Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

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Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology Dan Barbasch March 26 Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Transcript of Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Semisimple Lie Algebras

Math 649, 2013

Lie algebra cohomology

Dan Barbasch

March 26

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Generalities

For g a Lie algebra, let g′ be the dual vector space. Define

C n(g,V ) := Λng′ ⊗ V ' HomC(Λng,V ) (1)

where (π,V ) is a representation of g. There is a differential

d : C n → C n+1 (2)

given by

dω(x0 · · · xn) =∑

(−1)iπ(xi) · ω(x0, · · · , x̂i , · · · , xn) (3)

+∑

(−1)i+jω([xi , xj ], x0, · · · , x̂i , · · · , x̂j , · · · , xn).

NOTE: If (π,V ) is trivial, the first term is zero.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Generalities, continued

Suppose V is a (finite dimensional) vector space, V ′ its dual.We define

Λ∗V := T (V)/{x ⊗ y + y ⊗ x} x , y ∈ V . (4)

There is a pairing between Λ∗V and Λ∗V ′

〈f1 ∧ · · · ∧ fk , v1 ∧ · · · ∧ vk〉 := det[(fi , vj)]. (5)

Note that (Λ∗V)′ ' Λ∗V ′. ΛkV ′ is identified with alternatingfunctions. The exterior algebra has an algebra structure givenby ∧. If ω ∈ ΛkV ′, ν ∈ Λ`V ′ then define

(ω ∧ ν)(x1, · · · , xk+`) :=∑σ∈Sk+`

det(σ)1

k!`!ω(xσ(1), · · · , xσ(k))

· ν(xσ(k+1), · · · , xσ(k+`)).(6)

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Generalities, continued

We say α ∈ (End Λ∗V)i is of degree |α| = i , ifα(ΛjV) ⊆ Λj+iV . It is called a derivation if in addition

α(u ∧ v) = α(u) ∧ v + (−1)|u|·|α|u ∧ α(v). (7)

If f ∈ V ′, there is a map

⊗kV → Λk−1V (8)

x1 ⊗ · · · ⊗ xk 7→∑

(−1)i+1〈f , xi〉x1 ∧ · · · ∧ x̂i ∧ · · · ∧ xk (9)

which factors to ι(f ) : ∧kV −→ ∧k−1V . Soι(f ) ∈ (End ΛV)−1,

ι(f )(x1 ∧ · · · ∧ xk) :=∑

(−1)i+1〈f , xi〉x1 ∧ · · · ∧ x̂i ∧ · · · ∧ xk .

(10)Similarly, if u ∈ ΛV ′, let

ε(u)ω := u ∧ ω. (11)Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Proposition (2)

Given any linear map β : Λ1V → Λk+1V , there is a uniquederivation α ∈ (Der ΛV)k such that α|V = β. (α |K= 0).

Proof.

Exercise. Compare with the definition of the differential indifferential geometry, e.g. in Helgason.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Now suppose V = g is a Lie algebra. There is a linear map

∂ : Λ2g→ Λ1g ' g ∂(x ∧ y) := [x , y ] (12)

Then −∂t : Λ1g′ → Λ2g′ gives a derivation

d ∈ (Der Λg)1. (13)

We also define ∂ to be −d t .

Definition

For each x ∈ g, θ∗(x) ∈ (Der Λg)0, is defined to be the adjointaction, and θ(x) = −θ∗(x)t .

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Proposition

d =1

2

∑ε(fi)θ(xi)

∂ =1

2

∑θ∗(xi)ι(fi)

where f1 · · · fn, x1 · · · xn are dual bases;

Note: ∂ is not a derivation, but d is.d is the derivation from before with (π,V ) trivial.

Proposition

d2 = 0.

Proof.

Exercise. By the previous proposition you only need to checkd2f = 0 for f ∈ g′.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Properties

1 d ◦ θ(x) = θ(x) ◦ d , ∂ ◦ θ∗(x) = θ∗(x) ◦ ∂2 d ◦ ι(x) + ι(x) ◦ d = θ(x), ∂ ◦ ε(x) + ε(x) ◦ ∂ = θ∗(x)

3 θ∗(x) ◦ ι(y)− ι(y) ◦ θ∗(x) = ι([x , y ]),θ(x) ◦ ε(y)− ε(y) ◦ θ(x) = ε([x , y ])

4 d(u ∧ v) = du ∧ v + (−1)|u|u ∧ dv .

In particular, H i(g,K) := ker di/ im di . We get a cup product(wedge)

H i(g)⊗ H j(g)∧→ H i+j(g). (14)

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

If (π,V ) is a representation of g, we define

d = d0 + dV , ∂ = ∂0 + ∂V (15)

where d0 and ∂0 are as before, and

dV =∑

ε(fi)⊗ π(xi) ∂V =∑

ι(xi)⊗ π(fi). (16)

In these formulas we have changed to C i(g,V ) = Λig′ ⊗ V .Now suppose g is semisimple (actually all the results hold forreductive algebras). Then g and g′ are canonically identifiedvia B . We can assume x1 · · · xn, f1 · · · fn are all in g. Thenθ∗(x) = ad x and so we compute for f ∈ g′,f = fy (z) := B(z , y),θ(x)fy (z) = fy (− ad xz) = B(−[x , z ], y) = B(z , ad x(y)) i.e.θ(x)(y) = ad x(y).

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

The Casimir element

A semisimple Lie algebra has a distinguished invariant elementin U(g) corresponding to an invariant in S2(g). This is theanalog of the Laplace operator in Riemannian geometry.Suppose b is a symmetric nondegenerate form on g which isinvariant, i.e. it satisfies

b(ad x y , z) + b(y , ad x z) = 0. (17)

Let x1, . . . , xn be a basis and f1, . . . , fn be the dual basis withrespect to b.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Proposition

x1f1 + · · ·+ xnfn ∈ S2(g) and C = x1f1 + · · ·+ xnfn ∈ U(g) areinvariant.

Proof.

[x , x1f1 + · · ·+ xnfn] =∑

[x , xi ]fi +∑

xj [x , fj ]. But

[x , xi ] =∑

b([x , xi ], fj)xj =∑

b(x , [xi , fj ])xj (18)

[x , fj ] =∑

b([x , fj ], xi)fi =∑

b(x , [fj , xi ])fi

and we get∑i ,j

b(x , [xi , fj ])xj fi + b(x , [fj , xj ])xj fi = 0. (19)

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

If g is semisimple, g = a1 ⊕ · · · ⊕ ak , ai simple ideals.

Proposition

If g is simple, there is a unique symmetric bilinear form b( , )(up to a scalar) satisfying

b(ad x · y , z) + b(y , ad x · z) = 0. (20)

Proof.

Recall that g ' g′ even as representations. Then

Homg[C, g′ ⊗ g] = Homg[g, g]. (21)

Because g is simple, ad is an irreducible representation.Schur’s lemma implies that the dimension of Homg[g, g] equals1. Now g′ ⊗ g ≈ g⊗ g = S2(g)⊕ Λ2g. But S2(g) identifieswith symmetric bilinear forms and the Cartan-Killing form isinvariant.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Lemma

Let (π,V ) be an irreducible representation of g such thatker π = (0). Then

π(C ) =dim g

dim VId . (22)

Proof.

π(C ) is a scalar by Schur’s lemma. So π(C ) = λId . ThenTr π(C ) = λ · dim V . On the other hand considerb(x , y) := Tr(π(x)π(y)). This is a nondegenerate form on g.We can view g ↪→ End(V ) and consider a := rad b. It is anideal in g that satisfies b(x , x) = Tr(π(x)2) = 0 ∀ x ∈ a.By Cartan’s criterion, a is solvable, therefore 0. Choose a basisx1, . . . , xn and let f1, . . . fn be the dual basis with respect to b.

Then Tr π(C ) =∑

Tr[π(xi)π(fi)] =∑

b(xi , fi) = dim g.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

In general, there are as many invariants of the kind C as thereare simple factors in g.Now suppose that (π,V ) is still irreducible but has a kernel.Then g ' ker π ⊕ g̃, and the representation (π,V ) restrictedto g̃ satisfies the conditions of the lemma. We can extend theform b on g̃ to all of g so that it is nondegenerate and satisfies(17). For the Casimir element corresponding to this form b westill get

π(C ) =dim g

dim VId . (23)

The terms coming from ker π act by 0, do not contribute.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Proposition (Kuga’s lemma)

d∂V + ∂V d = 1⊗ Cπ, (24)

where Cπ is the Casimir operator for (π,V ).

Every general representation (π,V ) decomposes into a directsum of g−modules

V = V0 ⊕ V+ (25)

where π(C ) acts nilpotently on V0 and is an isomorphism onV+.

Corollary (1)

H i(g,V ) = H i(g,V0). The irreducible constituents of V0 areall isomorphic to the trivial representation. Furthermore,H1(g,V0) = 0 as well.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Proof.

Kuga’s lemma proves the first part; note that 1⊗ Cπcommutes with d , ∂ and θ. If f : g→ K is such thatf ([x , y ]) = 0 for all x , y ∈ g, it is identically zero becauseg = [g, g]. So H1(g,K) = (0).Assume V = V0, and write g = ker π + g̃. Using basesadapted to this decomposition, we find that π(CB) = λCπ fora λ 6= 0, where Cπ is the Casimir element constructed fromb(x , y) = Tr[π(x)π(y)]. The lemma implies thatm = dim g̃ = 0, so ker π = g.

Corollary (2)

H i(g,V ) ' H i(g,K)⊗ V g.

V g is another notation for V0.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Lie algebra cohomology with trivial coefficients

We consider the case of a general Lie algebra first. Theformulas are

d =1

2

∑ε(fi)θ(xi) ∂ =

1

2

∑θ∗(xi)ι(fi). (26)

If π : h→ g, is a Lie homomorphism, then it inducesπ∗ : Λh→ Λg so we get π∗ : H(g)→ H(h). Since Λg′ ispaired with Λg and d and ∂ are transpose of each other,

H∗(g) ' H∗(g)′. (27)

There is a distinguished element f 0 ∈ g′, f 0(x) := Tr ad x .

Definition

g is unimodular if Tr ad x = 0 for all x ∈ g.

A semisimple algebra is unimodular.Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Proposition

Hn(g) = K if and only if g is unimodular.

Proof.

θ(x)(f1 ∧ · · · ∧ fn) = (Tr ad∗ x)(f1 ∧ · · · ∧ fn),

θ∗(x)(x1 ∧ · · · ∧ xn) = (Tr ad x)(x1 ∧ · · · ∧ xn).

If g is not unimodular, there is x ∈ g such thatTr ad x = c 6= 0. Then

(dι(x) + ι(x)d)(f1 ∧ · · · ∧ fn) = cf1 ∧ · · · ∧ fn 6= 0. (28)

1

cd [ι(x)(f1 ∧ · · · ∧ fn)] = f1 ∧ · · · ∧ fn (29)

so Hn(g) = 0.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

proof continued.

Conversely, if f1 ∧ · · · ∧ fn = dω, then ∂(x1 ∧ · · · ∧ xn) 6= 0because

1 = 〈f1 ∧ · · · ∧ fn, x1 ∧ · · · ∧ xn〉 = 〈ω, ∂(x1 ∧ · · · ∧ xn)〉. (28)

Suppose θ(x)f1 ∧ · · · ∧ fn = 0 for all x ∈ g. Thendι(x)f1 ∧ · · · ∧ fn = 0 so ε(x)∂(x1 ∧ · · · ∧ xn) = 0 for all x .This implies ∂(x1 ∧ · · · ∧ xn) = 0 a contradiction. Thus thereis x such that

θ(x)f1 ∧ · · · ∧ fn = Tr ad x(f1 ∧ · · · ∧ fn) 6= 0. (29)

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Corollary

g is unimodular if and only if ∂(x1 ∧ · · · ∧ xn) = 0.

Now let u ∈ Λkg′, v ∈ Λn−kg′ be in ker d . Set

〈[u], [v ]〉 := 〈u ∧ v , ω〉. (30)

ω = x1 ∧ · · · ∧ xn.

Proposition

〈 , 〉 is nonsingular if g is unimodular. In this case, Hn−k(g) isdual to Hk(g).

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Proof.

Fix 0 6= ω ∈ Λng. Define a map Ψ : Λ∗g′ → Λ∗g byΨ(u) := ι(u)ω. Ψ is a bijection.

Claim: If g is unimodular, then Ψ carries coboundaries toboundaries and cocycles to cycles.Indeed,

dε(u)− (−1)|u|ε(u)d = ε(du) (31)

(−1)|u|∂ι(u)− ι(u)∂ = ι(du). (32)

If g is unimodular, evaluate on ω to get(−1)|u|∂ι(u)ω = ι(du)ω for all u. So du = 0 implies∂(ι(u)ω) = 0. It also says that coboundaries go to boundaries.Use the corollary.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

proof continued.

We claim ∂(ι(u)ω) = 0 if and only if du = 0. Let v bearbitrary. ∂(ι(u)ω) = ±(du). Then

〈du ∧ v , ω〉 = 〈v , ι(du)ω〉 = 0. (31)

But also

0 = 〈du ∧ v , ω〉 = ±〈du, ι(v)ω〉 for all v ∈ Λn−kg′. (32)

Since v 7→ ι(v)ω is an isomorphism, the claim follows.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Semisimple Case

Now assume g is semisimple, and use complete reducibility.Then g acts on Λg via the adjoint action and

d ◦ θ(x) = θ(x) ◦ d . (33)

The key is another version of Kuga’s lemma,∆ := d0∂0 + ∂0d0 = θ(CB).Thus

Λkg = (Λkg)g ⊕ g · Λkg = J (g)k ⊕ g · Λkg. (34)

The action · is via θ. Note that d(J (g)) = 0, so there is amap J (g)→ H∗(g) because of the formula (26).

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Semisimple Case

Lemma

If x ∈ g and ω ∈ ker d, then θ(x)ω ∈ Im d.

Proof.

θ(x)ω = (dι(x) + ι(x)d)ω = d(ι(x)ω). (33)

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Theorem

If g is semisimple, then the map J (g)→ H∗(g) is a ringisomorphism.

Proof.

The map is surjective because ker d is invariant under theaction of g and so also decomposes into (ker d)g ⊕ g · ker d .Then apply the previous lemma and observe thatJ (g) ∩ ker d = (ker d)g. Remains to show thatJ (g) ∩ im d = (0). Write

d =1

2

∑ε(fi)θ(xi) =

1

2

∑θ(xi)ε(fi)−

1

2

∑[θ(xi), ε(fi)].

But [θ(xi), ε(fi)] = ε(θ(xi)fi). Write f0 :=∑

θ(xi)fi . Then

〈f0, y〉 = −∑〈fi , [xi , y ]〉 = Tr ad y = 0 for all y . Thus the

second term is zero and the first term shows thatIm d ⊆ gΛ·g.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Exercise1 Compute the cohomology of sl(2).

2 Write bi = dim H i(g). Show that b1 = b2 = 0 but b3 6= 0.

3 Show that H(g⊕ h) = H(g)⊗ H(h).

In general, (g still semisimple)

Λg = Im ∂ ⊕ J (g)⊕ Im d . (34)

If ∆ = d∂ + ∂d , then ker ∆ = J (g).

Exercise

Read the section on bi-invariant forms in Helgason’s neweredition of the text on Lie groups, differential geometry andsymmetric spaces as well as the section on Hodge’s theorem inF. Warner.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Abstract homology theory

In an abelian category A the homology of a covariant functorF is defined as follows. Let V be an object in A. Take aresolution formed of projective objects

Pn → Pn−1 → · · · → P0 → V → 0, (35)

truncate V , and apply F :

→ F (Pn)→ · · · → F (P0)→ 0 (36)

Then RF∗(V ) is the homology of the above complex. Weassume that the functor F is right exact. This has the effectthat RF0(V ) = F (V ). For cohomology assume that thefunctor is contravariant and left exact.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

For Lie algebra cohomology we consider the functor V 7→ V g.For Lie algebra homology consider the functor V 7→ V /(gV ).We can construct a resolution by free objects as follows.If (π,V ) is a representation, let

P(V ) := U(g)⊗C V (37)

This is free as a module for the left action of g. There is ag−equivariant map P(V )→ V → 0, u ⊗ v 7→ π(u)v .

Dually one can take I (V ) := HomC(U(g),V ). Then0→ V → I (V ) via v 7→ fv (u) := π(u)v .

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

Koszul Complex

One of the ways to “compute” or identify the abstractcohomology theory with the specific theories is to construct anacyclic resolution of the trivial module C.

Kn := U(g)⊗C Λng, κ(x)(u⊗ω) = xu⊗ω+ u⊗ad x ω. (38)

The differential is

∂(u⊗x1∧· · ·∧xk) =∑

(−1)i+1uxi⊗x1∧· · ·∧x̂i∧· · ·∧xk . (39)

This induces a map

Homg(U(g)⊗ Λng,V )d−−−→ Homg(U(g)⊗ Λn+1g,V )∥∥∥ ∥∥∥

HomC(Λng,V )d−−−→ HomC(Λn+1g,V ).

(40)

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

(df )(1⊗ x0 ∧ · · · ∧ xn) = f (d(1⊗ x0 ∧ · · · ∧ xn))

= f(∑

(−1)ixi ⊗ x0 ∧ · · · ∧ x̂i ∧ · · · ∧ xn)

=∑

(−1)iπ(xi)f (x0 ∧ · · · ∧ x̂i ∧ · · · ∧ xn)

+∑

(−1)i f (ad xi(x0 ∧ · · · ∧ x̂i ∧ · · · ∧ xn)).

To see that d is exact, recall the filtration on U(g). We get

∂ : U(g)k ⊗ Λng→ U(g)k+1 ⊗ Λn−1g, (41)

which induces a map

∂ : S(g)⊗ Λng→ S(g)⊗ Λn−1g. (42)

Replace S(g) by P(g). It is enough to show∂ : K[x1 · · · xn]⊗ Λn(x1, . . . , xn)→ K[x1, . . . , xn]⊗ Λn−1(x1 · · · xn)is given by the formulas

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology

∂(p ⊗ ω) =n∑

i=1

xip ⊗ ι(xi)ω (43)

is exact. Define

d(p ⊗ ω) =n∑

i=1

∂ip ⊗ ε(xi)ω, (44)

Exercise

Compute d ∂ + ∂ d and show it is an isomorphism commutingwith d and ∂.

So (Kn, ∂n) is a resolution by free modules (therforeprojective). and we conclude that the abstract definitioncoincides with the explicit one.

Dan Barbasch Semisimple Lie Algebras Math 649, 2013 Lie algebra cohomology