Schur-Weyl dualities

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

Schur-Weyl dualities

Let $n \in ℤ_{> 0}$ and let $V$ be a vector space with basis $v_{1}, \dots, v_{n} .$ Then the tensor product $V^{\otimes k} = V \otimes V \otimes \dots V (k factors) has basis \{v_{i_{1}} \otimes \dots \otimes v_{i_{k}} | 1 \leq i_{1}, \dots, i_{k} \leq n\} .$ For $k \in ℤ_{> 0}$ let $V^{\otimes (k + \frac{1}{2})} = V^{\otimes k} \otimes v_{n}, a subspace of V^{\otimes (k + 1)}$ (which is isomorphic as a vector space to $V^{\otimes^{k}}$ ). If $b \in End (V^{\otimes k})$ let $b_{i_{1^{'}}, \dots, i_{k^{'}}}^{i_{1}, \dots, i_{k}} \in ℂ$ be the coefficients in the expansion $b (v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) = \sum_{1 \leq i_{1^{'}}, \dots, i_{k^{'}} \leq n} b_{i_{1^{'}}, \dots, i_{k^{'}}}^{i_{1}, \dots, i_{k}} v_{i_{1^{'}}} \otimes \dots \otimes v_{i_{k^{'}}} .$ For $d \in A_{k}$ and values $i_{1}, \dots, <_{k}, i_{1^{'}}, \dots i_{k^{'}} \in \{1 \dots n\}$ define

{(d)}_{i_{1}, \dots, i_{k}}^{i_{1^{'}}, \dots, i_{k^{'}}} = \{\begin{array}{rcl} 1, & if & i_{r} & = & i_{s} & when & r & and & s & are in the same block of & d & , \\ 0, & otherwise. \end{array}

For example, viewing

{(d)}_{i_{1}, \dots, i_{k}}^{i_{1^{'}}, \dots, i_{k^{'}}}

as the diagram

d

with vertices labeled by the values

i_{1}, \dots, i_{k}

and

i_{1^{'}}, \dots, i_{k^{'}},

we have

It follows from (???) and (???) that for all $d \in A_{k},$ ${(Φ_{k} (x_{d}))}_{i_{1^{'}}, \dots, i_{k^{'}}}^{i_{1}, \dots, i_{k}} = \{\begin{array}{rcl} 1, & if i_{r} = i_{s} iff r and s are in the same block of d, \\ 0, & otherwise. \end{array}$

The group ${GL}_{n} (ℂ)$ acts on the vector spaces $V$ and $V^{\otimes k}$ by $g v_{i} = \sum_{j = 1}^{n} g_{j i} v_{j}, and g (v_{i_{1}} \otimes v_{i_{2}} \otimes \dots \otimes v_{i_{k}}) = g v_{i_{1}} \otimes g v_{i_{2}} \otimes \dots \otimes g v_{i_{k}},$ for $g = (g_{j i}) \in {GL}_{n} (ℂ) .$ For any subgroup $G \subseteq {GL}_{n} (ℂ),$ ${End}_{G} (V^{\otimes k}) = \{b \in End (V^{\otimes k}) | b g v = g b v for all g \in G and v \in V^{\otimes k}\} .$

Let $n \in ℤ_{> 0}$ and let $\{x_{d} | d \in A_{k}\}$ be the basis of $ℂ A_{k} (n)$ defined in (???). Then the notation in (???) and (???) defines algebra homomorphisms $Φ_{k} : ℂ A_{k} (n) \to End (V^{\otimes k}) for k \in \frac{1}{2} ℤ_{> 0},$ giving a right action of the partition algebra $ℂ A_{k} (n)$ on $V^{\otimes k} .$ View the symmetric group $S_{n}$ as the subgroup of ${GL}_{n} (ℂ)$ of permutation matrices.

$Φ_{k} : ℂ A_{k} (n) \to End (V^{\otimes k})$ has image $im Φ_{k} = {End}_{S_{n}} (V^{\otimes k}) and \ker Φ_{k} = ℂ - span \{x_{d} | d has more than n blocks\},$ and
$Φ_{k + \frac{1}{2}} : ℂ A_{k + \frac{1}{2}} (n) \to End (V^{\otimes k})$ has image $im Φ_{k + \frac{1}{2}} = {End}_{S_{n - 1}} (V^{\otimes k}) and \ker Φ_{k + \frac{1}{2}} = ℂ - span \{x_{d} | d has more than n blocks\} .$

		Proof.
	As a subgroup of ${GL}_{n} (ℂ),$ $S_{n}$ acts on $V$ via its permutation representation and $S_{n}$ acts on $V^{\otimes k}$ by $σ (v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) = v_{σ (i_{1})} \otimes \dots \otimes v_{σ (i_{k})} .$ Then $b \in {End}_{S_{n}} (V^{\otimes k})$ iff $σ^{- 1} b σ = b$ (as endomorphisms on $V^{\otimes k}$ for all $σ \in S_{n} .$ Thus using the notation of (???), $b \in {End}_{S_{n}} (V^{\otimes k})$ iff $b_{i_{1^{'}}, \dots, i_{k^{'}}}^{i_{1}, \dots, i_{k}} = {σ^{- 1} b σ}_{i_{1^{'}}, \dots, i_{k^{'}}}^{i_{1}, \dots, i_{k}} = b_{σ (i_{1^{'}}), \dots, σ (i_{k^{'}})}^{σ (i_{1}), \dots, σ (i_{k})}, for all σ \in S_{n} .$ It follows that the matrix entries of $b$ are constant on the $S_{n}$ -orbits of its matrix coordinates. These orbits decompose $\{1 \dots k 1^{'} \dots k^{'}\}$ into subsets and thus correspond to set partitions $d \in A_{k} .$ Thus $Φ_{k} (x_{d})$ has 1s in the matrix position corresponding to $d$ and 0s elsewhere, and so $b$ is a linear combination of $Φ_{k} (x_{d}),$ $d \in A_{k} .$ Since $x_{d},$ $d \in A_{k},$ form a basis of $ℂ A_{k},$ $im Φ_{k} = {End}_{S_{n}} (V^{\otimes k}) .$ If $d$ has more than $n$ blocks, then by (???) the matrix entry ${(Φ_{k} (x_{d}))}_{i_{1^{'}}, \dots, i_{k^{'}}}^{i_{1}, \dots, i_{k}} = 0$ for all indices $i_{1}, \dots, i_{k}, i_{1^{'}}, \dots, i_{k^{'}},$ since we need a distinct $i_{j} \in \{1 \dots n\}$ for each block of $d .$ Thus $x_{d} \in \ker Φ_{k} .$ If $d$ has $\leq n$ blocks, then we can find an index set $i_{1}, \dots, i_{k}, i_{1^{'}}, \dots, i_{k^{'}}$ with ${(Φ_{k} (x_{d}))}_{i_{1^{'}}, \dots, i_{k^{'}}}^{i_{1}, \dots, i_{k}} = 1$ simply by choosing a distinct index from $\{1 \dots n\}$ for each block of $d .$ Thus, if $d$ has $\leq n$ blocks the $x_{d} \notin \ker Φ_{k},$ and so $\ker Φ_{k} = ℂ - span \{x_{d} \| d has more than n blocks\} .$ The vector space $V^{\otimes k} \otimes v_{n} \subseteq V^{\otimes (k + 1)}$ is a submodule both for $ℂ A_{k + \frac{1}{2}} \subseteq ℂ A_{k + 1}$ and $ℂ S_{n - 1} \subseteq ℂ S_{n} .$ If $σ \in S_{n - 1},$ then $σ (v_{i_{1}} \otimes \dots \otimes v_{i_{k}} \otimes v_{n}) = v_{σ (i_{1})} \otimes \dots \otimes v_{σ (i_{k})} \otimes v_{n} .$ Then as above $b \in {End}_{S_{n - 1}} (V^{\otimes k})$ iff $b_{i_{1^{'}}, \dots, i_{k^{'}}, n}^{i_{1}, \dots, i_{k}, n} = b_{σ (i_{1^{'}}), \dots, σ (i_{k^{'}}), n}^{σ (i_{1}), \dots, σ (i_{k}), n}, for all σ \in S_{n - 1} .$ The $S_{n - 1}$ orbits of the matrix coordinates of $b$ correspond to set partitions $d \in A_{k + \frac{1}{2}};$ that is vertices $i_{k + 1}$ and $i_{{(k + 1)}^{'}}$ must be in the same block of $d .$ The same argument as part (a) can be used to show that $\ker Φ_{k + \frac{1}{2}}$ is the span of $x_{d}$ with $d \in A_{k + \frac{1}{2}}$ having more than $n$ blocks. We always choose the index $n$ for the block containing $k + 1$ and ${(k + 1)}^{'} . □$

The multiplication in $ℂ A_{k} (n)$ is, in terms of the basis $\{x_{d}\},$ $x_{d_{1}} x_{d_{2}} = \{\begin{array}{rcl} 0, & if d_{1} \circ d_{2} match exactly in the middle, \\ \sum_{d} c_{d} x_{d}, & if d_{1} \circ d_{2} match exactly in the middle, \end{array}$ where the sum is over all coarsenings of $d_{1} \circ d_{2}$ obtained by merging a top horizontal block and a bottom horizontal block and $c_{d} = {(n - |d|)}_{[d_{1} \circ d_{2}]}, where {(l)}_{r} = l (l - 1) \dots (l - r + 1),$ $|d|$ is the number of blocks of $d,$ and $[d_{1} \circ d_{2}]$ is the number of internal blocks of $d_{1} \circ d_{2} .$

		Proof.
	Let $n > > k$ so that $ℂ A_{k} (n) ≅ {End}_{S_{n}} (V^{\otimes k}) .$ Then $(v_{i_{1}} \otimes \dots \otimes v_{i_{k}}) x_{d_{1}} x_{d_{2}} = \sum_{i_{1^{'}}, \dots, i_{k^{'}} i_{1^{"}}, \dots, i_{k^{"}}} {(x_{d_{1}})}_{i_{1^{'}}, \dots, i_{k^{'}}}^{i_{1}, \dots, i_{k}} {(x_{d_{2}})}_{i_{1^{"}}, \dots, i_{k^{"}}}^{i_{1^{'}}, \dots, i_{k^{'}}}$ and $\sum_{i_{1^{'}}, \dots, i_{k^{'}} i_{1^{"}}, \dots, i_{k^{"}}} {(x_{d_{1}})}_{i_{1^{'}}, \dots, i_{k^{'}}}^{i_{1}, \dots, i_{k}} {(x_{d_{2}})}_{i_{1^{"}}, \dots, i_{k^{"}}}^{i_{1^{'}}, \dots, i_{k^{'}}} = \sum_{interior blocks} 1 = (n - \|d\|) (n - \|d\| + 1) \dots (n - \|d\| - (l - 1)),$ where the sum is over labeled interior blocks of $d_{1} \otimes d_{2}$ such that the labels on these blocks are distinct and do not lie in $\{i_{1} \dots i_{k} i_{1^{'}} \dots i_{k^{'}}\} . □$

For $k \in ℤ_{> 0}$ the following result is due to Tanabe [Ta, Lemma 2.1].

Let $T$ be the subgroup of ${GL}_{n} (ℂ)$ of diagonal matrices in ${GL}_{n} (ℂ)$ and let $N = G_{ℂ^{*}, 1, n}$ be the normaliser of $T$ if ${GL}_{n} (ℂ) .$ Then $Φ_{k} (ℂ A_{k, \infty, 1} (n)) = {End}_{N} (V^{\otimes k}) and \ker Φ_{k} \cap ℂ A_{k, \infty, 1} (n) = ?$
Then $Φ_{k} (ℂ S_{k}) = {End}_{{𝔤 𝔩}_{n}} (V^{\otimes k}) and \ker Φ_{k} \cap ℂ S_{k} = the ideal generated by \sum_{w \in S_{k}} \det (w) w .$

Let $L_{n - 1} = (G_{r, 1, n} \times (ℤ / r ℤ)) \cap G_{r, p, n} .$ Then $im Φ_{k} = {End}_{G_{r, p, n}} (V^{\otimes k}) and \ker Φ_{k} = ℂ - span \{x_{d} | d has more than n blocks\},$ $im Φ_{k + \frac{1}{2}} = {End}_{L_{n - 1}} (V^{\otimes k}) and \ker Φ_{k + \frac{1}{2}} = ℂ - span \{x_{d} | d has more than n blocks\} .$

		Proof.
	In the case $r = p = 1$ this is a result of Jones [Jo] and Martin [Ma] (see [HR], Theorem ???). A direct computation (which we will not do here) shows that if $d \in A_{k} (r, p, n)$ then $x_{d}$ commutes with each of the generators $t_{1}^{p}, s_{1}, s_{2}, \dots, s_{n}$ of $G_{r, p, n} .$ Thus $im Φ_{k} \subseteq {End}_{G_{r, p, n}} (V^{\otimes k}) .$ If $a \in {End}_{G_{r, p, n}} (V^{\otimes k})$ then $a \in {End}_{S_{n}} (V^{\otimes k})$ and so by the Jones-Martin theorem, $a = \sum_{d \in A_{k}} c_{d} x_{d}, for some c_{d} \in ℂ .$ We shall show that if $d \notin A_{k} (r, p, n)$ then $c_{d} = 0 .$ Let $d \in A_{k}$ and let $v_{t} = v_{i_{1}} \otimes \dots \otimes v_{i_{k}} and v_{b} = v_{i_{1^{'}}} \otimes \dots \otimes v_{i_{k^{'}}}$ be such that $i_{r} = i_{s}$ iff $r$ and $s$ are in the same block of $d .$ Then $c_{d} = a v_{t} \|_{v_{b}},$ the coefficient of the basis element $v_{b}$ in the expansion of $a v_{t} .$ Choose a block $B$ of $d$ and let $l \in B .$ Then $c_{d} = (a v_{t}) \|_{v_{b}} = (t_{i_{l}}^{- p} a t_{i_{l}}^{p}) \|_{v_{b}} = ξ^{p κ (B)} c_{d} .$ Hence $c_{d} = 0$ unless $κ (B) = 0 (\mod r / p) .$ Now choose a pair of distinct blocks $B_{1}$ and $B_{2}$ in $d .$ Let $l \in B_{1}$ and $m \in B_{2} .$ Then $c_{d} = (a v_{t}) \|_{v_{b}} = (t_{i_{m}} t_{i_{l}}^{- 1} a t_{i_{l}} t_{i_{m}}^{- 1} v_{t}) \|_{v_{b}} = ξ^{κ (B_{1}) - κ (B_{2})} c_{d} .$ So $c_{d} = 0$ unles $κ (B_{1}) = κ (B_{2}) = 0 (\mod r) .$ The same argument with $v_{t} = v_{i_{1}} \otimes \dots \otimes v_{i_{k}} \otimes v_{n} and v_{b} = v_{i_{1^{'}}} \otimes \dots \otimes v_{i_{k^{'}}} \otimes v_{n}$ applies to establish case (b). $□$

If $d \in B_{k}$ is a diagram choose a labeling of the blocks of $d$ from ????? with $1, 2, \dots, k$ by marking one vertex in each block. An element $σ \in S_{k}$ permutes the marked vertices to produce a new diagram $σ d \in B_{k} .$

PICTURE INSERT HERE

Suppose $n + 1 \leq k .$ For a given element $d \in B_{k},$ the element of $ℂ B_{k} (n)$ given by $\sum_{σ \in S_{n + 1}} {(- 1)}^{l (σ)} σ d,$ depends on the choice of the labeling of $d,$ but the set $\{\sum_{σ \in S_{n + 1}} {(- 1)}^{l (σ)} σ d | d \in B_{k}\}$ does not???????/

If $k > n$ let $S_{n + 1}$ be the subgroup of $S_{k}$ which fixes $n + 2, \dots, k .$

(Schur-Weyl) Let $n \in ℤ_{> 0} .$

$Φ_{k} : ℂ S_{k} \to End (V^{\otimes k})$ has $im Φ_{k} = {End}_{{GL}_{n} (ℂ)} (V^{\otimes k})$ and $\ker Φ_{k}$ is the ideal of $ℂ S_{k}$ generated by $\{\sum_{σ \in S_{n + 1}} {(- 1)}^{l (σ)} σ d | d \in S_{k} has more than n blocks\} .$
$Φ_{k} : ℂ B_{k} \to End (V^{\otimes k})$ has $im Φ_{k} = {End}_{O_{n} (ℂ)} (V^{\otimes k})$ and $\ker Φ_{k}$ is the ideal generated by $\{\sum_{σ \in S_{n + 1}} {(- 1)}^{l (σ)} σ d | d \in S_{k} has more than n blocks\} .$

Need to define action of

σ

d .

Define linear maps $ε_{\frac{1}{2}} : End (V^{\otimes k}) \to End (V^{\otimes k})$ $ε^{\frac{1}{2}} : End (V^{\otimes k}) \to End (V^{\otimes (k - 1)})$ and $ε^{1} : End (V^{\otimes k}) \to End (V^{\otimes (k - 1)})$ by $ε_{\frac{1}{2}} {(b)}_{i_{1^{'}}, \dots, i_{k^{'}}}^{i_{1}, \dots, i_{k}} = b_{i_{1^{'}}, \dots, i_{k^{'}}}^{i_{1}, \dots, i_{k}} δ_{i_{k} i_{k^{'}}}$ $ε^{\frac{1}{2}} {(b)}_{i_{1^{'}}, \dots, i_{{(k - 1)}^{'}}}^{i_{1}, \dots, i_{k - 1}} = \sum_{j, l = 1}^{n} {(b)}_{i_{1^{'}}, \dots, i_{{(k - 1)}^{'}}, l}^{i_{1}, \dots, i_{k - 1}, j}$ and $ε_{1} {(b)}_{i_{1^{'}}, \dots, i_{{(k - 1)}^{'}}}^{i_{1}, \dots, i_{k - 1}} = \sum_{j = 1}^{n} b_{i_{1^{'}}, \dots, i_{{(k - 1)}^{'}}, j}^{i_{1}, \dots, i_{k - 1}, j} .$ Then $ε_{1} = ε_{\frac{1}{2}} \circ ε^{\frac{1}{2}} and Tr (b) = ε_{1}^{k} (b), for b \in End (V^{\otimes k}) .$ The relation between the maps $ε^{\frac{1}{2}}, ε_{\frac{1}{2}}$ and $ε_{1}$ in (???) and the maps $ε^{\frac{1}{2}}, ε_{\frac{1}{2}}$ and $ε_{1}$ in (???) is given by $Φ_{k - \frac{1}{2}} (ε_{\frac{1}{2}} (b)) = ε_{\frac{1}{2}} (Φ_{k} (b)) |_{V^{\otimes (k - \frac{1}{2})}}, Φ_{k - 1} (ε^{\frac{1}{2}} (b)) = \frac{1}{n} ε^{\frac{1}{2}} (Φ_{k} (b))$ and $Φ_{k - 1} (ε_{1} (b)) = ε_{1} (Φ_{k} (b)),$ where, on the right hand side of the middle inequality, $b$ is viewed as an element of $ℂ A_{k}$ via the natural inclusion $ℂ A_{k - \frac{1}{2}} (n) \subseteq ℂ A_{k} (n) .$ Then, for $k \in ℤ_{> 0},$ $Tr (Φ_{k} (b)) = {tr}_{k} (b), and Tr (Φ_{k - \frac{1}{2}} (b)) = \frac{1}{n} {tr}_{k - \frac{1}{2}} (b) .$

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)

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