“Garnir relations” and an analogue of Young’s natural basis

Last update: 3 March 2013

“Garnir relations” and an analogue of Young’s natural basis

Each of the modules ${\stackrel{\sim }{H}}^{\left(c,\lambda /\mu \right)}$ constructed in Theorem 4.1 has two natural bases:

1. The “seminormal basis” $\left\{{v}_{L} \mid L\in {ℱ}^{\lambda /\mu }\right\}$ which, up to multiplication of each basis vector by a constant, is given by $v∼L=τw vC, (5.1)$ where $C$ is the column reading tableau of shape $\lambda /\mu ,$ $w$ is the permutation such that $L=wC$ and ${\tau }_{w}$ is as in (3.3).
2. The “natural basis” $\left\{{n}_{L} \mid L\in {ℱ}^{\lambda /\mu }\right\}$ given by $nL=TwvC, (5.2)$ where $C$ is the column reading tableau of shape $\lambda /\mu ,$ $w$ is the permutation such that $L=wC$ and ${T}_{w}$ is as defined in (1.6).

Proposition 5.3. Let $\left(c,\lambda /\mu \right)$ be a placed skew shape with $n$ boxes and let ${\stackrel{\sim }{H}}^{\left(c,\lambda /\mu \right)}$ be the ${\stackrel{\sim }{H}}_{n}\text{-module}$ defined in Theorem 4.1. Let $C$ be the column reading tableau of shape $\lambda /\mu$ and let $wC$ denote the tableau $C$ with the entries permuted according to the permutation $w\text{.}$ For each standard tableau $L$ let ${n}_{L}$ be defined by formula (5.2). Then $\left\{{n}_{L} \mid L\in {ℱ}^{\lambda /\mu }\right\}$ is a basis of ${\stackrel{\sim }{H}}^{\left(c,\lambda /\mu \right)}\text{.}$

 Proof. If $L$ is a standard tableau of shape $\lambda /\mu$ let ${\stackrel{\sim }{v}}_{L}$ be as given in (5.1). It follows from the formulas defining the module ${\stackrel{\sim }{H}}^{\left(c,\lambda /\mu \right)}$ that the basis $\left\{{\stackrel{\sim }{v}}_{L} \mid L\in {ℱ}^{\lambda /\mu }\right\}$ is simply a renormalized version of the basis $\left\{{v}_{L} \mid L\in {ℱ}^{\lambda /\mu }\right\},$ i.e there are constants ${\kappa }_{L}\in {ℂ}^{*}$ such that ${\stackrel{\sim }{v}}_{L}={\kappa }_{L}{v}_{L}\text{.}$ Let ${s}_{{i}_{1}}\dots {s}_{{i}_{p}}=w$ be a reduced word for $w\text{.}$ Then, with notations as in Theorem 4.1, $v∼L= τi1…τip vC= ( Ti1- (Ti1)L2L2 ) ( Ti2- (Ti1)L3L3 ) … ( Tip- (Tip)LpLp ) nC,$ where ${L}_{j}={s}_{{i}_{j+1}}\dots {s}_{{i}_{p}}C\text{.}$ Expanding this expression yields $v∼L= ( Tw+∑u for some constants ${b}_{u}\in ℂ\text{.}$ The second equality is a consequence of the fact that, by Proposition 2.1, the tableaux $uC,u are always standard. Since $\left\{{\stackrel{\sim }{v}}_{L} \mid L\in {ℱ}^{\lambda /\mu }\right\}$ is a basis it follows from the triangular relation above that $\left\{{n}_{L} \mid L\in {ℱ}^{\lambda /\mu }\right\}$ is also a basis of ${\stackrel{\sim }{H}}^{\left(c,\lambda /\mu \right)}\text{.}$ $\square$

The construction of ${\stackrel{\sim }{H}}^{\left(c,\lambda /\mu \right)}$ in Theorem 4.1 makes the notational assumption that ${v}_{L}=0,$ whenever $L$ is not a standard tableau. Formula (5.1) can be used as a definition of ${\stackrel{\sim }{v}}_{L}$ even in the case when $L$ is not standard and, from the definition of the action in Theorem 4.1,

$v∼L=0,if L is not standard. (5.4)$

Theorem 5.5 proves that these relations, when expanded in terms of the basis $\left\{{n}_{L} \mid L\in {ℱ}^{\lambda /\mu }\right\}$ are exactly the classical Garnir relations!!

Let $\lambda /\mu$ be a skew shape. A pair of adjacent boxes in the same row of $\lambda /\mu$ determines a snake in $\lambda /\mu$ consisting of the boxes in the pair, all the boxes above the righthand box of this pair, and all the boxes below the lefthand box of the pair. See the picture in Theorem 5.5 (b).

Theorem 5.5. ("Garnir relations") Let $\left(c,\lambda /\mu \right)$ be a placed skew shape and let ${\stackrel{\sim }{H}}^{\left(c,\lambda /\mu \right)}$ be the ${\stackrel{\sim }{H}}_{n}\text{-module}$ defined in Theorem 4.1. Let $\left\{{n}_{L} \mid L\in {ℱ}^{\lambda /\mu }\right\}$ be the basis of ${\stackrel{\sim }{H}}^{\left(c,\lambda /\mu \right)}$ defined by Proposition 5.3 and let $C$ be the column reading tableau of shape $\lambda /\mu \text{.}$

1. If $i$ and $i+1$ are entries in the same column of $C$ then $TinC=-q-1 nC.$
2. Fix a snake in $\lambda /\mu \text{.}$ Let $P$ be the standard tableau which has all entries the same as $C$ except that the entries in the snake are entered in row reading order instead of in column reading order. $C= j+1 j+2 ⋮ ℓ-1 ℓ i i+1 ⋮ j-1 j P= i i+1 ⋮ k-1 k+1 k k+2 ⋮ ℓ-1 ℓ$ Let ${S}_{A},{S}_{B}$ and ${S}_{A\cup B}$ be the subgroups of ${S}_{n}$ consisting of the permutations of $A={i,i+1,…,j}, B= {j+1,…,ℓ-1,ℓ}$ and $A\cup B,$ respectively, and let ${S}_{A\cup B}/\left({S}_{A}×{S}_{B}\right)$ be the set of minimal length coset representatives of cosets of ${S}_{A}×{S}_{B}$ in ${S}_{A\cup B}\text{.}$ The elements of ${S}_{A\cup B}/\left({S}_{A}×{S}_{B}\right)$ are sometimes called the “shuffles” of $A$ and $B\text{.}$ Then $0 = ( ∑u∈SA∪B/(SA×SB) (-q) ℓ(wA∪B)- ℓ(wAwB)- ℓ(u) Tu ) nC = TknP+ ∑u≤P (-q) ℓ(wA∪B)- ℓ(wAwB)- ℓ(u) nuC,$ where $\ell \left({w}_{A\cup B}\right)=\left(\genfrac{}{}{0}{}{\ell -i+1}{2},\right)$ $\ell \left({w}_{A}{w}_{B}\right)=\left(\genfrac{}{}{0}{}{j-i+1}{2}\right)\left(\genfrac{}{}{0}{}{\ell -j}{2}\right)$ and the last sum is over all standard tableaux $uC$ which are obtained from $C$ by permuting entries which are in the snake.

 Proof. Part (a) follows immediately from the definition of ${\stackrel{\sim }{H}}^{\left(c,\lambda /\mu \right)}$ in Theorem 4.1. (b) The subgroups ${S}_{A},{S}_{B}$ and ${S}_{A\cup B}$ have longest elements $wA= ( ii+1…j-1j jj-1…i+1i ) ,wB= ( j+1j+2…ℓ-1ℓ ℓℓ-1…j+2j+1 ) ,$ $wA∪B= ( ii+1…ℓ-1ℓ ℓℓ-1…i+1i ) ,$ with lengths $\ell \left({w}_{A}\right)=\left(\frac{j-i+1}{2}\right),$ $\ell \left({w}_{B}\right)=\left(\frac{\ell -j}{2}\right),$ and $\ell \left({w}_{A\cup B}\right)=\left(\frac{\ell -i+1}{2}\right),$ respectively, and ${w}_{A}{w}_{B}$ is the longest element of ${S}_{A}×{S}_{B}\subseteq {S}_{A\cup B}\text{.}$ Let $t=\left({t}_{1},\dots ,{t}_{n}\right)=\left(a,a{q}^{2},a{q}^{4},\dots ,a{q}^{2\left(n-1\right)}\right)$ where $a={q}^{2\left(c\left(C\left(j\right)\right)-\left(i-1\right)\right)}\in {ℂ}^{*}\text{.}$ The positions of $C\left(i\right),\dots ,C\left(\ell \right)$ in $\lambda /\mu$ are such that $( q2c(C(i)) ,…, q2c(C(ℓ)) ) = wAwB ( q2c(C(j)), q2c(C(j-1)), …, q2c(C(i)), q2c(C(ℓ)), q2c(C(ℓ-1)), …, q2c(C(j+1)), ) = wAwB ( q2c(C(j)), q2c(C(j)) q2,…, q2c(C(j)) q2(ℓ-i+1) ) = wAwB (ti,…,tℓ).$ By Proposition 3.6 and the fact that ${T}_{s}{n}_{C}=-{q}^{-1}{n}_{C}$ for all $i\le s\le \ell ,s\ne j,$ $τwAwB ∣tnC= ( ∑w∈SA×SB (-q) ℓ(wAwB) -ℓ(w) Tw ) nC= [j-i+1]! [ℓ-j]!nC,$ where $\left[r\right]!=\left[r\right]\left[r-1\right]\dots \left[2\right]\left[1\right]$ and $\left[p\right]=\left({q}^{p}-{q}^{-p}\right)/\left(q-{q}^{-1}\right)\text{.}$ Let $\pi$ be the permutation such that $P=\pi C$ and let ${\stackrel{\sim }{v}}_{P}={\tau }_{\pi }{v}_{C}\text{.}$ Then ${s}_{i}\pi {w}_{A}{w}_{B}={w}_{A\cup B}$ and, by (3.5), $τiv∼P = τiτπvC= τiτπ ∣wAwBt nC = 1 [j-i+1]! [ℓ-j]! τiτπ ∣wAwBt τwAwB ∣tnC = 1 [j-i+1]! [ℓ-j]! τiτπ τwAwB ∣tnC = 1 [j-i+1]! [ℓ-j]! τwA∪B ∣tnC = 1 [j-i+1]! [ℓ-j]! ( ∑w∈SA∪B (-q) ℓ(wA∪B) -ℓ(w) Tw ) nC.$ Each element $w\in {S}_{A\cup B}$ has a unique expression $w=uv$ such that $v\in {S}_{A}×{S}_{B}$ and $\ell \left(w\right)=\ell \left(u\right)+\ell \left(v\right)\text{.}$ The left factor $u$ is the minimal length representative of the coset $w\left({S}_{A}×{S}_{B}\right)$ in ${S}_{A\cup B}\text{.}$ Then $∑w∈SA∪B (-q) ℓ(wA∪B) -ℓ(w) Tw = ( ∑u∈SA∪B/(SA×SB) (-q) ℓ(wA∪B)-ℓ (wAwB)-ℓ(u) Tu ) × ( ∑v∈SA×SB (-q) ℓ(wAwB)- ℓ(v) Tv ) .$ It follows from the formulas for the action on ${\stackrel{\sim }{H}}^{\left(c,\lambda /\mu \right)}$ that the element ${\stackrel{\sim }{v}}_{P}={\tau }_{\pi }{v}_{C}$ is a nonzero multiple of the basis element ${v}_{P}\text{.}$ Furthermore, by (5.4), ${\tau }_{k}{\stackrel{\sim }{v}}_{P}=0\text{.}$ So $0 = τiv∼P = 1 [j-i+1]! [ℓ-j]! ( ∑u∈SA∪B/(SA×SB) (-q) ℓ(wA∪B)-ℓ (wAwB)-ℓ(u) Tu ) × ( ∑v∈SA×SB (-q) ℓ(wAwB)- ℓ(v) Tv ) nC = ( ∑u∈SA∪B/(SA×SB) (-q) ℓ(wA∪B)-ℓ (wAwB)-ℓ(u) Tu ) nC$ For each $u\in {S}_{A\cup B}/\left({S}_{A}×{S}_{B}\right)$ except $u={w}_{A\cup B}{w}_{A}{w}_{B},$ the tableau $uC$ is standard. In fact these are exactly the standard tableaux $Q$ which are obtained by permuting entries of $C$ which are in the snake. The tableau ${w}_{A\cup B}{w}_{A}{w}_{B}C={s}_{k}P\text{.}$ Note that $\ell \left({w}_{A\cup B}{w}_{A}{w}_{B}\right)=\ell \left({w}_{A\cup B}\right)-\ell \left({w}_{A}{w}_{B}\right)\text{.}$ Thus we have $0=TknP+ ∑u≤P (-q) ℓ(wA∪B)- ℓ(wAwB)- ℓ(u) nuC,$ where the sum over all standard tableaux $uC$ which are obtained from $C$ by permuting entries which are in the snake. $\square$

Proposition 5.6. Let $\left(c,\lambda /\mu \right)$ be a placed skew shape and let $\left\{{n}_{L} \mid L \text{is a standard tableau of shape} \lambda /\mu \right\}$ be the basis of the ${\stackrel{\sim }{H}}_{n}\text{-module}$ ${\stackrel{\sim }{H}}^{\left(c,\lambda /\mu \right)}$ which is defined by Proposition 5.3.

1. If $w\in {S}_{n}$ and $L$ is a standard tableau of shape $\lambda /\mu$ then $TwnL=∑Q bQnQ, with coefficients bQ∈ ℤ[q,q-1].$
2. Assume that the content function $c$ takes values in $ℤ\text{.}$ If $1\le i\le n$ and $L$ is a standard tableau of shape $\lambda /\mu$ then $xinL=∑Q bQ′nQ, with coefficients bQ′∈ ℤ[q,q-1].$

 Proof. (a) It is sufficient to show that for all $1\le d\le n$ and all standard tableaux $L$ of shape $\lambda /\mu$ we have $TdnL= ∑QbQnQ,$ with coefficients ${b}_{Q}\in ℤ\left[q,{q}^{-1}\right]\text{.}$ For notational convenience let us identify each standard tableau $L$ of shape $\lambda /\mu$ with the permutation $w\in {S}_{n}$ such that $wC=L,$ where $C$ is the column reading tableau of shape $\lambda /\mu \text{.}$ From the definitions $TdnL = nsdL, if ℓ(sdL)= ℓ(L)+1 and sdL is standard, TdnL = (q-q-1)nL +nsdL, if ℓ(sdL)= ℓ(L)-1.$ Let $L$ be a standard tableau such that $\ell \left({s}_{d}L\right)=\ell \left(L\right)+1$ and ${s}_{d}L$ is not standard. The pair of boxes where the nonstandardness in ${s}_{d}L$ occurs in either a row or a column: $sdL= a ^ b ^ ⋮ ^ c ^ d > d+1 ^ e ^ ⋮ ^ f sdL= d+1 d Case (1) Case (2)$ Case (1): The two boxes in ${s}_{d}L$ where the nonstandardness occurs determine a snake in the shape $\lambda /\mu$ consisting of all the boxes above the entry $d$ and all the boxes below the entry $d+1\text{.}$ Use the same notation as in Theorem 5.5 so that $P$ is the standard tableau which is the same as the column reading tableau $C$ except that the entries in the snake are in row reading order. Then ${s}_{k}P$ is nonstandard and there exists a (unique) $v\in {S}_{n}$ such that ${s}_{d}L=v{s}_{k}P$ and $\ell \left({s}_{d}L\right)=\ell \left(v\right)+\ell \left({s}_{k}P\right)\text{.}$ The permutation $v$ is given by $v= ( ii+1…k-1 kk+1k+2…ℓ ab…cd d+1e…f )$ and $v\left(x\right)=L\left(x\right)$ for all $x\notin \left\{i,i+1,\dots ,\ell \right\}\text{.}$ Thus $TdnL=TdTL nC=TvTkTP nC=TvTknP.$ Let $\ell \left({w}_{A\cup B}\right)=\left(\genfrac{}{}{0}{}{\ell -i+1}{2}\right)\text{.}$ By Theorem 5.5 $TknP=-∑u≤P (-q) ℓ(wA∪B)- ℓ(wAwB)- ℓ(u) TunC,$ and so $TdnL=-Tv ( ∑u≤P (-q) ℓ(wA∪B)- ℓ(wAwB)- ℓ(u) TunC ) .$ The permutations $v$ and $u\le P$ are such that $\ell \left(vu\right)=\ell \left(v\right)+\ell \left(u\right)$ and so ${T}_{v}{T}_{u}={T}_{vu}$ for all $u\le P\text{.}$ Furthermore, the tableaux $vuP,$ for $u\le P,$ are exactly the standard tableaux which are obtained by permutations of the entries in $L$ which are in the snake. Thus $TdnL=-∑u≤ P (-q) ℓ(wA∪B)- ℓ(wAwB)- ℓ(u) nvuC.$ Case (2): Suppose that the boxes containing $d$ and $d+1$ in $L$ are the boxes containing k and $k+1$ in the column reading tableau $C\text{.}$ $C= k k+1$ Then $L{s}_{k}={s}_{d}L$ and $TdnL=TdTL nC=TLTknC =-q-1TLnC =-q-1nL,$ by Theorem 5.5 (a). (b) Let $L$ be a standard tableau of shape $\lambda /\mu$ and let $w$ be the permutation such that $L=wC,$ where $C$ is the column reading tableau of shape $\lambda /\mu \text{.}$ By repeatedly using the relations (1.4) we obtain $xinL=xiTw nC=Twxw-1(i) nC+∑v where the ${p}_{v}\left({x}_{1},\dots ,{x}_{n}\right)$ are polynomials in ${x}_{1},\dots ,{x}_{n}$ with coefficients in $ℤ\left[q,{q}^{-1}\right]\text{.}$ If ${x}_{i}$ acts on ${v}_{C}$ by integral powers of $q$ then we have $xinL=Tw q2c(C(w-1(i))) nC+∑v with coefficients ${b}_{v}^{\prime }$ in $ℤ\left[q,{q}^{-1}\right]\text{.}$ $\square$

Notes and References

This is an excerpt of the preprint entitled Skew shape representations are irreducible authored by Arun Ram in 1998.

Research supported in part by National Science Foundation grant DMS-9622985, and a Postdoctoral Fellowship at Mathematical Sciences Research Institute.