Last update: 3 March 2013
The following theorem classifies and constructs all irreducible calibrated representations of the affine Hecke algebra ${\stackrel{\sim}{H}}_{n}\text{.}$ It shows that the theory of standard Young tableaux plays an intrinsic role in the combinatorics of the representations of the affine Hecke algebra. The construction given in Theorem 4.1 is a direct generalization of A. Young’s classical “seminormal construction” of the irreducible representations of the symmetric group [You1931,You1934]. Young’s construction was generalized to Iwahori-Hecke algebras of type A by Hoefsmit [Hoe1974] and Wenzl [Wen1988] independently, to Iwahori-Hecke algebras of types B and D by Hoefsmit [Hoe1974] and to cyclotomic Hecke algebras by Ariki and Koike [AKo1994]. It can be shown that all of these previous generalizations are special cases of the construction for affine Hecke algebras given here. Recently, this construction has been generalized even further [Ram1998], to affine Hecke algebras of arbitrary Lie type. Some parts of Theorem 4.1 are due, originally, to I. Cherednik, and are stated in [Che1987, §3].
Garsia and Wachs [GWa1989] showed that the theory of standard Young tableaux and Young’s constructions play an important role in the combinatorics of the skew representations of the symmetric group. At that time it was not known that these representations are actually irreducible as representations of the affine Hecke algebra!!
Theorem 4.1. Let $(c,\lambda /\mu )$ be a placed skew shape with n boxes. Define an action of ${\stackrel{\sim}{H}}_{n}$ on the vector space
$${\stackrel{\sim}{H}}^{(c,\lambda /\mu )}=\u2102\text{-span}\{{v}_{L}\hspace{0.17em}\mid \hspace{0.17em}L\hspace{0.17em}\text{is a standard tableau of shape}\hspace{0.17em}\lambda /\mu \}$$by the formulas
$$\begin{array}{ccc}{x}_{i}{v}_{L}& =& {q}^{2c\left(L\left(i\right)\right)}{v}_{L},\\ {T}_{i}{v}_{L}& =& {\left({T}_{i}\right)}_{LL}{v}_{L}+({q}^{-1}+{\left({T}_{i}\right)}_{LL}){v}_{{s}_{i}L},\end{array}$$where ${s}_{i}L$ is the same as $L$ except that the entries $i$ and $i+1$ are interchanged,
$${\left({T}_{i}\right)}_{LL}=\frac{q-{q}^{-1}}{1-{q}^{2(c\left(L\left(i\right)\right)-c\left(L(i+1)\right))}},\phantom{\rule{2em}{0ex}}{v}_{{s}_{i}L}=0\hspace{0.17em}\text{if}\hspace{0.17em}{s}_{i}L\hspace{0.17em}\text{is not a standard tableau,}$$and $L\left(i\right)$ denotes the box of $L$ containing the entry $i\text{.}$
Step 1. The given formulas for the action of ${\stackrel{\sim}{H}}^{(c,\lambda /\mu )}$ define an ${\stackrel{\sim}{H}}_{n}\text{-module.}$
Proof. | |
If $L$ is a standard tableau then the entries $i$ and $i+1$ cannot appear in the same diagonal in $L\text{.}$ Thus, for all standard tableaux $L,$ $c\left(L\left(i\right)\right)\ne c\left(L(i+1)\right)$ and for this reason the constant ${\left({T}_{i}\right)}_{LL}$ is always well defined. Let $L$ be a standard tableau of shape $\lambda /\mu \text{.}$ Then ${\left({T}_{i}\right)}_{LL}+{\left({T}_{i}\right)}_{{s}_{i}L,{s}_{i}L}=q-{q}^{-1}$ and so $$\begin{array}{ccc}{T}_{i}^{2}{v}_{L}& =& ({\left({T}_{i}\right)}_{LL}^{2}+({q}^{-1}+{\left({T}_{i}\right)}_{LL})({q}^{-1}+{\left({T}_{i}\right)}_{{s}_{i}L,{s}_{i}L})){v}_{L}\\ & & +({q}^{-1}+{\left({T}_{i}\right)}_{LL})({\left({T}_{i}\right)}_{LL}+{\left({T}_{i}\right)}_{{s}_{i}L,{s}_{i}L}){v}_{{s}_{i}L}\\ & =& {\left({T}_{i}\right)}_{LL}({\left({T}_{i}\right)}_{LL}+{\left({T}_{i}\right)}_{{s}_{i}L,{s}_{i}L}){v}_{L}+{q}^{-1}({q}^{-1}+{\left({T}_{i}\right)}_{LL}+{\left({T}_{i}\right)}_{{s}_{i}L,{s}_{i}L}){v}_{L}\\ & & +({q}^{-1}+{\left({T}_{i}\right)}_{LL})(q-{q}^{-1}){v}_{{s}_{i}L}\\ & =& {\left({T}_{i}\right)}_{LL}(q-{q}^{-1}){v}_{L}+({q}^{-1}+{\left({T}_{i}\right)}_{LL})(q-{q}^{-1}){v}_{{s}_{i}L}+{q}^{-1}({q}^{-1}+q-{q}^{-1}){v}_{L}\\ & =& ((q-{q}^{-1}){T}_{i}+1){v}_{L}\text{.}\end{array}$$The calculations to check the identities (1.1a), (1.1f) and (1.5) are routine. Checking the identity ${T}_{i}{T}_{i+1}{T}_{i}={T}_{i+1}{T}_{i}{T}_{i+1}$ is more involved. One can proceed as follows. According to the formulas for the action, the operators ${T}_{i}$ and ${T}_{i+1}$ preserve the subspace $S$ spanned by the vectors ${v}_{Q}$ indexed by the standard tableaux $Q$ in the set $\{L,{s}_{i}L,{s}_{i+1}L,{s}_{i}{s}_{i+1}L,{s}_{i+1}{s}_{i}L,{s}_{i}{s}_{i+1}{s}_{i}L\}\text{.}$ Depending on the relative positions of the boxes containing $i,i+1,i+2$ in $L,$ this space is either 1, 2, 3 or 6 dimensional. Representative cases are when these boxes are positioned in the following ways. $$\begin{array}{cccc}\n\n\n\n\n\n& \n\n\n\n\n\n& \n\n\n\n\n\n& \n\n\n\n\n\n\\ \text{Case (1)}& \text{Case (2)}& \text{Case (3)}& \text{Case (4)}\end{array}$$In Case (1) the space $S$ is one dimensional and spanned by the vector ${v}_{Q}$ corresponding to the standard tableau $$\n\n\n\n\n\na\nb\nc\n\n$$where $a=i,$ $b=i+1,$ and $c=i+2\text{.}$ The action of ${T}_{i}$ and ${T}_{i+1}$ on $S$ is given by the matrices $${\varphi}_{S}\left({T}_{i}\right)=\left(\begin{array}{c}q\end{array}\right),\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{\varphi}_{S}\left({T}_{i+1}\right)=\left(\begin{array}{c}q\end{array}\right),$$respectively. In case (2) the space $S$ is two dimensional and spanned by the vectors ${v}_{Q}$ corresponding to the standard tableaux $$\n\n\n\n\n\na\nb\nc\n\n\phantom{\rule{4em}{0ex}}\n\n\n\n\n\na\nc\nb\n\n$$where $a=i,$ $b=i+1,$ and $c=i+2\text{.}$ The action of ${T}_{i}$ and ${T}_{i+1}$ on $S$ is given by the matrices $${\varphi}_{S}\left({T}_{i}\right)=\left(\begin{array}{cc}q& 0\\ 0& -{q}^{-1}\end{array}\right)\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{\varphi}_{S}\left({T}_{i+1}\right)=\left(\begin{array}{cc}\frac{q-{q}^{-1}}{1-{q}^{4}}& \frac{q-{q}^{-5}}{1-{q}^{-4}}\\ \frac{q-{q}^{3}}{1-{q}^{4}}& \frac{q-{q}^{-1}}{1-{q}^{-4}}\end{array}\right),$$In case (3) the space $S$ is three dimensional and spanned by the vectors ${v}_{Q}$ corresponding to the standard tableaux $$\n\n\n\n\n\na\nb\nc\n\n\phantom{\rule{4em}{0ex}}\n\n\n\n\n\na\nc\nb\n\n\phantom{\rule{4em}{0ex}}\n\n\n\n\n\nb\nc\na\n\n$$where $a=i,$ $b=i+1,$ and $c=i+2\text{.}$ The action of ${T}_{i}$ and ${T}_{i+1}$ on $S$ is given by the matrices $$\begin{array}{ccc}{\varphi}_{S}\left({T}_{i}\right)& =& \left(\begin{array}{ccc}q& 0& 0\\ 0& \frac{q-{q}^{-1}}{1-{q}^{2({c}_{1}-{c}_{3})}}& \frac{q-{q}^{2({c}_{3}-{c}_{1})-1}}{1-{q}^{2({c}_{3}-{c}_{1})}}\\ 0& \frac{q-{q}^{2({c}_{1}-{c}_{3})-1}}{1-{q}^{2({c}_{1}-{c}_{3})}}& \frac{q-{q}^{-1}}{1-{q}^{2({c}_{3}-{c}_{1})}}\end{array}\right)\phantom{\rule{2em}{0ex}}\text{and}\\ {\varphi}_{S}\left({T}_{i+1}\right)& =& \left(\begin{array}{ccc}\frac{q-{q}^{-1}}{1-{q}^{2({c}_{2}-{c}_{3})}}& \frac{q-{q}^{2({c}_{3}-{c}_{2})-1}}{1-{q}^{2({c}_{3}-{c}_{2})}}& 0\\ \frac{q-{q}^{2({c}_{2}-{c}_{3})-1}}{1-{q}^{2({c}_{2}-{c}_{3})}}& \frac{q-{q}^{-1}}{1-{q}^{2({c}_{3}-{c}_{2})}}& 0\\ 0& 0& q\end{array}\right)\end{array}$$respectively, where ${c}_{1}=c\left(L\left(i\right)\right),$ ${c}_{2}=c\left(L(i+1)\right)$ and ${c}_{3}=c\left(L(i+2)\right)\text{.}$ In case (4) the space $S$ is six dimensional and spanned by the vectors ${v}_{Q}$ corresponding to the standard tableaux $$\begin{array}{cccccc}\n\n\n\n\n\na\nb\nc\n\n& \n\n\n\n\n\nb\na\nc\n\n& \n\n\n\n\n\na\nc\nb\n\n& \n\n\n\n\n\nb\nc\na\n\n& \n\n\n\n\n\nc\na\nb\n\n& \n\n\n\n\n\nc\nb\na\n\n\end{array}$$where $a=i,$ $b=i+1,$ and $c=i+2\text{.}$ The action of ${T}_{i}$ and ${T}_{i+1}$ on $S$ is given by the matrices $$\begin{array}{ccc}{\varphi}_{S}\left({T}_{i}\right)& =& \left(\begin{array}{cccccc}\frac{q-{q}^{-1}}{1-{q}^{2{d}_{12}}}& \frac{q-{q}^{2{d}_{21}-1}}{1-{q}^{2{d}_{21}}}& 0& 0& 0& 0\\ \frac{q-{q}^{2{d}_{12}-1}}{1-{q}^{2{d}_{12}}}& \frac{q-{q}^{-1}}{1-{q}^{2{d}_{21}}}& 0& 0& 0& 0\\ 0& 0& \frac{q-{q}^{-1}}{1-{q}^{2{d}_{13}}}& \frac{q-{q}^{2{d}_{31}-1}}{1-{q}^{2{d}_{31}}}& 0& 0\\ 0& 0& \frac{q-{q}^{2{d}_{13}-1}}{1-{q}^{2{d}_{13}}}& \frac{q-{q}^{-1}}{1-{q}^{2{d}_{31}}}& 0& 0\\ 0& 0& 0& 0& \frac{q-{q}^{-1}}{1-{q}^{2{d}_{23}}}& \frac{q-{q}^{2{d}_{32}-1}}{1-{q}^{2{d}_{32}}}\\ 0& 0& 0& 0& \frac{q-{q}^{2{d}_{23}-1}}{1-{q}^{2{d}_{23}}}& \frac{q-{q}^{-1}}{1-{q}^{2{d}_{32}}}\end{array}\right)\end{array}$$and $$\begin{array}{ccc}{\varphi}_{S}\left({T}_{i+1}\right)& =& \left(\begin{array}{cccccc}\frac{q-{q}^{-1}}{1-{q}^{2{d}_{23}}}& 0& \frac{q-{q}^{2{d}_{32}-1}}{1-{q}^{2{d}_{32}}}& 0& 0& 0\\ 0& \frac{q-{q}^{-1}}{1-{q}^{2{d}_{13}}}& 0& 0& \frac{q-{q}^{2{d}_{31}-1}}{1-{q}^{2{d}_{31}}}& 0\\ \frac{q-{q}^{2{d}_{23}-1}}{1-{q}^{2{d}_{23}}}& 0& \frac{q-{q}^{-1}}{1-{q}^{2{d}_{32}}}& 0& 0& 0\\ 0& 0& 0& \frac{q-{q}^{-1}}{1-{q}^{2{d}_{12}}}& 0& \frac{q-{q}^{2{d}_{21}-1}}{1-{q}^{2{d}_{21}}}\\ 0& \frac{q-{q}^{2{d}_{13}-1}}{1-{q}^{2{d}_{13}}}& 0& 0& \frac{q-{q}^{-1}}{1-{q}^{2{d}_{31}}}& 0\\ 0& 0& 0& \frac{q-{q}^{2{d}_{12}-1}}{1-{q}^{2{d}_{12}}}& 0& \frac{q-{q}^{-1}}{1-{q}^{2{d}_{21}}}\end{array}\right)\end{array}$$where ${d}_{k\ell}=c\left(L(i+k-1)\right)-c\left(L(i+l-1)\right)\text{.}$ In each case we compute directly the products ${\varphi}_{S}\left({T}_{i}\right){\varphi}_{S}\left({T}_{i+1}\right){\varphi}_{S}\left({T}_{i}\right)$ and ${\varphi}_{S}\left({T}_{i+1}\right){\varphi}_{S}\left({T}_{i}\right){\varphi}_{S}\left({T}_{i+1}\right)$ and verify that they are equal. (This proof of the braid relation is, in all essential aspects, the same as that used by Hoefsmit [Hoe1974], Wenzl [Wen1988] and Ariki and Koike [AKo1994]. For a more elegant but less straightforward proof of this relation see the proof of Theorem 3.1 in [Ram1998].) $\square $ |
Step 2. The module ${\stackrel{\sim}{H}}^{(c,\lambda /\mu )}$ is irreducible.
Proof. | |
Let $L$ be a standard tableaux of shape $\lambda /\mu $ and define $${\pi}_{L}=\prod _{i=1}^{n}\prod _{P\ne L}\frac{{x}_{i}-{q}^{2c\left(P\left(i\right)\right)}}{{q}^{2c\left(L\left(i\right)\right)}-{q}^{2c\left(P\left(i\right)\right)}},$$where the second product is over all standard tableaux $P$ of shape $\lambda /\mu $ which are not equal to $L\text{.}$ Then ${\pi}_{L}$ is an element of ${\stackrel{\sim}{H}}_{n}$ such that $${\pi}_{L}{v}_{Q}={\delta}_{LQ}{v}_{L},$$for all standard tableaux $Q$ of shape $\lambda /\mu \text{.}$ This follows from the formula for the action of ${x}_{i}$ on ${\stackrel{\sim}{H}}^{(c,\lambda /\mu )}$ and the fact that the sequence $({q}^{2c\left(L\left(1\right)\right)},\dots ,{q}^{2c\left(L\left(n\right)\right)})$ completely determines the standard tableau $L$ (Lemma 2.2). Let $N$ be a nonzero submodule of ${\stackrel{\sim}{H}}^{(c,\lambda /\mu )}$ and let $v={\sum}_{Q}{a}_{Q}{v}_{Q}$ be a nonzero element of $N\text{.}$ Let $L$ be a standard tableau such that the coefficient ${v}_{L}$ is nonzero. Then ${\pi}_{L}v={a}_{L}{v}_{L}$ and so ${v}_{L}\in N\text{.}$ By Proposition 2.1 we may identify the set ${\mathcal{F}}^{\lambda /\mu}$ with an interval in ${S}_{n}$ (under Bruhat order). Under this identification the minimal element is the column reading tableau $C$ and there is a chain $C<{s}_{{i}_{1}}C<\dots <{s}_{{i}_{p}}\dots {s}_{{i}_{1}}C=L$ such that all elements of the chain are standard tableaux of shape $\lambda /\mu \text{.}$ Then, by the definition of the ${\tau}_{i}\text{-operators,}$ $${\tau}_{{i}_{1}}\dots {\tau}_{{i}_{p}}{v}_{L}=\kappa {v}_{C},$$for some constant $\kappa \in {\u2102}^{*}\text{.}$ It follows that ${v}_{C}\in N\text{.}$ Let $Q$ be an arbitrary standard tableau of shape $\lambda /\mu \text{.}$ Again, there is a chain $C<{s}_{{j}_{1}}C<\dots <{s}_{{j}_{p}}\dots {s}_{{j}_{1}}C=Q$ of standard tableaux in ${\mathcal{F}}^{\lambda /\mu}$ and we have $${\tau}_{{j}_{p}}\dots {\tau}_{{j}_{1}}{v}_{C}=\kappa \prime {v}_{Q},$$for some $\kappa \prime \in {\u2102}^{*}\text{.}$ Thus ${v}_{Q}\in N\text{.}$ It follows that $N={\stackrel{\sim}{H}}^{(c,\lambda /\mu )}\text{.}$ Thus ${\stackrel{\sim}{H}}^{(c,\lambda /\mu )}$ is irreducible. $\square $ |
Step 3. The modules ${\stackrel{\sim}{H}}^{(c,\lambda /\mu )}$ are nonisomorphic.
Proof. | |
Each of the modules ${\stackrel{\sim}{H}}^{(c,\lambda /\mu )}$ has a unique basis (up to multiplication of each basis vector by a constant) of simultaneous eigenvectors for the ${x}_{i}\text{.}$ Each basis vector is determined by its weight, the sequence of eigenvalues $({t}_{1},\dots ,{t}_{n})$ given by $${x}_{i}{v}_{t}={t}_{i}{v}_{t},\phantom{\rule{2em}{0ex}}\text{for}\hspace{0.17em}1\le i\le n\text{.}$$By the definition of the action of the ${x}_{i},$ a weight of ${\stackrel{\sim}{H}}^{(c,\lambda /\mu )}$ is equal to $({q}^{2c\left(L\left(1\right)\right)},\dots ,{q}^{2c\left(L\left(n\right)\right)})$ for some standard tableau $L\text{.}$ By Lemma 2.2, both the standard tableau $L$ and the placed skew shape $(c,\lambda /\mu )$ are determined uniquely by this weight. Thus no two of the modules ${\stackrel{\sim}{H}}^{(c,\lambda /\mu )}$ can be isomorphic. $\square $ |
Step 4. If $t=({t}_{1},\dots ,{t}_{n})$ is the weight of a calibrated ${\stackrel{\sim}{H}}_{n}\text{-module}$ $M$ then $t=({q}^{2c\left(L\left(1\right)\right)},\dots ,{q}^{2c\left(L\left(n\right)\right)})$ for some standard tableau $L$ of placed skew shape.
Proof. | |
Let ${m}_{t}$ be a weight vector in $M$ of weight $t=({t}_{1},\dots ,{t}_{n}),$ i.e. $${x}_{i}{m}_{t}={t}_{i}{m}_{t},\phantom{\rule{2em}{0ex}}\text{for all}\hspace{0.17em}1\le i\le n\text{.}$$We want $L$ such that $({q}^{2c\left(L\left(1\right)\right)},\dots ,{q}^{2c\left(L\left(n\right)\right)})=({t}_{1},\dots ,{t}_{n})\text{.}$ We shall show that if ${t}_{i}={t}_{j}$ for $i<j$ then there exist $k$ and $\ell $ such that $i<k<\ell <j,$ ${t}_{k}={q}^{\pm 2}{t}_{i}$ and ${t}_{\ell}={q}^{\mp 2}{t}_{i}\text{.}$ This will show that if there are two adjacent boxes of $L$ in the same diagonal then these boxes must be contained in a complete $2\times 2$ block, i.e. if there is a configuration in L$$ of the form $$\n\n\n\n\ni\nj\n\n\phantom{\rule{2em}{0ex}}\text{then}\hspace{0.17em}L\hspace{0.17em}\text{must contain}\phantom{\rule{2em}{0ex}}\n\n\n\n\n\n\ni\nk\n\u2113\nj\n\n\phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}\n\n\n\n\n\n\ni\n\u2113\nk\nj\n\n\text{.}$$This is sufficient to guarantee that $L$ is of skew shape. Let $j>i$ be such that ${t}_{j}={t}_{i}$ and $j-i$ is minimal. The argument is by induction on the value of $j-i\text{.}$ Case 1: $j-i=1\text{.}$ Then ${m}_{t}$ and ${T}_{i}{m}_{t}$ are linearly independent. If they were not we would have ${T}_{i}{m}_{t}=a{m}_{t}$ which would give $${t}_{i}a{m}_{t}={x}_{i}{T}_{i}{m}_{t}=({T}_{i}{x}_{i+1}-(q-{q}^{-1}){x}_{i+1}){m}_{t}=(a{t}_{i+1}-(q-{q}^{-1}){t}_{i+1}){m}_{t}=(a-(q-{q}^{-1})){t}_{i}{m}_{t}\text{.}$$Since $q-{q}^{-1}\ne 0,$ this equation implies that ${t}_{i}=0$ which is a contradiction. Now the relations (1.1d) and (1.4) show that $$\begin{array}{ccc}{x}_{i}{T}_{i}{m}_{t}& =& {t}_{i}({T}_{i}{m}_{t}-(q-{q}^{-1}){m}_{t}),\\ {x}_{i+1}{T}_{i}{m}_{t}& =& {t}_{i+1}({T}_{i}{m}_{t}-(q-{q}^{-1}){m}_{t}),\\ {x}_{j}{T}_{i}{m}_{t}& =& {t}_{j}{T}_{i}{m}_{t},\phantom{\rule{2em}{0ex}}\text{for all}\hspace{0.17em}j\ne i,i+1\text{.}\end{array}$$It follows that ${T}_{i}{m}_{t}$ is an element of ${M}_{t}^{\text{gen}}$ but not an element of ${M}_{t}\text{.}$ This is a contradiction to the fact that $M$ is calibrated. So this is not possible, i.e. ${t}_{i+1}$ cannot equal ${t}_{i}\text{.}$ Case 2: $j-i=2\text{.}$ Since ${t}_{i}\ne {t}_{i+1}$ and ${m}_{t}$ is a weight vector, the vector $${m}_{{s}_{i}t}=({T}_{i}-\frac{(q-{q}^{-1}){t}_{i+1}}{{t}_{i+1}-{t}_{i}}){m}_{t}$$is a weight vector of weight $t\prime ={s}_{i}t$ (see (3.7a)). Then ${t}_{i}^{\prime}={t}_{i+1}^{\prime}$ and so, by Case 1, ${m}_{{s}_{i}t}=0\text{.}$ This implies that $${T}_{i}{m}_{t}=\frac{(q-{q}^{-1}){t}_{i+1}}{{t}_{i+1}-{t}_{i}}{m}_{t}\text{.}$$By equation (1.3), all eigenvalues of ${T}_{i}$ are either $q$ or $-{q}^{-1}\text{.}$ Thus ${T}_{i}{m}_{t}=\pm {q}^{\pm 1}{m}_{t}$ and so ${t}_{i}={q}^{\pm 2}{t}_{i+1}\text{.}$ A similar argument shows that $${m}_{{s}_{i+1}t}=({T}_{i+1}-\frac{(q-{q}^{-1}){t}_{i+2}}{{t}_{i+2}-{t}_{i+1}}){m}_{t}$$must be 0 and thus that $${T}_{i+1}{m}_{t}=\frac{(q-{q}^{-1}){t}_{i+2}}{{t}_{i+2}-{t}_{i+1}}{m}_{t}=\frac{(q-{q}^{-1}){t}_{i}}{{t}_{i}-{t}_{i+1}}{m}_{t}=\mp {q}^{\mp 1}{m}_{t}\text{.}$$From ${T}_{i}{m}_{t}=\pm {q}^{\pm 1}{m}_{t}$ and ${T}_{i+1}{m}_{t}=\mp {q}^{\mp 1}{m}_{t}$ we get $$\pm {q}^{\pm 1}{m}_{t}={T}_{i}{T}_{i+1}{T}_{i}{m}_{t}={T}_{i+1}{T}_{i}{T}_{i+1}{m}_{t}=\mp {q}^{\mp 1}{m}_{t}\text{.}$$This is impossible since $q$ is not a root of unity. So this case is not possible, i.e. ${t}_{i+2}$ cannot equal ${t}_{i}\text{.}$ Induction step. Assume that $i$ and $j$ are such that ${t}_{i}={t}_{j}$ and the value $j-i$ is minimal such that this is true. If ${t}_{j-1}\ne {q}^{\pm 2}{t}_{j}$ then the vector $${m}_{{s}_{j}t}=({T}_{j}-\frac{(q-{q}^{-1}){t}_{j}}{{t}_{j-1}-{t}_{j}}){m}_{t}$$is a weight vector of weight $t\prime ={s}_{j}t$ and by (3.7b) this vector is nonzero. Since ${t}_{i}^{\prime}={t}_{i}={t}_{j}={t}_{j-1}^{\prime}$ we can apply the induction hypothesis to conclude that there are $k$ and $\ell $ with $i<k<\ell <j-1$ such that ${t}_{k}^{\prime}={q}^{\pm 2}{t}_{i}^{\prime}$ and ${t}_{\ell}^{\prime}={q}^{\mp 2}{t}_{i}^{\prime}\text{.}$ This implies that ${t}_{k}={q}^{\pm 2}{t}_{i}$ and ${t}_{\ell}={q}^{\mp 2}{t}_{i}\text{.}$ Similarly, if ${t}_{i}\ne {q}^{\pm 2}{t}_{i+1}$ then the vector $${m}_{{s}_{i}t}=({T}_{i}=\frac{(q-{q}^{-1}){t}_{i+1}}{{t}_{i+1}-{t}_{i}}){m}_{t}$$is a weight vector of weight $t\prime ={s}_{i}t$ and by (3.7b) this vector is nonzero. Since ${t}_{i+1}^{\prime}={t}_{i}={t}_{j}={t}_{j}^{\prime}$ we can apply the induction hypothesis to conclude that there are $k$ and $\ell $ with $i+1<k<\ell <j$ such that ${t}_{k}^{\prime}={q}^{\pm 2}{t}_{j}^{\prime}$ and ${t}_{\ell}^{\prime}={q}^{\mp 2}{t}_{j}^{\prime}\text{.}$ This implies that ${t}_{k}={q}^{\pm 2}{t}_{i}$ and ${t}_{\ell}={q}^{\mp 2}{t}_{i}\text{.}$ If we are not in either of the previous cases then ${t}_{i+1}={q}^{2}{t}_{i}$ or ${t}_{i+1}={q}^{-2}{t}_{i}$ and ${t}_{j-1}={q}^{2}{t}_{j}$ or ${t}_{j-1}={q}^{-2}{t}_{j}\text{.}$ We cannot have ${t}_{i+1}={t}_{j-1}$ since the $i$ and $j$ are such that $j-i$ is minimal such that ${t}_{i}={t}_{j}\text{.}$ Thus ${q}^{\pm 2}{t}_{i+1}={q}^{\mp 2}{t}_{j-1}={t}_{i}\text{.}$ $\square $ |
Step 5. Suppose that $M$ is an irreducible calibrated ${\stackrel{\sim}{H}}_{n}\text{-module}$ and that ${m}_{t}$ is a weight vector in $M$ with weight $t=({t}_{1},\dots ,{t}_{n})$ such that ${t}_{i}={q}^{\pm 2}{t}_{i+1}\text{.}$ Then ${\tau}_{i}{m}_{t}=0\text{.}$
Proof. | |
Assume that ${m}_{{s}_{i}t}={\tau}_{i}{m}_{t}\ne 0\text{.}$ Then, by the second identity in Proposition 3.2 (b), ${\tau}_{i}{m}_{{s}_{i}t}=0\text{.}$ Since $M$ is irreducible there must be some sequence of $\tau \text{-operators}$ such that $${\tau}_{{i}_{1}}\dots {\tau}_{{i}_{p}}{m}_{{s}_{i}t}=\kappa {m}_{t},$$with $\kappa \in {\u2102}^{*}\text{.}$ Assume that ${\tau}_{{i}_{1}}\dots {\tau}_{{i}_{p}}$ is a minimal length sequence such that this is true. We have ${s}_{{i}_{1}}\dots {s}_{{i}_{p}}{s}_{i}t=t\text{.}$ Assume that ${s}_{{i}_{1}}\dots {s}_{{i}_{p}}{s}_{i}\ne 1\text{.}$ Then there must be $1\le i<j\le n$ such that ${t}_{i}={t}_{j}$ (because some nontrivial permutation of the ${t}_{i}$ fixes $t\text{).}$ Since ${s}_{{i}_{1}}\dots {s}_{{i}_{p}}{s}_{i}$ is of minimal length such that it fixes $t$ it must be a transposition $(i,j)$ for some $i<j$ such that ${t}_{i}={t}_{j}\text{.}$ Furthermore there does not exist $i<k<j$ such that ${t}_{i}={t}_{k}\text{.}$ The element ${s}_{{i}_{1}}\dots {s}_{{i}_{p}}{s}_{i}$ switches the ${t}_{i}$ and the ${t}_{j}$ in $t\text{.}$ In the process of doing this switch by a sequence of simple transpositions there must be some point where ${t}_{i}$ and ${t}_{j}$ are adjacent and thus there must be some $\ell $ such that ${s}_{{i}_{\ell}}\left({s}_{{i}_{\ell +1}}\dots {s}_{{i}_{p}}{s}_{i}t\right)={s}_{{i}_{\ell +1}}\dots {s}_{{i}_{p}}{s}_{i}t\text{.}$ Then $${m}_{t\prime}={\tau}_{{i}_{\ell +1}}\dots {\tau}_{{i}_{\ell}}{\tau}_{i}{m}_{t}$$is a nonzero weight vector in $M$ of weight $t\prime ={s}_{{i}_{\ell +1}}\dots {s}_{{i}_{p}}{s}_{i}t\text{.}$ Since ${s}_{{i}_{\ell}}t\prime =t\prime $ it follows that ${t}_{{i}_{\ell}}={t}_{{i}_{\ell}+1}\text{.}$ Since $M$ is calibrated this is a contradiction to (Case 1 of) step 4. So ${s}_{{i}_{1}}\dots {s}_{{i}_{p}}{s}_{i}=1\text{.}$ Let $k$ be minimal such that ${s}_{{i}_{1}}\dots {s}_{{i}_{k}}$ is not reduced. Assume $p>1\text{.}$ Then we can use the braid relations (the third and fourth lines of Proposition 3.2 (b)) to write $$\kappa {m}_{t}={\tau}_{{j}_{1}}\dots {\tau}_{{j}_{k-2}}{\tau}_{{i}_{k}}{\tau}_{{i}_{k}}{\tau}_{{i}_{k+1}}\dots {\tau}_{{i}_{p}}{\tau}_{i}{m}_{t},$$for some ${j}_{1},\dots ,{j}_{k-2}\text{.}$ Then, by the second line of Proposition 3.2 (b), $$\kappa \prime {m}_{t}={\tau}_{{j}_{\prime}}\dots {\tau}_{{j}_{k-2}}{\tau}_{{i}_{k+1}}\dots {\tau}_{{i}_{p}}{\tau}_{i}{m}_{t},$$for some $\kappa \prime \in {\u2102}^{*}\text{.}$ This is a contradiction to the minimality of the length of the sequence ${\tau}_{{i}_{1}}\dots {\tau}_{{i}_{p}}\text{.}$ So $p=1,$ ${i}_{p}=i$ and ${\tau}_{i}{\tau}_{i}{m}_{t}=\kappa {m}_{t}\text{.}$ This is a contradiction since the second identity in Proposition 3.2 (b) and the assumption that ${t}_{i}={q}^{\pm 2}{t}_{i+1}$ imply that ${\tau}_{i}{\tau}_{i}=0\text{.}$ So ${\tau}_{i}{m}_{t}=0\text{.}$ $\square $ |
Step 6. An irreducible calibrated ${\stackrel{\sim}{H}}_{n}\text{-module}$ $M$ is isomorphic to ${\stackrel{\sim}{H}}^{(c,\lambda /\mu )}$ for some placed skew shape $(c,\lambda /\mu )\text{.}$
Proof. | |
Let ${m}_{t}$ be a nonzero weight vector in $M\text{.}$ Since $M$ is calibrated step 4 implies that there is a placed skew shape $(c,\lambda /\mu )$ and a standard tableau $L$ of shape $\lambda /\mu $ such that $t=({q}^{2c\left(L\left(1\right)\right)},\dots ,{q}^{2c\left(L\left(n\right)\right)}),$ Let us write ${m}_{L}$ in place of ${m}_{t}\text{.}$ Let $C$ be the column reading tableau of shape $\lambda /\mu \text{.}$ It follows from Proposition 2.1 that there is a chain $C,{s}_{{i}_{1}}C,\dots ,{s}_{{i}_{p}}\dots {s}_{{i}_{1}}C=L$ of standard tableaux of shape $\lambda /\mu \text{.}$ By (3.7b), all of the ${\tau}_{{i}_{j}}$ in this sequence are bijections and so $${m}_{C}={\tau}_{{i}_{1}}\dots {\tau}_{{i}_{p}}{m}_{L}$$is a nonzero weight vector in $M\text{.}$ Similarly, if $Q$ is any other standard tableau of shape $\lambda /\mu $ then there is a chain $C,{s}_{{j}_{1}}C,\dots ,{s}_{{j}_{p}}\dots {s}_{{j}_{1}}C=Q$ and so $${m}_{Q}={\tau}_{{j}_{p}}\dots {\tau}_{{j}_{1}}{m}_{C}$$is a nonzero weight vector in $M\text{.}$ Finally, by step 5, ${\tau}_{i}{m}_{Q}=0$ if ${s}_{i}Q$ is not standard (since ${q}^{2c\left(Q\left(i\right)\right)}={q}^{\pm 2}{q}^{2c\left(Q(i+1)\right)}\text{)}$ and so the span of the vectors $\{{m}_{Q}\hspace{0.17em}\mid \hspace{0.17em}Q\hspace{0.17em}\text{a standard tableau of shape}\hspace{0.17em}\lambda /\mu \}$ is a submodule of $M\text{.}$ Since $M$ is irreducible this must be all of $M$ and the map $$\begin{array}{ccc}M& \u27f6& {\stackrel{\sim}{H}}^{(c,\lambda /\mu )}\\ {\tau}_{v}{m}_{C}& \u27fc& {\tau}_{v}{v}_{C}\end{array}$$is an isomorphism of ${\stackrel{\sim}{H}}_{n}\text{-modules.}$ $\square $ |
This completes the proof of Theorem 4.1.
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.