Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2008 / B (Representation Theory of Symmetric Groups)

Use the convention that insertion returns the insertion tableau $P$ first and the recording tableau $Q$ second. Successive row insertions of $3,4,1,6,7,5,2$ giveFor example inserting five bumps six to the second row, and the final insertion of two bumps four from the first row and six from the second, creating the third row. The newly created boxes at these steps receive recording labels six and seven respectively.

The deletion input is this pair with the two tableaux interchanged. The symmetry property of RSK therefore gives the inverse permutation. One can also reverse-insert the boxes in descending recording-label order. The result isIndeed the displayed second row lists the positions of one through seven in the original permutation.

Let $f^{\lambda }$ be the number of standard tableaux of shape $\lambda$. First the associated polytabloids are linearly independent. Order tabloids lexicographically by reading each row after sorting its entries. For a column-increasing tableau $t$, the term $\{t\}$ in $e_t$ has coefficient one; every other tabloid term has larger row word. To see this, take the first row changed by a nontrivial column permutation. Earlier rows are fixed entry by entry, since each entry belongs to one original column. The changed row can only receive entries from lower positions of those increasing columns, so after sorting it is lexicographically larger. In a linear relation between standard polytabloids, choose the smallest row word with a nonzero coefficient. Its own tabloid coefficient cannot be cancelled by a larger standard polytabloid, a contradiction. Thus $\dim S^{\lambda }\ge f^{\lambda }$.

Work first over $\mathbb{C}$. The RSK bijection between permutations and pairs of standard tableaux of the same shape gives ${\sum }_{\lambda ⊢n}(f^{\lambda }{)}^2=n!$. The assumed pairwise nonisomorphic irreducible Specht modules, and the decomposition of the regular representation with irreducible multiplicities equal to their dimensions, give ${\sum }_{\lambda }(\dim S^{\lambda }{)}^2\le n!$. Combining the two inequalities forces equality term by term: $\dim S^{\lambda }=f^{\lambda }$. Hence the independent standard polytabloids are a basis.

The same conclusion holds over every characteristic-zero field. All polytabloid coordinates in the tabloid basis are integers, so the ranks of their coordinate matrices are the same over any characteristic-zero field as over $\mathbb{C}$. In particular the dimension equality and independence transfer without requiring an embedding of the whole field into $\mathbb{C}$.

The equality ${\sum }_{\lambda }(\dim S^{\lambda }{)}^2=n!$ exhausts the dimension of the regular representation. There is no room for another irreducible constituent, since it would contribute its positive squared dimension. ThusEach has the standard-polytabloid basis just proved. The usual split characteristic-zero statement follows over any characteristic-zero field because these integral Specht models remain irreducible after extension to an algebraic closure and already exhaust all dimensions there. In the literal wording of the source, polytabloids themselves are vectors; it is the Specht modules with those bases that form the family of representations.

If the cycle partition contains a part $r$, the Murnaghan–Nakayama rule isA border strip is a connected skew Young diagram with no $2\times 2$ block; its height is its number of occupied rows minus one. The remaining boxes must be a partition diagram. The recursion starts with character one for the empty diagram at the empty cycle type, and an empty sum is zero. Repeated removal for all cycle lengths computes the character value.

Write the hook cycle partition as $\mu =(k,1^{n-k})$, $1\le k\le n$. At the identity, the representation dimension is $n-1$: a standard tableau of shape $(2,1^{n-2})$ is determined by the entry in its top-right box, any one of $2,\dots ,n$.

For $2\le k\le n-2$, the only removable border strip of length $k$ is the vertical bottom strip. Its height is $k-1$, and the remaining hook has size $n-k$ and dimension $n-k-1$. A strip of length $n-1$ would leave the top-left box but disconnect the removed top-right box from the removed column, so none exists and the value is zero. For $k=n$, the whole hook is a border strip of height $n-2$ and contributes $(-1{)}^{n-2}$. These cases combine asIn particular the formula includes zero at $k=n-1$ and the full-cycle sign at $k=n$.

The two partition diagrams are transposes, so their characteristic-zero Specht modules differ by tensoring with the sign representation. A transposition has sign minus one. ConsequentlyThe middle character is $n-3$, since the standard representation is the point-permutation representation minus its trivial line, and a transposition fixes $n-2$ points.

The permutation module $M^{\lambda }$ has a basis consisting of row tabloids of shape $\lambda$: tableaux are identified when each row has the same set of entries. A permutation acts by applying it to every entry, and the action is extended linearly over $F$. Its dimension is $n!/{\prod }_i{\lambda }_i!$.

Here the two singleton rows form an ordered pair of distinct entries, soA standard tableau of the same shape must put one in its top-left corner. Choose two entries from $2,\dots ,n$ for the two remaining boxes of the first column, in increasing order; all other entries then fill the top row in increasing order. Thus the standard-polytabloid basis gives

Write $v_{ij}$ for the tabloid with second-row entry $i$ and third-row entry $j$. Equivariance gives $\varphi (v_{ij})=w_i+w_j$ for every distinct pair. For a tableau $t$, the only nontrivial column antisymmetrizer is on the three entries in its first column. In its alternating sum, pair each column permutation with the permutation obtained by interchanging the two singleton-row positions. Their images under $\varphi$ are the same unordered sum of two $w$'s, while their signs are opposite. Hence $\varphi (e_t)=0$ for every polytabloid, proving $\boxed{S^{(n-2,1,1)}\subseteq \ker \varphi }$.

For distinct $i,j,k$,Each summand is an image and two is invertible in characteristic zero. Thus every basis vector of $M^{(n-1,1)}$ is in the image, proving surjectivity. rank-nullity theorem then gives

The stabilizer of an ordered pair is the subgroup ${\Sigma }_{n-2}$ fixing both entries, so the permutation module is ${\mathrm{Ind}}_{{\Sigma }_{n-2}}^{{\Sigma }_n}1$. Induce in two stages through ${\Sigma }_{n-1}$. The first stage is $S^{(n-1)}\oplus S^{(n-2,1)}$. The characteristic-zero induction branching rule for a symmetric group adds one box in every possible partition position, with multiplicity one. Thus the two second-stage decompositions areBy transitivity of induction this yieldsThe dimensions are $1$, $2(n-1)$, $n(n-3)/2$ and $(n-1)(n-2)/2$, whose sum is $n(n-1)$, checking the decomposition also at $n=4$.

Give $M^{\lambda }$ the invariant symmetric bilinear form for which its tabloid basis is orthonormal. DefineIn characteristic $p>0$, this quotient is nonzero and simple when $\lambda$ is $p$-regular, meaning that no part occurs $p$ or more times. For other partitions it is zero, so the label “simple module” includes the nonzero/$p$-regular qualification.

The James submodule theorem states that for every $F{\Sigma }_n$-submodule $U\subseteq M^{\lambda }$, either $S^{\lambda }\subseteq U$ or $U\subseteq (S^{\lambda }{)}^{\perp }$. Applied to submodules of $S^{\lambda }$, it says that every proper submodule lies in the radical of the restricted form, giving the simplicity of the displayed nonzero quotient.

Choose the two standard tableauxand use their polytabloids $e_1,e_2$, with column-vector coordinates. Writing $[ab\mid cd]$ for the tabloid with those two row sets givesThe coefficients at $[12\mid 34]$ and $[13\mid 24]$ prove independence over every field. Applying the generators directly givesand $(34)$ acts in the same way as $(12)$. Their span is invariant under the generators, contains $e_1$, and contains its whole orbit, hence is the Specht module. Thus it is a basis without needing a characteristic-zero argument. The matrices areThese integral matrices may be reduced in the chosen characteristic.

For the characteristic-three Specht module for the two-by-two partition, reduce the matrices modulo three. The vector $e_1+e_2$ is fixed by every displayed transposition: $(12)$ and $(34)$ send it to $e_1-2e_2=e_1+e_2$, and $(23)$ interchanges its two terms. Thus its span is a proper nonzero trivial submodule.

On the quotient impose $e_2=-e_1$. Each generating transposition then acts as minus one, so the quotient is the sign representation. Consequentlyand the two composition factors are trivial and sign, each once. In the $D$ notation they are $D^{(4)}$ and $D^{(2,2)}$ respectively. Indeed the restricted tabloid form has Gram matrix $\left(\begin{array}{cc}4& 2\\ 2& 4\end{array}\right)$, whose characteristic-three radical is precisely the trivial line, so the simple quotient is $D^{(2,2)}$.

The extension does not split: the minus-one eigenspace of $\rho (12)$ is the line through $e_2$, but $\rho (23)$ sends that line to the line through $e_1$. Thus there is no sign submodule complementing the trivial line, consistent with the failure of semisimplicity.