A TASTE OF COMBINATORIAL REPRESENTATION THEORY. MATH B4900 5/02/2018

A TASTE OF COMBINATORIAL REPRESENTATION THEORY. MATH B4900 5/02/2018 Young s Lattice is an infinite leveled labeled graph with vertices and edges as follows. Vertices: Label vertices in label vertices on level k with partitions of k. Edges: See Figure 1. Draw and edge from a partition of k to a partition of k ` 1 if they differ by a box. Some combinatorial facts: (1) The representations of CS k are indexed by the partitions on level k. (2) The basis for the module corresponding to a partition λ is indexed by downward-moving paths from H to λ. (3) The action of CS k is encoded combinatorially as well: Define the content of a box b in row r and column c of a partition as cpbq c r, the diagonal number of b, where we start counting rows from the top (the top row is row 1), and start counting columns from the left (the left-most column is column 1). Label each edge in the diagram by the content of the box added. Then for the basis vector v corresponding to a particular path is a simultaneous eigenvector for all the Jucys-Murphy elements (defined below); specifically, X i acts on v by the contend of the box added at step i. The matrix entries for the transposition s i pi i ` 1q are functions of the values on the edges between levels i 1, i, and i ` 1 as well (you ll compute for S 4 ). (4) If S λ is the module indexed by λ, then Ind CS k`1 CS k ps λ q à µ$k`1 λ µ S µ and Res CS k CS k 1 ps λ q à µ$k 1 µ λ (where Res CS k CS k 1 ps λ q means forget the action of elements not in S k 1, and Ind CS k`1 CS k ps λ q CS k`1 b CSk S λ ). S µ 1

WORKSHEET 5/02/2018 2 Figure 1. Young s lattice, levels 0 5. Ŝ 3 : Ŝ 2 : Ŝ 0 : Ŝ 1 : H 0 1-1 2-1 1-2 3-1 2 0-2 1-3 Ŝ 4 : 4-1 3 0-2 2-2 2 0-3 1-4 Ŝ 5 :

WORKSHEET 5/02/2018 3 Define X i P CS n for i 1,..., n as follows: X 1 0, and X k p1 kq ` p2 kq ` pk 1 kq, for k 2,..., n. So, for example, X 2 p1 2q, X 3 p1 3q ` p2 3q, X 4 p1 4q ` p2 4q ` p3 4q, and so on. These elements are called the Jucys-Murphy elements of S n. (a) Show X 1 ` X 2 ` ` X n is an element of the center ZpCS n q. Proof. We have X 1 ` X 2 ` ` X n ř 1ďiăjďn pijq κ p12q P ZpCS n q. (b) For l ď n, using the natural inclusion S l ď S n, identify CS l as the subalgebra of CS n given by CS l Ctσ P S n σ fixes l ` 1, l ` 2,..., nu. Show that X l commutes with everything in CS l 1. [Hint: Show that for all σ P S l 1, we have σx l σ 1 X l, and conclude that X l commutes with everything in the span of S l 1.] Proof. If σ P S l 1, then σplq l. So So σpi lqσ 1 pσpiq σplqq pσpiq lq. σx l σ 1 lÿ pσpiq lq i 1 lÿ pi lq X l, since σ is a bijection on t1,..., l 1u. Thus σx l X l σ for all σ P S l 1 ; and so ax l X l a for all a P CS l 1. i 1 (c) Conclude that CxX 1,..., X n y is a commutative subalgebra of CS n. In fact, it is a quotient of the ring of polynomials in variables X 1,..., X n and coefficients in C. Proof. For i ă j, we have X i P CS j 1, so that X i X j X i X i. Thus CxX 1,..., X n y is a commutative subalgebra of CS n.

WORKSHEET 5/02/2018 4 (d) Now let S n act on C CxX 1,..., X n y by σx i X σpiq. We define C Sn tf P C σ f fu. In general, it is a theorem that ZpCS n q C Sn. Check that that the center ZpCS 3 q is the same vector space as the linear span of 1, X 1`X 2`X 3, and X 2 1 ` X2 2 ` X2 3, i.e. ZpCS 3 q Ct1, X 1 ` X 2 ` X 3, X 2 1 ` X 2 2 ` X 2 3u. Proof. Recall ZpCS 3 q Ct1, κ p12q, κ p123q u. As we saw, X 1 ` X 2 ` X 3 κ p12q. Finally, we have X1 2 ` X2 2 ` X3 2 0 2 ` p12q 2 ` pp13q ` p23qq 2 0 ` 1 ` 1 ` p13qp23q ` p23qp13q ` 1 3 ` p132q ` p123q 3 ` κ p123q. So Ct1, X 1 ` X 2 ` X 3, X 2 1 ` X 2 2 ` X 2 3u Ct1, κ p12q, 3 ` κ p123q u ZpCS 3 q. (e) Show that the action of X 1, X 2, and X 3 can be simultaneously diagonalized on each of the modules T, S, and R, and compute their eigenvalues on these modules (showing that they are what Young s lattice is told you they were). Proof. Recall the three simple CS 3 -modules are given by T where p12q ÞÑ 1 p23q ÞÑ 1 S where p12q ÞÑ 1 p23q ÞÑ 1 ˆ1 1 ˆa 0 1 b and ˆa b b R 1 where ˆ1 1 p12q ÞÑ 0 1 ˆ0. 1 p23q ÞÑ 1 0 For R, let s change basis to v 1 e 1 and v 2 e 1 ` 2e 2 before proceeding, so that ˆ1 0 ˆ 1{2 3{2 p12q ÞÑ and p23q ÞÑ. 0 1 1{2 1{2 Then in T, X 1 0 ÞÑ 0, X 2 p12q ÞÑ 1, X 3 p12qp23qp12q ` p23q ÞÑ 1 1 1 ` 1 2;

WORKSHEET 5/02/2018 5 in S, X 1 0 ÞÑ 0, X 2 p12q ÞÑ 1, and in R, ˆ0 0 X 1 0 ÞÑ, 0 0 ˆ1 0 X 2 p12q ÞÑ, 0 1 X 3 p12qp23qp12q ` p23q ÞÑ p 1q p 1q p 1q ` p 1q 2; X 3 p12qp23qp12q ` p23q ÞÑ ˆ1 0 0 1 ˆ 1{2 3{2 ˆ1 0 ˆ 1{2 3{2 ` 1{2 1{2 0 1 1{2 1{2 ˆ 1{2 3{2 ˆ 1{2 3{2 ` 1{2 1{2 1{2 1{2 These agree with the eigenvalues indicated on Young s lattice. ˆ 1 0 0 1. (f) With s i pi i ` 1q, show that s i X i X i`1 s i 1. (1) [Note that, together with part (b), this implies that where s i X j X si pjqs i ` fpi, jq, (2) $ & 1 if j i, fpi, jq 1 if j i ` 1, % 0 otherwise, This last form is called a divided difference operator.] X s i pjq X j X i`1 X i. Proof. We have s i X i s i pi i ` 1q ÿ i 1 ÿ pj iq pi i ` 1q pj i ` 1q X i`1 s i. i 1 j 1 j 1 So, multiplying both sides on the right by s i, we get s i X i X i`1 s i 1 as desired. (g) According to the conventions set by Young s lattice, which partition corresponds to the trivial module of CS n? Which partition corresponds to the sign module of CS n? Proof. There are two one-dimensional S n modules, one corresponding to and one corresponding to.... The first has one path, where X 2 p12q acts by 1, so must be the trivial

WORKSHEET 5/02/2018 6 module; and the second has one path, where X 2 module. p12q acts by 1, so must be the sign (h) Use the claim that the CS n -module indexed by the partition λ has a basis of downward-moving walks from H to λ in Young s lattice, with action of the Jucys-Murphy element X j given by the content of the box added at step j, together with the previous problem, to compute the 5 simple CS 4 -modules. For example, in the module indexed by, you know there are two basis vectors, call them u and v, where the action of the Jucys-Murphy elements is given by X 1 u X 1 v 0, X 2 u u, X 2 v v, X 3 u u, X 3 v v, X 4 u 0, X 4 v 0. Now, you also relations like p23q 2 1, so that p23q 2 u u and p23q 2 v v; also relations like p23qx 2 X 3 p23q 1, so that p23qu p23qx 2 u px 3 p23q 1qu X 3 p23qu u, and p23qv p23qx 2 v px 3 p23q 1qv X 3 p23qv ` v. Now suppose p23qu αu ` βv and p23qv γu ` δv. Plugging those in, what does that tell you about α, β, γ, δ? Proof. We already have checked that S is the trivial module, and S is the sign module. Now, let λ be a partition of 4, and let P λ be the set of downward-moving paths p ph p p0q Ñ p p1q Ñ Ñ p p4q λq. For p P P λ, let v p be the basis vector of S λ corresponding to p. And let c i ppq P Z be the content of the box added from p pi 1q Ñ p piq. If c i ppq c i`1 ppq 1, the it s possible to reverse the order of the boxes added from p pi 1q Ñ p piq and p piq Ñ p pi`1q. In this case, let s i p be that path, and let v si p be the corresponding vector. Otherwise, if c i ppq c i`1 ppq 1, define v si p 0. Define rs i s q p be the coefficient of v q in the expansion of s i v p. Now, if j i, i ` 1, then s i X j X j s i ; so s i X j v p c i ppqs i v p ÿ c j ppqrs i s q pv q qpp λ ÿ ÿ X j s i v p X j rs i s q pv q c j pqqrs i s q pv q. qpp λ qpp λ So, comparing coefficients of v q on either side (since the v q s form a basis), we get c j ppqrs i s q p c j pqqrs i s q p for all p, q P P λ. Therefore, either c j ppq c j pqq for all j i, i ` 1; or rs i s q p 0. Thus s i v p rs i s p pv p ` rs i s s ip p v si p;

where if v si p 0, then we ll define rs i s s ip p Next, we have s i X i X i`1 s i 1. So WORKSHEET 5/02/2018 7 0 as well. s i X i v p s i c i ppqv p c i ppqrs i s p pv p ` c i ppqrs i s s ip p v si p px i`1 s i 1qv p X i`1 prs i s p pv p ` rs i s s ip p v si pq v p pc i`1 ppqrs i s p p 1qv p ` c i`1 ps i pqrs i s s ip p v si p. Again, comparing coefficients of like basis vectors, this implies c i ppqrs i s p p c i`1 ppqrs i s p p 1 and c i ppqrs i s s ip p So we can solve for the diagonal entries, rs i s p p 1{pc i`1 ppq c i ppqq. c i`1 ps i pqrs i s s ip p. The second question is a direct consequence of the fact that if s i p is defined, so that v si p 0, we have that c i`1 ps i pq c i ppq by default. Note, in particular, that if v si p 0, then rs i s s ip s i p 1{pc i`1 ps i pq c i ps i pqq 1{pc i ppq c i`1 ppqq rs i s p p. Finally, we have s 2 i 1. So since s2 i p p, we have So v p s i v p s i prs i s p pv p ` rs i s s ip p v si pq rs i s p pprs i s p pv p ` rs i s s ip p v si pq ` rs i s s ip p prs i s p s i pv p ` rs i s s ip s i pv si pq pprs i s p pq 2 ` rs i s p s i prs i s s ip p qv p ` prs i s p prs i s s ip p ` rs i s s ip p rs i s s ip s i pqv si p. 1 prs i s p pq 2 ` rs i s p s i prs i s s ip p p1{pc i`1 ppq c i ppqqq 2 ` rs i s p s i prs i s s ip p ; and 0 rs i s s ip p prs i s p p ` rs i s s ip s i pq, which is a direct consequence of ( ). The first equation states that ( ) rs i s p s i prs i s s ip p 1 prs i s p pq 2. ( ) Now, up to a change of basis that keeps X 1,..., X 4 simultaneously diagonalized, this is the extent to what we can force. In other words, any choice of off-diagonal entries that yield ( ) will suffice! I like setting b rs i s s ip p 1 prs i s p pq 2 so to get symmetric entries. Now to the specifics! For λ, we have three paths p ph 0 ÝÑ 1 ÝÑ 2 ÝÑ 1 ÝÑ q ph 0 ÝÑ 1 ÝÑ 1 ÝÑ 2 ÝÑ q r ph 0 ÝÑ 1 ÝÑ 1 ÝÑ 2 ÝÑ q q

WORKSHEET 5/02/2018 8 Note s 2 q r, s 2 r q, s 3 p q, and s 3 q p ; and otherwise s i acts by 0. So with respect to the basis v p, v q, v r, we have s 1 ÞÑ 1 0 0 1 0 0 0 1 0, s 2 ÞÑ 0 1{2? 3{2 0 0 1 0?, 1{3?? 8{3 0 s 3 ÞÑ 8{3 1{3 0. 3{2 1{2 0 0 1 For λ, we also have three paths p ph 0 ÝÑ 1 ÝÑ 1 ÝÑ 2 ÝÑ q ph 0 ÝÑ 1 ÝÑ 1 ÝÑ 2 ÝÑ r ph 0 ÝÑ 1 ÝÑ 2 ÝÑ 1 ÝÑ q Note s 2 p q, s 2 q p, s 3 q r, and s 3 r q ; and otherwise s i acts by 0. So with respect to the basis v p, v q, v r, we have s 1 ÞÑ 1 0 0 0 1 0, 1{2?? 3{2 0 1 0 0 s 2 ÞÑ 3{2 1{2 0, s 3 ÞÑ 0 1{3?? 8{3. 0 0 1 0 0 1 0 8{3 1{3 Finally, for λ, we have two paths p ph 0 ÝÑ 1 ÝÑ 1 ÝÑ 0 ÝÑ q q ph 0 ÝÑ 1 ÝÑ 1 ÝÑ 0 ÝÑ q Here, s 2 swaps p and q, and s 1 and s 3 act by 0. So ˆ1 0 s 1 ÞÑ, s 0 1 2 ÞÑ? ˆ 1{2? 3{2, s 3{2 1{2 3 ÞÑ q q ˆ 1 0. 0 1 (i) Pick one of the simple CS 4 modules M that isn t one-dimensional, and compute Res CS 4 CS 3 pmq and check it is what Young s lattice tells you it should be. Proof. We have Res CS 4 CS 3 S Ctv p, v q, v r u as a vector space, with the action of CS 3 given by s 1 ÞÑ 1 0 0 1 0 0 0 1 0 and s 2 ÞÑ 0 1{2? 3{2 0 0 1 0?. 3{2 1{2 Note that in this representation, we automatically have that X 1 1 0 0 0 1 0 and X 2 2 0 0 0 1 0. 0 0 1 0 0 1

WORKSHEET 5/02/2018 9 This picks out Cv p S and Ctv q, v r u S. So Res CS 4 CS 3 S S S. (j) Compute Ind CS 4 CS 3 pt q, and verify that is what Young s lattice tells you it should be. Proof. Since 1, p14q, p24q, and p34q are a set of distinct left coset representatives of S 3 in S 4, if we let u be our favorite basis vector of T (so that σu u for all σ P S 3 ), then has basis V Ind CS 4 CS 3 pt q CS 4 b CS3 T v 1 1 b u, v 2 p14q b u, v 3 p24q b u, v 4 p34q b u. The most efficient way to compute the isomorphism type of V is to compute the action of X 2, X 3, and X 4, and simultaneously diagonalize them. To that end, we have p12qv 1 p12q b u 1 b p12qu 1 b u v 1, p12qv 2 p12qp14q b u p24qp12q b u p24q b u v 3, p12qv 3 p12qp24q b u p14qp12q b u p14q b u v 2, p12qv 4 p12qp34q b u p34qp12q b u p34q b u v 4, which has eigenspaces V 1 Ctv 1, v 2 ` v 3, v 4 u and V 1 Ctv 2 v 3 u; pp13q ` p23qqv 1 1 b pp13q ` p23qqu 2v 1, pp13q ` p23qqv 2 pp13q ` p23qqp14q b u pp34qp13q ` p14qp23qq b u p34q b u ` p14q b u v 2 ` v 4, pp13q ` p23qqv 3 pp13q ` p23qqp24q b u pp24qp13q ` p34qp23qq b u p24q b u ` p34q b u v 3 ` v 4, pp13q ` p23qqv 4 pp13q ` p23qqp34q b u pp14qp13q ` p24qp23qq b u p14q b u ` p24q b u v 2 ` v 3, which has eigenspaces V 2 Ctv 1, v 2 ` v 3 ` v 4 u, V 1 Ctv 2 v 3 u, and V 1 Ctv 2 ` v 3 2v 4 u ; and pp14q ` p24q ` p34qqv 1 v 2 ` v 3 ` v 4, pp14q ` p24q ` p34qqv 2 pp14q ` p24q ` p34qqp14q b u p1 ` p14qp12q ` p14qp13qq b u v 1 ` 2v 2, pp14q ` p24q ` p34qqv 3 pp14q ` p24q ` p34qqp24q b u pp24qp12q ` 1 ` p24qp23qq b u v 1 ` 2v 3, pp14q ` p24q ` p34qqv 4 pp14q ` p24q ` p34qqp34q b u pp34qp13q ` p34qp23q ` 1q b u v 1 ` 2v 4, which has eigenspaces V 3 Ctv 1 ` v 2 ` v 3 ` v 4, u, V 2 Ctv 2 v 3, v 3 v 4 u, and V 1 Ct 3v 1 ` v 2 ` v 3 ` v 4 u. So with respect to the basis u 1 v 1 ` v 2 ` v 3 ` v 4, u 2 3v 1 ` v 2 ` v 3 ` v 4, u 3 v 2 ` v 3 2v 4, and u 4 v 2 v 3,

WORKSHEET 5/02/2018 10 we have 1 0 0 0 2 0 0 0 3 0 0 0 X 2 0 1 0 0 0 0 1 0, X 3 0 2 0 0 0 0 1 0, X 4 0 1 0 0 0 0 2 0. 0 0 0 1 0 0 0 1 0 0 0 2 So V Ctu 1 u ` Ctu 2, u 3, u 4 u S S.