The vertical Pieri rule

Given: AC, $V={\mathbb{C}}^r$ with basis $e_1,\dots ,e_r$, a partition $\lambda$ with $\mathrm{\ell }(\lambda )\le r$, and an integer $d\ge 0$.

The one-column Specht module $S^{(1^d)}$ is the sign representation: its column stabilizer is all of $S_d$, and the signed sum of its distinct tabloids spans a line on which every permutation acts by its sign. Therefore $S_{(1^d)}(V)$ is the subspace of alternating tensors. It is isomorphic to the quotient exterior power in Symmetric and exterior powers over an arbitrary field via $v_1\wedge \cdots \wedge v_d\mapsto d{!}^{-1}{\sum }_{\sigma \in S_d}\mathrm{sgn}(\sigma )\sigma (v_1\otimes \cdots \otimes v_d)$. The displayed multilinear map vanishes when two inputs agree (pair permutations by their transposition), so it factors through the quotient. Conversely the quotient of this signed average is the original wedge, since a transposition changes a wedge's sign by expanding a repeated input $u+v$; the signed average fixes every alternating tensor. These are inverse equivariant maps (Schur modules and their characters, Column antisymmetrizers, polytabloids, and Specht modules). For $d=0$, $(1^0)$ means $\mathrm{\varnothing }$ and both spaces are $\mathbb{C}$; for $d>r$ both spaces vanish (If $k>\dim V$, then ${\mathrm{\Lambda }}^{k}V=0$).

Littlewood--Richardson rule: for partitions $\lambda ,\mu$ with $\mathrm{\ell }(\lambda ),\mathrm{\ell }(\mu )\le r$, $S_{\lambda }(V)\otimes S_{\mu }(V)\cong {⨁}_{\nu :\mathrm{\ell }(\nu )\le r}{S}_{\nu }(V{)}^{\oplus c_{\lambda \mu }^{\nu }}$, where $c_{\lambda \mu }^{\nu }$ is the number of LR tableaux of shape $\nu \mathrm{/}\lambda$ and content $\mu$, vanishing unless $\lambda \subseteq \nu$ and $\mathrm{\mid }\nu \mathrm{\mid }=\mathrm{\mid }\lambda \mathrm{\mid }+\mathrm{\mid }\mu \mathrm{\mid }$ (The Littlewood--Richardson tensor-product rule, Littlewood--Richardson tableaux and coefficients).

Let $U$ be a semistandard skew tableau of shape $\nu \mathrm{/}\lambda$ and content $(1^d)$, i.e. with entries $1,2,\dots ,d$ each occurring once. Its reading word $w(U)$ is a permutation of $1,\dots ,d$; it is a lattice word exactly when $w(U)=12\cdots d$, because the first letter of a lattice word of content $(1^d)$ must be $1$, and inductively the $k$-th letter must be $k$. If $\nu \mathrm{/}\lambda$ contains two boxes in the same row, at columns $c<c^{\mathrm{\prime }}$, then the right cell is read before the left cell in the reading order, while semistandardness gives the strictly smaller entry on the left, so in the reading word the larger entry $T(i,c^{\mathrm{\prime }})$ precedes the smaller entry $T(i,c)$ and the word is not $12\cdots d$. Hence an LR tableau of content $(1^d)$ exists only if $\nu \mathrm{/}\lambda$ is a vertical strip; conversely, if $\nu \mathrm{/}\lambda$ is a vertical strip, filling the boxes with $1,2,\dots ,d$ in the order in which they are read (equivalently, from top row to bottom row, since each row has at most one box) makes every column strictly increasing downward and gives the reading word $12\cdots d$, so the filling is the unique LR tableau of shape $\nu \mathrm{/}\lambda$ and content $(1^d)$ (Skew diagrams and semistandard skew tableaux, Semistandard tableaux and Kostka numbers, Littlewood--Richardson tableaux and coefficients).

The elementary symmetric polynomial $e_d={\sum }_{1\le i_1<\cdots <i_d\le r}{x}_{i_1}\cdots x_{i_d}$ equals $s_{(1^d)}$ by the one-column tableau expansion; $e_0=1$ and $e_d=0$ for $d>r$. It is the character of ${\mathrm{\Lambda }}^{d}V$, and the Schur characters $s_{\nu }$, $\mathrm{\ell }(\nu )\le r$, are linearly independent (Semistandard tableaux expand Schur characters, Stable Schur functions from bialternants, The Littlewood--Richardson tensor-product rule).