Representation Theory Seminar, Fall 2015
Sean McAfee
October 31, 2015
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Contents
1 9/1/15: Introduction to Verma Modules
3
2 9/8/15: Properties of Verma Modules
11
3 9/11/15: Properties of Verma Modules, Continued
16
4 9/15/15: Duality for Highest Weight Modules
22
5 9/22/15: Finite Length Modules, Revisited
24
6 9/25/15: Multiplicity of Irreducible Modules and the Grothendieck
Group
30
7 10/2/15: The Harish-Chandra Homomorphism
34
8 10/6/15: The Harish-Chandra Homomorphism, Continued 38
9 10/20/15: Another Approach to the Weyl Character Formula
42
10 10/23/15: Another Approach to the Weyl Character Formula, Continued
47
11 10/27/15: Another Approach to the Weyl Character Formula, Concluded
50
2
1
9/1/15: Introduction to Verma Modules
This lecture will use the following objects:
• g a complex, semisimple Lie algebra
• h a choice of Cartan subalgebra of g
• R the root system corresponding to (g, h) ⊂ h∗
• R+ a choice of positive roots in R
P
α
• ρ = 21
α∈R+
• n=
L
gα
α∈R+
• n̄ =
L
g−α (so both n and n̄ are nilpotent Lie subalgebras of g)
α∈R+
• b = h ⊕ n (a Borel subalgebra of g)
• U(g) the universal enveloping algebra of g
For future use, we note that
g = n̄ ⊕ h ⊕ n,
and thus, by Poincaré-Birkhoff-Witt,
U(g) = U(n̄) ⊗ U(b).
We start by considering the category M(g) of all g-modules. This is a
gigantic category, and we wish to find an interesting, useful subcategory. Our
first try is to look at the category Mfg (g) of finitely generated g-modules.
This is (apparently) still too general to be useful. On the other hand, the
category Mfd (g) of finite dimensional representations of g is (apparently) too
small to be interesting.
The categorical ”sweet spot” will turn out to be the category H of highest
weight modules.
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Definition 1.1. We call a g-module V a highest weight module if
i) V is a finitely generated U(g)-module, and
ii) any v ∈ V is U(b)-finite; i.e. dim(U(b)v) < ∞.
We let H be the category of highest weight modules. Since by definition any module in the category is finitely generated, and since any finite
dimensional representation is U(b)-finite, we have the following inclusions:
Mfd (g) ⊂ H ⊂ Mfg (g) ⊂ M(g).
We fix V ∈ H and λ ∈ h∗ for the following definition:
Definition 1.2. The weight subspace of V corresponding to λ is written V λ and is given by
V λ = v ∈ V | (ξ − λ(ξ))k v = 0 for some fixed k >> 0, ∀ξ ∈ h .
(Note: the space V λ is assumed to be nonzero.)
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We want to analyze the properties of V λ . First, a lemma:
Lemma 1.3. Let V be a finitely generated U(g)-module. The following are
equivalent:
i) V is a highest weight module.
ii) V satisfies
L λ
a) V =
V , and each V λ is finite dimensional.
λ
b) There exists a finite set S0 of weights of V such that, for any
weight ν of V , there exists a µ ∈ S0 such that µ − ν is a sum
of positive roots. (This is the origin of the terminology ”highest
weight module”.)
Proof.
(i) → (ii): Let V be finitely generated and U(b)-finite.
Let v1 , ..., vn be a U(g)-basis for V . Then we have that
U = U(b)v1 + · · · + U(b)vn
is a finite dimensional subspace of V which generates V over U(g). By
Poincareé-Birkhoff-Witt we have that
U(g) = U(n̄)U(b),
thus
V = U(g)U = U(n̄)U.
This gives us a map
φ0 : U(n̄) ⊗C U → V,
which is a restriction of the surjective map
φ : U(g) ⊗C U → V.
In fact, the map φ0 is an h-homomorphism. Observe that the action of
h on U(n̄) is adjoint; hence for H ∈ h, Y ∈ U(n̄),
(ad H)(Y ) = HY − Y H ∈ U(g).
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Let H ∈ h, and let Y ⊗ u ∈ U(n̄) ⊗C U . Then
H·(Y ⊗u) = (ad H(Y ))⊗u+Y ⊗(H·u) = (HY )⊗u−(Y H)⊗u+(Y H)⊗u = (HY )⊗u.
Now, we look at the weights of U(n̄) ⊗C U . First, observe that, for
Y = x−α1 · · · x−αk ∈ U(n̄) and H ∈ h we have
(ad H)(x−α1 · · · x−αk ) = −α1 (H) − · · · − αk (H),
thus weights of U(n̄) are of the form −α1 − · · · − αk , αi ∈ R+ .
Furthermore, since U is finite dimensional, it decomposes into a finite
sum of finite dimensional weight spaces:
M
U=
U µ.
Thus, the weights of U(n̄) ⊗C U are of the form
µ − α1 − · · · − αk ;
that is, there are a finite set of weights µ such that each weight is some
µ minus a postive sum of positive roots.
Now, since φ0 is a surjective h-homomorphism, the quotient
V = (U(n̄) ⊗C U )/ ker φ0
has the same property. This shows both a) and b) hold.
(ii) → (i): Suppose that V is a finitely generated U(g)-module, and satisfies both
properties a) and b). We need to show that any v ∈ V is U(b)-finite.
Since V is by assumption a direct sum of finite dimensional weight
spaces, it will suffice to show that, for a fixed weight space V λ , a given
v ∈ V λ is U(b)-finite.
So, let v ∈ V λ . We have that
U(b)v ⊂
M
V λ+α1 +···+αk .
(1)
αi ∈R+
We will show that there are a finite number of positive roots αi that may
appear in the direct sum above. We fix a finite set S0 guaranteed by
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property b). We claim (after possibly modifying our set S0 to another
finite set of weights S1 ) that any weight λ can be uniquely written in
the form
X
λ=µ−
mα α, (mα ∈ Z+ , µ ∈ S1 ).
α∈R+
Suppose that we have
X
X
λ = µ1 −
mα α = µ2 −
nα α, (µ1 , µ2 ∈ S0 ),
and consider
X
X
X
X
µ1 −µ2 =
(mα −nα )α =
kα α =
pα α −
qα α, (pα , qα ∈ Z+ ).
Thus we have
µ1 +
X
qα α = µ2 +
X
pα α.
If neither the qα ’s nor the pα ’s are all identically zero, we remove µ1
and µ2 from S0 and replace them by
X
X
µ = µ1 +
q α α = µ2 +
pα α.
Then our modified set (call it S00 ) has the same property as in b).
We can repeat this process for any two µi , µj ∈ S00 that differ by an
integral sum of roots; since S0 is finite, this process must terminate and
the result is a set S1 which has the same property as in b), and such
that each µ ∈ S1 can not be expressed the sum of another µ0 ∈ S1 plus
a nonzero sum of positive roots. Working in S1 , we follow the same
argument as above: if λ can be written in two ways as
X
X
λ = µ1 −
mα α = µ2 −
nα α (µ1 , µ2 ∈ S1 ),
then
µ1 +
X
qα α = µ2 +
X
pα α,
and by our construction of S1 , we must have that pα = qα = 0 for all
α. Thus, µ1 = µ2 and our decomposition is unique, as claimed.
Now that we have shown the unique decomposition of a weight λ in
terms of the weights contained in S0 , we will show that there are finitely
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many αi that may appear in ( 1 ). We let {βj } be a vector space basis
for the α’s, and we write (uniquely)
X
λ=µ−
mβ β, (mβ ≥ 0).
We may then write
λ + α1 + · · · + αk = λ +
X
X
mβ β +
kβ β
X
=µ−
(mβ − kβ )β.
kβ β, (kβ ≥ 0) = µ −
X
By the definition of S0 , each coefficient (mβ −kβ ) must be a nonnegative
integer. Since the mβ are fixed by uniqueness of the decomposition of
λ, that leaves only finitely many possible values for kβ . This implies
that there are only finitely many αi that may appear in the direct sum
(1), as claimed. This shows that our arbitrary v ∈ V was U(b)-finite.
So, to summarize, V is called a highest weight module since it is a U(g)module which is a direct sum of weight spaces V λ . Furthermore, these weight
spaces are determined by elements of the finite set S0 .
The category H has the following property: given a short exact sequence
0 → V 0 → V → V 00 → 0
of U(g)-modules, we have that
V 0 , V 00 ∈ H ⇐⇒ V ∈ H.
We have that H is an abelian category. Furthermore, the property above
makes it (by definition) a Serre subcategory of Mfg (g). It is closed under
extensions. So, we have created a category that contains all finite dimensional
representations of g; furthermore it is abelian and we can do homological
algebra with it.
We are now ready to construct Verma modules. Let λ ∈ h∗ be a weight
of a given U(g)-module V . We create a (one dimensional) b-module Cλ−ρ in
the following way: for X ∈ b = h ⊕ n, and for z ∈ Cλ−ρ ∼
= C, let
(
(λ − ρ)(X)z if X ∈ h
Xz =
,
0
if X ∈ n
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and extend by linearity to all of b. Furthermore, we may extend this
action to a (left) U(b)-action on Cλ−ρ . Now, viewing U(g) as a right U(b)module, we create the tensor product
U(g) ⊗U (b) Cλ−ρ .
This is a left U(g)-module in the obvious way; we call it the Verma
module corresponding to λ, and write M (λ).
So, given a U(g)-module V , we get an entire family of Verma modules
indexed by the weights of V . By construction, we have M (λ) ∈ H. An
interesting fact (which we will prove in the future) is that, for any w ∈ W ,
we have that M (λ) and M (wλ) have the same infinitesimal character.
We make a few observations, which will be used below. First, note that,
since
U(g) = U(n̄) ⊗C U(b),
we have that
M (λ) = U(g) ⊗U (b) Cλ−ρ = (U(n̄) ⊗C U(b)) ⊗U (b) Cλ−ρ
= U(n̄) ⊗C Cλ−ρ .
This tensor power has the natural structure of both a left and right h-module.
Note that the action of h on U(n̄) results in weights which are positive sums
of negative roots, and that the action of h on Cλ−ρ (by definition) results in
the weight λ − ρ. This means that the weights of M (λ) are all of the form
X
λ−ρ−
mα α, (mα ∈ Z+ , α ∈ R+ ).
In other words, the highest weight of M (λ) is λ−ρ. Furthermore, since M (λ)
is generated over U(g) by the element 1 ⊗ 1, we have that
dim M (λ)λ−ρ = 1.
Proposition 1.4. The Verma module M (λ) has a unique maximal proper
submodule N .
Proof. Since U(g) is Noetherian, we have that U(g) ⊗U (b) Cλ−ρ is Noetherian
as well. Therefore, M (λ) contains a (possibly non-unique) proper maximal
submodule N . Observe that, for any submodule N , either N has weight
λ − ρ or it does not. If N does have weight λ − ρ, then it must contain 1 ⊗ 1,
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since the one-dimensional term Cλ−ρ in M (λ) is responsible for the weight
λ − ρ. Since M (λ) is generated over U(g) by 1 ⊗ 1, this would imply that
N = M (λ). So, in the case that N is a maximal proper submodule of M (λ),
we must have that
N λ−ρ = 0.
Let P be an arbitrary submodule of M (λ), and suppose that P 6⊆ N . Then,
by maximality of N , we must have that N + P = M (λ). This means that
N + P has weight λ − ρ, which is impossible since by the above neither N
nor P has weight λ − ρ. Thus, we must have that P ⊆ N for any submodule
P of M (λ), completing the proof.
The above proposition implies that M (λ) has a unique irreducible quotient
L(λ) := M (λ)/N.
We thus have a map λ 7→ L(λ), which takes weights λ to irrreducible highest
weight modules L(λ). Next time, we will show that every irreducible highest
weight module is obtained in this way.
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2
9/8/15: Properties of Verma Modules
Last time, we defined the category H of highest weight U(g)-modules. Of
particular interest in this category are Verma modules, which are indexed by
weights λ:
M (λ) := U(g) ⊗U (b) Cλ−ρ .
We showed that each such M (λ) contains a unique maximal submodule N ,
thus each M (λ) has a unique irreducible quotient
L(λ) := M (λ)/N.
Furthermore, it is a consequence of Borel-Weil that, in the case of λ
dominant integral, we have that L(λ) is finite dimensional.
Proposition 2.1.
i) If V is an irreducible highest weight module, then V ∼
= L(λ) for some
highest weight λ.
ii) For two highest weights λ and µ,
L(λ) ∼
= L(µ) ⇐⇒ λ = µ.
Proof.
i) Suppose V is an irreducible highest weight module. We showed in the
last seminar that this implies there is a finite set of weights S0 such
that for λ ∈ S0 and for any weight ν, λ − ν is a positive sum of positive
roots. So, let λ ∈ S0 . Then we have V λ 6= 0. By the property of S0 ,
we also have that, for any v ∈ V λ and ξ ∈ gα (α ∈ R+ ),
ξv = 0.
This implies that
nv = 0.
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We call such a v a primitive vector of weight λ. Given such a
primitive vector v, we define a map
φ0 : U(g) ⊗C Cλ → V
Y ⊗ 1 7→ Y · v.
This map naturally extends to a map U(g)⊗U (b) Cλ → V in the following
way: for X ∈ U(b), we have
φ0 (Y X ⊗ 1) = Y Xv.
If X = n + h ∈ b, then
Xv = (n + h)v = nv + hv = 0 + λ(h)v = λ(h)v.
This shows that Y X ⊗ 1 = Y ⊗ (Xv) ∈ U(g) ⊗U (b) Cλ , hence the map
φ : U(g) ⊗U (b) Cλ → V
Y ⊗ (Xv) 7→ φ0 (Y X ⊗ 1)
is well defined. We have that φ is a morphism of g-modules, since for
X ∈ U(g) we have
X(Y ⊗ 1) = XY ⊗ 1 7→ X(Y v).
This map is also nontrivial, since φ(1 ⊗ 1) = v. Thus, since V is
irreducible, φ must be surjective. Since U(g) ⊗U (b) Cλ is just the Verma
module M (λ + ρ), this implies that V is an irreducible quotient of
M (λ + ρ). That is,
V = L(λ + ρ),
as claimed.
ii) One direction is trivial: if λ = µ then by construction L(λ) ∼
= L(µ). For
the other direction, suppose that L(λ) ∼
L(µ).
Consider
the
projection
=
map
π : M (λ) → L(λ).
This map takes 1 ⊗ 1 7→ 1 ⊗ 1. The image is a highest weight vector of
weight λ − ρ. Since L(λ) ∼
= L(µ), µ is another weight of L(λ) and thus
can be written as
X
µ−ρ=λ−ρ−
nβ β,
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where each nβ is a nonnegative integer and the β’s are a vector space
basis of R+ as in the previous seminar. This implies that
X
λ−µ=
nβ β.
We can make the same argument with the projection map M (µ) →
L(µ); thus
X
µ−λ=
pβ β.
Since λ and µ are both highest weights, uniqueness of expression in
the basis of β’s implies that all nβ ’s and pβ ’s are identically zero, thus
λ = µ, as needed.
We will be interested in calculating the infinitesimal character of these
irreducible modules L(λ). Let Z(g) denote the center of the universal enveloping algebra U(g), and recall the Harish-Chandra homomorphism:
γ : Z(g) → U(h).
Writing Z(g) = U(h) ⊕ U(g)n (this is a consequence of Poincaré-BirkoffWitt), this homomorphism is given by projection onto the U(h) factor of
Z(g). Note that this implies, for z ∈ Z(g),
z − γ(z) ∈ U(g)n.
We will need one more map in order to determine the infinitesimal character of L(λ). For ξ ∈ n and µ ∈ h∗ , we have the natural map
h→C
ξ 7→ µ(ξ).
This map can be extended by linearity to a map U(h) → C. We can identify
U(h) with the symmetric algebra S(h), which can in turn be identified with
the algebra P(h∗ ) of polynomials on the dual space h∗ . Thus, given µ ∈ h∗
and a polynomial Q ∈ P(h∗ ) we can interpret the map U(h) → C as a map
φµ : P(h∗ ) → C
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Q 7→ Q(µ).
We then have, for z ∈ Z(g) and 1 ⊗ 1 ∈ M (λ),
z(1 ⊗ 1) = γ(z)(1 ⊗ 1) = γ(z) ⊗ 1
= 1 ⊗ γ(z) · 1 = φλ−ρ (γ(z))(1 ⊗ 1).
Finally, we let
χλ (z) := φλ−ρ (γ(z)).
It is a fact (we will prove this later) that
χλ = χµ ⇐⇒ λ = wµ, w ∈ W.
Assuming this is true, we show that M (λ) has infinitesimal character χλ . Let
X ⊗ 1 ∈ M (λ), z ∈ Z(g). Then
z(X ⊗ 1) = zX ⊗ 1 = Xz ⊗ 1 = X(z(1 ⊗ 1))
= X(χλ (z)(1 ⊗ 1)) = χλ (z)(X(1 ⊗ 1)) = χλ (z)(X ⊗ 1).
So, we have that M (λ) has infinitesimal character χλ ; therefore, L(λ) has
infinitesimal character χλ as well.
Finally, observe that, given L(λ) and L(µ) with the same infinitesimal
character, we have that λ = wµ for some w in the Weyl group. Since the
Weyl group is finite, we have proven the following:
Proposition 2.2. For any infinitesimal character χ, there exists a finite set
λ1 , ..., λk ∈ h∗
such that L(λi ) has infinitesimal character χ.
We recall the following definitions: let V be an arbitrary module. A
finite filtration of V is a finite chain of proper inclusions of submodules of
V:
{0} ( U1 ( U2 ( · · · ( Un = V.
We say that V is of finite length if there exists a finite filtration such that
each quotient Ui /Ui−1 is irreducible. We call such a filtration a JordanHölder filtration. It can be shown that, although a Jordan-Hölder filtration
is not unique, the number of terms in any such filtration is constant. We call
this constant the length of the filtration. It can also be shown that (up to
shuffling) the quotients of a Jordan-Hölder filtration are isomorphic.
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Example 2.3. Let U and W be irreducible highest weight modules, and
consider the highest weight module V = U ⊕ W . Then we have a filtration
{0} ( U ( V.
The quotients of this filtration are V /U ∼
= W and U/0 ∼
= U , which are
irreducilbe; thus the filtration is Jordan-Hölder of length 2. Furthermore, we
see that the filtration
{0} ( W ( V
is also Jordan-Hölder of length 2, and has the same irreducible quotients.
It is a fact (which we will prove next time) that all Verma modules are
of finite length.
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3
9/11/15: Properties of Verma Modules, Continued
At the end of the last lecture, we defined an arbitrary module to be of finite
length if it admits a finite filtration such that each quotient of successive
submodules in the filtration is irreducible. We stated that all Verma modules
are of finite length; a fact which we will now prove. First, we make a few
observations.
Let V be a highest weight module. Then, by Lemma 1.3, V is a direct
sum of finite dimensional weight spaces:
M
V =
V λ.
We claim that this implies that this implies V is Z(g)-finite. Indeed, for a
given V λ , by definition we have that there is a fixed k ∈ Z+ such that, for
all v ∈ V λ and ξ ∈ h,
(ξ − λ(ξ))k v = 0.
The action of the Cartan subalgebra h commutes with the action of Z(g), so
for z ∈ Z(g) we have
z(ξ − λ(ξ))k v = (ξ − λ(ξ))k zv.
That is, we have that for all v ∈ V λ ,
Z(g)v ∈ V λ .
Therefore, since each V λ is finite dimensional, it follows that V is Z(g) finite.
By definition, V is finitely generated. The C-span of a choice of finite
generators forms a subspace U of V , hence
V = U(g)U.
We let I denote the kernel of the action of Z(g) on U . Since U is finite (say
n)-dimensional, this action can be represented by n×n matrices. This implies
that the quotient Z(g)/I is finite dimensional as well. Since I annihilates U ,
it annihilates V = U(g)U as well.
Thus, we have shown that there exists an ideal in Z(g) of finite codimension annihilating V . Notice that V does not neccessarily have infinitesimal
character; in fact, a finitely generated highest weight module has infinitesimal
character precisely when the ideal I above has codimension 1 in U(g) (there
is nothing mysterious here; this is precisely when Z(g) acts by scalars).
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Proposition 3.1. Any highest weight module V is of finite length.
Proof. Given a highest weight module V and arbitrary weight λ, define
V[λ] := v ∈ V | ∀z ∈ Z(g), (z − χλ (z))k v = 0 for some k >> 0 .
We say these subspaces V[λ] have generalized infinitesimal character.
We claim that V is a finite direct sum of such V[λ] ’s. To see this, first observe
that we have a natural finite dimensional representation
π : Z(g) → GL(Z(g)/I).
Since Z(g) is abelian, we can (by Dixmier’s Lemma) decompose Z(g)/I
into subspaces which are each annihilated by (z − χi (z))ki , where z is any
element of Z(g), ki is a fixed positive integer, and χi is a fixed homomorphism
Z(g) → C. Since Z(g)/I is finite dimensional, there are finitely many of these
χi . Consider the polynomial
n
Y
P (z) =
(z − χi (z))ki .
i=1
By construction, P (z) annihilates Z(g)/I; that is, P (z) ∈ I. Thus we have
that P (z) annihilates V = U(g)U , and we may therefore decompose as a
finite direct sum of spaces
V[λi ] = v ∈ V | (z − χλi (z))ki v = 0 ,
as claimed.
Since V is a finite direct sum of such subspaces, it will suffice to show
that each V[λ] is of finite length. For a fixed p, we let
Fp V[λ] = {v ∈ V | (z − χλ (z))p v = 0} .
Notice that the quotient
Fp V[λ] /Fp−1 V[λ]
has infinitesimal character. If we can show that each of these quotients has
finite length, then so will V[λ] and consequently so will V . Let F denote such
a quotient. Again, this is an (infinite dimensional) highest weight module
with infinitesimal character.
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Given an arbitrary highest weight module V and a filtration
0 ( W ( V,
we have a short exact sequence
0 → W → V → V /W → 0
which, given a weight µ, restricts to a short exact sequence of weight spaces
0 → W µ → V µ → (V /W )µ → 0.
This exact sequence splits (since the objects are vector spaces), so we have
dim V µ = dim W µ + dim(V /W )µ .
By induction on the length of the filtration, and using the above as a base
case, we see that given a filtration
0 = M0 ( M1 ( · · · ( Mn = V,
we have that
µ
dim V =
n
X
dim(Mi /Mi−1 )µ .
i=1
We use this fact to prove that F = Fp /Fp−i is of finite length. Let
M0 ( M1 ( · · · ( Mk = F
be a filtration of length k in F , where each Mi is a maximal proper submodule
of Mi+1 . By the above, for a given weight µ we have
µ
dim F =
k
X
dim(Mi /Mi−1 )µ + dim M0µ .
i=1
Each quotient (Mi /Mi−1 )µ is irreducible (except possibly M0µ ), thus each
quotient is isomorphic to L(wλ) for some weight λ. In particular, for a
highest weight µ = wλ − ρ, we have that L(wλ)wλ−ρ is one dimensional,
hence
dim F wλ−ρ ≥ #L(wλ) in Mi /Mi−1 .
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Since the Weyl group W is finite, and by our previous result that L(λ) =
L(µ) ⇐⇒ λ = wµ, we have that
X
dim F wλ−ρ ≥ #L(wλ), w ∈ W in Mi /Mi−1 .
w∈W
That is, the total sum of L(λ)’s appearing in all irreducible quotients in our
filtration is bounded by the finite sum of dimensions of finite dimensional
highest weight spaces of F . Thus F is of finite length, hence V is of finite
length as well.
So far we have shown that, if V is a highest weight module, then
• V is of finite length,
• V is Z(g)-finite, and
• any v ∈ V is annihilated by nk for some sufficiently large k.
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Proposition 3.2. Let V be a finitely generated U(g)-module. The following
are equivalent:
i) V is a highest weight module.
ii) V is Z(g)-finite, and any v ∈ V is annihilated by nk for some sufficiently large k.
Proof. By our earlier work, we have that i) implies ii). We prove the other
direction. Suppose that V is a finitely generated U(g)-module that satisfies
ii). Then, for any v ∈ V , U(n)v is finite dimensional. In particular, this
holds for any vector in a set of generators of V , hence we have that V is
generated by a finite dimensional n-invariant subspace U . We need to show
that any v ∈ V is U(b)-finite. We proceed by induction on the dimension of
U.
Suppose that U is one-dimensional, and let u ∈ U . By assumption there
exists a k such that nk u = 0. Since U is spanned by u, this implies that nu =
0, i.e. n annihilates U . Recall that for the Harish-Chandra homomorphism
γ we have
z − γ(z) ∈ U(g)n , ∀z ∈ Z(g),
thus
Z(g)u = γ(Z(g))u.
It is a fact (which we will not prove) that U(h) is finitely generated as a
γ(Z(g))-module. This implies that U(g)u is finite dimensional, hence (since
U(n)u = 0) U(b)u is finite dimensional as well. So, we have that V contains
a vector u which is U(b)-finite.
Consider the U(g)-bilinear map of U(g)-modules g × V → V given by
(X, v) 7→ X · v. This induces a U(g) homomorphism g ⊗ V → V which (since
it is a homomorphism) takes U(b)-finite vectors to U(b)-finite vectors. By the
universal property of tensor products, this means that the set of U(b)-finite
vectors in g ⊗ V is equivalent to the set of U(b)-finite vectors in g × V . Thus,
we have that the set VU (b) of U(b)-finite vectors in V are a g-submodule of
V . We have that u ∈ VU (b) , u generates U , and U generates V , thus V is
U(b)-finite, as claimed.
Now assume that this holds for any U as described above of dimension
less than n. By Lie’s Theorem, we may choose a vector u ∈ U that is
simultaneously annihilated by all elements of n. Let U 0 = Cu and let V 0 =
20
U(g)U 0 . We then have a short exact sequence of U(g)-modules
0 → V 0 → V → V /V 0 → 0.
The image of U in V /V 0 under this map has dimension dim U − 1, thus is
U(b)-finite by hypothesis. Therefore, we have that U is a direct sum of two
U(b)-finite modules, completing the induction.
So, to summarize, we can categorize the category of highest weight modules V by the properties
i) V is finitely generated as a U(g)-module,
ii) V is Z(g)-finite, and
iii) for any v ∈ V , there is a sufficiently large k such that nk v = 0.
Notice that property iii) is finer than the requirement that any v ∈ V is
U(n)-finite. If we replace this property by requiring any v ∈ V to be U(n)finite, we get a larger class, called the category of Whittaker modules.
21
4
9/15/15: Duality for Highest Weight Modules
Given a Lie algebra g and (π, V ) a finite-dimensional representation of g, we
can form the contragredient representation (π ∗ , V ∗ ), where V ∗ = {f : V → C}
and
(π ∗ (X)f )(v) := −f (π(X)v) , f ∈ V ∗ , v ∈ V, X ∈ g.
This map V
V ∗ is an exact, contravariant functor: if φ : V → U is a linear
map of finite dimensional vector space, then it induces a map φ∗ : U ∗ → V ∗
of the duals of these vector spaces. Furthermore, we have that
V
V∗
V ∗∗ = V.
A natural question follows: can we generalize this functoriality to a larger
category of (potentially infinite dimensional) representations? The problem
is that, if V is infinite dimensional, then V ∗ is much bigger than V ; that is,
V ∼
6= V ∗ . Hence we have that V is isomorphic to a submodule of V ∗∗ but
certainly not isomorphic to V ∗∗ (which is itself much larger than V ∗ ).
If V happens to be a highest
module, however, there is some hope.
Lweight
In this case we have that V =
V λ , where each of V λ are finite dimensional
weight spaces. Define
M
Vb :=
(V λ )∗ .
We have a natural embedding (V λ )∗ ,→ V ∗ . We call the direct sum Vb of these
embeddings the U(h)-finite linear forms of V . We remark that, since h is
abelian, we can describe the action of h on these summands by their Jordan
form.
Now, we claim that Vb is a g-invariant subspace of V ∗ : indeed, as in the
previous seminar we have that the map g⊗C Vb → V ∗ takes U(h)-finite vectors
to U(h)-finite vectors, hence this map factors through Vb . This implies that
Vb is a U(g)-module. We thus have a functor V
Vb which is ”better” than
∗
b
our original map V
V , since V is a more ”narrow” vector space.
We still have a problem, though: the vector spaces (V λ )∗ transform under
−λ by the definition of the contragredient representation at the beginning of
the lecture. This means that, instead of highest weight modules, these summands are ”lowest weight” modules. The fix will involve essentially ”twisting” by the negative identity.
22
More precisely, given h ⊂ g, look at
Auth (g) = {T ∈ Aut(g) | T (h) = h} .
∗
∗
∗
Any T ∈ Aut(g)
L with T (h) = h induces an automorphism T : h → h .
Since g = h ⊕ gα , we have that T (gα ) = gT ∗ α . Thus, we have a surjective
group homomorphism
φ : Auth (g) → Aut(R),
where R is the root space corresponding to (g, h). We have that −I ∈ Aut(R);
choose τ ∈ φ−1 (−I) (so we have that τ |h = −I). With the notation from
earlier, if (π, V ) is a finite-dimensional representation of g, then let
π̂ = π ∗ |Vb .
For X ∈ g, we have an action of X on Vb given by (π̂ ◦ τ )(X). So, letting
Ve = Vb and π̃ = π̂ ◦ τ , we have a functor V
Ve , with
M
M
Ve λ ,
V =
V λ and Ve =
as needed. Furthermore, we now have a g-compatible chain of functors
V
Ve
e
Ve ,
eλ
with each V λ ∼
= Ve , and with
f
dim V λ = dim Vfλ = dim Vfλ .
As we have seen before, this functor e· is contravariant; furthermore, it
preserves irreducibles. Indeed, if V is irreducible but Ve is not, then take
a nontrivial submodule W in Ve and apply e·. This will give a nontrivial
e
e
submodule in Ve , contradicting the fact that V ∼
= Ve .
] that is, our duality preserves irreducible
By the above, then, L(λ) = L(λ);
U(g)-modules.
23
5
9/22/15: Finite Length Modules, Revisited
The following discussion will hold in any arbitrary abelian categroy; we restrict our attention to the following set-up: let A be a ring, and let M(A)
be the category of A-modules. We recall the following definitions:
Definition 5.1. A nonzero V ∈ M(A) is irreducible if its only submodules
are 0 and V .
Definition 5.2. A finite filtration of V ∈ M(A) is a finite nested family
of proper submodules:
M0 = 0 ( M1 ( M2 ( · · · ( Mk = V.
Definition 5.3. We call the integer k in the previous definition the length
of the filtration.
We have a natural partial ordering on filtrations given by inclusion; that
is, if
F1 : 0 ( M1 ( · · · ( Mk
and
F2 : 0 ( N1 ( · · · ( N`
are two filtrations of V , then F1 F2 if each Mi can be included as a subset
in some Nj .
Definition 5.4. A maximal filtration of a given A-module V under the above
partial ordering is called a Jordan-Hölder filtration.
Lemma 5.5. Let V ∈ M(A), and let 0 ( · · · ( Mk = V be a finite filtration.
The following are equivalent:
i) The filtration is Jordan-Hölder.
ii) All successive quotients of the filtration are irreducible.
Proof. Follows directly from the definition of maximal.
We note that any irreducible V ∈ M(A) has a unique trivial filtration
0 ( V . It is Jordan-Hölder of length 1.
24
Proposition 5.6. Let V ∈ M(A) with a Jordan-Hölder filtration of length
n:
0 ( V0 ( V1 ( · · · ( Vn = V.
If
0 ( U0 ( U1 ( · · · ( Um = V
is another filtration of length m, then m ≤ n.
Proof. We proceed by induction on n. If n = 1, then V is irreducible and
the result is clear. Suppose that n > 1, and that the statement holds for all
W ∈ M(A) with a Jordan-Hölder filtration of length n − 1. Let V have a
Jordan-Hölder filtration of length n:
0 = V0 ( · · · ( Vn = V.
We quotient each term by V1 to get the filtration
0 = V1 /V1 ( V2 /V1 ( · · · ( Vn /V1 = V 0 .
Each quotient term in the above filtration is irreducible, thus it is JordanHölder of length n − 1. Let
0 = U0 ( U1 ( · · · ( Um = V
be another filtration of V , and let 0 ≤ k ≤ m be the smallest integer such
that
Us ∩ V1 = 0, 0 ≤ s ≤ k − 1 and Uk ∩ V1 6= 0.
Since V1 is irreducible, we have Uk ∩V1 = V1 . Furthermore, each of Uk+1 , Uk+2 , ...
contain V1 . Let p : V → V 0 = V /V1 be the projection map, and let
Ui0 = p(Ui ), 1 ≤ i ≤ m. The kernel of p is V1 , thus p|Ui is an injection
0
Ui ,→ Ui0 . We then have Ui−1
6= U for 1 ≤ i ≤ k − 1.
Since V1 ⊆ U` for k ≤ ` ≤ m, then Ui−1 6= Ui0 for k + 1 ≤ i ≤ m. Hence,
0
0
0
0 ( · · · ( Uk−1
( Uk+1
( · · · ( Um
=V0
is a finite filtration of V 0 of length m − 1. By our induction hypothesis, then,
m − 1 ≤ n − 1, therefore m ≤ n, as needed.
As an immediate consequence of the above, we have
25
Corollary 5.7. Let V ∈ M(A) have a Jordan-Hölder filtration of length n.
Then
i) Any other Jordan-Hölder filtration of V has length n, and
ii) any finite filtration of V is a subfiltration of some Jordan-Hölder filtration.
By part i) of the corollary, we can make the following definition:
Definition 5.8. If V ∈ M(A) admits a Jordan-Hölder filtration, say of
length n, then we say V is a module of finite length. We write `(V ) = n
to denote this length.
Not all A-modules admit a Jordan-Hölder filtration. For example, any
infinite dimensional vector space has an arbitrarily long filtration with irreducible quotients: let {vi } be a sequence of linearly independent vectors;
then the filtration
V ) V /hv1 i ) V /hv1 , v2 i ) V /hv1 , v2 , v3 i ) ...
has each quotient irreducible but does not have finite length. Even finitely
generated modules may not admit a Jordan-Hölder filtration: let V = C[x]
and view V as a C[x] module. Then the filtration
V ) V /hxi ) V /hx2 i ) V /hx3 i ) ...
has irreducible quotients but does not terminate.
Lemma 5.9. Let
0 → V 0 → V → V 00 → 0
be an exact sequence of A-modules. Then
i) `(V ) < ∞ ⇐⇒ `(V 0 ), `(V 00 ) < ∞.
ii) If `(V ), `(V 0 ), `(V 00 ) < ∞, then `(V ) = `(V 0 ) + `(V 00 ).
Proof. We begin by proving ii). Suppose that `(V ), `(V 0 ), `(V 00 ) < ∞, and
consider the filtration
0 ( V 0 ( V.
26
This has finite length, thus is contained in some Jordan-Hölder filtration
0 ( V1 ( · · · ( Vn = V.
This implies that Vp = V 0 for some 1 ≤ p ≤ n − 1, hence the filtration
0 ( V1 ( · · · ( Vp = V 0
is Jordan-Hölder of length p, i.e. `(V 0 ) = p. We quotient the higher terms
(those with index p + 1 and higher) of our original filtration of V :
0 = Vp /Vp ( Vp+1 /Vp ( · · · ( Vn /Vp = V 00 .
This has irreducible quotients, hence is Jordan-Hölder of length n − p, i.e
`(V 00 ) = n − p. Thus,
`(V ) = n = `(V 0 ) + `(V 00 ),
proving part ii) of the lemma and the right direction of part i). To prove the
converse of part i), suppose that `(V 0 ), `(V 00 ) < ∞. We then have JordanHölder filtrations
0 = v00 ( · · · Vq0 = V 0
and
0 = V000 ( · · · Vs00 = V 00
. Let p denote the projection V → V 00 , let Vi = Vi0 for 1 ≤ i ≤ q, and let
Vq+j := p−1 (Vj00 ) for 1 ≤ j ≤ s. Then we have a Jordan-Hölder filtration of
V given by
0 = V0 ( V1 ( · · · ( Vq+s = V,
completing the proof.
We denote by Mf l (A) the category of A-modules of finite length. This
is an abelian category, and a full subcategory of M(A).
Definition 5.10. Let V ∈ Mf l (A) have Jordan-H´’older filtration
0 = V0 ( · · · ( Vn = V.
We call the irreducible quotients Vi /Vi−1 the composition factors of this
filtration.
27
Theorem 5.11. Let V ∈ Mf l (A), and let {Li , 1 ≤ i ≤ n}, {Mi , 1 ≤ i ≤ n}
be the composition factors for two Jordan-Hölder filtrations. Then there exists
a permutation σ ∈ Sn such that Mj = Lσ(i) for all i.
Example 5.12. Let U, V be irreducible A-modules, and consider the JordanHölder filtrations
0 ( U ( U ⊕ V and 0 ( V ( U ⊕ V.
Then both filtrations have composition factors isomorphic to U and V , but
their orders are permuted.
Proof. We again proceed by induction on n. If n = 1, then V is irreducible
and we are done. Suppose n > 1, and suppose the theorem holds for all
modules of length ≤ n − 1. Let V ∈ Mf l (A) with `(V ) = n, and with two
Jordan-Hölder filtrations
F1 : 0 = V0 ( · · · ( Vn = V
and
F2 : 0 = U0 ( · · · ( Un = V.
Then we have that
F10 : 0 = V1 /V1 ( · · · ( V /V1 = V 0
is Jordan-Hölder of length n − 1. We create another filtration of length n − 1:
let 1 ≤ k ≤ n be such that
Us ∩ V1 = 0 , 1 ≤ s ≤ k − 1 and Uk ∩ V1 6= 0.
Note that this implies Uk ∩ V1 = V1 . Letting p be the projection V → V 0 =
V /V1 , and letting Ui0 = p(Ui ), a similar argument as in Proposition 5.6 gives
us a Jordan-Hölder filtration
0
0
( Uk+1
( · · · ( V 0 = V /V1 .
F20 : 0 = U00 ( · · · ( Uk−1
Let Li , Mi denote the composition factors of F1 , F2 , respectively. Then the
composition factors of F10 have the form
(Vi /V1 )/(Vi−1 /V1 ) = Li , 2 ≤ i ≤ n.
28
Similarly, the composition factors of F20 have the form
0
(Ui0 /U1 )/(Ui−1
/U1 ) = Mi , 1 ≤ i ≤ k − 1 , k + 1 ≤ i ≤ n.
By our induction hypothesis, then, there is a bijection
π : {1, ..., k − 1, k + 1, ...n} → {2, ..., n}
such that Mi = Lπ(i) . Since we have observed earlier that Uk ∩ V1 = V1 ,
this bijection can be extended to a permutation σ ∈ Sn with Mi = Lσ(i) , as
required.
29
6
9/25/15: Multiplicity of Irreducible Modules and the Grothendieck Group
We again consider the category Mf l (A) of finite length A-modules, where A
is an arbitrary ring. Let V ∈ Mf l (A), with Jordan-Hölder filtration
0 = V0 ( · · · ( Vn = V
and composition factors Li = Vi /Vi−1 .
Definition 6.1. With V as above and M ∈ Mf l (A) an irreducible A-module,
the cardinality of the subset
{i | Li ∼
= M } ⊂ {1, ..., n}
is called the multiplicity of M in V . We denote this multiplicity by m(V ; M ).
Lemma 6.2. Let
0 → V 0 → V → V 00 → 0
be a short exact sequence of finite-length A-modules. Then, for any irreducible
A-module M , we have
m(V ; M ) = m(V 0 ; M ) + m(V 00 ; M ).
Proof. We have that 0 ( V 0 ( V is a filtration of V , thus it admits an
inclusion in some Jordan-Hölder filtration
0 ( · · · ( Vp = V 0 ( · · · ( Vn = V.
The terms leading up to and including Vp are then a Jordan-Hölder filtration
of Vp = V 0 . If we take the quotient of the terms to the right of Vp by Vp ,
we also have a Jordan-Hölder filtration of V 00 . The composition factors of
each of these two filtrations exhaust the composition factors of V (see our
discussion from the previous seminar), thus we see that
m(V ; M ) = m(V 0 ; M ) + m(V 00 ; M ),
as needed.
30
We now relate our discussion of multiplicity to algebraic objects called
Grothendick Groups. Warning: we will blatantly ignore any potential settheoretic issues in our construction of these groups. We have been assured
that everything to follow is kosher in this context...
Let A be an abelian category (so in particular it makes sense to talk
about short exact sequences). Let G(A) be the free abelian group generated
by objects in A. Given a short exact sequence
0→X→Y →Z→0
of such objects, we say that ”Y is an extension of X by Z”. We consider
the subgroup N ≤ G(A) generated by elements of the form Y − X − Z
representing such short exact sequences.
Definition 6.3. The quotient K(A) = G(A)/N is called the Grothendieck
Group (or K-group) of A.
In particular, for a given ring A, we have that Mf l (A) is an abelian
category; we write K = K(Mf l (A)) to denote the K-group of Mf l (A).
We make a few simple observations of K. Let V, U be two arbitrary
A-modules in Mf l (A). We have a short exact sequence
0 → U → U ⊕ V → V → 0,
thus [U ⊕ V ] − [U ] − [V ] = 0 in K, hence [U ⊕ V ] = [U ] + [V ] in K. In
particular, if U = 0, we see that [0] acts as the identity under addition in K.
Also, observe that if U ∼
= V , we have the short exact sequence
0 → V → U → 0 → 0,
hence in this case [V ] = [U ] in K. Finally, we remark that as the quotient
of a free abelian group, generic elements of K are of the form
[U1 ] + · · · + [Un ] − [V1 ] − · · · − [Vm ].
Now, the multiplicity function m(−; M ) described earlier defines a homomorphism from K to Z for any fixed irreducible M :
m(−; M ) : Mf l (A) → Z
V 7→ m(V ; M ).
31
We have seen earlier that m(Y ; M ) = m(X; M ) + m(Z; M ) for any short
exact sequence
0 → X → Y → Z → 0,
thus N is in the kernel of m(−; M ), hence the map above factors through
K. We then have a well-defined homomorphism m([−]; M ) : K → Z. Furthermore, this map will be constant on any isomorphism class of a given
irreducible M , so we have that m([−]; M ) = m([−]; [M ]).
We let I denote the family of equivalence classes of irreducible M in
Mf l (A), and write Z(I) to denote the free abelian group generated by I.
We then have a homomorphism
φ : K → Z(I)
X
[V ] 7→
m([V ]; [M ])[M ].
[M ]∈I
Lemma 6.4. Let V ∈ Mf l (A), with `(V ) = n and composition factors
n
P
{M1 , ..., Mn }. Then [V ] = [Mi ].
i=1
Proof. Induction on n.
Theorem 6.5. The map φ described above gives an isomorphism K ∼
= Z(I).
Proof. To see that φ is surjective, observe that for M irreducible we have
φ([M ]) = [M ]. To prove injectivity, recall that an arbitrary element of K is
of the form
[U1 ] + · · · [Un ] − [V1 ] − · · · − [Vm ].
By applying the property [U ⊕ V ] = [U ] + [V ], we can then write any element
of K in the form [U ] − [V ]. In particular, elements of ker φ have this form.
For such an element in the kernel, we have
0 = φ([U ] − [V ]) = φ([U ]) − φ([V ]),
thus φ([U ]) = φ([V ]). Now, we claim that if φ([U ]) = φ([V ]), then [U ] = [V ],
and injectivity will follow. Indeed, if φ([U ]) = φ([V ]), then m([U ]; M ) =
m([V ]; M ) for any irreducible M . This implies that U and V have the same
composition factors; by Lemma 6.4, then, [U ] and [V ] have the same decomposition into classes of irreducibles in K, hence [U ] = [V ], as claimed.
32
As a final remark, we give an example of the sometimes unexpected behavior of K-groups. Let A be the category of finite dimensional vector spaces
over a field k. Then, by the previous theorem, we have that K(A) ∼
= Z. Let B
be the category of all vector spaces. The natural inclusion (injection) A ,→ B
induces a map F : K(A) → K(B). This map is not injective, however! In
fact, its image in K(B) is trivial. To see this, let V be a finite dimensional
vector space and let U be infinite dimensional. We then have that V ⊕U ∼
= U.
Thus, the short exact sequence
0→V →U ⊕V →U →0
becomes
0 → V → U → U → 0,
hence f ([V ]) = [V ] = [U ] − [U ] = 0.
33
7
10/2/15: The Harish-Chandra Homomorphism
Recall that given a semisimple, complex lie algebra g, we can construct the
universal enveloping algebra as the quotient
U(g) = T (g)/(x ⊗ y − y ⊗ x − [x, y]),
where T (g) is the tensor algebra of g. This construction is the smallest
associative algebra which contains g (which is very far from associative). We
wish to understand the center
Z(g) = {z ∈ U(g) | zu = uz ∀u ∈ U(g)}
of U(g). In particular, we would like to show that Z(g) is ”big” in an appropriate sense and use this fact to analyze U(g)-modules. We can do this by
using the Harish-Chandra homomorphism which has appeared earlier in
our seminar series. Recall that as a consequence of the Poincaré-Birkoff-Witt
Theorem we may write
U(g) = U(h) ⊕ (n− U(g) + U(g)n+ ).
Viewing elements of U(g) as weight vectors of the adjoint representation, it
can be shown that
Z(g) ⊆ U(h) ⊕ U(g)n+ .
The Harish-Chandra homomorphism γ 0 is given by projection onto the U(h)
factor of an element of the center; for z ∈ Z(g), write z = h + p in the above
decomposition. Then define
γ 0 : Z(g) → U(h)
z = h + p 7→ h.
We can slightly modify γ 0 to arrive at a more useful homomorphism P
in the
1
+
α.
following way: choose a system ∆ of positive roots, and let ρ = 2
α∈∆+
Define a map
τ : h → U(h)
H 7→ H − ρ(H)
34
and extend this to a map τ : U(h) → U(h) by letting τ act distributively on
each term of U(h). For example,
τ (H1 ⊗ H2 ) = τ (H1 ) ⊗ τ (H2 ) = (H1 − ρ(H1 )) ⊗ (H2 − ρ(H2 ))
= H1 ⊗ H2 − ρ(H1 )H2 − ρ(H2 )H1 + ρ(H1 )ρ(H2 ).
Now, define γ to be the composition
γ : Z(g) → U(h)
z 7→ τ (γ 0 (z)).
We refer to this construction as ”twisting” the map γ 0 by ρ.
Theorem 7.1. The map γ induces an isomorphism Z(g) → U(h)W , where
U(h)W is the set of elements of U(h) which are invariant under the action of
the Weyl group W .
We will not prove this theorem in its entirety; instead we will show an
example and an application of the theorem in this lecture, then prove the
following in the next lecture:
i) Im γ ∈ U(h)W
ii) γ is a homomorphism
iii) γ does not depend on the choice of positive roots
We give a quick example of the homomorphism: let g = sl(2, C) and
consider the Casimir element (described last semester)
1
Ω = H 2 + XY + Y X ∈ Z(g).
2
We have that H = [X, Y ] = XY − Y X, thus we may rewrite
1
Ω = ( H 2 + H) + 2Y X ∈ U(h) ⊕ U(g)n+ .
2
We then have
1
γ 0 (Ω) = H 2 + H.
2
35
The Weyl group of g is isomorphic to Z/2Z; its nontrivial element w acts on
H by
w · H = −H.
This extends to an action on U(h) (W just acts on each term distributively)
, hence
1
1
w · γ 0 (Ω) = w · ( H 2 + H) = H 2 − H.
2
2
0
Thus we see that γ (Ω) is not W -invariant. Applying the twist by ρ, however,
we have
1
1
γ(Ω) = τ (γ 0 (Ω)) = τ ( H 2 + H) = (H − 1)2 + (H − 1)
2
2
1
1
1
= (H 2 − 2H + 1) + H − 1 = H 2 − H + + H − 1
2
2
2
1
1
= H2 − .
2
2
Then the result is invariant under the action of W ; letting w ∈ W be the
nontrivial element again, we have
1
1
1
1
1
1
w · γ(Ω) = w · ( H 2 − ) = (−H)2 − = H 2 − .
2
2
2
2
2
2
To conclude this lecture, we give an application of the Harish-Chandra
homomorphism. Recall that a U(g)-module M is said to have infinitesimal
character if Z(g) acts on M by scalars. In such a case, these scalars are
given by χλ (z) := λ(γ(z)) for some fixed λ ∈ h∗ , where γ is the HarishChandra homomorphism. We stated without proof earlier in the semester
that, for λ, µ ∈ h∗ ,
χλ = χµ ⇐⇒ λ = wµ
for some w ∈ W . Assume that we have proven that γ is an isomorphism
between Z(g) and U(h)W .
If λ = wµ for some w ∈ W , then we have, for any z ∈ Z(g),
χλ (z) = λ(γ(z)) = wµ(γ(z))
= µ(w−1 γ(z)) = µ(γ(z)) = χµ (z),
36
where the second to last equality comes from the W -invariance of the image
of γ.
For the other direction, suppose that χλ = χµ , but λ 6= wµ for any
w ∈ W . Since W is finite, we may use interpolation and appropriate scaling
to choose a polynomial p on h∗ such that p(wλ) = 1 and p(wµ) = 0 for any
w ∈ W . Now, write
1 X
wp.
p̃ =
|W | w∈W
This polynomial p̃ has the same properties as p, and is clearly W -invariant.
By the Harish-Chandra isomorphism, then, we can find a z ∈ Z(g) such that
γ(z) = p̃. Then we have
χλ (z) = λ(γ(z)) = p̃(λ) = 1,
χµ (z) = µ(γ(z)) = p̃(µ) = 0,
thus χλ 6= χµ , contradicting our assumption and proving the claim.
37
8
10/6/15: The Harish-Chandra Homomorphism, Continued
Last time, we described the Harish-Chandra map γ, claimed it induced an
isomorphism between Z(g) and U(h)W , and gave and example and an application. In this lecture we will prove the following:
i) Im γ ∈ U(h)W
ii) γ is a homomorphism
iii) γ does not depend on the choice of positive roots
Assume for the moment that we have proven i); we will begin by proving
ii) and iii).
Proof that γ is a homomorphism: Let z1 , z2 ∈ Z(g). We show that
γ(z1 z2 ) = γ(z1 )γ(z2 ).
Recall that as a consequence of Poincaré-Birkhoff-Witt we have that
Z(g) ⊆ U(h) ⊕ U(g)n+ .
We claim that γ(z1 )γ(z2 ) is the entire U(h)-component of z1 z2 , i.e.
z1 z2 − γ(z1 )γ(z2 ) ∈ U(g)n+ .
Then, by the uniqueness of the direct sum decomposition, we will have
γ(z1 )γ(z2 ) = γ(z1 z2 ).
Indeed, since z1 is in the center of U(g) and since U(h) is abelian we have:
z1 z2 − γ(z1 )γ(z2 ) = z1 z2 − z1 γ(z2 ) + z1 γ(z2 ) − γ(z1 )γ(z2 )
= z1 (z2 − γ(z2 )) + γ(z2 )z1 − γ(z2 )γ(z1 )
= z1 (z2 − γ(z2 )) + γ(z2 )(z1 − γ(z1 )).
38
Since the terms (z2 − γ(z2 )) and (z1 − γ(z1 )) both lie in U(g)n+ , we see that
z1 (z2 − γ(z2 )) + γ(z2 )(z1 − γ(z1 )) ∈ U(g)n+ ,
hence
z1 z2 − γ(z1 )γ(z2 ) ∈ U(g)n+ ,
as claimed.
Proof that γ does not depend on a choice of positive root system: Let ∆1 , ∆2
be two choices of positive roots system, with corresponding γ1 , γ2 , ρ1 , ρ2 , and
+
U(g)n+
1 , U(g)n2 . Basic theory gives us an element w ∈ W = W (g, h) such
that w∆1 = ∆2 .
We make the following observation: let γ given by a choice of positive
root system with corresponding ρ, and let z = h + p ∈ U(h) ⊕ U(g)n+ . Then,
λ(γ(z)) = λ(h − ρ(h)) = λ(h) − ρ(h) = (λ − ρ)(γ 0 (z)).
Since g is assumed to be complex semisimple in our lecture series, there
exists a compact connected G with g as the complexification of its Lie algebra.
Thus we have that the Weyl groups W (G, H), W (g, h) correspond and we
may choose an element x ∈ G such that
w = Ad(x)|h .
This element x ∈ G acts on weight spaces by
x · gα = gwα ,
thus we have
+
x · U(g)n+
1 = U(g)n2 and x · U(h) = U(h).
Therefore, we see that wγ10 = x · γ10 = γ20 . We also have that wρ1 = ρ2 .
Putting this all together, we see that
λ(γ2 (z)) = (λ − ρ2 )(γ20 (z) = (λ − wρ1 )(wγ10 (z))
= w−1 (λ − wρ1 )(γ10 (z)) = (w−1 λ − ρ1 )(γ10 (z))
39
= w−1 λ(γ1 (z)) = λ(wγ1 (z))
= λ(γ1 (z)).
The last equality follows from the W -invariance of γ1 ; thus we have
γ2 (z) = γ1 (z),
as needed.
We now prove the proposition that the image of the Harish-Chandra map
lies in U(h)W . There will be a few technical details to show, but the main
idea is to prove that, for any w ∈ W and dominant integral λ ∈ h∗ , the Verma
module M (wλ) is contained in M (λ). We will discuss how this implies that
w(γ(z)) = γ(z) for any λ ∈ h∗ and z ∈ Z(g); thus we will have shown that
the image of γ is W -invariant.
Lemma 8.1. Let the (ρ-twisted) Harish-Chandra map γ be defined as above.
In order to show that the image of γ lies in U(h)W , it suffices to show that,
for all w ∈ W and for all dominant integral λ ∈ h∗ ,
λ(w(γ(z)) = λ(γ(z)).
Proof. First, observe that, since h is abelian, there is a natural isomorophism
between U(h) and S(h), the symmetric tensors in h. There is also a natural
ismorphism between S(h) and P(h∗ ), the space of polynomials on h∗ . With
this in mind, fix some z ∈ Z(g) and w ∈ W , and consider the polynomial
w(γ(z)) − γ(z) ∈ P(h∗ ).
The dual space h∗ is generated by the dominant integral weights. Since these
weights form an unbounded, countable subset of h∗ , a polynomial which
vanishes on each dominant integral weight must be identically zero. That is,
if
λ · (w(γ(z)) − γ(z)) = 0
for all dominant integral λ, then
wγ(z) − γ(z) = 0,
40
hence
wγ(z) = γ(z).
Since w and z were arbitrary, this shows that under the assumptions of the
lemma the image of γ is W -invariant, completing the proof.
We can refine the lemma above by observing that, since the Weyl group
is generated by simple reflections, it will suffice to show that
λ(sα (γ(z))) = λ(γ(z))
for all simple reflections sα . Now recall the Verma module
M (λ) = U(g) ⊗U (b) Cλ−ρ
constructed earlier this semester. This has infinitesimal character χλ : Z(g) →
C given by
χλ (z) = λ(γ(z)).
Thus, in order to prove the proposition it will be enough to show that the
Verma modules M (λ) and M (sα λ) have the same infinitesimal character;
then we will have
χλ (z) = λ(γ(z)) = λ(sα γ(z)) = χsα λ .
To do this, we show that M (sα λ) can be viewed as a submodule of M (λ),
hence has the same infinitesimal character.
41
9
10/20/15: Another Approach to the Weyl
Character Formula
The goal of the next three lectures is as follows:
i) Introduce a partial ordering on h
ii) Describe a basis of Verma modules for the Grothendieck (aka K-)
Group
iii) Define the notion of character for g-modules
iv) Use the above language to offer an alternate description/proof of the
Weyl Character Formula
In this lecture, we use the following objects:
• g a semisimple, complex Lie algebra
• h a Cartan subalgebra, with dual h∗
• R = R(g, h) a root system in h∗
• R+ a choice of positive roots
• B a (vector space) basis composed of simple (positive) roots
• H the category of highest weight modules
• W = W (g, h) the Weyl group
We first give a quick description of a partial ordering on h∗ and make a
few observations. For λ, µ ∈ h∗ , we say λ ≤ µ if
X
µ−λ=
mα α , mα ∈ Z≥0 .
α∈B
Definition 9.1. We call λ, µ ∈ h∗ integrally equivalent if they differ by a
sum of (not necessarily positive) roots.
Note that λ ≤ µ implies λ, µ are integrally equivalent. We state the
following two lemmata without proof (they are both an easy combinatorial
exercise).
42
Lemma 9.2. Let λ1 , ..., λm be integrally equivalent elements of h∗ . Then
there exists a µ ∈ h∗ such that λi ≤ µ for 1 ≤ i ≤ m.
Lemma 9.3. Let λ, µ ∈ h∗ be such that λ ≤ µ. Then the set
{ν ∈ h∗ | λ ≤ ν ≤ µ}
is finite.
For a given λ ∈ h∗ , consider the Weyl group orbit θ = W λ. We have that
θ determines an infinitesimal character χλ : Z(g) → C. Consider Hθ , the full
subcategory of H composed of modules with infinitesimal character χλ . We
have seen that its Grothendieck group K = K(Hθ is the free abelian group
with basis [L(wλ)], w ∈ W . The partial ordering on h∗ introduced above
induces a partial ordering on θ.
Lemma 9.4. Let λ ∈ θ. Let L(µ) be a composition factor of the Verma
module M (λ). Then µ ∈ θ and µ ≤ λ.
Proof. A composition factor of M (λ) has the same infinitesimal character
χλ as M (λ). Thus, for some w ∈ W , L(µ) = L(wλ); that is, µ ∈ θ. We have
that weights of M (λ) are all of the form
X
mα α , mα ∈ Z≥0 .
λ−ρ−
α∈B
Since µ − ρ is the highest weight of L(µ), it must be a weight of M (λ), so
X
µ−ρ=λ−ρ−
mα α,
αnB
hence
λ−µ=
X
mα α,
α∈B
i.e. µ ≤ λ.
Proposition 9.5. Let N be the unique maximal submodule of M (λ). Then
the composition factorws of N are isomorphic to L(µ) for µ ≤ λ , µ ∈ θ, , µ 6=
λ.
43
Proof. Given N , we have a Jordan-Hölder filtration of M (λ):
0 ( · · · ( N1 ( N ( M (λ).
Thus we see that composition factors L(µ) of N are composition factors of
M (λ). By the previous lemma, then, we have that all µ are in θ, with µ ≤ λ.
We have seen previously that the weight space M (λ)λ−ρ is one-dimensional,
hence L(λ)λ−ρ is one-dimensional as well. The short exact sequence
0 → N → M (λ) → L(λ) → 0
induces a short exact sequence
0 → N λ−ρ → M (λ)λ−ρ → L(λ)λ−ρ → 0,
hence we must have N λ−ρ ∼
= M (λ)λ−ρ /L(λ)λ−ρ = 0. We have also seen
previously that the multiplicity of L(λ) in M (λ) is 1, with M (λ)/N ∼
= L(λ),
thus L(λ) is not a composition factor of N , i.e. µ 6= λ, completing the
proof.
Corollary 9.6. Let λ be a minimal (with respect to the partial ordering
above) element of θ. Then M (λ) is irreducible, and M (λ) = L(λ).
Proof. Since λ is minimal, N has no composition factor by the proposition,
so N = 0. Thus, M (λ) is irreducible, hence M (λ) = L(λ).
We let
Sλ = {µ ∈ θ | µ ≤ λ, µ 6= λ} .
Note that Sλ is finite. We have that, for [V ] ∈ K,
X
[V ] =
m([V ]; [M ])[M ].
M irreducible
Thus,
[M (λ)] = [L(λ)] +
X
m([M (λ)]; [L(µ)])[L(µ)].
µ∈Sλ
Theorem 9.7. Let λ ∈ θ. Then
[L(λ)] = [M (λ)] +
X
mλ,µ [M (µ)] , mλ,µ ∈ Z.
µ∈Sλ
44
Before proving the theorem, we give a quick example. Let g = sl(2, C),
and let λ = ρ = 12 α. We have W ∼
= Z/2Z, so θ = W ρ = {ρ, −ρ}. We have
that −ρ ≤ ρ, since ρ − (−ρ) = α, hence Sρ = −ρ. By the previous corollary,
since −ρ is minimal we have M (−ρ) = L(−ρ). By the discussion preceding
the above theorem, then, we may write
[M (ρ)] = [L(ρ)] + m([M (ρ)]; [L(−ρ)])[L(−ρ)]
= [L(ρ)] + m([M (ρ)]; [M (−ρ)])[M (−ρ)].
Thus we have
[L(ρ)] = M [(ρ)] − m([M (ρ)]; [M (−ρ)])[M (−ρ)].
Furthermore, by our discussion of sl(2, C) leading up to the W -invariance
of the Harish-Chandra map, we see that M (ρ) has M (−ρ) as a maximal
submodule, thus M (ρ)/M (−ρ) ∼
= L(ρ). The short exact sequence induced
by this isomorphism shows that
[L(ρ)] = [M (ρ)] − [M (−ρ)],
which agrees with the statement of the theorem.
Proof. We proceed by induction on the partially ordered set θ. Let S ⊆ θ
be the elements of θ for which
X
[L(λ)] = [M (λ)] +
mλ,µ [M (µ)] , mλ,µ ∈ Z
µ∈Sλ
holds for all λ ∈ S. By the corollary above, this trivially holds for a minimal
element λ, since in this case M (λ) = L(λ) and Sλ = ∅. Thus we have S 6= ∅.
Let T = θ − S. We assume T 6= ∅ and derive a contradiction. Let λ ∈ T be
minimal. We have observed earlier that
X
[L(λ)] = [M (λ)] −
m([M (λ)]; [L(µ)])[L(µ)].
µ∈Sλ
By the minimality of λ in T , we have Sλ ⊆ S. So, for all µ ∈ Sλ we have
X
[L(µ)] = [M (µ)] +
mµ,ν [M (ν)].
ν∈Sµ
45
For ν ∈ Sµ , we have ν ≤ µ ≤ λ with ν 6= λ, thus ν ∈ Sλ . That is, Sµ ⊆ Sλ .
For some ν ∈
/ Sµ , we have mµ,ν = 0, thus we may rewrite the above as
X
[L(µ)] = [M (µ)] +
mµ,ν [M (ν)].
ν∈Sλ
We then have
!
[L(λ)] = [M (λ)] −
X
m([M (λ)]; [L(µ)]) [M (µ)] +
µ∈Sλ
X
mµ,ν [M (ν)]
ν∈Sλ
= [M (λ)] +
X
nλ,µ [M (µ)].
µ∈Sλ
This is of the form presented in the theorem, contradicting the fact that
λ ∈ T . Thus we have T = ∅, i.e. the theorem holds for all λ ∈ θ.
The theorem tells us that the isomorphism classes [M (λ)] , λ ∈ θ generate
the Grothendieck group K. Now, consider the free abelian group G with
formal basis {[[M (λ)]] , λ ∈ θ}. There is a natural homomorphism
φ:G→K
[[M (λ)]] 7→ [M (λ)].
By our theorem, this map is surjective. Let H = Ker(φ). Since θ is finite,
this is a free abelian group of finite rank. We have that G is generated by
[[M (λ)]], so the rank of G is just the cardinality of θ. Thus,
#(θ) = Rank(K) = Rank(G) − Rank(H) = #(θ) − Rank(H).
Therefore, Rank(H) = 0 and the homomorphism φ is an isomorphism. We
have then proven the following:
Theorem 9.8. The elements [M (λ)] , λ ∈ θ form a basis for K.
46
10
10/23/15: Another Approach to the Weyl
Character Formula, Continued
Last time, we defined a partial order on h∗ and described a basis of the
Grothendieck group K in terms of Verma modules M (λ). In this lecture,
we continue by revisiting the idea of an algebraic character for an arbitrary
g-module, where g is as always a complex, semisimple Lie algebra. We introduce the following notation, which we will use for this lecture and the
next.
∗
• Zh := the set of additive mappings h∗ → Z
∗
• For f ∈ Zh , define the support of f
Supp(f ) := {λ ∈ h∗ | f (λ) 6= 0}
∗
• Z[h∗ ] := the subgroup of Zh of elements of finite support
P
+
∗
• Q := µ ∈ h | µ =
mα α , mα ∈ Z≥0
α∈R+
S
∗
h∗
+
∗
• Zhh i := f ∈ Z | Supp(f ) ⊆
(ν − Q ) , ν ∈ h
finite
(i.e. this is the set of f whose support lies in a finite union of integral cones in h∗ )
• eλ : h∗ → Z , eλ (λ) = 1 , eλ (µ) = 0 for µ 6= λ (note that eλ ∈ Zhh∗ i)
We start by making a few observations. First, notice that we can express
∗
any (by definition additive) f ∈ Zh as
X
f=
f (λ)eλ .
λ∈h∗
Also notice that we have the following inclusions:
∗
Z[h∗ ] ( Zhh∗ i ( Zh .
∗
Finally, observe that we can give a ring structure to Zhh∗ i; write any f ∈ Zh
in the summation form shown above. Since eλ eµ = eλ+ mu for any λ, µ ∈ h∗
we can then define the product of
X
X
f=
f (λ)eλ , g =
g(µ)eµ ∈ Zhh∗ i
µ∈h∗
λ∈h∗
47
by
!
f ·g =
X
f (λ)eλ
!
X
g(µ)eµ
!
=
µ∈h∗
λ∈h∗
X
X
ν∈h∗
λ+µ=ν
f (λ)g(µ) eν .
To make sure this is well defined, we check that each coefficient of eν on the
right is finite; by defintion f and g both have support in a finite union of
integral cones in h∗ . For a fixed ν ∈ h∗ , if λ+µ = ν, then obviously λ = ν −µ.
This is to say that, if λ ∈ Supp(f ), then λ and ν are integrally equivalent.
By a lemma proved in the previous lecture, we have only finitely weights η
such that λ ≤ η ≤ ν. We may make the same argument for µ by writing
µ = ν − λ; thus we have only finitely many available λ, µ such that λ + µ = ν
and the multiplication is well defined.
We remark that the ring structure on Zhh∗ i induces a ring structure on
(the subring) Z[h∗ ].
Definition 10.1. Let V be a g-module. We say V has character if
M
Vµ ,
V =
µ∈h∗
with each generalized eigenspace Vµ finite dimensional. In such a case, the
map
ch(V ) : h∗ → Z
µ 7→ dim Vµ
∗
is an element of Zh called the character of V .
We compare this notion of character with that of the classical trace character. Let V be a finite dimensional g-module. Then there exists a compact
Lie group G which has a Lie algebra g0 whose complexification is g. Let
H ≤ G be a maximal torus. In this case we have that finite dimensional
representations of G corresponed to finite dimensional representations of g0 ,
so we may consider V as a representation (π, V ) of G. Furthermore, H acts
semisimply in this representation, i.e. for h ∈ H we may choose a basis of V
such that


eµ1 (h)
0


...
π(h) = 
,
µn (h)
0
e
48
where eµ : H → C is the function whose differential is µ. The classical
character χV is then
X
χV (h) = Tr(π(h)) =
(dim V µ )eµ (h).
µ a weight of V
Compare this with our new definition of character:
X
ch(V ) =
(dim V µ )eµ .
µ∈h∗
Our new definition of character has two easily verified properties: first,
given a g-module V and g-submodule V 0 , we have that
ch(V ) = ch(V 0 ) + ch(V /V 0 ).
(This can be seen by applying the definition of ch to the short exact sequence
induced by V, V 0 ). A second useful property is that, given highest weight
modules V, W , the tensor produce V ⊗ W has character given by
ch(V ⊗ W ) = ch(V ) · ch(W ),
where the multiplication on the right side of the equality is multiplication in
Zhh∗ i.
In the next lecture, we will use the objects and character map defined
earlier to describe the Weyl character formula in a different language.
49
11
10/27/15: Another Approach to the Weyl
Character Formula, Concluded
The goal of this talk is to show that the character map ch defined in the
previous lecture is in fact the Weyl character formula. Recall that, for an
irreducible g-module V of highest weight λ, the Weyl character formula is
given by
P
(−1)`(w) ew(λ+ρ)
w∈W
.
ch(V ) := Q
(eα/2 − e−α/2 )
α∈R+
We begin with a few definitions. First, we introduce the Kostant partition function P ; let γ ∈ Q+ (recall Q+ is the set of elements of h∗ which
can be expressed as a nonnegative sum of positive roots), and define P (γ)
to be the number of ways γ can be written as a sum of positive roots. For
example, for g = sl(3, C) we may choose a positive root system composed of
α = e1 − e2 , β = e2 − e3 , and γ = α + β = e1 − e3 . In this case, P (γ) = 2.
We let
X
K :=
P (γ)e−γ ,
γ∈Q+
and let
d :=
Y
eα/2 − e−α/2 .
α∈R+
Note that d is denominator in the Weyl character formula above.
Now, let λ be a dominant weight, and let F λ be the irreducible finite
dimensional representation of g with highest weight λ. Then F λ is the irreducible quotient L(λ + ρ) of the Verma module M (λ + ρ). We remark that,
for any dominant weight λ, the (dominant) weight λ + ρ is regular, i.e. the
Weyl group acts transitively on λ + ρ. This proves
Lemma 11.1.
[F λ ] = [M (λ + ρ)] +
X
mw [M (w(λ + ρ))].
w6=1∈W
We saw in the last lecture that the classes highest weight modules [M (λ)]
form a basis for the Grothendieck group K. Thus we have that the ”new”
character map ch factors through the K-group:
ch : Hθ → K(Hθ ) → Zhh∗ i
50
M (λ) 7→ [M (λ)] 7→ ch(M (λ)).
So, we see that
X
ch(F λ ) = ch(M (λ + ρ)) +
mw ch(M (w(λ + ρ))).
w6=1∈W
Lemma 11.2. In the ring Zhh∗ i, d is invertible, with
d−1 = Ke−ρ .
Proof. Let α ∈ R+ . We have
(1 − e
−α
)
∞
X
e−nα = (1 − e−α )(1 + e−α + e−2α + · · · ) = 1.
n=0
We also have that
e−ρ d =
Y
(1 − e−α ).
α∈R+
Thus,

Ke−ρ d = 

X
P (γ)e−γ 
γ∈Q+
"
=
Y
∞
X
α∈R+
n=0
=
(1 − e−α )
α∈R+
!
!#
Y
e−nα
(1 − e−α )
α∈R+
"
Y
Y
(1 − e−α )
∞
X
n=0
α∈R+
proving the lemma.
51
#
e−nα = 1,