Murphy elements <math> <mo>ℂ</mo><msub> <mi>A</mi><mi>k</mi> </msub><mfenced> <mi>r</mi><mi>p</mi><mi>n</mi> </mfenced> </math>

Murphy elements $ℂ A_{k} (r, p, n)$

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 20 June 2010

Murphy elements $ℂ A_{k} (r, p, n)$

Let $S \subseteq \{12 \dots k\}$ and let $I \subseteq S \cup S^{'} .$ Define $b_{S}, d_{I} \in A_{k}$ by $b_{S} = \{S \cup S^{'}, {\{l, l^{'}\}}_{l \notin S}\} and d_{I \subseteq S} = \{I, I^{c} {\{l, l^{'}\}}_{l \notin S}\} .$ For example, in $A_{9},$ if $S = \{2, 4, 5, 8\}$ and $I = \{2, 4, 4^{'}, 5, 8\}$ then

and

Note that $d_{I} = d_{I^{c}}, d_{S \cup S^{'}} = d_{\emptyset} = b_{S}, d_{\{l, l^{'}\}} = d_{{\{l, l^{'}\}}^{c}} = b_{S - \{l\}} .$ For $k \in \frac{1}{2} ℤ_{\geq 0}$ and $r \in ℤ_{> 0},$ $Z_{k, r} = (\begin{matrix} n \\ 2 \end{matrix}) + \sum_{|S| \geq 1} {(- 1)}^{|S|} ((n - 1) b_{S} + \sum_{κ (I) = 0 (\mod r)} {(- 1)}^{κ (I)} (d_{I} - b_{S})),$ where the outer sum is over $S \subseteq \{12 \dots k\}$ such that $S \neq \emptyset$ and the inner sum is over $I \subseteq S \cup S^{'}$ such that $d_{I} \in A_{k} (r, 1, n)$ and $d_{I} \neq b_{S} .$

For $n, r \in ℤ_{\geq 0},$ $κ_{r, n} = \frac{1}{r} \sum_{m = 0}^{r - 1} \sum_{1 \leq i < j \leq n} t_{i}^{m} t_{j}^{- m} s_{i j}$ is a central element of $ℂ G_{r, p, n} .$ $κ_{r, n} = \sum_{b \in λ} c (b), as operators on G_{r, p, n}^{(λ, j)},$ the irreducible $G_{r, p, n}$ -module indexed by $(λ, j),$ where $λ = (λ^{(0)} \dots λ^{(r - 1)})$ is a multipartition with $n$ boxes.
Let $n, k \in ℤ_{\geq 0} .$ Then, $κ_{r, n} = Z_{k, r} and κ_{n - 1} = Z_{k + \frac{1}{2} r}, as operators on V^{\otimes k} .$
Let $n \in ℂ, k \in \frac{1}{2} ℤ_{\geq 0} .$ Then $Z_{k} (r)$ is a central element of $ℂ A_{k} (r, p, n),$ and, if $n \in ℂ$ is such that $ℂ A_{k} (r, p, n)$ is semisimple then $Z_{k} (r) = \sum_{b \in λ} c (b), as operators on A_{k}^{λ},$ where $A_{k}^{λ}$ is the irreducible $ℂ A_{k} (r, p, n)$ -module indexed by $λ .$

Proof.

The element $κ_{n}$ is the class sum corresponding to the conjugacy class of the element $s_{12}$ in $G r, p, n$ and thus $κ r, n$ is a central element of $ℂ G_{r, p, n} .$ The constant by which $κ r, n$ acts on $G_{r, p, n}^{(λ, j)}$ follows from theorem???.
The first statement follows from parts (a) and (b) and Theorems 3.6 and 3.22 as follows. By theorem 3.6, $ℂ A_{k} (n) ≅ {End}_{S_{n}} (V^{\otimes k})$ if $n \geq 2 k .$ Thus, by Theorem 3.22, if $n \geq 2 k$ then $Z_{k}$ acts on the irreducible $ℂ A_{k} (n)$ -module $A_{k}^{λ} (n)$ by the constant given in the statement. This means that $Z_{k}$ is a central element of $ℂ A_{k} (n)$ for all $n \geq k .$ Thus, for $n \geq 2 k, d Z_{k} = Z_{k} d$ for all diagrams $d \in A_{k} .$ Since the coefficients in $d Z_{k}$ (in terms of the basis of diagrams) are polynomials in $n,$ it follows that $d Z_{k} = Z_{k} d$ for all $n \in ℂ .$
If $n \in ℂ$ is such that $ℂ A_{k} (n)$ is semisimple let $χ_{ℂ A_{k} (n)}^{λ}$ be the irreducible characters. Then $Z_{k}$ acts on $A_{k}^{λ} (n)$ by the statement, therefore it is a polynomial in $n,$ determined by its values for $n \geq 2 k .$
The proof of the second statement is completely analogous using $ℂ A_{k + \frac{1}{2}}, S_{n - 1},$ and the second statement in part (b).
Then $2 κ_{n} (v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) = \frac{1}{r} \sum_{m = 0}^{r - 1} \sum_{i, j = 1, i \neq j}^{n} t_{i}^{m} t_{j}^{- m} s_{i j} v_{i_{1}} \otimes \dots \otimes t_{i}^{m} t_{j}^{- m} s_{i j} v_{i_{k}}$ $= \frac{1}{r} \sum_{m = 0}^{r - 1} \sum_{i, j = 1, i \neq j}^{n} (1 - E_{i i} - E_{j j} + t_{i}^{m} E_{i j} + t_{j}^{- m} E_{j i}) v_{i_{1}} \otimes \dots \otimes (1 - E_{i i} - E_{j j} + t_{i}^{m} E_{i j} + t_{j}^{- m} E_{j i}) v_{i k} .$ Expanding this sum, let $c_{S, I} = (\prod_{l \in S^{c}} δ_{i_{l} i_{l^{'}}}) {(- 1)}^{# (\{l, l^{'}\} \subseteq I) + # (\{l l^{'} \subseteq I^{c}\})} ξ^{m (\{l \in I, l^{'} \in I^{c}\} - # \{l \in I, l \in I^{c}\})} (\prod_{l \in I} δ_{i_{l} i}) (\prod_{l \in I^{c}} δ_{i_{l} j})$ and $c_{S, I} = (\prod_{l \in S^{c}} δ_{i_{l} i_{l^{'}}}) {(- 1)}^{# (\{l, l^{'}\} \subseteq I) + # (\{l l^{'} \subseteq I^{c}\})} ξ^{m (\{l \in I, l^{'} \in I^{c}\} - # \{l \in I, l \in I^{c}\})} (\prod_{l \in I} δ_{i_{l} i}) (\prod_{l \in I^{c}} δ_{i_{l} j})$ so that $2 κ_{n} (v_{i_{1}} \otimes \dots \otimes v_{i_{k}})$ is equal to $\frac{1}{r} \sum_{S \subseteq \{1 \dots k\}} \sum_{i_{1^{'}}, \dots, i_{k^{'}}} \sum_{m = 0}^{r - 1} \sum_{i, j = 1, i \neq j}^{n} \sum_{I \subseteq S \cup S^{'}} c_{S, I} (v_{i_{1^{'}}} \otimes \dots \otimes v_{i_{k^{'}}})$ $= \frac{1}{r} \sum_{S \subseteq \{1 \dots k\}} \sum_{i_{1^{'}}, \dots, i_{k^{'}}} \sum_{m = 0}^{r - 1} \sum_{I \subseteq S \cup S^{'}} (\sum_{i, j = 1}^{n} c_{S, I} (v_{i_{1^{'}}} \otimes \dots \otimes v_{i_{k^{'}}}) - \sum_{i, j = 1}^{n} {c_{S, I}}^{'} (v_{i_{1^{'}}} \otimes \dots \otimes v_{i_{k^{'}}})) .$ Here $S^{c} \subseteq \{1 \dots k\}$ corresponds to the tensor products where 1 is acting, $I \subseteq S \cup S^{'}$ corresponds to the tensor position that must equal $i,$ and $I^{c}$ corresponds to the tensor positions that must equal $j .$
When $|S| = 0$ the set $I$ is empty and the sum in (???) is equal to $(n^{2} - n) (v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) since c_{S, I} = c_{S, I}^{'} = (\prod_{l \in \{1 \dots k\}} δ_{i_{l} i_{l^{'}}}) .$ Assume $|S| \geq 1$ and separate the sum according to the cardinality of $I .$ Note that the sum for $I$ is equal to the sum for $I^{c}$ since the whole sum is symmetric in $i$ and $j .$ When $I = S \cup S^{'},$ $c_{S, I} = c_{S, I}^{'} = (\prod_{l \in S^{c}} δ_{i_{l} i_{l^{'}}}) {(- 1)}^{|S|} (\prod_{< S \cup S^{'}} δ_{i_{l} i})$ and the sum in (???) is equal to $\frac{1}{r} \sum_{m = 0}^{r - 1} \sum_{i_{1^{'}}, \dots, i_{k^{'}}} (n - 1) \sum_{i = 1}^{n} c_{S, I} (v_{i_{1^{'}}} \otimes \dots \otimes v_{i_{k^{'}}}) = (n - 1) {- 1}^{|S|} b_{S} (v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) .$ We get a similar contribution from the sum of the terms with $I = \emptyset .$
For each of the remaining subsets $I \subseteq S \cup S^{'}$ the sum in (???) contributes 0 when $κ (I) \neq 0 (\mod r)$ and ${(- 1)}^{# (\{l, l^{'}\} \subseteq I) + # (\{l l^{'} \subseteq I^{c}\})} (d_{I \subseteq S}) (v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) = {(- 1)}^{|S| + κ (I)} (d_{I \subseteq S} - b_{S}) (v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) .$ when $κ (I) = 0 (\mod r) .$
For the second statement, since $(1 - E_{i i} - E_{j j} + E_{i i} E_{j j}) v_{n} = \{\begin{array}{rcl} 0, & if & i = n or j = n, \\ v_{n}, & if & otherwise, \end{array}$ $\begin{array}{rcl} 2 κ_{n - 1} (v_{i_{1}} \otimes \dots \otimes v_{i_{k}} \otimes v_{n}) & = & \frac{1}{r} (\sum_{m = 0}^{r - 1} \sum_{i, j = 1, i \neq j}^{n - 1} t_{i}^{m} t_{j}^{- m} s_{i j}) (v_{i_{1}} \otimes \dots \otimes v_{i_{k}} \otimes v_{n}) \\ = & \frac{1}{r} \sum_{m = 0}^{r - 1} \sum_{i, j = 1, i \neq j}^{n} t_{i}^{m} t_{j}^{- m} s_{i j} v_{i_{1}} \otimes \dots \otimes t_{i}^{m} t_{j}^{- m} s_{i j} v_{i_{k}} \otimes (1 - E_{i i} - E_{j j} + E_{i i} E_{j j}) v_{n} \\ = & (\frac{1}{r} \sum_{m = 0}^{r - 1} \sum_{i \neq j} s_{i j}) (v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) \otimes v_{n} \\ + \frac{1}{r} \sum_{m = 0}^{r - 1} \sum_{i, j = 1, i \neq j}^{n} (1 - E_{i i} - E_{j j} + ξ^{m} E_{i j} + ξ^{- m} E_{j i}) v_{i_{1}} \otimes \dots \\ \dots \otimes (1 - E_{i i} - E_{j j} + ξ^{m} E_{i j} + ξ^{- m} E_{j i}) v_{i_{k}} \otimes (- E_{i i} - E_{j j}) v_{n} \\ = \frac{1}{r} \sum_{m = 0}^{r - 1} \sum_{i, j = 1}^{n} t_{i}^{m} t_{j}^{- m} s_{i j} v_{i_{1}} \otimes \dots \otimes t_{i}^{m} t_{j}^{- m} s_{i j} v_{i_{k}} \otimes E_{i i} E_{j j} v_{n} . \end{array}$ The first sum is known to equal $2 κ_{n} (v_{i_{1}} \otimes \dots \otimes v_{i_{k}})$ and is known by the computation proving the first statement, and the last is 0 since $i \neq j .$ The middle sum is treated exactly as in (???) except that now the sum is over $S$ such that $k + 1 \in S$ and $I$ such that $\{k + 1, {(k + 1)}^{'}\} \subseteq I$ or $\{k + 1, {(k + 1)}^{'}\} \subseteq I^{c} . □$

$\begin{array}{rcl} (\sum_{i = 1}^{n} t_{i}^{m}) (v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) & = & \sum_{i = 1} (t_{i}^{m} v_{i_{1}} \otimes \dots \otimes t_{i}^{m} v_{i_{k}}) \\ = & \sum_{i = 1} (1 - E_{i i} + ξ^{m} E_{i i}) v_{i_{1}} \otimes \dots \otimes (1 - E_{i i} + ξ^{m} E_{i i} \\ = & \sum_{i_{1}^{'}, \dots, i_{k}^{'}} \sum_{I \subseteq \{1 \dots k\}} {(ξ^{m} - 1)}^{|I|} (\prod_{l \in I^{c}} δ_{i_{l} i_{l^{'}}}) (\prod_{l \in I^{c}} δ_{i_{l} i} δ_{i_{l^{'}} i}) (v_{i_{1^{'}}} \otimes \dots \otimes v_{i_{k^{'}}}) \\ = & \sum_{I \subseteq \{1 \dots k\}} {(ξ^{m} - 1)}^{|I|} b_{I} \\ = & n + k (ξ^{m} - 1) + \sum_{|I| \geq 2} {(ξ^{m} - 1)}^{|I|} b_{I} \\ = & n - k + k ξ^{m} + \sum_{|I| \geq 2} {(ξ^{m} - 1)}^{|I|} b_{I} . \end{array}$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

page history

Murphy elements ℂ Ak rpn

Murphy elements ℂ Ak rpn

References [PLACEHOLDER]

Murphy elements $ℂ A_{k} (r, p, n)$

Murphy elements $ℂ A_{k} (r, p, n)$