Representation theory of the symmetric group

The representation theory of $S_n$ is organized by a partition of an integer. Over the complex numbers, partitions label the irreducible Specht modules; over positive-characteristic fields, the simple modules are labelled by regular partitions.

The sign representation of $S_n$ is the one-dimensional representation in which a permutation acts by its sign. Tensoring a representation by it multiplies its character value at $g$ by $\mathrm{sgn}(g)$.

For a subgroup $H\le G$, a character $\chi$ of $G$ and a character $\varphi$ of $H$,It follows directly from the induced-character formula because $\chi (g)=\chi (x^{-1}gx)$.

Let $n$ be prime. If the Kronecker product ${\chi }^{\alpha }{\chi }^{\beta }$ of two irreducible characters of $S_n$ is irreducible, then one of $\alpha ,\beta$ is $(n)$ or $(1^n)$. Comparing the two self-products shows that only one may contain the standard character; the restriction branching rule then makes one partition rectangular, and primality makes that rectangle a single row or column.

For the standard inclusions $S_2,S_3\le S_4$,

A partition $\lambda ⊢n$ is a weakly decreasing sequence ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge 0$ with sum $n$.

The generating function for partitions whose parts are at most $e$ isConjugating Young diagrams shows that it also counts partitions with at most $e$ parts. Multiplication by $x^e$ gives the generating function for partitions with exactly $e$ positive parts.

A self-conjugate partition is determined by the hook lengths of its diagonal cells. These are distinct odd positive integers, and every partition into distinct odd parts arises uniquely this way. Hence their generating function is

The Young diagram of $\lambda$ is the set of cells $(i,j)$ with $1\le j\le {\lambda }_i$.

A removable node is a cell whose deletion leaves a Young diagram. The set ${\lambda }^{-}$ consists of the distinct partitions obtained by deleting one removable node from $\lambda$.

The conjugate partition ${\lambda }^{\prime }$ is obtained by reflecting the Young diagram across its main diagonal; thus ${\lambda }_j^{\prime }$ is the number of parts of $\lambda$ at least $j$.

The hook $H_{i,j}(\lambda )$ consists of $(i,j)$, the cells to its right in row $i$, and the cells below it in column $j$. Its hook length is

For $(i,j)\in Y(\lambda )$, the integers from one through $h_{i,j}$ split as the disjoint unionTracing the southeast boundary of the hook records the first set at horizontal steps and the second at vertical steps.

If a partition has a hook of length $ef$, it has a hook of length $e$. In a beta set, the first hook is a bead-gap pair at distance $ef$; along the arithmetic progression with step $e$, some consecutive pair changes from a bead to a gap.

A rim hook, or border strip, is a connected skew Young diagram containing no $2\times 2$ square. Its leg length is one less than its number of occupied rows.

The content of the cell $(i,j)$ is $c_{i,j}=j-i$. Transposing the diagram negates every content and preserves every hook length.

The principal hook lengths are $h_{1,1},h_{2,2},\dots$ along the main diagonal. They are strictly decreasing positive integers, and their sum is the size of the partition.

For $m\ge \ell (\lambda )$, a beta set isA hook of length $h$ corresponds to a bead $b$ and an empty position $b-h$; removing the hook moves the bead to that gap.

Let $X=\{h_1,\dots ,h_m\}$ be a beta set and let $H_i(\lambda )$ be the hook lengths in row $i$. ThenThus hooks are exactly bead-gap pairs on the partition abacus.

An $e$-runner abacus arranges the nonnegative integers by their residues modulo $e$ and places beads at the positions in a beta set. Moving a bead up one place on its runner removes an $e$-hook.

The $e$-core is obtained by repeatedly removing hooks of length $e$. On an $e$-runner abacus it is obtained by sliding every bead as high as possible, which proves independence of the order of removals.

The $e$-weight is the number of $e$-hooks removed to reach the $e$-core. It satisfiesand equals the total size of the partitions in the $e$-quotient.

Removing a rim hook of length two preserves the number of odd hook lengths minus the number of even hook lengths. If the 2-core is the staircase $(r,r-1,\dots ,1)$, every one of its $r(r+1)/2$ hooks is odd. Hence the difference for the original partition is

The $e$-quotient $Q_e(\lambda )=({\lambda }^{(0)},\dots ,{\lambda }^{(e-1)})$ is the tuple of partitions represented by the individual runners after their positions are divided by $e$.

Hooks of $\lambda$ whose lengths are divisible by $e$ correspond bijectively to hooks in the $e$-quotient. A bead-gap pair on one runner with distance $eh$ becomes a bead-gap pair of distance $h$ in that runner partition, and hook removal commutes with this correspondence.

The $e$-quotient tower recursively takes the $e$-quotient of every partition at the preceding level. For $e>1$, the sum of the sizes at each new level is at most $1/e$ times the preceding sum, so every partition has finite depth. For $e=1$, the quotient is the original partition and the tower has finite depth only for the empty partition.

With runners ordered by residues $0,1,2,3$, an abacus for $(3,1)$ gives

The nonempty levels of the 2-quotient tower of $(3,1)$ areand

For every $e,k\ge 1$, the $e^k$-quotient $Q_{e^k}(\lambda )$ is a permutation of level $k$ of the $e$-quotient tower. Writing a runner residue modulo $e^k$ in base $e$ shows that taking one quotient chooses one digit at a time; iteration may reverse the order of those digits but selects the same runner partitions.

The $p$-core tower places at level $r$ the $p$-cores of all partitions at level $r$ of the quotient tower of a partition. If $q_r$ is the sum of the sizes at quotient level $r$ and $c_r$ the sum at core level $r$, then

If $n=|\lambda |$ and $d_p(n)$ is its base-$p$ digit sum, thenThis combines the Hook-length formula, Legendre formula, the abacus divisible-hook correspondence, and the recurrence $q_r=c_r+p{q}_{r+1}$.

For every partition $\lambda$,The core-tower formula reduces this to subadditivity of the base-$p$ digit sum across the sizes of the quotient partitions.

The $e$-residue of a cell $(i,j)$ is $j-i$ modulo $e$, and the $e$-content is the multiset of cell residues. Removing an $e$-hook removes exactly one cell of each residue. Among partitions of a fixed size, two partitions have the same $e$-content exactly when they have the same $e$-core.

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.

For partitions of the same integer, $\lambda$ dominates $\mu$ whenfor every $r$.

If an irreducible character has $p$-defect zero, it vanishes on every element whose order is divisible by $p$. Consequently, $\chi (g)\ne 0$ implies that the square-free part of the order of $g$ divides $|G|/\chi (1)$.

For $\chi \in \mathrm{Irr}(S_n)$, if the order of $g\in S_n$ does not divide $|S_n|/\chi (1)$, then $\chi (g)=0$. The Hook-length formula turns the co-degree into a product of hook lengths, while the Murnaghan–Nakayama rule detects a cycle whose required prime-power hook cannot be removed.