Young tableau

A Young tableau of shape $\lambda$ is a bijective filling of the cells of its Young diagram by $1,\dots ,n$.

The row stabilizer $R(t)$ permutes entries within each row, while the column stabilizer $C(t)$ permutes entries within each column. Their intersection is trivial.

A tabloid is an equivalence class of tableaux under permutations within rows. The tabloids of shape $\lambda$ form a transitive $S_n$-set.

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

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 $\lambda$.

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.

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.

For a tableau $t$, its polytabloid is

The column antisymmetrizer of a Young tableau $t$ is the group algebra elementOn the Young permutation module it has one-dimensional image $\mathbb{F}e(t)$.

For tableaux $v,w$ of the same shape, $b_v\{w\}\ne 0$ exactly when no row of $w$ contains two entries from one column of $v$. In that case a column permutation $h\in C(v)$ satisfies $h\{v\}=\{w\}$, and $b_v\{w\}=\pm e(v)$.

The Specht module $S^{\lambda }$ is the span of the polytabloids of shape $\lambda$. It is cyclic, generated by any one polytabloid, and over $\mathbb{C}$ it is irreducible.

Let $U$ be a submodule of the Young permutation module $M^{\lambda }$ over any field. Then eitherfor the tabloid bilinear form. The key identity is $b_{t}u=⟨u,e(t)⟩e(t)$: if one pairing is nonzero, the cyclic generator $e(t)$ and hence the whole Specht module lies in $U$.

Over a field of characteristic zero,The map from $M^{{\lambda }^{\prime }}$ sending the tabloid of $g{t}^{\prime }$ to $ge(t)\otimes ge(u)$ is a surjection whose kernel is $(S^{{\lambda }^{\prime }}{)}^{\perp }$.

Over the complex numbers, restriction from $S_n$ to $S_{n-1}$ is multiplicity-free:Equivalently, one removes one Removable node of a Young diagram in every possible distinct way.

For $\lambda ⊢n$ with $n\ge 2$,Indeed, adjoining the trivial character to the standard character gives the point-permutation character, and Frobenius reciprocity turns its multiplicity in the self-product into the norm of the multiplicity-free restriction.

Order tabloids lexicographically by the row containing $1$, then the row containing $2$, and so on. For a standard tableau $t$, the tabloid $\{t\}$ occurs with coefficient one in $e(t)$ and precedes every other tabloid occurring in it. Distinct standard tableaux have distinct leading tabloids, so their polytabloids are linearly independent.

The degree of the complex irreducible character labelled by $\lambda ⊢n$ is

If a permutation has a $k$-cycle and remaining cycle type $\rho$, thenwhere the sum is over removable rim hooks of length $k$.

The parity of the sum of the leg lengths in any sequence that removes all $e$-hooks from $\lambda$ is independent of the sequence. Its sign ${\epsilon }_e(\lambda )$ is therefore well defined and supplies the common sign in repeated applications of the Murnaghan–Nakayama rule.

For even $n$, the virtual charactervanishes on every permutation having an odd cycle. Under the Frobenius characteristic map, the Jacobi–Trudi identity identifies its characteristic with the degree-$n$ part of $H(t)H(-t)$, which contains only products of even-indexed power sums.

At a permutation whose cycle lengths are the principal hook lengths of $\lambda$, the Murnaghan–Nakayama rule has a unique complete removal sequence. Consequently ${\chi }^{\lambda }$ has value $1$ or $-1$ there.

For a partition $\lambda$, the irreducible character ${\chi }^{\lambda }$ vanishes on every cycle type containing an even part exactly when $\lambda$ is a staircase $(m,m-1,\dots ,1)$. A staircase has only odd hook lengths. Conversely, vanishing first forces $\lambda$ to be self-conjugate; applying the Murnaghan–Nakayama rule at the largest even hooks then forces consecutive row lengths.

Conjugating a Young diagram twists its complex Specht module by the sign representation:

For the alternating expression ${\psi }^{\lambda }$ indexed by an integer composition, swapping two adjacent entries of $\lambda -\mathrm{id}$ negates ${\psi }^{\lambda }$. If two entries coincide the expression vanishes; otherwise sorting produces, up to sign, the irreducible character indexed by the resulting partition.

Over a field of characteristic $p$, the invariant tabloid form on $S^{\lambda }$ may be degenerate. Its radical is $S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$.

The tabloid bilinear form makes the tabloid basis orthonormal and restricts to an invariant bilinear form on each Specht module. In positive characteristic of a field, its radical controls the corresponding simple quotient.

A partition is $p$-regular when no part occurs $p$ or more times.

For a $p$-regular partition,is nonzero and absolutely irreducible. Distinct $p$-regular partitions label nonisomorphic simple modules.

If $\lambda =(n^{a_n},\dots ,1^{a_1})$ is $p$-regular, thenFor a tableau $t$ and its row reversal $t^{\ast }$,which is nonzero in characteristic $p$. Applying a column antisymmetrizer to an endomorphism at $e(t^{\ast })$ therefore forces its value on the cyclic generator $e(t)$ to be scalar.