Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2007 / B

A tabloid records the sets of entries in each row of a tableau. Let $C_t$ be the subgroup permuting entries independently within each column. The polytabloid isand the Specht module $S^{\lambda }$ is spanned by these vectors. A standard polytabloid is one whose tableau increases along rows and down columns.

Order tabloids by first sorting each row increasingly, then comparing the resulting row-reading words lexicographically. For a standard tableau $t$, every nonidentity column permutation produces a strictly larger word. In the first row affected by that permutation, each replacement comes from farther down the same column and is larger than the entry it replaces; all earlier rows remain unchanged. Sorting the affected row therefore increases its word. Also no nontrivial column permutation fixes its row sets, since a cell is the unique intersection of its row and column. Hence the least tabloid in $e_t$ is $\{t\}$ with coefficient one.

Distinct standard tableaux have distinct row tabloids. In a nonzero linear relation among their polytabloids, choose the least leading tabloid with a nonzero coefficient. None of the vectors with larger leading tabloids can contribute to it, so its coefficient cannot cancel. This contradiction proves linear independence of the standard polytabloids.

The hook length of a cell is one plus the number of cells to its right in the same row and below it in the same column. The hook-length formula isFor $\lambda =(n-k,k)$, with $0\le k\le n/2$, the second-row hook product is $k!$. In the first row, the hooks are $n-k-j+2$ for $1\le j\le k$ and $n-k-j+1$ for $k<j\le n-k$, so their product is $(n-k+1)!/(n-2k+1)$. ConsequentlyAt $k=0$, interpret the second binomial coefficient as zero.

For a cycle type $\mu =(m,\nu )$, the Murnaghan–Nakayama rule isThe sum is over partitions $\rho$ whose diagrams remain after removing the specified border strip; $\nu$ is the remaining cycle type, and the recursion ends with the empty character equal to one. At the identity, ${\chi }^{\rho }(1^{|\rho |})=f^{\rho }$. Incompatible total sizes contribute zero.

For shape $(2,2)$ and a transposition, a horizontal strip in the bottom row leaves $(2)$ with positive sign; a vertical strip in the right column leaves $(1,1)$ with negative sign. Both remaining dimensions are one. Thus

For shape $(n-2,2)$ with $n\ge 6$, there are exactly two removable rim hooks of size two: the bottom row, leaving $(n-2)$, and the final two cells of the first row, leaving $(n-4,2)$. Both have height zero. A vertical pair cannot be removed while leaving a Young diagram because the first row is longer than the second by at least two.

By the Murnaghan–Nakayama rule and part (b),

For an integer partition $\lambda =({\lambda }_1,\dots ,{\lambda }_r)$, its Young diagram has ${\lambda }_i$ left-justified cells in row $i$. A Young tableau of shape $\lambda$ is a filling of its cells. Under the usual standard numbering convention, a standard Young tableau uses each integer $1,\dots ,n$ once and increases strictly along every row and down every column. For a different fixed ordered set of distinct labels, the same inequalities define the corresponding ordered-label version, which becomes standard after replacing its labels by their ranks.

First convert the printed cycle to one-line notation:Apply row insertion to these values, recording the insertion time in each new cell. The successive insertion-tableau row lists areTherefore the Robinson-Schensted-Knuth correspondence givesBoth have shape $(4,2,1,1)$.

First prove preservation of the row and column inequalities for distinct ordered labels. If $b>c_l$, row insertion appends it to the first row, and the inequalities remain valid. Otherwise let $j$ be the first position with $c_j>b$. Replace $c_j$ by $b$ and insert the bumped $c_j$ in the second row. Its first larger entry is some $d_k$ with $k\le j$, because $d_j>c_j$. Replace $d_k$ by $c_j$ and put the bumped $d_k$ in the new third-row cell.

The first-larger-entry choices preserve strict row order. At column $k$, if $k<j$ the first-row entry is $c_k<c_j$, while if $k=j$ it is the new $b<c_j$. At column $j$ when $k<j$, the unchanged lower entry satisfies $d_j>c_j>b$. Every other old column keeps its previous inequalities. The new third-row entry is larger than the second-row first entry: for $k=1$ this is $d_1>c_j$, while for $k>1$ it follows from $d_k>d_1$. Thus the new shape $(l,l,1)$ is a Young diagram with strictly increasing rows and columns. This proves the insertion assertion for the ordered-label convention.

There is a numbering qualification in the literal standard convention of part (a). If $T$ uses exactly $1,\dots ,2l$, an arbitrary missing natural number $b$ need not be $2l+1$. For example, inserting six into rows $(1,2);(3,4)$ produces $(1,2,6);(3,4)$, which has the order inequalities but not the labels $1,\dots ,5$. Thus standardness with consecutive labels requires $b=2l+1$, or increasing relabelling afterward. The printed wording is retained in the private transcription; row insertion itself preserves the needed order structure.

The Robinson-Schensted-Knuth correspondence is a bijection between permutations and ordered pairs of standard Young tableaux of the same shape. Hence the class with shape $\lambda$ has size $(f^{\lambda }{)}^2$. For $\lambda =(4,2,1,1)$, the hook lengths arewith product $448$. The hook-length formula gives $f^{\lambda }=8!/448=90$. Therefore

The symmetric group acts on tabloids by permuting their entries; extending linearly gives the permutation module $M^{\mu }$. Its dimension isA tableau of type $\mu$ contains exactly ${\mu }_i$ copies of the label $i$. Fix a numbering of the $\lambda$-diagram's cells by $1,\dots ,n$. For a filling $T$, let the $i$th row set of an associated $\mu$-tabloid be the set of cell numbers whose labels in $T$ equal $i$. This is a bijection from type-$\mu$ fillings to $\mu$-tabloids.

Define $\sigma T$ by moving labels along the numbered positions: its label at cell $j$ is the old label at cell ${\sigma }^{-1}j$. The associated row set is then the image under $\sigma$ of the old row set. Thus the bijection is equivariant, and its linear extension makes $F[{\mathcal{T}}_{\lambda ,\mu }]$ and $M^{\mu }$ isomorphic representations. Extend from group elements to the group algebra by linearity.

Young's rule states that in characteristic zero,where the Kostka number $K_{\lambda \mu }$ counts Semistandard Young tableaux of shape $\lambda$ and type $\mu$: rows are weakly increasing, columns strictly increasing, and label $i$ occurs ${\mu }_i$ times. The summands are irreducible Specht modules, and the coefficients are nonnegative integer multiplicities.

For type $(1^n)$, all labels are distinct, so a semistandard tableau is precisely a standard Young tableau. Thus $K_{\lambda ,(1^n)}=f^{\lambda }$, and Young's rule givesThis is also the decomposition of the regular representation, since tabloids with one entry per row are naturally indexed by all permutations.

For shape and type both $\mu$, column strictness makes every entry in row $r$ at least $r$. All ${\mu }_1$ entries equal to one must therefore be in the first row, which has exactly that length; it is filled with ones. Remove that row and repeat after subtracting one from the remaining labels. Induction gives the unique filling whose $i$th row consists of $i$'s. Thus $K_{\mu \mu }=1$, and Young's rule proves that $S^{\mu }$ occurs exactly once.

More generally, if a semistandard filling of shape $\lambda$ and type $\mu$ exists, its entries at most $k$ must lie in the first $k$ rows. Counting them givesThese are the inequalities defining the dominance order on partitions. Hence $K_{\lambda \mu }>0$ implies $\boxed{\lambda \ge \mu }$, and the asserted restriction on summands follows from Young's rule.

We calculate the exact multiplicity without using Young's rule, via a two-subset incidence realization of a two-row Specht module. Let $V_2$ be the permutation representation on unordered pairs from a six-element set, and $V_1$ the natural permutation representation on its singletons. The equivariant incidence mapis surjective: the transpose map sends $\sum c_i{e}_i$ to the pair coefficients $c_i+c_j$, and all such sums can be zero only when every $c_i=0$ in characteristic zero. Thus $\dim \ker D=15-6=9$.

Identify the tabloids of shape $(4,2)$ with their two-element bottom rows. A polytabloid is a signed four-edge vector $e_{cd}-e_{ad}-e_{cb}+e_{ab}$, whose incidence at every vertex is zero. Hence $S^{(4,2)}\subseteq \ker D$. The standard polytabloid basis theorem and the hook-length formula give $\dim S^{(4,2)}=\left(\genfrac{}{}{0ex}{}{6}{2}\right)-\left(\genfrac{}{}{0ex}{}{6}{1}\right)=9$, so equality holds. By Maschke's theorem,Take $H={\Sigma }_3\times {\Sigma }_2\times {\Sigma }_1$, preserving blocks of sizes three, two and one. It has three orbits on singletons and five on pairs: pairs inside the first block, inside the second, and between each of the three distinct pairs of blocks. Thus $\dim V_1^H=3$ and $\dim V_2^H=5$. Averaging over $H$ makes taking fixed vectors exact in characteristic zero, so $\dim (S^{(4,2)}{)}^H=2$.

Finally $M^{(3,2,1)}={\mathrm{Ind}}_H^{{\Sigma }_6}1$. Frobenius reciprocity, and equality of invariant and coinvariant dimensions in characteristic zero, identify the required multiplicity with this two-dimensional fixed space. Thereforewhich in particular satisfies the requested lower and upper bounds.

The simple modules are indexed by the p-regular partitions $\lambda$ of $n$, meaning no part is repeated $p$ or more times. Equip the tabloid permutation module with the bilinear form invariant under the symmetric group making its tabloid basis orthonormal. The construction isThe standard modular Specht theorem says this quotient is nonzero precisely for p-regular $\lambda$, is then simple, and these quotients give all simple modules.

To verify distinctness of labels, use the standard dominance triangularity theorem: if $D^{\nu }$ occurs as a composition factor of $S^{\lambda }$, then $\nu \ge \lambda$; for a p-regular label the diagonal multiplicity is one. If $D^{\lambda }\cong D^{\mu }$, then $D^{\lambda }$ occurs in $S^{\mu }$, so $\lambda \ge \mu$, and the symmetric argument gives $\mu \ge \lambda$. Antisymmetry of the dominance order on partitions yields $\boxed{\lambda =\mu }$. These are the general representation-theoretic results allowed in this part.

Realize $S^{(n-1,1)}$ as the augmentation submoduleof the natural permutation module: its polytabloids are the differences $e_i-e_j$, which span exactly $W$. Let $U$ be a submodule. If it contains a vector $v$ with two unequal coordinates $v_i,v_j$, thenThe coefficient is nonzero, so $e_i-e_j\in U$; its symmetric-group orbit spans $W$, forcing $U=W$. Therefore every proper nonzero submodule consists only of constant-coordinate vectors. Those form at most a one-dimensional trivial submodule. If such a trivial submodule is absent, no proper nonzero submodule exists, so the Specht module is simple.

On a one-dimensional representation, $s_i$ acts by a nonzero scalar ${\epsilon }_i$. Its square is one. The braid relation becomes ${\epsilon }_i^2{\epsilon }_{i+1}={\epsilon }_{i+1}^2{\epsilon }_i$, and hence ${\epsilon }_{i+1}={\epsilon }_i$. Thus all adjacent transpositions act by one common scalar $\epsilon$ with ${\epsilon }^2=1$.

In characteristic different from two, $\epsilon =1$ or $-1$. Since adjacent transpositions generate the group, these choices give exactly the trivial representation and sign representation. For $n\le 1$ they coincide, leaving only the trivial representation.

In characteristic three with $n=5$, a constant vector $(c,c,c,c,c)$ belongs to the augmentation submodule only if $5c=0$, which forces $c=0$. Thus $S^{(4,1)}$ has no trivial submodule. Part (b) proves that it is simple, of dimension four, and therefore equals $D^{(4,1)}$.

The standard polytabloid basis is defined over the integers and remains a basis after reduction modulo three. Thus $\dim S^{(3,2)}=f^{(3,2)}=\left(\genfrac{}{}{0ex}{}{5}{2}\right)-\left(\genfrac{}{}{0ex}{}{5}{1}\right)=5$. The given submodule has dimension four by part (d), so the quotient is one-dimensional and part (c) leaves only the trivial or sign action.

A transposition has ordinary trace one on $S^{(3,2)}$: the Murnaghan–Nakayama rule has only the horizontal bottom-row two-strip, leaving shape $(3)$ with dimension one. Reduction of its integral action matrix preserves this trace in $F$. Its trace on $D^{(4,1)}=S^{(4,1)}$ is $3-1=2$, since the natural five-point permutation representation has three fixed coordinates and contains the trivial line. Therefore its trace on the one-dimensional quotient is $1-2=-1$, distinct from one in characteristic three. This gives the characteristic-three sign quotient of the Specht module of shape three two. The quotient is the sign representation: