UC Berkeley Summer Undergraduate Research Program 2015 July 9 Lecture

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1 UC Berkeley Summer Undergraduate Research Program 205 July 9 Lecture We will introduce the basic structure and representation theory of the symplectic group Sp(V ). Basics Fix a nondegenerate, alternating bilinear form ω : V V C, where V is a finite dimensional C-vector space. This means that. ω(u, v) = ω(v, u) 2. ω(u, V ) = 0 = ω(v, u) = u = 0 V, 3. ω(au + bv, w) = aω(u, w) + bω(v, w). Let e 0. Then, by (2) there is some f / span(e ) such that ω(f, e ) 0; scaling we may assume that ω(f, e ) =. Let U = span(e, f ) V. The subspace U = {v V ω(u, v) = 0, for all u U } V satisfies U U = {0} (why?). Nondegeneracy ensures an isomorphism V V ; v ω(, v), so that we can identify the annihilator U V with some dim V 2 subspace V V via this isomorphism. Moreover, U V = {0}, and ω V is nondegenerate: if ω(v, V ) = 0 then ω(v, V ) = 0 = v = 0. Hence, by induction we can assume that V admits a basis (e 2,..., e n, f n,..., f 2 ) such that ω(f i, e j ) = δ ij. Then, with respect to the basis (e,..., e n, f n,..., f ), the matrix of ω is Remark: J = J = J t J def = [ω] =... The symplectic group (with respect to ω) Sp(V, ω) is the group of all operators on V preserving ω... Sp(V, ω) = {g GL(V ) ω(gu, gv) = ω(u, v), for all u, v V }. We have just seen above that any nondegenerate alternating form ω admits a basis such that the matrix of ω is J: this means that all such pairs (V, ω) (such a pair is defined as a symplectic vector space; a corresponding basis is called a symplectic basis) are equivalent, in the sense that, for any two such pairs (V, ω), (V, ω ), there is an isomorphism T : V V, such that ω (Tu, Tv) = ω(u, v). In particular, we may as well choose V = C 2n, and ω to be the nondegenerate, alternating form ω 0 : C 2n C 2n C ; (u, v) u t Jv.

2 Hence, we can restrict ourselves to the group Sp(C 2n, ω 0 ) = {g GL 2n g t Jg = J}. We will denote this group Sp 2n : it is a subvariety of GL 2n defined by the equations g t Jg = J. Hence, Sp 2n is an affine algebraic variety such that its group operations are given by morphisms of algebraic varieties. Moreover, we see that Exercise:. If g Sp 2n then g t Sp 2n. det(g t Jg) = det J = det(g) 2 = = det(g) = ±. 2. Let g Sp 2n, v V, such that gv = λv. Show that there exists w V such that gw = λ w. Deduce that Sp 2n SL 2n. Tori, Weyl Group It can be checked that the intersection T = Sp 2n T 2n, with the standard maximal torus in GL 2n, is t... T = t n tn t,..., t n C = (C ) n... t Hence, T is a complex torus. Moreover, T is maximal. You can think of T as being those operators on C 2n that preserve ω 0 and for which the symplectic basis is a common eigenbasis. Let s guess what we expect the Weyl group W = N Sp2n (T )/T to be: - For GL n we saw that the normaliser N G (T ) consists of all those operators that preserved the set of lines defined by the common eigenbasis for T GL n. When we divided out by the action of T we forgot about the scaling that we could have within each line, so that N G (T )/T was the group acting on the set of n eigenlines. - You might guess that we simply permute all elements of the symplectic basis but this is not correct: if we swapped e and e 2 then we would no longer preserve the form ω 0 as = ω 0 (f, e ) ω 0 (f, e 2 ) = 0. A moment s thoughtand we see that permuting e i and e j means that we must also permute f i and f j. Hence, there should be a copy of S n sitting inside W : it consists of those symmetries that preserve the pairs of lines (Ce i, Cf i ). - We are allowed further symmetry: within each pair of lines (Ce i, Cf i ), we can swap the lines. However, when we swap the lines we must map Ce i to Cf i, in order that the symmetry preserve ω 0. - Hence, we should think of the Weyl group of Sp 2n as being the symmetry we have in any ordering of the symplectic planes we choose. We expect there to be subgroups of W isomorphic to S n - corresponding to swapping the pairs of lines (Ce i, Cf i ) - and (Z/2Z) n - swapping the lines within each pair (with the minus twist ). 2

3 Fact: the Weyl group of Sp 2n is W (C n ) def = W = S n (Z/2Z) n, where σ S n W acts on (a,..., a n ) (Z/2Z) n by permuting entries. For example, the representatives of the symmetric group appearing in W are of the form [ ] σ 0 0 σ where σ is the permutation matrix obtained by reflecting σ in the antidiagonal. For example, if σ = (2) S 3 then we have. The element (a,..., a n ) (Z/2Z) n has representative a 2n 2n matrix [c ij ] with c ii = c 2n+ i,2n+ i =, c 2n+ i,i = = c i,2n+ i, whenever a i = 0 Z/2Z, whenever a i = Z/2Z. For example, in Weyl group W (C 2 ) we have the following representatives for (2) S 2, (, 0) (Z/2Z) 2, (0, ) (Z/2Z) , , Exercise: W (C 2 ) = D 8. However, W (C n ) = D4n in general (why?). Roots etc The character lattice of T is generated by the projections onto the i th diagonal entry. Denote these projections χ i, so that X (T ) = Zχ i. We are going to determine the root system - for this we need to know the Lie algebra of Sp 2n. It is a fact that sp 2n def = {X M 2n X t J + JX = 0} = {[ ] A B C D where t M denotes the transpose across the antidiagonal. For example, we have a b e f sp 4 = c d g e h i d b j h c a a b c x x 2 z d e f x 2 y x 2 sp 6 = g h i x x 2 x y y 2 w i f c y 2 v y 2 h e b u y 2 y g d a 3 } t C = C, t B = B, D = t A

4 Then, Sp 2n acts linearly on sp 2n by conjugation, and T admits a common eigenbasis: namely, they are the same basis vectors as we see for the above examples. For example, for sp 4 we have eigenbasis {E E 44, E 22 E 33, E 2 E 34, E 2 E 43, E 3 + E 24, E 4, E 23, E 32, E 4, E 3 + E 42 } and the weights are 0, ±χ χ 2, ±χ + χ 2, ±2χ, ±2χ 2 If we let α = χ χ 2, α 2 = 2χ 2, then we can write the nonzero elements above as these are the roots of sp 4. In the sp 6 case we have weights If we let then we can write the above nonzero weights ±α, ±(α + α 2 ), ±(2α + α 2 ), ±α 2 ; 0, ±χ i χ j, ±χ k + χ l, i < j, k l. α = χ χ 2, α 2 = χ 2 χ 3, α 3 = 2χ 3, ±α i, ±(α + α 2 ), ±(α 2 + α 3 ), ±(2α 2 + α 3 ) ± (α + α 2 + α 3 ), ±(α + 2α 2 + α 3 ), these are the roots of sp 6. ±(α + 2α 2 + α 3 ), ±(2α + 2α 2 + α 3 ); In general, the roots of Sp 2n (with respect to our choice of T ) are and a set of simple roots are R def = {χ i χ j i j} {χ i + χ j i j} X (T ) This allows us to define the positive roots and the negative roots R +. For α R, a root subgroup is a subgroup α = χ χ 2, α 2 = χ 2 χ 3,..., α n = 2χ n. R + def = {α R α = n i α i, n i Z 0 }, U α = I 2n + ae α, a C where E α is an eigenvector with T -eigenvalue α. Proposition/Definition: The standard Borel B Sp 2n is the subgroup generated by T and U α, α R +. It is a Borel subgroup. It is the intersection Sp 2n B 2n of Sp 2n with the upper triangular matrices in GL 2n. We define the fundamental weights ω = χ, ω 2 = χ + χ 2,..., ω n = χ χ n, ω n = χ χ n. 4

5 The corresponding Dynkin diagram is = This signifies that the angle between α i and α i+ is 2π/3, for i =,..., n ; the angle between α n and α n is 3π/4; and all other angles between simple roots is π/2. All of these angles are measured in the Euclidean space Rχ i, with respect to the standard inner (= dot) product. The set of roots R is a root system: the reflections defined by each simple root α i are s i : v v 2 (v, α i) (α i, α i ) α i. The group generated by the s i in GL(X (T )) is isomorphic to the Weyl group. Remark: Observe, that rankzr = n = dim T, in contrast to the GL n case: this is because Sp 2n is (semi)simple. Representation Theory The main Theorem is Theorem:. Let W be a finite dimensional irreducible representation of Sp 2n. Then, there is a unique line L W that is B-invariant, and on which T acts by λ X (T ). Moreover, if µ Λ(W ) is a weight then λ µ; we call λ the highest weight in W. The weight λ = i a iχ i satisfies a... a n To any weight λ = i a iχ i, a... a n 0, there is an irreducible representation of Sp 2n with λ as its highest weight. We will focus on the example Sp 4. 5

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