Young permutation module ($M^{\lambda }$)

The Young permutation module is the permutation module on the $\lambda$-tabloids.

For $n\ge 3$, let $S_n$ permute the coordinates of $M=F^n$. Write $T$ for the constant vectors and $U=S^{(n-1,1)}$ for the augmentation submodule. These and zero and $M$ are all the submodules. Indeed a vector with two unequal coordinates yields a nonzero multiple of $e_i-e_j$ by subtracting its image under $(i,j)$; the orbit of such a difference spans $U$. A submodule without such a vector lies in $T$. If the characteristic does not divide $n$, then $M=T\oplus U$ and $U$ is simple. If it divides $n$, the lattice is the chain $0<T<U<M$, with $U/T$ simple of dimension $n-2$ and trivial end factors. The direct-sum projection in the first case is $x\mapsto x-({\sum }_i{x}_i)1/n$.

A tabloid is fixed by a permutation exactly when each of its cycles lies wholly within one row. Assigning the distinct length-$q$ cycles to rows gives the displayed coefficient. More explicitly, sum ${\prod }_q{m}_q!/{\prod }_j{a}_{qj}!$ over nonnegative integers $a_{qj}$ with ${\sum }_j{a}_{qj}=m_q$ and ${\sum }_{q}q{a}_{qj}={\lambda }_j$. Equal-length rows remain distinguished. Thus the character counts fixed ordered row sets, rather than unordered set partitions.

Over the complex numbers, the signed Young permutation module isThe isomorphism is the tensor identity for induced representations. The partition index is unchanged by this twist; conjugation instead changes the labels of its irreducible constituents. For the conjugate partition ${\lambda }^{\prime }$, the character inner product of $M^{\lambda }$ and ${\tilde{M}}^{{\lambda }^{\prime }}$ is one: Mackey restriction formula reduces it to the unique zero-one matrix with row margins $\lambda$ and column margins ${\lambda }^{\prime }$.

Over the complex numbers,where the Kostka number $K_{\alpha \lambda }$ counts semistandard tableaux of shape $\alpha$ and content $\lambda$. It is nonzero only when $\alpha$ dominates $\lambda$.

Let $M(\mu ,\lambda )$ be the multiplicity of $V^{\mu }$ in the complex Young permutation module $M^{\lambda }$. For $\lambda ⊢n$ and $\rho ⊢n-1$,Here $\nearrow$ means adding one cell, and zero parts are omitted. Equal row lengths contribute separately on the right. Restrict $M^{\lambda }$ to $S_{n-1}$ and separate the tabloid orbits by the row containing $n$ to obtain the right side. Decompose into irreducibles and use the restriction branching rule for a symmetric group to obtain the left side. Equating multiplicities proves the formula.

For $0\le k\le n/2$, over the complex numbers,By Young's rule, the multiplicity is the Kostka number for content $(n-k,k)$. A semistandard Young tableau on just the entries $1,2$ has at most two rows. For shape $(n-j,j)$, every lower entry is $2$ and every entry above it is $1$; the remaining top row is uniquely fixed by the content. Such a filling exists precisely for $j\le k$. Thus each displayed Specht module occurs once. A zero second part is omitted.

The Kostka number $K_{\alpha \lambda }$ is the number of semistandard Young tableaux of shape $\alpha$ and content $\lambda$: rows are weakly increasing and columns are strictly increasing.

Over the complex numbers, the multiplicity of $S^{\mu }$ in the Young permutation module $M^{\lambda }$ equals $K_{\mu \lambda }$. Its character is ${\mathrm{Ind}}_{S_{\lambda }}^{S_n}1$, so Frobenius reciprocity expresses this multiplicity as the average of the restricted irreducible character over the Young subgroup. For $\lambda =(2,1^{n-2})$ that subgroup contains only the identity and one transposition, giving $({\chi }_{\mu }(1)+{\chi }_{\mu }((12)))/2$. For $\lambda =(1^n)$ it is trivial, giving the character degree. This supplies a calculation from ordinary character values without enumerating semistandard tableaux. An indexing convention writing the multiplicity as $m_{\lambda ,\mu }$ reverses the conventional Kostka indices.

For $n\ge 4$, identify $M^{(n-2,1^2)}$ with the permutation module on ordered pairs $(i,j)$ with $i\ne j$. Let $p_1(i,j)=e_i$, $p_2(i,j)=e_j$, and $q(i,j)=\{i,j\}$. With $U=\ker p_1$ and $V=U\cap \ker p_2$, one has a Specht filtrationwhose successive quotients areThe same construction for $n=3$ omits the zero $S^{(n-2,2)}$ factor.