Theorem: (Weyl's Theorem)  If L is semisimple and V is a finite dimensional L-module then V is completely reducible (i.e. V = V1(+)V2(+)..(+)Vk (where (+) represents the direct sum), Vi irreducible).

Remark:  It is enough to show that if W is a submodule of V then there is a complementary submodule so V = W + W'.

Note: As a module W' should be isomorphic to V/W.  Equivalently we need to show that every exact sequence of form
0 ~> W ~> V ~> V/W ~> 0 splits (i.e. there exists s: V/W ~> V so that s(p(v)) is the identity) as a sequence of L modules.

Lemma: (Schur's Lemma)  Suppose L is a Lie Algebra, V is an irreducible L-module, and T is an endomorphism which commutes with the action (i.e. x(Tv) = T(xv) for all x in L, v in V) then T is of the form cI for some c in C.

Proof:  Let c be an eigenvalue of T and let W be the c-eigenspace.  Then for x in L, w in W, c(xw) = x(Tw) = T(xw) and so xw is also an eigenvector for eigenvalue c.  This implies W is L invariant.  Since W is nontrivial and V is irreducible W=V and hence Tv=cv for all v in V.  Thus T=cI. n

Def:  If L is a Lie Algebra and f: L ~ gl(V) is a representation, we define the trace form of f to be the symmetric bilinear form given by (x, y)f = trace(f(x)f(y)).

Lemma:  If f is a representation of a Lie algebra L then Rad f is and ideal of L.

Proof:  Let y be in Rad f, x in L.  Then for any z in L, trace(f([xy])f(z)) = trace ([f(x)f(y)]f(z)) = trace ([f(y)f(-x)]f(z)) = trace (f(y)[f(-x)f(z)]) = trace (f(y)f([zx])) = 0.  Thus for x in L, y in Rad f, [xy] is in Rad f. n

Lemma:  Suppose L is a semisimple Lie Algebra and f: L ~> gl(V) is injective then (x, y)f is non-degenerate.

Proof:  R = Rad f = {x in L : trace f(x)f(y)=0 for all y in L}is an ideal of L, thus f(R) is an ideal of f(L).  As f(L) is a subalgebra of gl(V) we can apply Cartan's solvability criterion (the general form not the ad one) to f(R).  Since trace(f(x)f(y)) = 0 for f(x) in [f(R)f(R)], f(y) in f(R) (in fact we know something stronger) we know f(R) is solvable.  Since f is injective R is isomorphic to f(R) and a solvable ideal of L.  Thus R is zero and (x, y)f is non-degenerate. n

Note:  To prove Weyl's theorem we can assume that V is a faithful L module.  Otherwise we can mod out and look at what we have on the quotient space).

Note:  Suppose V is a vector space and ( , ) is a non-degenerate bilinear form on V.  Then for and basis u1, u2, ..., un of V we can define a dual basis v1, v2, ..., vn of V satisfying (ui, vj) = d(i, j).

Def:  Given a semisimple Lie algebra L and a faithful f: L ~> gl(V) we define the Casmir element cf in gl(V) as follows:  Choose a basis x1, x2, ..., xn and let y1, y2, ..., yn be the dual basis with respect to the trace form.  Set cf = f(x1)f(y1)+f(x2)f(y2)+...+f(xn)f(yn).

Note:  cf is invariant under change of basis.

Lemma:  For any trace form (x, y)f, ([x,y],z)f = (x,[y,z])f .

Proof: ([x,y],z)f = trace(f([xy])f(z)) = trace([f(x)f(y)]f(z)) =  trace(f(x)f(y)f(z) - f(y)f(x)f(z)) = trace(f(x)f(y)f(z)) - trace(f(y)f(x)f(z)) = trace(f(x)f(y)f(z)) - trace(f(x)f(z)f(y)) =  trace(f(x)f(y)f(z) - f(x)f(z)f(y)) = trace(f(x)[f(y)f(z)]) = trace(f(x)f([yz])) =  (x,[y,z])f .

Lemma:  For all x in L, v in V cf(f(x)v) = f(x)(cfv).

Proof:  We want to show f(x)(Sf(xi)f(yi)) = (Sf(xi)f(yi))f(x).  We can manipulate this by subtracting (Sf(xi)f(x)f(yi)) from both sides to get (Sf(x)f(xi)f(yi)) - (Sf(xi)f(x)f(yi)) = (Sf(xi)f(yi)f(x)) - (Sf(xi)f(x)f(yi)) or
(Sf(x)f(xi)f(yi) - f(xi)f(x)f(yi)) = (Sf(xi)f(yi)f(x) - f(xi)f(x)f(yi)) equivalent to
(S(f(x)f(xi) - f(xi)f(x))f(yi)) = (Sf(xi)(f(yi)f(x) - f(x)f(yi)) or (S[f(x)f(xi)]f(yi)) = (Sf(xi)[f(yi)f(x)]) or finally
Sf[x,xi]f(yi) = - Sf(xi)[f(x)f(yi)].  Thus this is eqivalent to what we want to show.
            Now we know that the trace form is invariant.  Thus ([xi,x],yj)f = (xi,[x,yj])f .  We can write [xi, x] = Sjaijxj and [x, yj] = Sibjiyi and get aik = Sjaij(xj ,yk)f  = (Sjaijxj ,yk)f = ([xi,x],yk)f = (xi,[x,yk])f = (xi, Sjbkjyj)f = Sjbkj(xi, yj)f = bki, so aij = bji .
            Finally we return to our equation and write Sif[x,xi]f(yi) = Sif(Sjaijxj)f(yi) = Sif(-Sjaijxj)f(yi) = SiSj(-aij)f(xj)f(yi) = SiSj(-bji)f(xj)f(yi) = SjSi(-bji)f(xj)f(yi) = Sjf(xj)Si(-bji)f(yi) = Sjf(xj)f(-Sibjiyi) = Sjf(xj)f(-Sibjiyi) = Sjf(xj)f(-Sibjiyi) = -Sjf(xj)f([x, yj]) = -Sjf(xj)[f(x)f(yj)] = - Sif(xi)[f(x)f(yi)].  Thus we have proved the equality above. n

Note:  Suppose V and W are L-modules.  Then the space HomC(V,W) of all linear transformations from V to W is naturally an L-module via the action (xf)(v) = x(f(v)) - f(xv).
 
2/25/99

Note:  I still have a problem with a part of the proof of Weyl's Theorem, which occupies this entire lecture.  In good conscience, I cannot write it up as I do not fully understand it.  Hopefully I will find time to get back to this proof soon.

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