Contents
1. Day 3
• (a) Representations of sl(2)
• (b) Enveloping algebra
• (c) Killing form and Casimir operator
1.1. Representations of sl(2). The structure theory of semisimple Lie algebras is based on the classification of irreducible representations ofsl(2).
In this section we assumeK to be algebraically closed of characteristic zero.
Recall the basis e, f, h ofg=sl(2) with relations [h, e] = 2e; [h, f] =−2f; [e, f] =h.
Let V be a finite-dimensional irreducible representation of g. Since K is algebraically closed we can find a non-zero eigenvector v ∈ V for h with hv =λv. Considerv =v0,vi =eiv =evi−1. Suppose by induction that vj is an eigenvector forh with eigenvalue λj,j≤i. We compute:
hvi+1 =hevi =ehvi+ [h, e]vi =e(λi)vi+ 2evi = (λi+ 2)vi+1. It follows that each non-zeroviis an eigenvector, with eigenvalueλi =λ+2i.
In particular, the eigenvalues are all distinct – we are using that K is of characteristic 0! – hence the non-zero vi are linearly independent. Since dimV <∞ there must be a smallest i such that vi+1 = 0, in other words evi= 0. Write w=vi,µ=λi, so that
hw =µw, ew= 0.
Since V is irreducible,w generates V as a representation of g; that is, V is the smallestg-invariant subspace ofV containing w.
So letw0 =w,wi=fiw. The above argument shows that each non-zero wi is an eigenvector for h with eigenvalueµi =µ−2i. Again, there is ann such thatwn6= 0,f wn=wn+1 = 0. LetWibe the span ofwj, j ≤i. I claim that {wi, i= 0, . . . , n} is a basis forV. Because they are eigenvectors for h with distinct eigenvalues, they are linearly independent; thus it suffices to show that Wn=V, in particular thatWn is stable underg. We will in fact show that eachWi is stable under b, the Lie algebra generated by h and e,
1
and indeed thateWi ⊂Wi−1. We know this to be true fori= 0. Assume it forj≤i−1. Now
ewi=ef wi−1 =f ewi−1+hwi−1=µi−1wi−1 (modf(Wi−2)⊂Wi−1. by induction.
Thus dimV =n+ 1 andhis diagonalizable with eigenvalues µi =µ−2i, i = 0, . . . , n. But h = [e, f] and therefore T r(h) = 0. We compute: 0 = Pn
i=0µ−2i = (n+ 1)µ−2n(n+1)2 and it follows that µ= n = dimV −1.
We have proved the uniqueness part of
Theorem 1.1. For everyn≥0there is a unique irreducible representation Vn of sl(2), up to isomorphism, of dimensionn+ 1. The action of h onVn is diagonalizable, with eigenvalues{−n,−n+ 2, . . . , n−2, n}.
Proof. We have to show existence. For n= 0 we take the trivial represen- tation; for n= 1 we take the tautological representation onK2. In general, we can let
Vn=Symn(V) = (V ⊗V ⊗ · · · ⊗V)Sn
(invariant n-fold tensors). Alternatively we can write down the matrices explicitly in a basisv1, . . . , vn+1(new indices): h(vi) = (n−2i+ 2)vie(vi) = (n−i+ 2)vi−1,f(vi) =ivi+1 (with v0=vn+2 = 0). Then we can check that these matrices in M(n+ 1, K) satisfy the relations
[h, e] = 2e; [h, f] =−2f; [e, f] =h
and therefore define a representation of sl(2). For example, ef(vi) = (n− i+ 1)ivi, f e(vi) = (i−1)(n−i+ 2)vi, so
[e, f]vi= [(n−i+ 1)i−(i−1)(n−i+ 2)]vi= (n−2i+ 2)vi =h(vi) as required. Any non-trivial subrepresentation would contain an eigenvector for h, and since none of the relevant coefficients of eand f vanish, we see that even one h-eigenvector is enough to generate all the others.
1.2. Enveloping algebra. Any vector spaceW overK defines two graded associative algebras overK. Thetensor algebraT(W) =⊕i≥0W⊗i has mul- tiplication that is bilinear over K (distributive) and defined on generators by
v1⊗v2⊗ · · · ⊗vn·w1⊗w2⊗ · · · ⊗wm =v1⊗v2⊗ · · · ⊗vn⊗w1⊗w2⊗ · · · ⊗wm (taking Tn(W)×Tm(W) to Tn+m(W) for all n, m). This is the universal object for K-linear maps to associative K-algebras: for all K-algebras A, the restriction map
HomK−alg(T(W), A) → HomK(W, A)
is bijective. The symmetric algebra S(W) is the quotient of T(W) by the two-sided ideal generated byv⊗w−w⊗vforv, w∈T(W); it is the universal object for K-linear maps to commutativeK-algebras.
In particular, ifV is a second vector space then any linear mapW → End(V) factors through a map T(W)→ End(V). This just says that the elements of any subspace of End(V) generate a subalgebra by (matrix) multiplica- tion. Now suppose W =g is a Lie algebra, and suppose g →End(V) is a Lie algebra representation. Then by definition, it factors through a homo- morphism
T(g)/I → End(V)
where (I) is the two-sided ideal generated by expressions of the form X⊗ Y −Y ⊗X−[X, Y]. We denote by U(g) the quotient T(g)/I, and call it theuniversal enveloping algebra ofg. It is the universal associative algebra for representations ofg; in other words, the universal associative algebra in which the Lie bracket becomes the canonical bracket XY −Y X. Alterna- tively
Proposition 1.2. There is a canonical equivalence between the category of representations of g and the category of U(g)-modules, given by restriction of aU(g)-module to g.
The algebra U(g) is no longer graded: the element XY −Y X of degree 2 is identified with the element [X, Y] of degree 1. However, the subspace Un(g) of elements of degree ≤nis well defined for alln≥0; it is the image of ⊕ni=0g⊗i inU(g). Moreover,Un(g)⊂Un+1(g) for alln, so we can define grU(g) = ⊕i≥0Ui(g)/Ui−1(g) and this is a commutative algebra. The map g 7→ grU(g) therefore factors canonically through a map S(g)→ grU(g), and it is easy to see that it is surjective.
Theorem 1.3 (Poincar´e-Birkhoff-Witt theorem). For any Lie algebra g, the map S(g) → grU(g) is an isomorphism of graded algebras.
The proof can be found in any textbook on Lie algebras. We only need U(g) for a single element of degree 2, the Casimir operator (defined in the next section).
1.3. Killing form and Casimir operator. As always,gis a finite-dimensional Lie algebra over K.
Definition 1.4. A bilinear form B : g⊗g 7→ K is invariant if, for all X, Y, Z ∈g
B(adZ(X), Y) +B(X, adZ(Y)) = 0.
Equivalently, B is an element ofHom(g⊗g, K)g.
Example 1.5. If g=gl(n), let B(X, Y) =T r(XY). Then B is invariant.
Proof. . We need to show that, for allX, Y, Z
T r((ZX−XZ)Y) +T r(X(ZY −Y Z)) = 0.
which comes down toT r(ZXY) =T r(XY Z).
Remark 1.6. This cancellation, and others like it, may appear to have come from nowhere. It’s easier to see what is happening in terms of the action of the groupGL(n). The formB is invariant under the adjoint representation of GL(n):
B(gXg−1, gY g−1) =B(X, Y).
This is easier to see:
T r(gXg−1·gY g−1) =T r(gXY g−1) =T r(XY g−1g) =T r(XY).
The magical cancellation for the Lie algebra action is the derivative of the cancellation for the group action.
Corollary 1.7. Let φ:g → gl(V) be a representation. Then Bφ(X, Y) = T r(φ(X)φ(Y))is an invariant bilinear form.
In particular, every g has a canonical invariant symmetric bilinear form, namely theKilling form
Bad(X, Y) =T r(adX ◦adY).
Exercise 1.8. (a) Show thatX⊗Y 7→T r(XY) is nondegenerate on gl(n).
(b) Suppose K = R or C. Show that, if g ⊂ gl(n) is invariant under transpose, then the restriction to g of T r(XY) is nondegenerate.
Lemma 1.9. Leth⊂gbe an ideal. Then the restriction of the Killing form of g to his the Killing form of h.
Proof. For any X, Y ∈h, adX ◦adY(g) ⊂h. So the trace of adX ◦adY on g equals its trace on h. (Write down a matrix with a basis of g consisting first of elements of h, then of a complement of hing).
In the rest of this section, we assume we are given φ such that Bφ is nondegenerate. (This will imply thatgis reductive; ifBad is nondegenerate then g is semisimple.) The form B =Bφ is an element of Hom(g⊗g, K), and as such defines a homomorphism of vector spaces
g7→g∗;X7→[(Y 7→B(X, Y)],
which is an isomorphism because B is assumed nondegenerate. But B is also ad(g) invariant, which implies that this homomorphism is also ad(g)- invariant. Thus the canonical representations ad and ad∗ on g and g∗ are isomorphic and B defines an explicit isomorphism.
ThusB defines isomorphisms
Hom(g,g)−→g∼ ⊗g∗ −→g∼ ⊗g
of representations ofg. There is a canonicalg-invariant element ofHom(g,g), namely the identity, which in any matrix representation is given by a multi- ple of the trace when identified with ag-invariant element ofHom(Hom(g,g), K).
The corresponding element of g⊗g∗ is written in the basis X1, . . . , XN of g
asPN
i=1Xi⊗Yi, whereYi is the dual basis ofg∗. With respect toB, we let Xi∗ be the dual basis of g, and define the Casimir element
CB =
N
X
i=1
Xi⊗Xi∗ ∈g⊗g;
we use the same notation to denote the image of CB inU(g).
Proposition 1.10. This expression doesn’t depend on the basis, and CB
belongs to the center of U(g).
Proof. The first assertion is a formal consequence of the invariance of the trace, and the second is a formal consequence of theg-invariance ofCB, but we give a computational proof.
So let Zi be a second basis of g. ThenZi =P
kB(Zi, Xk∗)Xk and Zi∗ = P
jB(Zi∗, Xj)Xj∗ (basic properties of nondegenerate bilinear forms). Thus X
i
Zi⊗Zi∗=X
i,k,j
B(Zi, Xk∗)B(Zi∗, Xj)Xk⊗Xj∗
=X
k,j
[B(X
i
B(Zi∗, Xj)Zi, Xk∗)]Xk⊗Xj∗
=X
k,j
[B(Xj, Xk∗)]Xk⊗Xj∗=X
k
Xk⊗Xk∗
That proves the first statement. As for the second, letX ∈g; then X⊗X
i
Xi⊗Xi∗−X
i
Xi⊗Xi∗⊗X = X
i
(X⊗Xi⊗Xi∗−Xi⊗X⊗Xi∗) +X
i
(Xi⊗X⊗Xi∗−Xi⊗Xi∗⊗X)
=X
i
[X, Xi]⊗Xi∗−Xi⊗[Xi∗, X]
=X
i
X
k
(B([X, Xi], Xk∗)Xk⊗Xi∗−Xi⊗B(Xk,[Xi∗, X])Xk∗)
=X
i,k
(B([X, Xk], Xi∗)Xi⊗Xk∗+B(Xk,[X, Xi∗])Xi⊗Xk∗) by reindexing. But for all i, k,
B([X, Xk], Xi∗) +B(Xk,[X, Xi∗]) = 0
by invariance of B, so we are done.
Schur’s Lemma then implies
Corollary 1.11. Suppose B is an invariant bilinear form on g. Suppose K is algebraically closed. Let (φ, V) be any irreducible representation of g.
Then φ(CB) is a scalar.
Lemma 1.12. Let (φ, V) be a representation of g such that B = Bφ is nondegenerate. Then T r(φ(CB)) = dimg. On the other hand, CB acts as0 on the trivial representation.
Proof. The second statement is clear, because anyXi∗acts as 0 on the trivial representation. On the other hand,
T r(φ(CB) =X
i
T r(φ(Xi)φ(Xi∗)) =X
i
Bφ(Xi, Xi∗) =X
i
1 = dimg.
