The irreducible quotients of the Verma modules

Combining Observations 2’ and 3’ and the fact that the polynomial function h^∗ →C, λ 7→det

(·,·)^◦_λ,n is not identically zero, we conclude that the polynomial function

h^∗ →C, λ 7→det

(·,·)^◦_λ,n is the leading term of the polynomial function

h^∗ →C, λ 7→det

(·,·)_λ,n . This proves Proposition 2.6.17.

Now that Proposition 2.6.17 is proven, the proof of Theorem 2.6.6 is also complete (because we have already proven Theorem 2.6.6 using Proposition 2.6.17).

2.7. The irreducible quotients of the Verma modules

We will now use the form (·,·)_λ to develop the representation theory of g. In the following, we assume thatg is nondegenerate.

Definition 2.7.1. Let (·,·) denote the form (·,·)_λ. Let J_λ^± be the kernel of (·,·) on M_λ^±. This is a graded g-submodule of M_λ^± (since the form (·,·) is g-invariant).

Let L^±_λ be the quotient module M_λ^±J_λ^±. Then, (·,·) descends to a nondegenerate pairing L⁺_λ ×L⁻_−λ →C.

Remark 2.7.2. For Weil-genericλ(away from a countable union of hypersurfaces), we have J_λ^± = 0 (by Theorem 2.6.6) and thus L^±_λ =M_λ^±.

Theorem 2.7.3. (i) The g-moduleL^±_λ is irreducible.

(ii) The g-module J_λ^± is the maximal proper graded submodule of M_λ^±. (This means that J_λ^± contains all proper graded submodules in M_λ^±.)

(iii)Assume that there exists some L∈g₀ such that every n ∈Z satisfies (adL)|_gn=n·id|_gn .

(In this case it is said thatthe grading on g is internal, i. e., comes from bracketing with some L∈g₀.) Then J_λ^± is the maximal proper submodule of M_λ^±.

Remark 2.7.4. Here are two examples of cases when the grading on g is internal:

(a)Ifgis a simple finite-dimensional Lie algebra, then we know (from Proposition 2.5.6) that choosing a Cartan subalgebra h and corresponding Chevalley generators e₁, e₂, ..., e_m, f₁, f₂, ..., f_m, h₁, h₂, ..., h_m of g endows g with a grading. This grading is internal. In fact, in this case, we can take L=ρ^∨, whereρ^∨ is defined as the element of h satisfying α_i(ρ^∨) = 1 for all i (where α_i are the simple roots ofg).

Since the actions of the α_i onh are a basis of h^∗, this ρ^∨ is well-defined and unique.

(But it depends on the choice of h and the Chevalley generators, of course.)

(b)Ifg= Vir, then the grading ong is internal. In fact, in this case, we can take L=−L₀.

On the other hand, if g is the affine Kac-Moody algebra bg_ω of Definition 1.7.6, then the grading on g is not internal.

Proof of Theorem 2.7.3. (i)Let us show thatL⁻_λ is irreducible (the proof for L⁺_λ will be similar).

In fact, assume the contrary. Then, there exists a nonzero w∈L⁻_λ such thatU(g)· w 6= L⁻_λ. Since L⁻_λ is graded by nonnegative integers, we can choose w to have the smallest possible degree m (without necessarily being homogeneous). Clearly, m >0.

Thus we can write w = w₀+w₁ +...+w_m, where each w_i is homogeneous of degree degwi =i and wm 6= 0.

Let a ∈ g_j for some j < 0. Then aw = 0 (since deg (aw) < degw, but still U(g)·aw6=L⁻_λ (since U(g)·aw⊆U(g)·wand U(g)·w6=L⁻_λ), and we have chosen w to have the smallest possible degree). By homogeneity, this yields awm = 0 (since aw_m is the (m+j)-th homogeneous component of aw).

For every u ∈L⁺_−λ[−m−j], the term (au, w_m) is well-defined (since au ∈L⁺_−λ and w_m ∈L⁻_λ). Since the form (·,·) isg-invariant, it satisfies (au, w_m) = −



u, aw_m

|{z}=0



= 0.

But sincem >0, we haveL⁺_−λ[−m] = P

j<0

g_j·L⁺_−λ[−m−j] (because Proposition 2.5.15 (a) yields M_−λ⁺ =U(n−)v_λ⁺, so that L⁺_−λ =U(n−)v_λ⁺, thus

L⁺_−λ[−m] = U(n−) [−m]

| {z }

=P

j<0

(n−)[j]·U(n−)[−m−j]

v_λ⁺ =X

j<0

(n−) [j]

| {z }

=g[j]=gj

·U(n−) [−m−j]v_λ⁺

| {z }

=L⁺_−λ[−m−j]

(sinceU(n−)v⁺_λ=L⁺_−λ)

=X

j<0

gj·L⁺_−λ[−m−j]

). Hence, any element of L⁺_−λ[−m] is a linear combination of elements of the form au with a∈g_j (for j <0) andu∈L⁺_−λ[−m−j]. Thus, since we know that (au, w_m) = 0

for every a ∈ gj and u ∈ L⁺_−λ[−m−j], we conclude that L⁺_−λ[−m], w_m is nondegenerate, this yields w_m = 0. This is a contradiction to w_m 6= 0. This contradiction shows that our assumption was wrong. Thus,L⁻_λ is irreducible. Similarly, L⁺_λ is irreducible.

(ii)First let us prove that theg-moduleJ_λ⁺ is the maximal proper graded submodule of M_λ⁺.

LetK ⊆M_λ⁺be a proper graded submodule, and letK be its image inL⁺_λ. Then, K lives in strictly negative degrees (because it is graded, so if it would have a component in degrees ≥ 0, it would contain v_λ⁺ and thus contain everything, and thus not be proper). Hence, K also lives in strictly negative degrees, and thus is proper. Hence, by (i), we have K = 0, thus K ⊆ J_λ⁺. This shows that J_λ⁺ is the maximal proper graded submodule of M_λ⁺. The proof of the corresponding statement for J_λ⁻ and M_λ⁻ is similar.

(iii) Assume that there exists someL∈g0 such that every n ∈Z satisfies (adL)|_gn=n·id|_gn .

Consider this L. It is easy to prove (by induction) that [L, a] = na for every a ∈ U(g) [n].

We are now going to show that all g-submodules of M_λ⁺ are automatically graded.

In fact, it is easy to see thatM_λ⁺[n]⊆Ker L|_M⁺

λ −(λ(L) +n) id

for everyn∈Z.

68 In other words, for everyn ∈Z, then-th homogeneous componentM_λ⁺[n] of M_λ⁺ is contained in the eigenspace of the operator L|_M⁺

λ for the eigenvalue λ(L) +n. Now,

eigenspace of the operatorL|

M+

eigenspace of the operator L|_M⁺

λ for the eigenvalue λ(L) +n

Since all eigenspaces of L|_M⁺

λ are clearly contained inM_λ⁺, this rewrites as M_λ⁺ =X

n∈Z

eigenspace of the operator L|_M⁺

λ for the eigenvalue λ(L) +n .

Since eigenspaces of an operator corresponding to distinct eigenvalues are linearly disjoint, the sum P

n∈Z

eigenspace of the operator L|_M⁺

λ for the eigenvalue λ(L) +n must be a direct sum, so this becomes

M_λ⁺ =M

n∈Z

eigenspace of the operator L|_M⁺

λ for the eigenvalue λ(L) +n

. (70) As a consequence of this, the map L |_M⁺

λ is diagonalizable, and all of its eigenvalues belong to the set {λ(L) +n | n∈Z}.

eigenspace of the operator L|_M⁺

λ for the eigenvalue λ(L) +n , but the direct sum of these inclusions over alln ∈Z is an equality (since

M

n∈Z

M_λ⁺[n] =M_λ⁺ =M

n∈Z

eigenspace of the operator L|_M⁺

λ for the eigenvalue λ(L) +n by (70)). Hence, each of these inclusions must be an equality. In other words,

M_λ⁺[n] =

eigenspace of the operator L|_M⁺

λ for the eigenvalue λ(L) +n

for every n∈Z. (71)

Now, let K be a g-submodule of M_λ⁺. Then, L |K is a restriction of L |_M⁺

λ to K. Hence, map L |_K is diagonalizable, and all of its eigenvalues belong to the set {λ(L) +n | n ∈Z} (because we know that the map L|_M⁺

λ is diagonalizable, and all of its eigenvalues belong to the set{λ(L) +n | n∈Z}). In other words,

K =M

n∈Z

(eigenspace of the operator L|K for the eigenvalue λ(L) +n)

| {z }

=K∩

eigenspace of the operatorL|

M+

eigenspace of the operator L|_M⁺

λ for the eigenvalue λ(L) +n

Hence, K is graded. We thus have shown that every g-submodule of M_λ⁺ is graded.

Similarly, every g-submodule ofM_λ⁻ is graded. Thus, Theorem 2.7.3(iii)follows from Theorem 2.7.3 (ii).

Remark 2.7.5. Theorem 2.7.3 (ii) does not hold if the word “graded” is removed.

In fact, here is a counterexample: Let g be the 3-dimensional Heisenberg algebra.

(This is the Lie algebra with vector-space basis (x, K, y) and with Lie bracket given by [y, x] =K, [x, K] = 0 and [y, K] = 0. It can be considered as a Lie subalgebra of the oscillator algebra Adefined in Definition 1.1.4.) It is easy to see that gbecomes a nondegenerate Z-graded Lie algebra by setting g−1 = hxi, g₀ = hKi, g₁ = hyi and g_i = 0 for everyi ∈Z{−1,0,1}. Then, on the Verma highest-weight module M₀⁺ =C[x]v⁺₀, bothK and yact as 0 (andx acts as multiplication withx), so that Iv₀⁺ is ag-submodule of M₀⁺ for every idealI ⊆C[x], but not all of these ideals are graded, and not all of them are contained in J₀⁺ (as can be easily checked).

Corollary 2.7.6. For Weil-generic λ (this means a λ outside of countably many hypersurfaces in h^∗), theg-modules M_λ⁺ and M_λ⁻ are irreducible.

Definition 2.7.7. Let Y be a g-module. A vector w∈Y is called a singular vector of weight µ∈h^∗ (here, recall thath =g₀) if it satisfies

hw=µ(h)w for every h∈h and

aw= 0 for every a∈g_i for every i >0.

We denote by Sing_µ(Y) the space of singular vectors of Y of weight µ.

When people talk about “singular vectors”, they usually mean nonzero singular vectors in negative degrees. We are not going to adhere to this convention, though.

Lemma 2.7.8. Let Y be ag-module. Then there is a canonical isomorphism Hom_g M_λ⁺, Y

→Sing_λY, φ7→φ v_λ⁺

.

Proof of Lemma 2.7.8. We have M_λ⁺=U(g)⊗U(h⊕n+)C^λ = Ind^g_h⊕n+C^λ, so that Hom_g M_λ⁺, Y

= Hom_g Ind^g_h⊕n

+C^λ, Y

∼= Homh⊕n+(C^λ, Y) (by Frobenius reciprocity). But Homh⊕n+(Cλ, Y)∼= Sing_λY (because every C-linear map Cλ →Y is uniquely

de-termined by the image ofv_λ⁺, and this map is a (h⊕n+)-module map if and only if this image is a singular vector ofY of weightλ). Thus, Hom_g M_λ⁺, Y∼= Homh⊕n₊(Cλ, Y)∼= Sing_λY. If we make this isomorphism explicit, we notice that it sends everyφtoφ v⁺_λ

, so that Lemma 2.7.8 is proven.

Corollary 2.7.9. The representation M_λ⁺ is irreducible if and only if it does not have nonzero singular vectors in negative degrees. Here, a vector inM_λ⁺is said to be

“in negative degrees” if its projection on the 0-th homogeneous component M_λ⁺[0]

is zero.

Proof of Corollary 2.7.9. ⇐=: Assume that M_λ⁺ does not have nonzero singular vectors in negative degrees.

We must then show that M_λ⁺ is irreducible.

In fact, assume the contrary. Then, M_λ⁺ is not irreducible. Hence, there exists a nonzerohomogeneous v ∈ M_λ⁺ such that U(g)·v 6=M_λ⁺. ⁶⁹ Consider this v. Then, U(g)·v is a proper graded submodule of M_λ⁺, and thus is contained in J_λ⁺. Hence, J_λ⁺ 6= 0.

There exist some d∈Z such thatJ_λ⁺[d]6= 0 (since J_λ⁺6= 0 and since J_λ⁺ is graded).

All such d are nonpositive (since J_λ⁺ is nonpositively graded). Thus, there exists a highest integer d such that J_λ⁺[d] 6= 0. Consider this d. Clearly, d < 0 (since the bilinear form (·,·) :M_λ⁺×M_−λ⁻ is obviously nondegenerate on M_λ⁺[0]×M_−λ⁻ [0], so that J_λ⁺[0] = 0).

Every i >0 satisfies g_i· J_λ⁺[d]

⊆J_λ⁺[i+d] since J_λ⁺ is a graded g-module

= 0 since i+d > d, but d was the highest integer such that J_λ⁺[d]6= 0 . By Conditions(1)and(2)of Definition 2.5.4, the Lie algebrag₀ is abelian and

finite-dimensional. Hence, every nonzero g₀-module has a one-dimensional submodule⁷⁰.

69Proof. Notice thatM_λ⁺ is a gradedU(g)-module (sinceM_λ⁺ is a gradedg-module).

Since M_λ⁺ is not irreducible, there exists a nonzero w∈M_λ⁺ such thatU(g)·w6=M_λ⁺. Since M_λ⁺ is graded bynonpositive integers, we can write win the form w=

m

70Proof. This is because of the following fact:

Thus, the nonzero g0-module J_λ⁺[d] has a one-dimensional submodule. Let w be the generator of this submodule. Then, this submodule ishwi.

For everyh∈h, the vectorhwis a scalar multiple ofw(sinceh∈h=g₀, so thathw lies in the g0-submodule of J_λ⁺[d] generated by w, but this submodule is hwi). Thus, we can writehw=λ_hwfor someλ_h ∈C. Thisλ_h is uniquely determined (sincew6= 0), so we can define a mapµ:h→C such thatµ(h) = λ_h for everyh∈h. This mapµis easily seen to be C-linear, so that we have found a µ∈h^∗ such that

hw=µ(h)w for every h∈h.

Also,

aw= 0 for every a∈g_i for every i >0 (since a

|{z}∈g_i

w

|{z}

∈J_λ⁺[d]

∈ g_i · J_λ⁺[d]

⊆ 0). Thus, w is a nonzero singular vector. Since w ∈ J_λ⁺[d] and d < 0, this vector w is in negative degrees. This contradicts to the assumption thatM_λ⁺ does not have nonzero singular vectors in negative degrees. This contradiction shows that our assumption was wrong, so that M_λ⁺ is irreducible. This proves the⇐= direction of Corollary 2.7.9.

=⇒: Assume that M_λ⁺ is irreducible.

We must then show that M_λ⁺ does not have nonzero singular vectors in negative degrees.

Let v be a singular vector of M_λ⁺ in negative degrees. Let it be a singular vector of weightµ for some µ∈h^∗.

By Lemma 2.7.8 (applied toµandM_λ⁺instead ofλandY), we have an isomorphism Hom_g M_µ⁺, M_λ⁺

→Sing_µ M_λ⁺ , φ7→φ v_µ⁺

.

Letφ be the preimage of v under this isomorphism. Then, v =φ v⁺_µ . Since v is in negative degrees, we have v ∈ P

n<0

M_λ⁺[n]. Now, M_µ⁺ = U(n−)v⁺_µ = P

m≤0

U(n−) [m]v_µ⁺ (sinceM_µ⁺ is nonpositively graded), so that

φ M_µ⁺

=φ X

m≤0

U(n−) [m]v_µ⁺

!

= X

m≤0

U(n−) [m] φ v_µ⁺

| {z }

=v∈P

n<0

M_λ⁺[n]

since φ ∈Hom_g M_µ⁺, M_λ⁺

∈ X

m≤0

U(n−) [m]X

n<0

M_λ⁺[n] =X

m≤0

X

n<0

U(n−) [m]·M_λ⁺[n]

| {z }

⊆M_λ⁺[m+n]

(sinceM_λ⁺ is a gradedg-module)

⊆ X

m≤0

X

n<0

M_λ⁺[m+n]⊆X

r<0

M_λ⁺[r].

Every nonzero finite-dimensional module over an abelian finite-dimensional Lie algebra has a one-dimensional submodule. (This is just a restatement of the fact that a finite set of pairwise commuting matrices on a finite-dimensional nonzeroC-vector space has a common nonzero eigen-vector.)

Thus, the projection of φ M_µ⁺

onto the 0-th degree of M_λ⁺ is 0. Hence, φ M_µ⁺ is a proper g-submodule of M_λ⁺. Therefore, φ M_µ⁺

= 0 (since M_λ⁺ is irreducible). Thus, v =φ v⁺_µ

∈φ M_µ⁺

= 0, so that v = 0.

We have thus proven: Whenever v is a singular vector of M_λ⁺ in negative degrees, we havev = 0. In other words, M_λ⁺ does not have nonzero singular vectors in negative degrees. This proves the =⇒ direction of Corollary 2.7.9.

Here is a variation on Corollary 2.7.9:

Corollary 2.7.10. The representation M_λ⁺ is irreducible if and only if it does not have nonzero homogeneous singular vectors in negative degrees.

Proof of Corollary 2.7.10. =⇒: This follows from the =⇒ direction of Corollary 2.7.9.

⇐=: Repeat the proof of the ⇐= direction of Corollary 2.7.9, noticing that w is homogeneous (since w∈J_λ⁺[d]).

Corollary 2.7.10 is thus proven.

Im Dokument Pavel Etingof: 18.747: Infinite-dimensional Lie algebras (Spring term 2012 at MIT) (Seite 118-125)