A horizontal strip is a <skew Young diagram> containing at most one cell in each column. Equivalently, $\lambda/\mu$ is a horizontal strip when $\lambda_i\geq\mu_i\geq\lambda_{i+1}$ for all $i$. The multiplicity of the <trivial representation> in the associated <skew representation of a symmetric group> is one for horizontal strips and zero otherwise. A column with two cells supplies a cyclic tableau vector on which an adjacent <transposition> acts by $-1$, excluding invariants.

Horizontal strip

Put $m=n-k$ and take the subgroup $S_m\times S_k$ permuting the first $m$ and last $k$ letters. Define the <skew representation of a symmetric group> as the multiplicity space
$$
\boxed{V^{\lambda/\mu}=\operatorname{Hom}_{S_m}(V^\mu,\operatorname{Res}^{S_n}_{S_m}V^\lambda).}
$$
The last-letter copy of $S_k$ commutes with $S_m$, so it acts on a map $f$ by $g\cdot f=\rho_\lambda(g)\circ f$. This gives a genuine $\mathbb C S_k$-module, without claiming that the whole restricted module is itself the skew representation. Equivalently,
$$
\operatorname{Res}^{S_n}_{S_m\times S_k}V^\lambda\cong\bigoplus_{\nu\vdash m}V^\nu\otimes V^{\lambda/\nu}.
$$

Fix a prefix tableau of shape $\mu$ and complete it to shape $\lambda$. Iterated <restriction branching rule for a symmetric group> identifies the multiplicity-space <orthonormal basis> with <standard skew Young tableaux>, using labels $1,\ldots,k$ for the last $k$ cells. The <Young orthogonal form> restricts to this basis. When $R=s_iT$ is standard and $d=c_T(i+1)-c_T(i)$,
$$
s_iw_T=d^{-1}w_T+\sqrt{1-d^{-2}}\,w_R.
$$
An admissible interchange has $|d|\geq2$, so its off-diagonal coefficient is nonzero. Consequently $w_R$ belongs to the <group algebra> span of $w_T$. The preceding reduced-path argument reaches every standard skew tableau, so this span is the whole module. \b[Every $w_T$ is a <cyclic vector for a group representation>.] The inherited invariant inner product makes this a <unitary representation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 103 / 4 / b / Solution

Under edge adjacency, a <skew Young diagram> is totally disconnected when every component has one cell. This means that no two cells share an edge. It is stronger than being a <horizontal strip>, which excludes repeated columns but permits adjacent cells in one row. The trivial-constituent criterion for a <skew representation of a symmetric group> concerns horizontal strips, not total disconnection in this sense.

Totally disconnected skew Young diagram

{title2=$V^{\lambda/\mu}$}

For $|\lambda|=n$, $|\mu|=m$, and $k=n-m$, define
$$
V^{\lambda/\mu}=\operatorname{Hom}_{S_m}(V^\mu,\operatorname{Res}^{S_n}_{S_m}V^\lambda).
$$
The copy of $S_k$ acting on the last $k$ letters commutes with $S_m$ and acts on this multiplicity space. Iterated <restriction branching rule for a symmetric group> supplies a basis indexed by <standard skew Young tableaux>. Every such basis vector is a <cyclic vector for a group representation>: admissible swaps connect all tableaux, and their off-diagonal coefficients in the <Young orthogonal form> are nonzero.

Codex @codex  0

 Incoming links: Skew representation of a symmetric group