UC Berkeley Summer Undergraduate Research Program 2015 July 8 Lecture

Transcription

1 UC Berkeley Summer Undergraduate Research Program 25 July 8 Lecture This lecture is intended to tie up some (potential) loose ends that we have encountered on the road during the past couple of weeks We will focus on root subgroups and some connections with Lie algebras Root Subgroups Fix G = GL n and let T B G be the diagonal matrices sitting inside the upper triangular matrices Let R = {χ i χ j i j} be the set of roots, R + = {χ i χ j i < j} the positive roots (relative to B) For α R, we define the unipotent subgroup U α = {I n + ae ij a C} This is a subgroup of G isomorphic to C - there exists a group isomorphism C U α that is also a morphism of algebraic varieties Let s write Exercise: u α (a) = exp(ae ij ) T normalises U α and u α (a) = I n + ae ij U α t(i n + ae ij )t = I n + α(t)ae ij = tu(a)t = u(α(t)a) Hence, identifying U α with C, we find that conjugation by T induces an action of T on C via the root α: t a = α(t)a Now, suppose that W is a representation of G and w W (µ) is a (nonzero) weight vector of weight µ: this means that tw = µ(t)w, for t T Fix α R, some root of G Important Proposition: u α (a)w = i ai w i, where w = w, w i W (µ + iα), and the sum is finite Proof: Restrict the representation of G on W to U α = C: this amounts to giving a morphism of varieties that is also a group homomorphism F : C = U α G GL(W ) Such a morphism of affine varieties is determined by the induced morphism on coordinate rings F : A GL(W ) A C = C[t] Fix a weight basis B = (w, u 2,, u m ) of W (relative to T ), such that w is a basis vector Then, we get GL(W ) = GL m, and A GL(W ) = C[xij, det ], so that any C-algebra morphism is determined by F (x ij ) = f ij (t) C[t]: we have [u α (a)] B = [f ij (a)] with respect to this basis Hence, we have u α (a)w = f i (a)u i = a k v k, v k W i k Note that the v k s above are not weight vector a priori We will show that, in fact, they are! Let t T Then, t(u α (a)w) = a k tv k k

2 Also, we have t(u α (a)w) = ( tu α (a)t ) (tw), = u α (α(t)a)µ(t)w, = µ(t) f i (α(t)a)u i, i = µ(t) k α(t) k a k v k Hence, for every a C, t T, µ(t) k α(t) k a k v k = k a k tv k = k a k (µ(t)α(t) k v k tv k ) = W Remark: a nicer proof of the following discussion was provided by Mark during lecture; I ll keep what I had for prosperity Suppose that k ak w k is a polynomial with coefficients in W (ie w k W ) Fix any basis of W, (x,, x m ) Then, we have, for every a C, = k a k w k = m f j (a)u j = f j (a) = = f j = C[t] j= Note that the coefficients of f j = n j i= c ijt i are such that w k = m j= c kju j, so that we must have w k = W, for all k In fact, we have then that Otherwise, let µ(t)α(t) k v k = tv k, for all t T, and each k such that v k Then, for k J we observe that J = {k µ(t)α(t) k v k tv k, for some t T } v k = j c j u j, where c j is the coefficient of t k in f j (t), and tv k = j c j tu j = j c j µ j (t)u j (u j W (µ j )) Moreover, µ(t)α(t) k v k = j c j µ(t)α(t) k u j, so that we must have µ(t )α(t ) k µ j (t ), for some t T, some j such that c j, and any k J Hence, for every a C, W = k J a k j c j (µ(t )α(t ) k µ j (t ))u j = = k J a k c j (µ(t )α(t ) k µ j (t )), for every j, a C 2

3 In particular, for every a C, = c j a k (µ(t )α(t ) k µ j (t )) = c j =, k J which is absurd The result follows! (Can you think of a nicer proof? I feel like that there is a more elegant way to prove this, without the juggling of quantifiers) Why do we care? We have just shown that the root subgroups move weight vectors in the direction of α Namely, if w W (µ) then u α (a)w i W (µ + iα) Recall that we have defined a partial order on weights: for λ, µ X (T ) we define λ µ λ µ = i n i α i, n i Z In particular, if α R + so that α = j m jα j, with m j Z, then for w W (µ), U α w w + λ>µ W (λ) Similarly, if α R +, and w W (µ), then U α w w + λ<µ W (λ) Draw some A 2 examples: fundamental represenations, Ad, decompose some symmetric products The above results tell us more If we choose a weight basis B = (w,, w m ), w i W (µ i ), for a representation W and order the basis elements so that they (linearly) refine the above partial ordering - this means µ i µ j = i < j - then the matrix of an element u U α, for any α R +, is upper triangular, unipotent: we must always have [ ] a [u α (a)] B =, with powers of a appearing above the diagonal Similarly, the matrix of an element u U α, for any α R +, is lower triangular, unipotent For example, fix the basis (e 2, e 3, e 4, e 23, e 24, e 34 ) of 2 C 4 We have the set of weights of this representation 2 Λ( C 4 ) = {χ i + χ j i < j} Note that µ 2 = χ + χ 2 = ω 2, µ 3 = χ + χ 3 = ω 2 α 2, µ 4 = χ + χ 4 = ω 2 α 2 α 3, µ 23 = χ 2 + χ 3 = ω 2 α α 2, 3

4 The poset (Λ( 2 C 4 ), ) is We see that, for example, µ 24 = χ 2 + χ 4 = ω 2 α α 2 α 3, µ 34 = χ 3 + χ 4 = ω 2 α 2α 2 α 3 µ 2 µ 3 µ 4 µ 23 µ 24 µ 34 [u α3 (a)] = a a a [u α +α 2 (a)] = The Weyl Group Suppose that w W (µ) is a weight vector of weight µ (with respect to T ) Recall that the Weyl group of G (relative to T ) is defined to be W = N G (T )/T = S n For σ W, let x σ N G (T ) be a representative Then, W acts on X (T ) as follows: for σ W, α X (T ), σ α : t α(x σ tx σ ) 2 if α R and s α W is the reflection in the hyperplane orthogonal to α, then t(x sα w) = x sα (x s α tx sα )w = x sα µ(x s α tx sα )w = (σ µ)(t)x sα w = x sα w W (σ µ) Hence, W permutes the weight spaces; in particular, for any σ W, dim W (µ) = dim W (σ µ) The Lie algebra The Lie algebra of G is defined as g def = T e G, the tangent space at the identity For us, g = M n We can think of tangent vectors as derivatives of curves passing through e G at t = For example, γ α : [, ] U α ; t u α (t), and we have, assuming α = χ i χ j, γ () = E ij 4

5 Given a rational representation ρ : G GL(W ), we get an induced map of tangent spaces: dρ : T e G T id GL(W ) ; γ () (ρ γ) () For example, if we choose a weight basis B for W = 2 C 4 refining, as above, then we can identify GL(W ) with GL 6, and for any u U α we have seen that the i th column of [u α (a)] B is of the form c a i c 2 a i 2 c i a, c i {, } Hence, if we restrict ρ to U α then we find that (ρ γ α ) () is a matrix with s everywhere away from the main superdiagonal, where there may appear s For example, in the above examples, we see that Observing that dρ(γ α 3 ()) = dρ(γ α +α 2 ()) = γ α 3 () = E 34 M 4, γ α +α 2 () = E 3 M 4, we can think of the above matrices as showing us how the root operators E 34, E 3 act on the induced Lie algebra representation (whatever this means) For example, E 34 does nothing to any weight space W (µ ij ), except for W (µ 4 ) and W (µ 24 ), and E 34 W (µ 4 ) W (µ 3 ), E 34 W (µ 24 ) W (µ 23 ) Observe that we have µ 3 = µ 4 + α 3, µ 23 = µ 24 + α 3 ; this is no accident Fact: If α R, then E α W (µ) W (µ + α) 5