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

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.

In Case (1) the space $S$ is one dimensional and spanned by the vector $v_Q$ corresponding to the standard tableau

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

respectively. In case (2) the space $S$ is two dimensional and spanned by the vectors $v_Q$ corresponding to the standard tableaux

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

In case (3) the space $S$ is three dimensional and spanned by the vectors $v_Q$ corresponding to the standard tableaux

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

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

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

and

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].)

$\square$

Let $L$ be a standard tableaux of shape $\lambda /\mu$ and define

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

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 ${ℱ}^{\lambda /\mu }$ 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 ${\tau }_i\text{-operators,}$

for some constant $\kappa \in {ℂ}^{*}\text{.}$ 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 ${ℱ}^{\lambda /\mu }$ and we have

for some $\kappa \prime \in {ℂ}^{*}\text{.}$ 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.

$\square$

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

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.

$\square$

Let $m_t$ be a weight vector in $M$ of weight $t=(t_1,\dots ,t_n),$ i.e.

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

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

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

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

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

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

must be 0 and thus that

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

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

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

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 $q^{\pm 2}{t}_{i+1}=q^{\mp 2}{t}_{j-1}=t_i\text{.}$

$\square$

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

with $\kappa \in {ℂ}^{*}\text{.}$ 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

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

for some $j_1,\dots ,j_{k-2}\text{.}$ Then, by the second line of Proposition 3.2 (b),

for some $\kappa \prime \in {ℂ}^{*}\text{.}$ 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 ${\tau }_i{m}_t=0\text{.}$

$\square$

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

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

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

is an isomorphism of ${\stackrel{\sim }{H}}_n\text{-modules.}$

$\square$