in
counting the index 2 subgroups up to conjugation in π1 (Σg ) where Σg is the g-holed torus for g ≥ 0. Recall
that all index 2 subgroups are normal, hence we may remove “up to conjugation” in the previous sentence.
Such subgroups induce natural quotient homomorphism π1 (Σg ) → Z/2, and conversely every such surjective
homomorphism induces an index 2 subgroup, namely, its kernel. But Z/2 is abelian, hence a homomorphism
ϕ : π1 (Σg ) → Z/2 factors through
π1 (Σg )/[π1 (Σg ), π1 (Σg )] ∼
= H1 (Σg ) ∼
= Z2g .
In the first of the two isomorphisms above, we applied the Hurewicz Theorem. Our problem now is reduced to
counting the surjective homomorphisms Z2g → Z/2. Note that Z2g is the free abelian group of 2g generators,
say e1 , . . . , e2g . Therefore, homomorphisms ϕ
e : Z2g → Z/2 are uniquely determined by ϕ(e
e i ), and conversely,
every such choice yields a homomorphism. The only non-surjective homomorphism corresponds to the
assignment of [0] ∈ Z/2 to every generator ei . In conclusion, Σg has exactly
22g − 1
2-sheeted covering spaces up to isomorphism.
1.2. Day 2.
Problem 1
TODO
Problem 2
Let a be an arbitrary real number and b a positive real number. Evaluate the integral
Z ∞
cos(ax)
dx.
cosh(bx)
0
Solution. Start by noting that
Z
Z
Z ∞
1 ∞ cos(ax)
1 ∞
eiax
cos(ax)
dx =
dx =
dx.
I=
cosh(bx)
2 −∞ cosh(bx)
2 −∞ cosh(bx)
0
Shifting the line of integration up by πi/b we get
Z ∞
Z ∞
eia(x+πi/b)
eiax
dx = −e−πa/b
dx = −2Ie−πa/b .
−∞ cosh(b(x + πi/b))
−∞ cosh(bx)
In doing so, we picked only one pole at πi/2b with residue
resπi/2b
z − πi
eiax
1
2b
= eiaπi/2b resπi/2b
= e−aπ/2b lim
cosh(bx)
cosh(bx)
z→πi/2b cosh(bx)
1
1
= e−aπ/2b lim
= e−aπ/2b .
ib
z→πi/2b b sinh(bx)
1. SPRING 2010
65
We used two statements worth mentioning: (1) if f has a simple pole at z0 and g is holomorphic at z0 ,
then resz0 f g = g(z0 ) resz0 f ; (2) a complex version of L’Hôpital’s rule. Putting the collected information
together, we obtain
1
2I − −2Ieaπ/b = 2πi e−aπ/2b .
ib
Rearranging, we conclude
πe−aπ/2b
.
I=
b(1 + e−aπ/b )
Problem 3
Let Λ1 and Λ2 ⊂ R4 be complementary 2-planes, and let X = R4 \ (Λ1 ∪ Λ2 ) be the complement of their
union. Find the homology and cohomology groups of X with integer coefficients. j
Solution. Since Λ1 and Λ2 are complimentary, there exists an isomorphism T : R4 → R4 such that
R(Λ1 ) = R2 × {0}2 and T (Λ2 ) = {0}2 × R2 . Such R-linear isomorphisms are homeomorphisms, hence we
may assume that Λi by its image under T . Then
X = R4 \ (Λ1 ∪ Λ2 ) = (R2 \ {0})2 .
The space R2 \ {0} deformation retracts to S 1 , so
H• (R2 \ {0}) = Z(0) ⊕ Z(1) .
The homology in all degrees is finitely generated and free, hence the Künneth formula implies
∼ H• (R2 \ {0})⊗2 ∼
H• (X) =
= Z(0) ⊕ Z2(1) ⊕ Z(2) .
All groups are free, hence any Ext(−, Z) group involving them would vanish. The Universal Coefficients
Theorem enables us to compute the cohomology
H• (X) ∼
= Z(0) ⊕ Z2(1) ⊕ Z(2) .
Problem 4
TODO
Problem 5
TODO
Problem 6
Let p be a prime, and let G be the group Z/p2 Z ⊕ Z/p2 Z.
(a) How many subgroups of order p does G have?
(b) How many subgroups of order p2 does G have? How many of these are cyclic?
Solution. (a) Every subgroup of order p is generated by an element of order p, and each subgroup
possesses ϕ(p) = p − 1 such generators. Therefore, it suffices to count the number of elements of order p and
divide by p − 1. Let | − | denote the order of an element. The order of an element (a, b) ∈ G is given by
|(a, b)| = lcm(|a|, |b|). If |(a, b)| = p, then the possibilities are (|a|, |b|) = (1, p), (p, 1), (p, p). Since Z/p2 Z has
exactly p elements of order dividing p, and exactly one of oder 1, it follows it has p − 1 elements of order p.
Putting all information together, the number of subgroups of order p is
1 · (p − 1) + (p − 1) · 1 + (p − 1) · (p − 1)
= (1 + 1 + p − 1) = p + 1.
p−1
(b) Let us start by counting the cyclic subgroups of order p2 . Each such corresponds to ϕ(p2 ) = p(p − 1)
generators. The group Z/p2 Z has ϕ(p) = p(p − 1) generators, hence p(p − 1) elements of order p2 . If
66
2. PAST EXAMS
|(a, b)| = p2 , then (|a|, |b|) = (1, p2 ), (p2 , 1), (p, p2 ), (p2 , p), (p2 , p2 ). The number of cyclic subgroups of order
p2 is
1 · p(p − 1) + p(p − 1) · 1 + (p − 1) · p(p − 1) + p(p − 1) · (p − 1) + p(p − 1) · p(p − 1)
= p(p + 1).
p(p − 1)
Next, let us consider the non-cyclic subgroups of order p2 . It is easy to see each of these is isomorphic
to Z/pZ ⊕ Z/pZ. There are exactly p2 elements of order dividing p in G, and also p2 such elements in
Z/pZ ⊕ Z/pZ, hence there is exactly one non-cyclic subgroup of order p2 . In total, there are
p(p + 1) + 1 = p2 + p + 1
subgroups of order p2 in G.
1.3. Day 3.
Problem 1
TODO
Problem 2
Let f be holomorphic on a domain containing the closed disk {z | |z| ≤ 3}, and suppose that
f (1) = f (i) = f (−1) = f (−i) = 0.
Show that
1
max |f (z)|.
80 |z|=3
and find all such functions for which equality holds.
|f (0)| ≤
Solution. Consider the holomorphic functions
h(z) = (z − 1)(z + 1)(z − i)(z + i) = z 4 − 1
and


f (z)/h(z)
if z 6= ±1, ±i,



0


f
(1)/4
if z = 1,

0
g(z) = if (i)/4
if z = i,


0

−f
(−1)/4
if
z = −1,



−if 0 (−i)/4 if z = −i
defined on the same domain as f . Let us compute the minimum of h on the circle C = {z | |z| = 3}. We
have
|h(3eiθ )|2 = 34 e4iθ − 1 34 e−4iθ − 1 = (38 − 1) − 2 · 34 cos 4θ,
hence the required minimum of |h| is
p
38 − 1 − 2 · 34 = 34 − 1 = 80.
Setting c = max|z|=3 |f (z)|, it follows that
|g| ≤ c/80
on C, hence on |z| ≤ 3 by the maximum modulus principle. But then |f | ≤ c|z 4 − 1|/80, so
|f (0)| ≤
c|04 − 1|
c
=
.
80
80
If equality is attained, then
c
|f (0)|
= |f (0)| =
.
4
|0 − 1|
80
Combined with the fact |g| ≤ c/80 on C and the maximum modulus principle, we conclude g = cc0 /80 is
constant, where c0 is a constant of absolute value 1. It follows that f = cc0 h/80.
|g(0)| =
1. SPRING 2010
67
Problem 3
TODO
Problem 4
TODO
Problem 5
Let X = RP2 × RP4 .
(a) Find the homology groups H• (X, Z/2).
(b) Find the homology groups H• (X, Z).
(c) Find the cohomology groups H • (X, Z).
Solution.
(a) Recall that
H• (RPn ; Z/2) ∼
=
M
Z/2(i) .
0≤i≤n
Note that Z/2 is a field, and apply the Künneth formula to compute
H• (X; Z/2) ∼
= H• (RP2 ; Z/2) ⊗ H• (RP4 ; Z/2)
∼
= Z/2(0) ⊕ Z/2(1) ⊕ Z/2(2) ⊗ Z/2(0) ⊕ Z/2(1) ⊕ Z/2(2) ⊕ Z/2(3) ⊕ Z/2(4)
∼
= Z/2(0) ⊕ (Z/2)2(1) ⊕ (Z/2)3(2) ⊕ (Z/2)3(3) ⊕ (Z/2)3(4) ⊕ (Z/2)2(5) ⊕ Z/2(6) .
(b) We have
H• (RP2 ) ∼
= Z(0) ⊕ Z/2(1) ,
H• (RP4 ) ∼
= Z(0) ⊕ Z/2(1) ⊕ Z/2(3) .
The general Künneth formula tells us we have split short exact sequences
0
/
L
i
Hi (RP2 ) ⊗ Hn−i (RP4 )
/ Hn (X)
/
L
i
Tor(Hi (RP2 ), Hn−i−1 (RP4 ))
/ 0.
Using this, we compute
H0 (X) ∼
= Z,
H4 (X) ∼
= Z/2,
H1 (X) ∼
= (Z/2)2 ,
H5 (X) ∼
= Z/2,
H2 (X) ∼
= Z/2,
H6 (X) ∼
= 0.
H3 (X) ∼
= (Z/2)2 ,
(c) The universal coefficients for cohomology implies we have split short exact sequences
0
/ Ext(Hn−1 (X), Z)
/ H n (X)
/ Hom(Hn (X), Z)
/0
for all n. We use these to compute
H0 (X) ∼
= Z,
H4 (X) ∼
= (Z/2)2 ,
Problem 6
TODO
H1 (X) ∼
= 0,
H5 (X) ∼
= Z/2,
H2 (X) ∼
= (Z/2)2 ,
H6 (X) ∼
= Z/2.
H3 (X) ∼
= Z/2,
68
2. PAST EXAMS
2. Fall 2009
2.1. Day 1.
Problem 1
TODO
Problem 2
Let CPn be complex projective n-space.
(a) Describe the cohomology ring H • (CPn , Z) and, using the Künneth formula, the cohomology ring H • (CPn ×
CPn , Z).
(b) Let ∆ ⊂ CPn × CPn be the diagonal, and δ = i∗ [∆] ∈ H2n (CPn × CPn , Z) the image of the fundamental
class of ∆ under the inclusion i : ∆ ,→ CPn × CPn . In terms of your description of H • (CPn × CPn , Z)
above, find the Poincaré dual δ ∗ ∈ H 2n (CPn × CPn , Z) of δ.
Solution.
(a) We know that H • (CPn ) ∼
= Z[α]/(αn+1 ) with |α| = 2, so the Künneth formula implies
∼ Z[α]/(αn+1 ) ⊗Z Z[β]/(β n+1 ) ∼
H • (CPn × CPn ) =
= Z[α, β]/(αn+1 , β n+1 ).
(b) TODO
Problem 3
TODO
Problem 4
Let Ω ⊂ C the the open set
Ω = {z | |z| < 2 and |z − 1| > 1}.
Give a conformal isomorphism between Ω and the unit disc ∆ = {z | |z| < 1}.
Solution. Set Ω1 = Ω. We proceed to give a sequence of conformal isomorphisms fi : Ωi → Ωi+1 .
f1 (z) = z − 2
1
f2 (z) =
z
f3 (z) = −4πi z +
Ω2 = {z | |z + 2| < 2 and |z + 1| > 1}
Ω3 = {z | − 1/2 < Re z < −1/4}
1
4
Ω4 = {z | 0 < Im z < π}
f4 (z) = ez
i−z
f5 (z) =
i+z
The composition f = f5 ◦ · · · ◦ f1 is given by
Ω5 = {z | Im z > 0}
Ω6 = {z | |z| < 1}
πi(6 − z)
i − exp −
z−2
.
f (z) =
πi(6 − z)
i + exp −
z−2
Problem 5
TODO
Problem 6
TODO
2. FALL 2009
69
2.2. Day 2.
Problem 1
Let ∆ = {z | |z| < 1} be the unit disk, and ∆∗ = ∆ \ {0} the punctured disk. A holomorphic function f on
∆∗ is said to have an essential singularity at 0 if z n f (z) does not extend to a holomorphic function on ∆
for any n.
Show that if f has an essential singularity at 0, then f assumes values arbitrarily close to every complex
number in any neighbourhood of 0 – that is, for any w ∈ C, ε > 0, and δ > 0, there exists z ∈ ∆∗ with
|z| < δ
|f (z) − w| < ε.
and
Solution. For contradiction assume the opposite – there exists some w ∈ C, ε > 0, and δ > 0, such
that for all z ∈ B(0, δ) \ {0}, we have |f (z) − w| ≥ ε. Consider the function g(z) = 1/(f (z) − w) which is
well-defined on B(0, δ) \ {0} and bounded by 1/ε. Riemann’s Theorem on removable singularities implies
that g extends holomorphically over 0. By abusing notation, we call this extension g again. Suppose that g
has a zero of order n at 0, in other words, there exists a holomorphic function h such that h(z) = g(z)/z n
away from 0. But then
h(z)
1
+w = n +w
f (z) =
g(z)
z
has a pole of order n at 0, contradicting the hypothesis there is an essential singularity there. This completes
our claim.
Remark. This statement is known as the Casorati-Weierstrass Theorem (see Ahlfors, page 129 or Stein &
Shakarchi, page 86).
Problem 2
Let X = S 1 ∨ S 1 be a figure 8, p ∈ X the point of attachment, and let α and β : [0, 1] → X be loops with
base point p (that is, such that α(0) = α(1) = β(0) = β(1) = p) tracing out the two halves of X. Let Y be
the CW complex formed by attaching two 2-disks to X, with attaching maps homotopic to
α2 β
and
αβ 2 .
(a) Find the homology groups Hi (Y, Z).
(b) Find the homology groups Hi (Y, Z/3).
Solution.
(a) We associate a cellular chain complex
···
/0
/ Z2
(2)
( 21 12 )
/ Z2
(1)
(0 0)
/ Z(1)
/0
/ ··· ,
whose homology is
H• (Y, Z) ∼
= Z(0) ⊕ Z/3(1) .
(b) The universal coefficients for homology yields split short exact sequences
0
/ Hn (Y ) ⊗ Z/3
/ Hn (Y, Z/3)
/ Tor(Hn−1 (Y ), Z/3)
/ 0.
We obtain
H• (Y, Z/3) ∼
= Z/3(0) ⊕ Z/3(1) ⊕ Z/3(2) .
Problem 3
TODO
Problem 4
70
2. PAST EXAMS
TODO
Problem 5
TODO
Problem 6
TODO
2.3. Day 3.
Problem 1
TODO
Problem 2
Let τ1 and τ2 ∈ C be a pair of complex numbers, independent over R, and Λ = Zhτ1 , τ2 i ⊂ C the lattice of
integral linear combinations of τ1 and τ2 . An entire meromorphic function is said to be doubly periodic with
respect to Λ if
f (z + τ1 ) = f (z + τ2 ) = f (z)
∀z ∈ C.
(a) Show that an entire holomorphic function doubly periodic with respect to Λ is constant.
(b) Suppose now that f is an entire meromorphic function doubly periodic with respect to Λ, and that f is
either holomorphic or has one simple pole in the closed parallelogram
{aτ1 + bτ2 | a, b ∈ [0, 1] ⊂ R}.
Show that f is constant.
Solution. (a) Let P denote the closed parallelogram defined above. It is clear that P is compact, its
image under f is also compact, hence closed and bounded. On the other hand, f (C) = f (P ) since f is
doubly periodic, so f is bounded. Liouville’s Theorem implies f is constant as required.
(b) Without loss of generality (by shifting P slightly), we may assume P has no poles on ∂P . The Cauchy
Residue Theorem implies combined with the double periodicity yields
Z
X
1
f = 0.
resz f =
2πi ∂P
z
The sum above is taken over all poles of f occurring inside P . If f is holomorphic then f is constant by part
(a). If f had a simple pole, then the above equality is contradiction.
Problem 3
TODO
Problem 4
TODO
Problem 5
Find the fundamental groups of the following spaces:
(a) SL(2, R)
(b) SL(2, C)
(c) SO(3, R)
Solution.
(a) TODO
3. SPRING 2009
71
(b) TODO
(c) TODO
Problem 6
TODO
3. Spring 2009
3.1. Day 1.
Problem 1
TODO
Problem 2
TODO
Problem 3
TODO
Problem 4
Let X = S 1 ∨ S 1 be a figure 8.
(a) Exhibit two three-sheeted covering spaces f : Y → X and g : Z → X such that Y and Z are not
homeomorphic.
(b) Exhibit two three-sheeted covering spaces f : Y → X and g : Z → X such that Y and Z are homeomorphic, but not as covering spaces of X.
(c) Exhibit a normal three-sheeted covering space of X.
(d) Exhibit a non-normal three-sheeted covering space of X.
(e) Which of the above would still be possible if we consider two-sheeted covering spaces instead of theesheeted ones?
Solution. We start by treating π1 (X) = ha, bi where we use the wedge-point as a basepoint.
(a) Consider the two covering spaces corresponding the subgroups
ha3 , b3 , ab−1 , a−1 bi,
ha3 , b, aba−1 , a−1 bai.
The former has the property that removing any single point from the covering space leaves it connected,
while the latter does not. Hence the two covers are not homeomorphic.
(b) Consider the two covering spaces corresponding to subgroups
ha, b3 , bab−1 , b−1 abi,
ha3 , b, aba−1 , a−1 bai.
Drawing the respective spaces it is clear they are homeomorphic. To see they are not isomorphic as covers
consider the preimage of one of the circles in X, say the one given by the image of a. If the covers are
isomorphic, then these preimages would be homeomorphic. In the former case, the preimage is three disjoint
circles and in the latter case a triangle. Counting connected components, we conclude these two spaces
cannot be homeomorphic, hence the two covers are not isomorphic.
(c) Both covers we used in part (b) were normal. To be specific we consider the first one, namely corresponding to the subgroup H = ha, b3 , bab−1 , b−1 abi. It suffices to show H ⊂ π1 (X) is normal. To do this, it
suffices to show H is closed under conjugation. Since π1 (X) is generated by a and b, it suffices to show H
is closed under conjugation by a, a−1 , b, b−1 . Since a ∈ H, we are left to consider b and b−1 only which is a
routine exercise.
72
2. PAST EXAMS
(d) The cover corresponding to the subgroup
ha, b3 , bab, ba−1 bi
is not normal.
(e) Considering two-sheeted covering spaces, we could have answered in the following manner.
(a) Such covers exist for the same reason – for example, consider
ha2 , b2 , abi,
ha, b2 , babi.
(b) The answer is also in the affirmative – consider
ha, b2 , babi,
hb, a2 , abai.
(c) All two-sheeted covers are normal since all index 2 subgroups are normal. Any of the above examples
would suffice.
(d) As explained in part (c) above, no non-normal covers exist.
Problem 5
TODO
Problem 6
TODO
3.2. Day 2.
Problem 1
TODO
Problem 2
Show that the function defined by
f (z) =
∞
X
n
z2
n=0
is analytic in the open disk |z| < 1, but has no analytic continuation to any large domain.
Solution. For the first part, it suffices to compute the radius of convergence of the series defining f .
Set ak = 1 if k = 2n for an integer n ≥ 0, and ak = 0 otherwise. Then the radius of convergence is given by
(
!−1
−1
p
1 if k is a power of 2,
k
lim sup
= 1.
R = lim sup |ak |
0 otherwise
k→∞
k→∞
For the second part, we first claim that it suffices to show limr→1 f (r) = ∞. Assume for contradiction that
f extends to a region Ω strictly larger than the unit disk. It follows that Ω contains some point on the unit
circle and a small disk around it. Since points of the form exp(2πik/2` ) for k, ` ≥ 0 are dense in the unit
circle, it follows that Ω contains such a point, say z0 = exp(2πik/2` ). Since f extends holomorphically over
z0 , it follows that limr→1 f (rz0 ) is bounded. Then
∞
∞
∞
X
X
X
k n 2n
2n
n−` 2n |f (rz0 )| = r exp 2πi ` 2 ≥ C + r exp 2πik2
r ,
=C +
2
n=0
n=`
n=`
where C can be chosen to be a constant independent of r. If the above expression were bounded, this would
contradict the unboundedness of limr→1 f (r). It suffices to provide an argument for our last point, namely
f (r) =
∞
X
n=0
n
r2 ≥
N
X
n=0
n
N
r2 ≥ N r2 .
3. SPRING 2009
73
N +1
Above, we could have chosen N to be any positive integer. Given such a choice of N , consider r = N −1/2
so that
√
f (r) ≥ N .
Since N could have been chosen arbitrarily large, this furnished the necessary claim.
,
Nasko: Is there a way to solve the second part of the given problem with techniques from complex analysis
rather than brute force?
Problem 3
TODO
Problem 4
TODO
Problem 5
TODO
Problem 6
Let X = S 2 × RP3 and Y = S 3 × RP2 .
(1) Find the homology groups Hn (X, Z) and Hn (Y, Z) for all n.
(2) Find the homology groups Hn (X, Z/2) and Hn (Y, Z/2) for all n.
(3) Find the homotopy groups π1 (X) and π1 (Y ).
Solution.
(a) Recall that
H• (S 2 ) = Z(0) ⊕ Z(2) ,
H• (RP3 ) = Z(0) ⊕ Z/2(1) ⊕ Z(3) ,
H• (S 3 ) = Z(0) ⊕ Z(3) ,
H• (RP2 ) = Z(0) ⊕ Z/2(1) .
Since the homology of S n is free and finitely generated in all degrees, the Künneth formula in its simplest
form implies
H• (X) ∼
= H• (S 2 ) ⊗ H• (RP3 ) ∼
= Z(0) ⊕ Z/2(1) ⊕ Z(2) ⊕ (Z ⊕ Z/2)(3) ⊕ Z(5) ,
3
2 ∼
∼
H• (Y ) = H• (S ) ⊗ H• (RP ) = Z(0) ⊕ Z/2(1) ⊕ Z(3) ⊕ Z/2(4) .
(b) Universal coefficient for homology implies
H0 (X, Z/2) ∼
= Z/2,
H3 (X, Z/2) ∼
= (Z/2)2 ,
H1 (X, Z/2) ∼
= Z/2,
H4 (X, Z/2) ∼
= Z/2,
H2 (X, Z/2) ∼
= (Z/2)2 ,
H5 (X, Z/2) ∼
= Z/2,
H0 (Y, Z/2) ∼
= Z/2,
H3 (Y, Z/2) ∼
= Z/2,
H1 (Y, Z/2) ∼
= Z/2,
∼
H4 (Y, Z/2) = Z/2,
H2 (Y, Z/2) ∼
= Z/2,
∼
H5 (Y, Z/2) = Z/2.
(c) We compute
π1 (X) ∼
= π1 (S 2 ) × π1 (RP3 ) ∼
= 1 × Z/2 ∼
= Z/2,
3
2 ∼
∼
π1 (Y ) = π1 (S ) × π1 (RP ) = 1 × Z/2 ∼
= Z/2.
74
2. PAST EXAMS
3.3. Day 3.
Problem 1
TODO
Problem 2
TODO
Problem 3
TODO
Problem 4
TODO
Problem 5
TODO
Problem 6
Let X and Y be two CW complexes.
(a) Show that χ(X × Y ) = χ(X)χ(Y ).
(b) Let A and B be two subcomplexes of X such that X = A∪B. Show that χ(X) = χ(A)+χ(B)−χ(A∩B).
Solution. In order to define χ(X) and χ(Y ), we need to assume they are finite.
(a) For all n ∈ Z, let the number of n-cells of X and Y be xn and yn respectively. It is a trivial observation
that
xn = rank Hn (X n , X n−1 ),
yn = rank Hn (Y n , Y n−1 ).
It is an easy algebraic exercise to show that the Euler characteristic of a finite chain complex does not change
when we take its homology. Therefore cellular homology implies
X
X
X
χ(X) =
(−1)n rank Hn (X) =
(−1)n rank Hn (X n , X n−1 ) =
(−1)n xn
n
n
n
and similarly for Y . On the otherP
hand, the CW structures on X and Y induce such a structure on X × Y
in which the number of n-cells is i xi yn−i . Therefore,

!
X
X
X
X
X
χ(X × Y ) =
(−1)n
xi yn−i =
(−1)i+j xi yj =
(−1)i xi  (−1)j yj  = χ(X)χ(Y ).
n
i
i,j
i
j
(b) Since A and B are subcomplexes of X, there exists neighbourhoods UA and UB of A and B respectively
which deformation retract for the respective subcomplexes. It follows that there exists a Mayer-Vietoris long
exact sequence relative H• (A ∩ B), H• (A) ⊕ H• (B), and H• (X). The required equality follows directly by
counting ranks in this sequence.
4. Fall 2008
4.1. Day 1.
Problem 1
TODO
Problem 2
4. FALL 2008
Evaluate the integral
Z
0
∞
75
√
t
dt.
(1 + t)2
√
Solution. We start with a change of variable t = x2 or x = t (the square root is taken to be positive
here), which yields
Z ∞
Z ∞
Z ∞ √
x2
x2
t
dt
=
2
dx
=
dx.
I=
2 2
(1 + t)2
(1 + x2 )2
0
−∞ (1 + x )
0
We are lead to consider the meromorphic function f (z) = z 2 /(1 + z 2 )2 which has two double poles at i and
−i respectively. The residue at i is
1 d
z
1
z2
= 2i lim
= .
(z − i)2
2
2
3
z→i 1! dz
z→i (z + i)
(z − i) (z + i)
4i
resi f = lim
Consider a semicircular contour in the upper half-plane spanning the interval [−R, R] for R > 0. Since f
is rational and the degree of the numerator is 2 less than the degree of the denominator, it follows that the
integral over the semicircle converges to 0 as R → ∞. Therefore
π
I = 2πi resi f = .
2
TODO
Problem 3
TODO
Problem 4
Let X = (S 1 × S 1 ) \ {p} be a once-punctured torus.
(a) How many connected, 3-sheeted covering spaces f : Y → X are there?
(b) Show that for any of these covering spaces, Y is either a 3-times punctured torus or a once-punctured
surface of genus 2.
Solution.
(a) TODO
(b) TODO
Problem 5
TODO
Problem 6
TODO
4.2. Day 2.
Problem 1
TODO
Problem 2
Let Ω ⊂ C be the open region
Ω = {z | |z − 1| < 1 and |z − i| < 1}.
Find a conformal map f : Ω → ∆ of Ω onto the unit disk ∆ = {z | |z| < 1}.
76
2. PAST EXAMS
Solution. Set Ω1 = Ω. We proceed to give a sequence of conformal isomorphisms fi : Ωi → Ωi+1 .
1
z
−1 + i
f2 (z) = i z +
2
Ω2 = {z | Re z > 1/2 and Im z < −1/2}
f1 (z) =
Ω3 = {z | Re z > 0 and Im z > 0}
f3 (z) = z 2
i−z
f4 (z) =
i+z
Ω4 = {z | Im z > 0}
Ω5 = {z | |z| < 1}
The composition f = f4 ◦ · · · ◦ f1 is given by
f (z) =
z 2 − 2i(i − 1)z − 2i
.
3z 2 + 2i(i − 1)z + 2i
Problem 3
TODO
Problem 4
TODO
Problem 5
TODO
Problem 6
Let X = S 2 ∧ RP2 be the wedge of the 2-sphere and the real projective plane. (This is the space obtained
from the disjoint union of the 2-sphere and the real projective plane by the equivalence relation that identifies
a given point in S 2 with a given point in RP2 , with the quotient topology.)
(a) Find the homology groups Hn (X, Z) for all n.
(b) Describe the universal covering space of X.
(c) Find the fundamental group π1 (X).
Solution.
e • (S 2 ) ⊕ H
e • (RP2 ) and
e • (X) ∼
(a) Recall that H
=H
e • (S 2 ) ∼
H
= Z(2) ,
e • (RP2 ) ∼
H
= Z/2(1) ,
so
e • (X) ∼
H
= Z/2(1) ⊕ Z(2) .
Since X is connected, we have
e • (X) ⊕ Z(0) ∼
H• (X) ∼
=H
= Z(0) ⊕ Z/2(1) ⊕ Z(2) .
e of X can be described as the union of three 2-spheres of radius 1
(b) The universal covering space X
centered at (−2, 0, 0), (0, 0, 0), and (2, 0, 0) sitting inside R3 (the first two touch at (−1, 0, 0) and the latter
e → X is the quotient under the identification x ∼ −x.
two at (1, 0, 0)). The covering map q : X
(c) Observing that the covering space described above is two-sheeted, we obtain π1 (X) ∼
= Z/2. Alternatively, one could also derive this from the van Kampen Theorem which implies that
π1 (X) ∼
= π1 (S 2 ) ∗ π1 (RP2 ∼
= 1 ∗ (Z/2) ∼
= Z/2.
4. FALL 2008
77
4.3. Day 3.
Problem 1
TODO
Problem 2
TODO
Problem 3
Let X and Y be compact, connected, oriented 3-manifolds, with
π1 (X) = (Z/3) ⊕ Z ⊕ Z,
π1 (Y ) = (Z/6) ⊕ Z ⊕ Z ⊕ Z.
(a) Find Hn (X, Z) and Hn (Y, Z) for all n.
(b) Find Hn (X × Y, Q) for all n.
Solution. (a) Let us start with X and we will later handle Y analogously. We expect non-trivial homology
only in degrees 0 to 3 since X is a 3-manifold. Connectedness implies H0 (X) ∼
= H 0 (X) ∼
= Z. Orientedness
3
∼
implies H3 (X) ∼
H
(X)
Z.
The
Hurewicz
Theorem
implies
=
=
H1 (X) ∼
= π1 (X)/[π1 (X), π1 (X)] ∼
= (Z/3) ⊕ Z2 .
Finally, we are left to determine H2 (X). Poincaré duality implies H 2 (X) ∼
= H1 (X). Since X is a closed
manifold, there is a finite CW complex structure. Since the double dual of a free abelian group is naturally
isomorphic to itself, and the groups in the cellular complex are such, a version of universal coefficients from
cohomology from homology holds. We compute
H2 (X) ∼
= Hom(H 2 (X), Z) ⊕ Ext(H 3 (X), Z) ∼
= Z2 .
Similarly one can compute the homology of Y . We summarize
H• (X) ∼
= Z(0) ⊕ (Z/3 ⊕ Z2 )(1) ⊕ Z2(2) ⊕ Z(3) ,
∼ Z(0) ⊕ (Z/6 ⊕ Z3 )(1) ⊕ Z3 ⊕ Z(3) .
H• (Y ) =
(2)
(b) Using the fact all Tor(−, Q) groups vanish, then universal coefficients implies
H• (X; Q) ∼
= H• (X; Z) ⊗ Q ∼
= Q(0) ⊕ Q2(2) ⊕ Q2(2) ⊕ Q(3) ,
H• (Y ; Q) ∼
= H• (Y ; Z) ⊗ Q ∼
= Q(0) ⊕ Q3 ⊕ Q3 ⊕ Q(3) .
(2)
(2)
Since Q is a field, a simple version of the Künneth formula implies
14
11
5
H• (X × Y ; Q) ∼
= H• (X; Q) ⊗ H• (Y ; Q) ∼
= Q(0) ⊕ Q5(1) ⊕ Q11
(2) ⊕ Q(3) ⊕ Q(4) ⊕ Q(5) ⊕ Q(6) .
Problem 4
TODO
Problem 5
TODO
Problem 6
TODO
78
2. PAST EXAMS
5. Spring 2008
5.1. Day 1.
Problem 1
Let K = C(x) be the field of rational functions in one variable over C, and consider the polynomial
f (y) = y 4 + x · y 2 + x ∈ K[y].
(a) Show that f is irreducible in K[y].
(b) Let L = K[y]/(f ). Is L a Galois extension of K?
(c) Let L0 be the splitting field of f over K. Find the Galois group of L0 /K.
Solution. (a) This follows from applying the following generalized version of Eisenstein’s criterion to the
UFD K = C(x) and the prime ideal (x) ⊆ K:
(Generalized Eisenstein’s Criterion) Let D be a UFD and let f (x) = an xn + an−1 xn−1 + · · · + a0 ∈ D[x]
be a polynomial. Suppose there exists a prime ideal P such that
(1) an 6∈ P ,
(2) an−1 , . . . , a0 ∈ P ,
(3) a0 6∈ P 2 .
Then f is irreducible over F [x], where F is the field of fractions of D. If f is primitive, then it is also
irreducible in D[x].
(b) TO DO: From the wording of part (c) it seems that f doesn’t split completely in L but I can’t figure
out how to show it. Part (c) is done assuming that part (b) has been done.
(c) Let α be a root of f (y) = 0, so that L = K(α). Then L0 = L(β), where β is another root of f (y) = 0
such that β 6∈ L. Note that −α is also a root of f (y) = 0, so that the irreducible polynomial for β over L has
degree 2. Hence [L0 : K] = [L0 : L][L : K] = 2 · 4 = 8, and the Galois group of L0 /K has order 8. Now, the
Galois group associated to the splitting field of an irreducible polynomial of degree 4 must be a transitive
subgroup of S4 . The transitive subgroups of S4 are S4 , A4 , D4 , C4 and the Klein four-group. The only one
of these with order 8 is D4 , so Gal(K/Q) = D4 .
Problem 2
Let f be a holomorphic function on the unit disk ∆ = {z | |z| < 1}. Suppose |f (z)| < 1 for all z ∈ ∆, and
that
1
1
f
=f −
= 0.
2
2
Show that |f (0)| ≤ 1/3.
Solution. We consider the function g : ∆ → C given by

f (z)
1


 (z − 1/2)(z + 1/2) if z 6= ± 2 ,
g(z) = f 0 (1/2)
if z = 1/2,


 0
−f (−1/2)
if z = −1/2,
which is holomorphic on ∆. Next pick an arbitrary 1/2 < r < 1, and consider the function h(z) =
(z − 1/2)(z + 1/2) = z 2 − 1/4. We would like to find the minimal value of |h| on the circle Cr given by
|z| = r. This is the square of the minimal value of
|h(reiθ )|2 = r2 −
r cos 2θ
1
+
2
16
5. SPRING 2008
79
p
which is r2 − r/2 + 1/16, hence |h| ≤
r2 − r/2 + 1/16. Since g = f /h and |f | < 1, it follows that
2
−1/2
|g| ≤ (r − r/2 + 1/16)
on Cr . By the maximum modulus principle |g| ≤ (r2 − r/2 + 1/16)−1/2 on the
disk ∆r given by |z| ≤ r. But |g| = |f |/|h|, so
|f | ≤ p
and
|z 2 − 1/4|
r2 − r/2 + 1/16
,
1
|f (0)| ≤ p
.
2
4 r − r/2 + 1/16
Taking the limit as r → 1, we obtain
|f (0)| ≤
1
.
3
Problem 3
Let CPn be complex projective n-space.
(a) Describe (without proof) the cohomology ring H • (CPn , Z).
(b) Let i : CPn → CPn+1 be the inclusion of CPn as a hyperplane in CPn+1 . Show that there does not exist
a continuous map f : CPn+1 → CPn such that the composition f ◦ i is the identity on CPn .
Solution.
(a) We have H • (CPn , Z) ∼
= Z[x]/(xn+1 ) where deg(x) = 2.
(b) For contradiction, assume such f exists. Say H • (CPn , Z) ∼
=
= Z[x]/(xn+1 ) and H • (CPn+1 , Z) ∼
n+2
∗
Z[y]/(y
). For degree reasons, we have f (x) = ky for some k ∈ Z. Then
0 = f ∗ (0) = f ∗ (xn+1 ) = f ∗ (x)n+1 = k n+1 y n+1 ,
hence k n+1 = 0 and k = 0. It follows that f ∗ = 0, but this is impossible since
i∗ ◦ f ∗ = (f ◦ i)∗ = id∗CPn = idH • (CPn ,Z) .
Problem 4
TODO
Problem 5
TODO
Problem 6
TODO
5.2. Day 2.
Problem 1
TODO
Problem 2
Let V be an n-dimensional vector space over a field K, and Q : V × V → K a symmetric bilinear form. By
the kernel of Q we mean the subspace of V of vectors v such that Q(v, w) = 0 for all w ∈ V , and by the
rank of Q we mean n minus the dimension of the kernel of Q.
Let W ⊂ V be a subspace of dimension n − k, and let Q0 be the restriction of Q to W . Show that
rank(Q) − 2k ≤ rank(Q0 ) ≤ rank(Q).
80
2. PAST EXAMS
Solution. Recall that for linear morphisms f : U → V , g : V → W , we have dim Ker(g ◦ f ) ≤
dim Ker f + dim Ker g.
Let Q0 : W × W → K be the symmetric bilinear form on W induced by Q. Suppose Q and Q0 correspond to
e : V → V ∗ and Q
f0 : W → W . If i : W ,→ V is the natural inclusion, then consider the following
the maps Q
diagram.
W
f0
Q
i∗
i
V
/ W∗
O
e
Q
/ V∗
TODO
Nasko: The second inequality is easy. I have some ideas about the latter but still haven’t fully realized
them.
Problem 3
TODO
Problem 4
Let S be a compact orientable 2-manifold of genus g, and let S2 be its symmetric square, that is, the quotient
of the ordinary product S × S by the involution exchanging factors.
(1) Show that S2 is a manifold.
(2) Find the Euler characteristic χ(S2 ).
(3) Find the Betti numbers of S2 .
Solution. (a) It is clear that S2 is Hausdorff and second countable, therefore it suffices to furnish a
cover by charts. Let q : S × S → S2 be the quotient map, and ∆ ⊂ S × S the diagonal. Away from ∆
the map q is a 2-sheeted cover, and therefore, S2 \ q(∆) is a manifold. It suffices to furnish charts around
points along q(∆). Therefore, it suffices to take S = R2 and prove S2 is a manifold in that case. The key
point is to note S = R2 ∼
= C. Therefore, we can think of a point in S2 as a pair of complex numbers,
disregarding ordering. Such complex pairs {α, β} can always be arranged to be the root of a quadratic
equation z 2 − (α + β)z + αβ = 0. Note that the space of monic quadratic polynomials, properly topologized,
is homeomorphic to C2 (take as coordinates the coefficients of 1 and z). We have furnished a map S2 → C2 .
Since each monic quadratic polynomial has precisely two roots (by the fundamental theorem of algebra), it
can be shown this map is a homeomorphism, and we conclude S2 ∼
= C2 is a manifold.
(b) TODO
(c) TODO
Nasko: I know how to do (a) but not (b) and (c).
Problem 5
TODO
Problem 6
TODO
5.3. Day 3.
Problem 1
5. SPRING 2008
For c a non-zero real number, evaluate the integral
Z ∞
0
81
log z
dz.
+ c2
z2
Solution. Note that the projected answers for c and −c would necessarily agree. Therefore, we may
assume c is positive, carry out the computation, and in the very last step replace c with |c|. Start by choosing
a branch of the logarithm well-defined on C \ (−∞, 0]i, and define f (z) = log z/(z 2 + c2 ). Make a choice
of two constants R and ε satisfying 0 < ε < R. We choose a contour which traverses the semicircle from
R to −R in the upper half-plane, the interval from −R to −ε, followed by the semicircle from −ε to ε in
the upper half-plane, and finally the interval from ε to R. It is not hard to show that letting R → ∞ and
ε → 0, the integral of f along the two semicircles approaches 0. The integral over the negative line segment
is related to the integral over the positive one:
Z R
Z R
Z R
Z R
Z −ε
log(−z)
log z + πi
log z
πi
log z
dz
=
dz
=
dz
=
dz
+
πi
dz.
2
2
2
2
2
2
2
2
2
2
z +c
z +c
ε
ε
ε z +c
ε z +c
−R z + c
The residue of f at ic is
resic f = log(ic) resic
Also recall the computation
1
log c + πi/2
=
.
z 2 + c2
2ic
∞
Z
1
1 ∞ 1
π
dz
=
dz = .
2 + c2
2+1
z
c
z
2c
0
0
Let I denote the value of the integral we are seeking. Putting all pieces of information we listed so far,
applying the Cauchy Residue Theorem, and letting R → ∞, ε → 0, we obtain
Z
π(log c + πi/2)
π2 i
= 2πi resic f =
.
2c
c
Simplifying the previous equation and replacing c with |c|, we have shown that
2I +
I=
π log |c|
.
2|c|
Problem 2
TODO
Problem 3
Let S be a compact orientable 2-manifold of genus 2 (that is, a 2-holed torus) and let f : S → S be any
orientation-preserving homeomorphism of finite order.
(a) Show that f must have a fixed point.
(b) Is this statement still true if we drop the hypothesis that f is orientation reversing? Prove or give a
counterexample.
(c) Is this statement still true if we replace S by a compact orientable 2-manifold of genus 3? Again, prove
or give a counterexample.
Solution.
(a) TODO
(b) The statement is false. If one positions symmetrically a 2-holed torus around the origin in R3 , then
the map x 7→ −x is an orientation-reversing homeomorphism without fixed points.
(c) The statement is false if the genus is 3. One can position a 3-holed torus symmetrically around the
origin in R3 , such that rotation by π along an axis of symmetry is a orientation-preserving homeomorphism
without fixed points.
82
2. PAST EXAMS
Nasko: Any ideas about this problem?
Problem 4
(a) State Fermat’s Little Theorem on powers in the field F37 with 37 elements.
(b) Let k be any natural number not divisible by 2 or 3, and let a ∈ F37 be any element. Show that there
exists a unique solution to the equation
xk = a
in F37 .
(c) Solve the equation
x5 = 2
in F37 .
Solution. (a) Fermat’s Little Theorem in this specific case would state that every a ∈ F37 satisfies a37 = a.
36
Alternatively, we may state that every a ∈ F×
= 1.
37 satisfies a
(b) If a = 0, then it is easy to see the unique solution is x = 0. Next, assume a 6= 0. Note that since k is not
divisible by 2 or 3, it is then coprime to 36 = 22 · 32 , and there exist integers s and t satisfying ks + 36t = 1.
Let us start by showing uniqueness. If x is some solution, then raising xk = a to the power s, we obtain
as = xks = xks+36t = x1 = x.
For the existence claim, we note that x = as is always a solution.
(c) Note that (−7) · 5 + 1 · 36 = 1, so the desired solution is
x = 2−7 = 17−1 = 24.
Problem 5
TODO
Problem 6
TODO
6. Fall 2007
6.1. Day 1.
Problem 1
Let f (x) = x4 − 7 ∈ Q(x).
(a) Show that f is irreducible in Q[x].
(b) Let K be the splitting field of f over Q. Find the Galois group of K/Q.
(c) How many subfields L ⊂ K have degree 4 over Q? How many of them are Galois over Q?
Solution.
(a) This follows from Eisenstein’s criterion with the prime p = 7.
√
√
√
(b) By observation, K = Q( 4 7, i). This has degree 8 over Q since [Q( 4 7) : Q] = 4 and i 6 inQ( 4 7. Hence
the Galois group of K/Q has order 8. Now, we know that the Galois group associated to the splitting field
of an irreducible polynomial of degree 4 must be a transitive subgroup of S4 . The transitive subgroups of S4
are S4 , A4 , D4 , C4 and the Klein four-group. The only one of these with order 8 is D4 , so Gal(K/Q) = D4 .
(c) The subfields L ⊂ K of degree 4 over Q are the subfields Q ⊂ L ⊂ K such that [K : L] = 2. These are
in bijective correspondence with the subgroups of Gal(K/Q) of order 2. The group D4 has 5 subgroups of
6. FALL 2007
83
order 2 (if D4 has the presentation hx, y; x4 , y 2 , xyxyi, then the subgroups are hx2 i, hyi, hxyi, hx2 yi, hx3 yi),
so there are 5 such subfields L. The normal subgroups correspond to Galois extensions L/Q. Only the
subgroup hx2 i is a normal subgroup of D4 , so only one of the subfields is Galois over Q.
Problem 2
Lemma. Let f be a convex function on (a, b). Let a < s < t < u < b. Then
f (t) − f (s)
f (u) − f (s)
f (u) − f (t)
≤
≤
.
t−s
u−s
u−t
(†)
Proof. Let t = λs + (1 − λ)u, then λ =
u−t
u−s .
f (t) ≤
The convexity of f implies that
u−t
t−s
f (s) +
f (u).
u−s
u−s
Rearranging, we obtain
f (u) − f (s)
f (t) − f (s)
≤
.
t−s
u−s
t−s
, and the convexity of f implies that
Similarly, if we let t = λu + (1 − λ)s, then λ = u−s
f (t) ≤
u−t
t−s
f (u) +
f (s).
u−s
u−s
Rearranging, we obtain
f (u) − f (s)
f (u) − f (t)
≤
.
u−s
u−t
Now we return to the problem. For x ∈ (a, b), choose δ > 0 such that (x − δ, x + δ) ∈ (a, b). First suppose
that z ∈ (x − δ, x). Applying the second inequality in (†) to x − δ < z < x, we obtain
f (x) − f (x − δ)
f (z) − f (x)
≤
,
δ
z−x
and applying the outer inequality in (†) to z < x < x + δ, we obtain
f (z) − f (x)
f (x + δ) − f (x)
≤
.
z−x
δ
We can obtain similar inequalities if z ∈ (x, x + δ).
Hence |f (x) − f (z)| ≤ C|x − z| for all z ∈ (x − δ, x + δ). This proves the continuity of f .
Problem 3
Let τn : S n → S n be the antipodal map, and let X be the quotient S n × S m by the involution (τn , τm ) –
that is,
X = S n × S m /(x, y) ∼ (−x, −y).
(a) What is the Euler characteristic of X?
(b) Find the homology groups of X in case n = 1.
Solution.
(a) Recall that H∗ (S n ) ∼
= Z(0) ⊕ Z(n) , so
(
0 if n is odd,
χ(S ) = 1 + (−1) =
2 if n is even.
n
n
Künneth formula implies χ(S n ×S m ) = χ(S n )χ(S m ). It is easy to see that the projection map p : S n ×S m →
X is a 2-fold cover, hence
(
0 if m or n is odd,
1
n
m
χ(X) = χ(S )χ(S ) =
2
2 if m and n are both even.
84
2. PAST EXAMS
(b) In the case m = 1, a cut and paste argument enables us to identify X with the 2-torus, hence
H• (X) ∼
= Z(0) ⊕ Z2(1) ⊕ Z(2) . We will later see this calculation fits with the result for m > 1. In what follows,
assume m > 1. Consider the following open cover of X:
A = exp(i(0, π/2) ∪ i(π, 3π/2))/∼,
B = exp(i(π/2, π) ∪ i(3π/2, 2π))/∼ .
m
It is not hard to check that A ' B ' S and A ∩ B ' S m t S m . The relevant parts of the associated
Mayer-Vietoris sequence and the associated maps are as follows.
0
/ H1 (X)
/ H0 (A ∩ B)
/ H1 (X)
/ Z2
/ H0 (A) ⊕ H0 (B)
∼
=
0
0
/ Hm+1 (X)
/ Hm (A ∩ B)
/ Hm+1 (X)
/ Z2
We conclude
(
H• (X) ∼
=
/0
/ H0 (X)
/0
∼
=
/ Z2
( 11 11 )
∼
=
0
/ H0 (X)
/ Hm (A) ⊕ Hm (B)
1
1
1 (−1)m+1
/ Hm (X)
/0
/ Hm (X)
/0
∼
=
/ Z2
Z(0) ⊕ Z(1) Z/2(m)
Z(0) ⊕ Z(1) ⊕ Z(m) ⊕ Z(m+1)
if n is even,
if n is odd.
Note that this computation extends to the case m = 1 handled above.
Problem 4
Construct a surjective conformal mapping from the pie wedge
Ω = {z = reiθ | θ ∈ (0, π/4), r < 1}
to the unit disk
∆ = {z | |z| < 1}.
Solution. Set Ω1 = Ω. We proceed to give a sequence of conformal isomorphisms fi : Ωi → Ωi+1 .
f1 (z) = z 4
Ω2 = {z | |z| < 1 and Im z > 0}
f2 (z) = z + 1
1
f3 (z) =
z
Ω3 = {z | |z − 1| < 1 and Im z > 0}
f4 (z) = i z −
Ω4 = {z | Re z > 1/2 and Im z < 0}
1
2
Ω5 = {z | Rez > 0 and Im z > 0}
f5 (z) = z 2
i−z
f6 (z) =
i+z
The composition f = f6 ◦ · · · ◦ f1 is given by
Ω6 = {z | Im z > 0}
Ω7 = {z | |z| < 1}
f (z) = −i
z 8 + 2iz 4 + 1
.
z 8 − 2iz 4 + 1
Problem 5
TODO
Problem 6 Compute the curvature and the torsion of the curve
ρ(t) = (t, t2 , t3 )
6. FALL 2007
85
in R3 .
Solution. The curvature κ of the curve is given by the formula
|ρ0 × ρ00 |
.
κ=
|ρ0 |3
Substituting in the parametrization ρ(t) = (t, t2 , t3 ), we obtain
√
36t4 + 36t2 + 4
κ=
.
(1 + 4t2 + 9t4 )3/2 )
The torsion of the curve is given by the formula
(ρ0 × ρ00 ) · ρ000
τ=
,
|ρ0 × ρ00 |2
and substituting in the given parametrization, we obtain
3
.
κ= 4
9t + 9t2 + 1
6.2. Day 2.
Problem 1
Evaluate the integral
Z
∞
0
x2
dx.
x4 + 5x2 + 4
Solution. Taking
Z
I=
0
∞
1
x2
dx =
4
2
x + 5x + 4
2
Z
∞
−∞
x2
dx,
(x2 + 1)(x2 + 4)
we are lead to consider the meromorphic function f (z) = z 2 /((z 2 + 1)(z 2 + 4)). It has four simple poles at
±i and ±2i respectively. We are interested in
1
z2
1
lim (z − i) 2
=− ,
0! z→i
(z + 1)(z 2 + 4)
6i
z2
1
1
lim (z − 2i) 2
= .
res2i f =
0! z→2i
(z + 1)(z 2 + 4)
3i
resi f =
A simple estimation leads to
I=
1
π
2πi (resi f + res2i f ) = .
2
6
Problem 2
TODO
Problem 3
TODO
Problem 4
Consider the following three topological spaces
A = CP3 ,
B = S2 × S4,
C = S2 ∧ S4 ∧ S6,
where CP3 is the complex projective 3-space, S n is the n-sphere, and ∧ denotes the wedge product.
(a) Calculate the cohomology groups (with integer coefficients) of all three.
(b) Show that A and B are not homotopy equivalent.
(c) Show that C is not homotopy equivalent to any compact manifold.
86
2. PAST EXAMS
(a) We know that H • (CPn ) ∼
= Z[α]/(αn+1 ) with |α| = 2, so the relative cohomology groups
Solution.
are
H • (CP3 ) ∼
= Z(0) ⊕ Z(2) ⊕ Z(4) ⊕ Z(6) .
•
2 ∼
For the second space, we have H (S ) = Z(0) ⊕ Z(2) and H • (S 4 ) ∼
= Z(0) ⊕ Z(4) . Since all groups are finitely
generated and free, a simple version of the Künneth formula implies
H • (S 2 × S 4 ) ∼
= H • (S 2 ) ⊗ H • (S 4 ) ∼
= Z(0) ⊕ Z(2) ⊕ Z(4) ⊕ Z(6) .
Finally, reduced cohomology is additive with respect to the wedge product. To pass from reduced to nonreduced for a connected space such as S 2 ∧ S 4 ∧ S 6 , we only need to add a factor of Z in degree 0. We
compute
e • (S 2 ∧ S 4 ∧ S 6 ) ∼
e • (S 2 ) ⊕ H
e • (S 4 ) ⊕ H
e • (S 6 )
H • (S 2 ∧ S 4 ∧ S 6 ) ∼
= Z(0) ⊕ H
= Z(0) ⊕ H
∼
= Z(0) ⊕ Z(2) ⊕ Z(4) ⊕ Z(6) .
(b) Note that α ` α = α2 6= 0 in H • (CP3 ). On the other hand, since H • (S 2 × S 4 ) ∼
= H • (S 2 ) ⊗ H • (S 4 )
as rings, the product of any two degree 2 classes is trivial there. We conclude that the cohomology rings of
CP3 and S 2 × S 4 are not isomorphic, hence the spaces are not homotopy equivalent.
(c) For contradiction, assume C is homotopy equivalent to a compact manifold M . Since C is connected,
so is M . Van Kampen’s Theorem implies
π1 (M ) ∼
= π1 (C) ∼
= π1 (S 2 ) ∗ π1 (S 4 ) ∗ π1 (S 6 ) ∼
=1∗1∗1∼
= 1,
hence M is simply-connected. If M were non-orientable, it would admit a connected 2-sheeted cover. This
is however impossible in view of π1 (M ) = 1, hence M is orientable. A simple computation with cohomology
rings reveals that the ring H • (M ) ∼
= H • (C) is trivial. In other words α ` β = 0 if one of the degrees |α|
•
or |β| is positive for α, β ∈ H (M ). If α denotes the generator of H 2 (M ), then Poincaré duality implies
there exists β ∈ H 4 (M ) such that (α ` β)[M ] = 1, in particular α ` β 6= 0. This yields the necessary
contradiction.
Problem 5
TODO
Problem 6
TODO
6.3. Day 3.
Problem 1
TODO
Problem 2
TODO
Problem 3
(1) Show that any continuous map from the 2-sphere S 2 to a compact orientable manifold of genus
g ≥ 1 is homotopic to a constant map.
(2) Recall that if f : X → Y is a map between compact, oriented n-manifolds, the induced map
f∗ : Hn (X) → Hn (Y ) is a multiplication by some integer d, called the degree of the map f . Now let
S and T be compact oriented 2-manifolds of genus g and h respectively, and f : S → T a continuous
map. Show that if g < h, then the degree of f is zero.
6. FALL 2007
87
Solution. (a) Let f : S 2 → Σg be a continuous map. Since g ≥ 1, the universal cover of Σg is R2 , say
via p : R2 → Σg . Note that π1 (S 2 ) = 1 ⊂ p∗ (π1 (R2 )) = p∗ (1) = 1, and S 2 is path-connected and locally
path-connected, hence the lifting criterion furnished a map fe: S 2 → R2 lifting f . Since R2 is contractible,
the map fe is nullhomotopic. Composing this homotopy with p, we conclude that f is also nullhomotopic.
(b) Let us start by describing the cohomology ring of Σg , a compact orientable surface of genus g. We have
H 0 (Σg ) ∼
= Z which are generated respectively by the unit 1 ∈ H 0 (Σg ) and the fundamental
= Z and H 2 (Σg ) ∼
2
class [Σg ] ∈ H (Σg ). The degree 1 part H 1 (Σg ) is generated freely by 2g elements α1 , β1 , . . . , αg , βg which
satisfy αi ` βi = [Σg ] and all other products are trivial. The ring H • (Σg ; Q) is analogously presented but
with coefficients in Q.
For convenience set S = Σg and T = Σh . Consider a map f : Σg → Σh for g < h of degree d = deg f .
A dimension count implies the map f ∗ : H 1 (Σh ; Q) → H 1 (Σg ; Q) has nontrivial kernel. Pick a non-trivial
element γ in the kernel of this map. Up to scaling and reordering of the generators, we may write γ = α1 + γ 0
where γ 0 is a linear combination of all generators except α1 . Note that
γ ` β1 = (α1 + γ 0 ) ` β1 = [Σh ].
We have
d[Σg ] = f ∗ [Σh ] = f ∗ (γ ` β1 ) = f ∗ (γ) ` f ∗ (β1 ) = 0 ` f ∗ (β1 ) = 0,
hence d = 0 as desired.
Problem 4
TODO
Problem 5
TODO
Problem 6
TODO