Semistandard maps span the Hom space in characteristic zero

[construct] Fix $f\in {\mathrm{Hom}}_{S_n}(S^{\lambda },M^{\mu })$ and expand $f(e_{t_0})={\sum }_{T\in T_{\lambda ,\mu }}{c}_{T}T$ in the filling basis of [F1]; the sum is finite by [F2]. Since $e_{t_0}$ generates $S^{\lambda }$ as an $S_n$-module by [F3], the map $f$ is determined by $f(e_{t_0})$: if $f(e_{t_0})=0$ then $f(\sigma \cdot e_{t_0})=\sigma \cdot f(e_{t_0})=0$ for every $\sigma \in S_n$, and these elements span $S^{\lambda }$. It therefore suffices to express $f(e_{t_0})$ as a $\mathbb{C}$-linear combination of the vectors ${\theta }_T(e_{t_0})$ with $T$ semistandard: if $f(e_{t_0})={\sum }_T{a}_T{\theta }_T(e_{t_0})$ then the $S_n$-linear maps $f$ and ${\sum }_T{a}_T{\theta }_T$ agree on the generator $e_{t_0}$, hence are equal.

For $\tau \in C_{t_0}$ and every filling $T$ one has $c_{\tau \cdot T}=\mathrm{sgn}(\tau )c_T$: by [F3] and $S_n$-linearity of $f$, $\tau \cdot f(e_{t_0})=f(\tau \cdot e_{t_0})=\mathrm{sgn}(\tau )f(e_{t_0})$, while expanding the left side in the filling basis gives $\tau \cdot f(e_{t_0})={\sum }_T{c}_T(\tau \cdot T)$, so comparing coefficients of the basis element $\tau \cdot T$ yields $c_T=\mathrm{sgn}(\tau )c_{\tau \cdot T}$ and hence $c_{\tau \cdot T}=\mathrm{sgn}(\tau )c_T$ by [F7].

(Cross swaps increase $N$.) Let $x$ be a box in a column $j$, let $y$ be a box in the column $j+1$, let $f$ be a filling with $f(x)>f(y)$, and let $\tau :=(t_0(x)t_0(y))$, so that $\tau \cdot f$ is obtained from $f$ by exchanging the values at $x$ and $y$ by [F1]. Then $N_{\tau \cdot f}(i,k)\ge N_f(i,k)$ for all $i,k$, with strict inequality for $i=f(y)$, $k=j$: for $k<j$ or $k\ge j+1$ the swap changes no count of entries in the first $k$ columns, while for $k=j$ the two counts differ by $[f(y)\le i]-[f(x)\le i]$, which is $0$ or $1$ because $f(y)<f(x)$, namely $1$ exactly when $f(y)\le i<f(x)$.

(A transverse transversal.) Let $X,Y$ be disjoint nonempty label sets carried by a set of boxes of column $j$ and column $j+1$ respectively, put $Z:=X\cup Y$ and $p:=\mathrm{\mid }X\mathrm{\mid }$, and for every $p$-element subset $A\subseteq Z$ write $X\setminus A=\{a_1<\cdots <a_r\}$ and $A\setminus X=\{b_1<\cdots <b_r\}$, with $r:=\mathrm{\mid }X\setminus A\mathrm{\mid }=\mathrm{\mid }A\setminus X\mathrm{\mid }$, and put $g_A:=(a_1{b}_1)\cdots (a_r{b}_r)$, the empty product for $A=X$ giving $g_X=1$. Then $g_A(X)=A$ and $g_A$ maps each element of $X\setminus A$ to an element of $Y$ and conversely, so $g_A^{-1}=(a_r{b}_r)\cdots (a_1{b}_1)$ is a product of $r$ transpositions each exchanging an element of $X$ with an element of $Y$; and $\{g_A:\mathrm{\mid }A\mathrm{\mid }=p\}$ is a set of representatives for the left cosets of $H=S_X\times S_Y$ in $S_Z$ containing $1$, because $H$ is exactly the setwise stabilizer of $X$ in $S_Z$ so the coset $gH$ is determined by $g(X)\in \{A:\mathrm{\mid }A\mathrm{\mid }=p\}$ and $g_A(X)=A$ realizes each value; consequently this is a transversal of the kind required in [F5].

(Finite descending induction.) Since $T_{\lambda ,\mu }$ is finite by [F2], the set $\mathcal{N}:=\{N_T:T\in T_{\lambda ,\mu }\}$ is finite, and we fix a total order $⪯$ on $\mathcal{N}$ extending the componentwise order in the sense that $N⊑N^{\mathrm{\prime }}$ implies $N⪯N^{\mathrm{\prime }}$; such an order exists because a finite partial order is listed by repeatedly removing a maximal element. For a nonzero $v\in M^{\mu }$ let $m(v)$ be the $⪯$-greatest element of $\{N_T:\text{the coefficient of }T\text{ in }v\text{ is nonzero}\}$. Claim: if $v=f(e_{t_0})$ for some $f\in {\mathrm{Hom}}_{S_n}(S^{\lambda },M^{\mu })$ and $v\mathrm{\ne }0$, then there is a semistandard $S$ such that $v-c_S{\theta }_S(e_{t_0})=0$ or $m(v-c_S{\theta }_S(e_{t_0}))\prec m(v)$, where $c_S$ is the coefficient of $S$ in $v$.

If a filling $T$ has two boxes $x\mathrm{\ne }y$ of the same column carrying the same entry, then $c_T=0$. Indeed, let $a:=t_0(x)\mathrm{\ne }b:=t_0(y)$ be the labels of those boxes and let $\tau :=(ab)$; then $\tau \in C_{t_0}$ by [F6] and, by [F1], $\tau$ exchanges the entries in the boxes labelled $a$ and $b$, so $\tau \cdot T=T$; step 1.2 gives $c_T=c_{\tau \cdot T}=\mathrm{sgn}(\tau )c_T=-c_T$ by [F7], hence $2c_T=0$ and $c_T=0$ in $\mathbb{C}$. Thus every filling with $c_T\mathrm{\ne }0$ has pairwise distinct entries in each of its columns.

(Expansion of a semistandard ${\theta }_S$.) Let $S\in T_{\lambda ,\mu }$ be semistandard and write ${\theta }_S(e_{t_0})={\sum }_g{d}_{g}g$ in the filling basis. Then $d_g=0$ unless $N_g⊑N_S$ componentwise, and for every $g$ with $N_g=N_S$ there is a unique ${\tau }_g\in C_{t_0}$ with ${\tau }_g\cdot S=g$, and $d_g=\mathrm{sgn}({\tau }_g)$. Indeed ${\theta }_S(e_{t_0})={\sum }_{u\in R_{t_0}\cdot S}{\kappa }_{t_0}\cdot u$, and the coefficient of $g$ in ${\kappa }_{t_0}\cdot u$ vanishes unless $N_g=N_u$ by [F4], while $N_u⊑N_S$ for $u$ in the row orbit; so $d_g=0$ unless $N_g=N_u$ for some such $u$, whence $N_g⊑N_S$, and if $N_g=N_S$ then $N_u=N_S$ forces $u=S$ by [F4], so $d_g$ is the coefficient of $g$ in ${\kappa }_{t_0}\cdot S$, namely ${\sum }_{\gamma \in C_{t_0}:\gamma \cdot S=g}\mathrm{sgn}(\gamma )$; the column-sorted semistandard $S$ has distinct entries in each column by [F2], so $\gamma \cdot S=S$ only for $\gamma =1$ and there is exactly one $\gamma \in C_{t_0}$ with $\gamma \cdot S=g$, giving $d_g=\mathrm{sgn}(\gamma )$ for that unique $\gamma$, which we call ${\tau }_g$.

Let us record the standing choice of a maximal level. For the fixed $f$ of step 1.1 with support $\{T:c_T\mathrm{\ne }0\}$ nonempty put $v:=f(e_{t_0})$ and choose $T_1$ in the support with $N_{T_1}=m(v)$, the $⪯$-greatest support level of step 1.5; then every support filling $T$ satisfies $N_T⪯N_{T_1}$, and no support filling has $N_T$ strictly above $N_{T_1}$ componentwise, because $N_T⊐N_{T_1}$ would imply $N_T\succ N_{T_1}$ in the total order by the extension property of $⪯$. Replace $T_1$ by the unique filling obtained from it by sorting each column increasingly; $N_{T_1}=m(v)$ is unchanged by [F4] because column sorting only rearranges entries within columns, the coefficient still satisfies $c_{T_1}\mathrm{\ne }0$ (it is multiplied by a sign by step 1.2), the entries of $T_1$ are pairwise distinct in each column by step 2.1, and the columns of $T_1$ are strictly increasing by construction. So we may assume: $c_{T_1}\mathrm{\ne }0$, the columns of $T_1$ strictly increase, $N_{T_1}=m(v)$, and no filling $T$ with $c_T\mathrm{\ne }0$ has $N_T$ strictly above $N_{T_1}$ in the componentwise order.

(Descent contradicts maximality.) Suppose the filling $T_1$ of step 3.1 is not semistandard. Since its columns are strictly increasing and it is not semistandard, some row $q$ of $T_1$ descends between adjacent columns $j,j+1$: with $a:=T_1(q,j)$ and $b:=T_1(q,j+1)$ one has $a>b$. Let $X$ be the set of labels of the boxes $(r,j)$ with $q\le r\le {\lambda }_j^{\mathrm{\prime }}$ and $Y$ the set of labels of the boxes $(r,j+1)$ with $1\le r\le q$; these are label sets of column $j$ and column $j+1$ of $t_0$, and $\mathrm{\mid }X\mathrm{\mid }+\mathrm{\mid }Y\mathrm{\mid }=({\lambda }_j^{\mathrm{\prime }}-q+1)+q={\lambda }_j^{\mathrm{\prime }}+1>{\lambda }_j^{\mathrm{\prime }}$. Because the columns of $T_1$ strictly increase, $T_1(x)\ge a$ for every box $x\in X$ and $T_1(y)\le b$ for every box $y\in Y$, and $a>b$, so $T_1(x)>T_1(y)$ for all such boxes.

With $X,Y,Z$ as in step 4.1 and the transversal $\{g_A\}$ of step 1.4, the Garnir relation [F5] gives ${\sum }_A\mathrm{sgn}(g_A)g_A\cdot e_{t_0}=0$ in $M^{\lambda }$, where $A$ ranges over the $p$-element subsets of $Z$. Applying the $S_n$-linear map $f$ and expanding yields the identity $0={\sum }_A{c}_{g_A^{-1}{T}_1}\mathrm{sgn}(g_A)$ in $\mathbb{C}$: the coefficient of the basis element $T_1$ in ${\sum }_A\mathrm{sgn}(g_A)g_A\cdot f(e_{t_0})$ is ${\sum }_A\mathrm{sgn}(g_A)c_{g_A^{-1}\cdot T_1}$, because within the $A$-th summand exactly the filling $g_A^{-1}\cdot T_1$ is transported to $T_1$ by $g_A$.

In the identity of step 5.1, every term with $A\mathrm{\ne }X$ vanishes. Indeed, for $A\mathrm{\ne }X$ the permutation $g_A^{-1}$ is a product of $r=\mathrm{\mid }X\setminus A\mathrm{\mid }\ge 1$ transpositions $(a_i{b}_i)$ with $a_i\in X$ and $b_i\in Y$ by step 1.4, and these act on disjoint pairs of boxes labelled by elements of $Z$; applying them one after another to $T_1$, each step exchanges the value at the box labelled $a_i$, which is still $T_1$'s value there and is $>b$, with the value at the box labelled $b_i$, which is still $\le b$, and hence, by step 1.3 and $T_1(x)>T_1(y)$ for all boxes $x\in X$, $y\in Y$ from step 4.1, strictly increases the count vector $N$ at each step. So $N_{g_A^{-1}{T}_1}(i,k)\ge N_{T_1}(i,k)$ for all $i,k$ with strict inequality somewhere, and $c_{g_A^{-1}{T}_1}=0$ by the maximality of $N_{T_1}$ in step 3.1.

Therefore the identity of step 5.1 reduces to $c_{T_1}\mathrm{sgn}(g_X)=c_{T_1}=0$, contradicting $c_{T_1}\mathrm{\ne }0$ from step 3.1. Hence the filling $T_1$ is semistandard.

(Subtraction kills a whole level.) Keep $T_1$ semistandard with $c_{T_1}\mathrm{\ne }0$ and $N_{T_1}=m(v)$ from step 7.1 and step 3.1 and put $v^{\mathrm{\prime }}:=f(e_{t_0})-c_{T_1}{\theta }_{T_1}(e_{t_0})$, with coefficients $c_g^{\mathrm{\prime }}=c_g-c_{T_1}{d}_g$ in the filling basis. If $N_g=N_{T_1}$ then $g$ is a column rearrangement of $T_1$ by [F4], so $g={\tau }_g\cdot T_1$ for the unique ${\tau }_g\in C_{t_0}$ of step 2.2; step 1.2 gives $c_g=\mathrm{sgn}({\tau }_g)c_{T_1}$ and step 2.2 gives $d_g=\mathrm{sgn}({\tau }_g)$, so $c_g^{\mathrm{\prime }}=\mathrm{sgn}({\tau }_g)c_{T_1}-\mathrm{sgn}({\tau }_g)c_{T_1}=0$. Every $g$ with $c_g^{\mathrm{\prime }}\mathrm{\ne }0$ satisfies $N_g⪯N_{T_1}$: if $c_g\mathrm{\ne }0$ then this is the maximality of $N_{T_1}=m(v)$ among the support levels in the total order of step 1.5, and if $d_g\mathrm{\ne }0$ then $N_g⊑N_{T_1}$ by step 2.2, hence again $N_g⪯N_{T_1}$. Since no such $g$ has $N_g=N_{T_1}$, every nonzero coefficient of $v^{\mathrm{\prime }}$ sits at a level strictly below $N_{T_1}$ in the total order.

This proves the claim of step 1.5: take $S:=T_1$, which is semistandard by step 7.1. Step 8.1 says that the residual $v^{\mathrm{\prime }}:=v-c_S{\theta }_S(e_{t_0})$ either is zero or has every support level strictly below $m(v)$, so in the nonzero case $m(v^{\mathrm{\prime }})\prec m(v)$.

Set $f_0:=f$ and $v_0:=f(e_{t_0})$. Whenever $v_k\mathrm{\ne }0$, apply step 9.1 to $v_k=f_k(e_{t_0})$, choose the resulting semistandard $S_k$ and its coefficient $a_k$ in $v_k$, and put $f_{k+1}:=f_k-a_k{\theta }_{S_k}{\mathrm{\mid }}_{S^{\lambda }}$ and $v_{k+1}:=f_{k+1}(e_{t_0})$. Each $f_{k+1}$ is $S_n$-linear by [F1]. Either $v_{k+1}=0$ and we stop, or $m(v_{k+1})\prec m(v_k)$. The nonzero residuals thus have strictly decreasing levels in the finite set $\mathcal{N}$, so after at most $\mathrm{\mid }\mathcal{N}\mathrm{\mid }$ subtractions we reach $v_k=0$. Then $f(e_{t_0})={\sum }_{j<k}{a}_j{\theta }_{S_j}(e_{t_0})$, and the $S_n$-linear maps agree on a module generator, hence $f={\sum }_{j<k}{a}_j{\theta }_{S_j}{\mathrm{\mid }}_{S^{\lambda }}$ by step 1.1. If $v_0=0$, the same conclusion holds with the empty sum.

The remaining cases are the empty ones: for $n=0$ one has $\lambda =\mu =\mathrm{\varnothing }$, the set $T_{\mathrm{\varnothing },\mathrm{\varnothing }}$ consists of the single empty filling, which is semistandard because its row and column conditions are vacuous, and ${\theta }_{\mathrm{\varnothing }}$ is the identity of $M^{\mathrm{\varnothing }}=\mathbb{C}$, so $\mathrm{Hom}(S^{\mathrm{\varnothing }},M^{\mathrm{\varnothing }})=\mathbb{C}$ is spanned by that restriction; here step 1.1 applies with $e_{t_0}$ generating $S^{\mathrm{\varnothing }}$, and the argument of the descent and subtraction steps is either vacuous or terminates at once, since the support of $f(e_{t_0})$ is empty or consists of the semistandard empty filling. No division is used anywhere, only the fact that $2\mathrm{\ne }0$ in $\mathbb{C}$ in step 2.1, and all choices made are selections of a maximal element or a unique column-sorted filling from finite explicitly given sets, so no choice principle is invoked.