Alternating hook induction cancels disconnected rim removals

If $\nu \subseteq ̸\lambda$, every coefficient is zero by its definition [F1], so the claims about coefficients hold and there is no strip. Hence assume $\nu \subseteq \lambda$. Let $T$ be a semistandard skew tableau of shape $\lambda \mathrm{/}\nu$ and content $(x,1^{r-x})$ whose reading word is a lattice word, and put $s=r-x+1$. Then: (a) in each row of $\lambda \mathrm{/}\nu$ at most one entry exceeds $1$, and such an entry is the rightmost entry of its row, all other entries being $1$; (b) the labels $2,3,\dots ,s$ lie in strictly increasing rows; (c) the first (topmost) nonempty row of $\lambda \mathrm{/}\nu$ consists entirely of $1$'s. For (a): a lattice word contains an occurrence of $j$ before any occurrence of $j+1$, because at the prefix ending at the first occurrence of $j+1$ one has $\mathrm{\#}j\ge \mathrm{\#}(j+1)=1$; so $2$ precedes $3$ precedes $\dots$ precedes $s$ in the reading word. Two entries $b<c$ greater than $1$ in one row would have $b$ to the left of $c$ (weak increase), so the right-to-left reading word would contain $c$ before $b$, contradicting $b$ before $c$; hence at most one entry of a row exceeds $1$. If a row contains $b>1$, every entry strictly to its right is $\ge b>1$ and distinct from $b$, so one would be a second entry greater than $1$: impossible; so $b$ is rightmost, and the entries to its left are $<b$ and hence equal to $1$. For (b): if $j$ and $j+1$ lay in one row, weak increase would put $j$ to the left of $j+1$ and the reading word would contain $j+1$ before $j$, contradicting $j$ before $j+1$. For (c): the rightmost entry of the first nonempty row is the first letter of the reading word; a first letter $j\ge 2$ would violate $\mathrm{\#}(j-1)\ge \mathrm{\#}j$ in the one-letter prefix, so the first letter is $1$; but by (a) any entry greater than $1$ in that row would be its rightmost entry, so the row has none.

Consequences and the inverse construction. (a) If $\lambda \mathrm{/}\nu$ contained a $2\times 2$ block $\{(u,j),(u,j+1),(u+1,j),(u+1,j+1)\}$, then the entries at $(u,j)$ and $(u+1,j)$ are each not the rightmost entry of their row, hence equal $1$ by step 1.1(a), contradicting strict increase down column $j$; so $\lambda \mathrm{/}\nu$ has no $2\times 2$ block and $\mathrm{\mid }{I}_{t-1}\cap I_t\mathrm{\mid }\le 1$ for every $t\ge 2$. (b) Let $L:=\{t\in \{1,\dots ,w\}:\text{row }t\text{ of }\lambda \mathrm{/}\nu \text{ contains an entry different from }1\}$. By step 1.1(a,b) each row in $L$ carries exactly one label, these labels are $2,3,\dots ,s$ in increasing row order, $1\mathrm{\notin }L$ by step 1.1(c), and the content forces $\mathrm{\mid }L\mathrm{\mid }=r-x$; all other entries are $1$. Moreover, for every column $j$ and rows $t<t^{\mathrm{\prime }}$ with $j\in I_t\cap I_{t^{\mathrm{\prime }}}$ one must have $t^{\mathrm{\prime }}\in L$: the entries at $(u_{t^{\mathrm{\prime }}},j)$ lying below a non-label entry $1$ at $(u_t,j)$ would have to be $>1$, i.e. equal to the label of row $t^{\mathrm{\prime }}$, so a non-label row of a column must precede every label row of that same column, and at most one row of a column can be non-label (two $1$'s in a column violate strict increase); hence among the rows of any column all but possibly the first are in $L$. (c) Conversely, assume the shape has no $2\times 2$ block. If column $j$ meets an earlier row and row $t$, column-convexity makes it meet row $t-1$. Their interval endpoints weakly decrease, so their overlap is $({\nu }_{u_{t-1}},{\lambda }_{u_t}]$; since it has exactly one cell, $j={\lambda }_{u_t}$. Thus every cell below the top of its column is rightmost in its row. Let $L\subseteq \{2,\dots ,w\}$ satisfy $\mathrm{\mid }L\mathrm{\mid }=r-x$ and: whenever $j\in I_t$ and $j\in I_{t^{\mathrm{\prime }}}$ with $t^{\mathrm{\prime }}<t$, then $t\in L$. Define $T$ by filling each row $t\in L$ with $1$'s except for the label $k+1$ when $t$ is the $k$-th element in the increasing enumeration of $L$ at the rightmost cell $(u_t,{\lambda }_{u_t})$, and each row $t\mathrm{\notin }L$ with $1$'s. Then rows are weakly increasing and columns strictly increasing: in a column, the rows below the first all lie in $L$ by hypothesis, so their entries are labels increasing with the row, and the possible first entry $1$ is smaller than all of them. The content is correct because there are $s-1=r-x$ labels and $r-(r-x)=x$ ones, and the reading word is a lattice word: it begins with the $1$'s of the first row and the labels occur in increasing row order, so $j$ precedes $j+1$ and at least one $1$ precedes the $2$. The two constructions are inverse to each other, so the LR tableaux of shape $\lambda \mathrm{/}\nu$ and content $(x,1^{r-x})$ are in bijection with these admissible sets $L$.

Count the admissible sets. Let $f:=\mathrm{\#}\{t\in \{2,\dots ,w\}:I_{t-1}\cap I_t\mathrm{\ne }\mathrm{\varnothing }\}$ and ${\epsilon }_t:=1$ if $I_{t-1}\cap I_t\mathrm{\ne }\mathrm{\varnothing }$ and $0$ otherwise (so $f={\sum }_{t=2}^w{\epsilon }_t$). A row $t\ge 2$ is forced into $L$ exactly when ${\epsilon }_t=1$: if $j\in I_{t-1}\cap I_t$ then $j$ lies in the rows $t-1<t$ of column $j$, so step 2.1(b) gives $t\in L$; and if ${\epsilon }_t=0$ then for every $j\in I_t$ the row $t$ is the first row of column $j$ (a row $t^{\mathrm{\prime }}<t$ with $j\in I_{t^{\mathrm{\prime }}}$ would force $j\in I_{t-1}$ by column-convexity), so no constraint from step 2.1(b) applies to $t$. Hence the admissible sets are exactly the subsets of the $w-1-f$ free rows $\{2,\dots ,w\}\setminus \{t:{\epsilon }_t=1\}$ of size $r-x-f$; their number is $\left(\genfrac{}{}{0px}{}{w-1-f}{r-x-f}\right)$ when $0\le r-x-f\le w-1-f$ and $0$ otherwise. Now count incidences $(t,j)$ with $j\in I_t$: on one hand $r={\sum }_t\mathrm{\mid }{I}_t\mathrm{\mid }$; on the other hand ${\sum }_j\mathrm{\mid }\{t:j\in I_t\}\mathrm{\mid }=c+f$ because each column contributes $1$ for its first row and one further unit for each $t\ge 2$ with $t-1,t$ both rows of the column, and $\mathrm{\mid }{I}_{t-1}\cap I_t\mathrm{\mid }\le 1$ by step 2.1(a). Hence $f=r-c$. For the components: if $I_{t-1}\cap I_t$ contains $j$, then $u_t=u_{t-1}+1$, because an intermediate partition row $u_{t-1}<i<u_t$ would be a row of $[\lambda ]$ with ${\nu }_i={\lambda }_i\ge j$ (as $i<u_t$ and ${\lambda }_{u_t}\ge j$) while ${\nu }_i\le {\nu }_{u_{t-1}}<j$ (as $i>u_{t-1}$ and $j\in I_{t-1}$), a contradiction; so the cells $(u_{t-1},j)$ and $(u_t,j)$ are vertically adjacent and rows $t-1$ and $t$ lie in one component. Conversely, any path in $\lambda \mathrm{/}\nu$ joining rows $t-1$ and $t$ consists of horizontal steps within rows and vertical steps within a column between consecutive partition rows, and the only consecutive partition rows between the indices $t-1$ and $t$ are these two, so such a path exists only if ${\epsilon }_t=1$. Thus the components are the maximal blocks of consecutive indices joined by overlaps, and their number is $m=1+\mathrm{\#}\{t\ge 2:{\epsilon }_t=0\}=w-f$. Therefore the number of LR tableaux of shape $\lambda \mathrm{/}\nu$ and content $(x,1^{r-x})$ is $\left(\genfrac{}{}{0px}{}{m-1}{c-x}\right)$ (zero outside the range $0\le c-x\le m-1$), and it is $0$ when $\lambda \mathrm{/}\nu$ contains a $2\times 2$ block. This is claim 1, in the form in which the $m$ connected components of a $2\times 2$-free $\lambda \mathrm{/}\nu$ are the skew hooks.

For the alternating sum, first suppose that $\lambda \mathrm{/}\nu$ contains a $2\times 2$ block: then every coefficient is $0$ by step 3.1 and the left-hand side of claim 2 is $0$, while $\lambda \mathrm{/}\nu$ is not a connected border strip, so the right-hand side is also $0$. Otherwise $\lambda \mathrm{/}\nu$ is $2\times 2$-free with $m$ components, $c$ columns and $r$ nodes, and by step 3.1 the sum equals ${\sum }_{x=1}^r(-1{)}^{r-x}\left(\genfrac{}{}{0px}{}{m-1}{c-x}\right)$. The nonzero terms are exactly those with $0\le c-x\le m-1$, that is $x\in \{c-m+1,\dots ,c\}$; here $c\ge m$ because column-convexity puts all cells of each column in one component, so distinct components have disjoint columns and each occupies at least one column, and $c\le r$ because $r\ge 1$ and each column is nonempty, so all these $x$ lie in $\{1,\dots ,r\}$. Substituting $k=c-x$ gives ${\sum }_{k=0}^{m-1}(-1{)}^{r-c+k}\left(\genfrac{}{}{0px}{}{m-1}{k}\right)=(-1{)}^{r-c}{\sum }_{k=0}^{m-1}(-1{)}^k\left(\genfrac{}{}{0px}{}{m-1}{k}\right)$, which is $(-1{)}^{r-c}$ when $m=1$ and $0$ when $m\ge 2$ by the alternating row sum [F4]. A connected $2\times 2$-free $\lambda \mathrm{/}\nu$ with $r$ nodes and $c$ columns is exactly a connected border strip with those data [F5]; so the last display is claim 2.

For claim 3, apply the outer Littlewood–Richardson rule [F3] to $\mu =(x,1^{r-x})$ and multiply by $(-1{)}^{r-x}$: ${\sum }_{x=1}^r(-1{)}^{r-x}{\chi }^{\nu }\circ {\chi }^{(x,1^{r-x})}={\sum }_{x=1}^r(-1{)}^{r-x}{\sum }_{\lambda ⊢n}{c}_{\nu ,(x,1^{r-x})}^{\lambda }{\chi }^{\lambda }={\sum }_{\lambda ⊢n}[{\sum }_{x=1}^r(-1{)}^{r-x}{c}_{\nu ,(x,1^{r-x})}^{\lambda }]{\chi }^{\lambda }$, the exchange of the two finite sums being legitimate because each is finite and the character ring is abelian. By step 4.1 the bracket equals $(-1{)}^{r-c(\lambda \mathrm{/}\nu )}$ when $\lambda \mathrm{/}\nu$ is a connected border strip of size $r$ and $0$ for every other $\lambda$, so the double sum collapses to ${\sum }_{\lambda \mathrm{/}\nu }(-1{)}^{r-c(\lambda \mathrm{/}\nu )}{\chi }^{\lambda }$ over the removable border strips of size $r$; since $\mathrm{ht}(\lambda \mathrm{/}\nu )=r-c(\lambda \mathrm{/}\nu )$ by [F5], this is the displayed character identity of claim 3.