Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 103 (Representation Theory)

Work over the complex numbers. The chain $S_0\subset S_1\subset \cdots \subset S_n$ has multiplicity-free restriction: each irreducible representation restricts to a direct sum of pairwise inequivalent irreducible representations. This structural fact can be proved before identifying the branching diagram. Indeed, the permitted Olshanskii centralizer lemma makes $C_{\mathbb{C}{S}_r}(\mathbb{C}{S}_{r-1})$ commutative, since it is generated by the center at the previous level and the commuting element $X_r$. In each irreducible block this is the endomorphism algebra of the restricted module; a repeated summand would give a noncommutative matrix algebra factor. Thus no identification of the branching graph with Young diagrams is being assumed here.

Successively restricting an irreducible representation $V$ gives one-dimensional spaces $V_T$ indexed by paths of irreducible representations from the trivial $S_0$-module to $V$. Choosing one nonzero vector in each $V_T$ gives a Gelfand–Tsetlin basis. Define the Gelfand–Tsetlin algebra as the subalgebra of $\mathbb{C}{S}_n$ acting diagonally in all these bases. The Artin–Wedderburn theorem identifiesUnder this identification the Gelfand–Tsetlin algebra is the direct sum of the full diagonal matrix algebras in the indicated bases, so it is commutative.

To see that it is actually available inside the group algebra, let $e_{\alpha }^{(r)}$ be the central primitive idempotent selecting the irreducible representation $\alpha$ of $S_r$. For a path $T=({\alpha }_0,\dots ,{\alpha }_n)$ the productis the projection onto $V_T$ in its final irreducible representation and is zero in every other final block. These products commute: a center at a higher level commutes with every element at a lower level. The $P_T$ are precisely the diagonal matrix units, and therefore span the proposed algebra.

If an element of the group algebra commutes with every $P_T$, its matrix preserves every $V_T$ and is diagonal in every block. It already belongs to the Gelfand–Tsetlin algebra. Thuswhich proves that it is a maximal commutative subalgebra.

The projections $P_T$ in the Gelfand–Tsetlin algebra satisfyConsequently the map sending a diagonal element to its scalar on each Gelfand–Tsetlin basis vector is a ring isomorphismEach factor is a simple module over this algebra, and the algebra is their direct sum as a module over itself. This proves that $G{Z}_n$ is a semisimple algebra. The argument uses its explicit diagonal structure; being a subalgebra of a semisimple algebra by itself would not suffice.

The Young–Jucys–Murphy elements arewith products of permutations composed from right to left. Fix $i<j$. In $[X_i,X_j]$, all pairs of disjoint transpositions commute. The only terms left, for each $a<i$, areIndeed $(ai)(ij)=(aj)(ai)$ and $(ij)(ai)=(ai)(aj)$. Summing directly provesThus all the Young–Jucys–Murphy elements commute.

The primitive idempotent products used above show that the Gelfand–Tsetlin algebra is generated by the centers at all levels: those products give every diagonal matrix unit, whereas each center acts diagonally. Hence

For the second generation assertion use the permitted Olshanskii centralizer lemma, in the precise formIn particular $Z_r\subseteq \mathrm{alg}(Z_{r-1},X_r)$. Induction puts every $Z_r$ in $\mathrm{alg}(1,X_1,\dots ,X_r)$. Conversely, if $A_r$ is the sum of all transpositions in $S_r$, then $A_r\in Z_r$ and $X_r=A_r-A_{r-1}\in G{Z}_n$. ThereforeHere generation is unital; the zero generator $X_1$ causes no problem.

The deletion projection for a symmetric group removes $n$ from the cycle containing it. A cycle $(a_1\cdots a_{r}n)$ becomes $(a_1\cdots a_r)$, with a singleton understood as a fixed point; all other cycles remain unchanged. This describes a genuine permutation of $\{1,\dots ,n-1\}$ and immediately proves the two identity assertions.

Put $H=S_{n-1}$. For $\sigma ,\theta \in H$ and $k<n$, if $\pi (\theta (k))\ne n$, deletion gives $\sigma \pi (\theta (k))$. If $\pi (\theta (k))=n$, it gives $(\sigma \pi \theta )(n)=\sigma \pi (n)$. These are exactly the two cases in $\sigma {\pi }_n\theta (k)$. Thus

The printed uniqueness assertion has an exceptional case at $n=4$. To locate it, use the two double cosetsAn $H$-bimodule-equivariant set map is determined by $u=\Phi (1)$ and $v=\Phi (t)$. Equivariance requires $u\in Z(H)$ andbecause $H\cap tHt=S_{n-2}$ and $t$ commutes with this subgroup. These conditions are also sufficient: define $\Phi (h)=hu$ and $\Phi (ht{h}^{\prime })=hv{h}^{\prime }$. Changing the double-coset expression only inserts an element of $S_{n-2}$, which commutes with $v$.

For $n\ge 5$, the center of $H$ and this centralizer are both trivial. For the latter, a commuting permutation must fix the unique point outside $\{1,\dots ,n-2\}$ and lie in the center of $S_{n-2}$; that center is trivial when $n-2\ge 3$. Therefore $u=v=1$, giving deletion. For $n=4$, however, $C_{S_3}(S_2)=\{1,(12)\}$. The choiceis well defined and satisfies all the printed equivariance conditions, but sends $(34)$ to $(12)$ rather than to the identity. Uniqueness holds for $n\ge 5$; it is false at $n=4$ as printed. Even requiring the map to restrict to the identity on $H$ does not remove this counterexample.

There is a separate problem with the printed linear-span assertion, when $Z_{((n-1),1)}$ denotes the usual centralizer of a subalgebra. Let $A_{n-1}$ be the sum of all transpositions in $H$. Direct expansion in the group algebra givesThus $Y=X_n^2-2A_{n-1}$ commutes with $H$ and satisfies ${\Pi }_n(Y)=(n-1)1$. For $n\ge 3$ it contains nonzero coefficients on three-cycles involving $n$, whereas every element of $\mathrm{span}\{1,X_n\}$ is supported only on the identity and transpositions. Therefore $Y$ is an explicit counterexample to the asserted equality.

One correct description of the entire preimage is obtained by putting $C=C_{\mathbb{C}{S}_n}(\mathbb{C}H)$. Equivariance gives ${\Pi }_n(C)\subseteq Z(\mathbb{C}H)$, and ${\Pi }_n$ is the identity on that center. It follows thatA useful corrected two-dimensional statement restricts the support to the identity and the transpositions $(in)$: invariance under conjugation by $H$ then forces their coefficients to be equal, yielding exactly $\mathrm{span}\{1,X_n\}$. This additional support restriction is absent from the PDF.

Use the Young–Jucys–Murphy elements in the group algebra $\mathbb{C}{S}_n$. Their joint spectrum isEquivalently one can use the regular representation, which contains every irreducible representation. The commuting elements are self-adjoint in a unitary representation, so there is a simultaneous eigenbasis.

A standard Young tableau has entries increasing along each row and down each column. If entry $r$ occupies cell $(i_r,j_r)$, its Content of a Young-diagram cell is $c_T(r)=j_r-i_r$. Define the set of content vectors of standard Young tableaux byThe shape here is a partition of an integer; it is not itself the vector of eigenvalues. For $n=1$ both sets consist of $(0)$.

The first Young–Jucys–Murphy element is the empty sum $X_1=0$. Applying it to the nonzero simultaneous eigenvector defining a point of the joint spectrum gives $0=a_{1}v$. HenceFor the remaining spectral arguments use the following local facts, without assuming the restriction branching rule for a symmetric group: consecutive eigenvalues are distinct; a difference $a_{r+1}-a_r=\pm 1$ makes $s_r=(rr+1)$ act on the corresponding line as $\pm 1$; every other adjacent interchange gives another point of the joint spectrum in the same irreducible representation. These are the local spectral rules for Young–Jucys–Murphy elements. They follow from $s_r{X}_r{s}_r+s_r=X_{r+1}$ and the one- or two-dimensional analysis of the two neighboring eigenspaces. They are the appropriate local results used below.

First exclude the short patterns $(a,a+1,a)$ and $(a,a-1,a)$ in the joint spectrum. In the first pattern, the local spectral rules for Young–Jucys–Murphy elements make $s_r$ act by $+1$ and $s_{r+1}$ by $-1$ on the same simultaneous eigenvector. The two sides of the braid relation in a Coxeter group then act by opposite signs. The second pattern gives the same contradiction with signs reversed.

Now induct on the length of a spectral vector. A prefix of length $i-1$ is spectral for $S_{i-1}$: decompose the restricted group representation and retain a nonzero component of the simultaneous eigenvector. If $a_i$ differed by neither $+1$ nor $-1$ from every earlier entry, move it left using the allowed adjacent interchanges. Encountering an equal entry would contradict the distinctness of consecutive eigenvalues. Otherwise it reaches the first position, forcing $a_i=0$. For $i>1$ the original first entry was also zero, so an equal entry would indeed have been encountered. ThusThis argument uses only the stated local rules. In particular, iterating the result from $a_1=0$ also shows that all coordinates are integers.

Induct on the distance between equal entries $a_i=a_j=a$ of a point of the joint spectrum. Consecutive equal entries are impossible by the local spectral rules for Young–Jucys–Murphy elements. It suffices to consider consecutive occurrences of $a$: proving the assertion for each such pair proves it for any wider pair.

Suppose $a-1$ is missing between them. There can be at most one occurrence of $a+1$ in the interval. Otherwise a pair of consecutive $a+1$ entries has a shorter gap, and the induction hypothesis forces an intervening $a$, contrary to our choice of the pair. Move the two endpoint entries $a$ inward past entries different from $a,a-1,a+1$. Every move is an admissible spectral interchange. If there is no $a+1$, this creates consecutive equal entries. If there is one, this creates $(a,a+1,a)$, already excluded by the braid relation in a Coxeter group. Both are impossible.

Interchanging the roles of $a-1$ and $a+1$ excludes the omission of $a+1$ as well. Therefore

Necessity is a property of standard Young tableaux alone. The first entry occupies $(1,1)$, with Content of a Young-diagram cell zero. Every subsequent cell has a cell immediately above or to its left, of content one larger or one smaller, already present. If two entries occupy the same diagonal, the later cell lies strictly southeast of the earlier. The cells immediately right of and immediately below the earlier one exist and have contents $a+1,a-1$; their entries lie strictly between the two given entries. This proves all three conditions for content vectors of standard Young tableaux.

For sufficiency, build the Young diagram one cell at a time. All coordinates are integers, because each new coordinate differs by one from an earlier coordinate, starting at zero. Cells of any fixed content $a$ are ordered northwest to southeast. For $a\ge 0$ their positions are $(r,r+a)$, $r=1,2,\dots$; for $a<0$ they are $(r-a,r)$.

At the first occurrence of a positive $a$, the candidate is $(1,a+1)$. The occurrence of $a+1$ earlier would already force a cell of content $a$ in its row, so the neighbor condition must instead supply the content $a-1$. This supplies the left predecessor. The negative case is symmetric, supplying the upper predecessor, and the first zero gives $(1,1)$.

At any later occurrence of $a$, take the next position on its diagonal. The preceding occurrence already supplied all but the next necessary predecessor on each neighboring diagonal. The repeated-entry condition supplies both $a-1$ and $a+1$ after that preceding occurrence, so those next predecessors are present. More explicitly, for $a>0$ and the $r$th occurrence, the left predecessor is the $r$th cell of content $a-1$, and the upper predecessor is the $(r-1)$st cell of content $a+1$. For $a=0$ both are the $(r-1)$st cells on their respective diagonals; for $a<0$ the roles are reversed. These are exactly the predecessors supplied by the two new neighboring occurrences.

The new cell is therefore an addable node of a Young diagram. Induction produces a partition of an integer at every stage, and filling the added cell by its insertion time gives a standard Young tableau. No choice was possible because addable nodes of a Young diagram have distinct contents. Thusand the construction also proves uniqueness of the tableau with a specified vector.

The local spectral interchanges preserve an irreducible representation, and standard Young tableaux of a fixed shape are connected by admissible adjacent interchanges. Different shapes have different multisets of Young-diagram cell contents. Their elementary symmetric functions in the Young–Jucys–Murphy elements are central: the identityproves this coefficient by coefficient. By the Schur lemma, one irreducible representation cannot therefore contain two different shape classes. The multiset determines a diagram because the counts on its positive and negative diagonals determine its arm and leg lengths at the diagonal cells.

Different irreducible representations cannot share a joint eigenvalue vector: the Gelfand–Tsetlin algebra is generated by the Young–Jucys–Murphy elements, so all its elements would act identically on that vector in both representations, whereas a central primitive idempotent distinguishes their two blocks. Thus distinct irreducibles give distinct shape classes.

The number of irreducible representations is the number $p(n)$ of conjugacy classes, each indexed by a partition of an integer. Every spectral vector gives a tableau by the preceding construction. Closure under admissible interchanges makes every shape class that occurs occur in full. The two counts $p(n)$ then force every shape to occur, with exactly one irreducible representation per shape. This supplies $\mathrm{Spec}(n)=\mathrm{Cont}(n)$.

On restricting to $S_{n-1}$, delete the last coordinate. For tableaux of shape $\lambda$, the entry $n$ occupies one Removable node of a Young diagram. Grouping by that node gives one copy of the corresponding module; the path decomposition has simple multiplicities. Hence the restriction branching rule for a symmetric group isThe different removable nodes produce distinct partitions. The multiplicity-free structural input defining the Gelfand–Tsetlin basis is distinct from identifying this graph with the graph of Young diagrams.

Use a Gelfand–Tsetlin basis in Young seminormal form.

For a standard Young tableau $T$, put $d=c_T(i+1)-c_T(i)$, using the Content of a Young-diagram cell. Choose the row-reading tableau $T_0$, and let $\ell (T)$ be the Coxeter length of the unique permutation sending $T_0$ to $T$. A Gelfand–Tsetlin basis can be chosen so that, when $R=s_{i}T$ is standard and $\ell (R)=\ell (T)+1$,If $R$ is not standard, the action is $+v_T$ for two consecutive entries in one row, and $-v_T$ for two in one column. This is one usual normalization of the Young seminormal form.

Here is a construction and proof of the normalization. Fix $v_{T_0}\ne 0$, let $P_T$ be the projection onto the tableau line, and defineThe permutation ${\pi }_T$ has a reduced expression consisting entirely of admissible swaps, by the reduced adjacent-swap path between linear extensions. At each swap the off-diagonal coefficient is nonzero. In its expansion, the only term that can reach a tableau at distance $\ell (T)$ uses all $\ell (T)$ swaps; omitting a swap gives a shorter path. Thus $v_T\ne 0$. This also makes its definition independent of a chosen reduced expression, because ${\pi }_T$ itself is fixed.

The relation $s_i{X}_i{s}_i+s_i=X_{i+1}$ forces the coefficient of $v_T$ in $s_i{v}_T$ to be $1/d$, and forces every other component to lie on the swapped line. If length increases, project the identity ${\pi }_R=s_i{\pi }_T$ onto the line for $R$. The shorter terms of ${\pi }_T{v}_{T_0}$ cannot reach $R$ in one step, so the coefficient of $v_R$ is exactly one. Applying $s_i^2=1$ then gives the reverse coefficient $1-d^{-2}$ and diagonal coefficient $-1/d$. In the nonstandard cases the same relation gives $d=\pm 1$ and the asserted scalar action. This proves the theorem rather than merely specifying pairwise scalings.

Assume $n\ge 2$. The place-permutation action on ${\mathbb{C}}^n$ preserves the coefficient-sum kernelIt is the augmentation subrepresentation of a permutation representation. The point action of $S_n$ is two-transitive, so the irreducible augmentation criterion for a transitive group action makes $W$ irreducible. The two-row Young permutation module decomposition of $M^{(n-1,1)}\cong {\mathbb{C}}^n$ identifies its nontrivial summand as $V^{(n-1,1)}$. Hence $W\cong V^{(n-1,1)}$.

All standard Young tableaux of shape $(n-1,1)$ are $T_j$, $2\le j\le n$, with $j$ below the first cell and the remaining entries increasing along the first row. Their content vectors of standard Young tableaux areAn explicit orthonormal basis realizing these tableau lines isThe sums of their coordinates vanish. Their norms are one, and the inner product of $w_j$ with $w_l$, $j<l$, is zero because the coefficients of $w_j$ sum to zero. Directly summing the action of $(ar)$ gives $X_r{w}_j=c_{T_j}(r)w_j$.

For $s_i=(ii+1)$, $s_1{w}_2=-w_2$. For $2\le i\le n-1$, its only nontrivial two-dimensional block isEvery other $w_j$ is fixed, including all $w_j$ with $j>2$ when $i=1$. These formulas follow by swapping coordinates $i,i+1$ in the displayed vectors, and are the Young orthogonal form with axial distance $i$ for $T_i$.

For the cycle-sum identity for Young–Jucys–Murphy elements, let $C_m$ be the sum of all $m$-cycles on $\{1,\dots ,m\}$. Multiplying such a cycle by $(jm+1)$ inserts $m+1$ immediately after $j$ in that cycle, with our composition convention. Every $(m+1)$-cycle has a unique predecessor $j$ of $m+1$, so deleting $m+1$ recovers a unique term of $C_m{X}_{m+1}$. Since $X_2=(12)$, induction givesFor $n=1$ the empty product is the identity, also the unique one-cycle.

For the product assertion in a non-hook shape, ${\lambda }_2\ge 2$, so cell $(2,2)$ is present. Its Content of a Young-diagram cell is zero and its label is at least four. Since $X_r{w}_T=c_T(r)w_T$, that factor in the product vanishes. Therefore

A hook partition $\lambda =(n-k,1^k)$ has cell contentsIn a standard Young tableau, the unique zero-content cell is $(1,1)$ and contains $1$. The other eigenvalues are precisely the remaining contents, in whatever order the tableau supplies. Their product is independent of that order:The scalar is nonzero over the complex numbers. For the single row or single column, one of the factorials is $0!=1$; the formula still gives $(n-1)!$ or $(-1{)}^{n-1}(n-1)!$, respectively.

The Gelfand–Tsetlin basis spans $V^{\lambda }$, and the scalar computed on its vectors depends only on the shape. Thusfor every $w\in V^{\lambda }$. The product is the sum of the $(n-1)!$ permutations in the conjugacy class of an $n$-cycle. Taking traces therefore gives $(n-1)!{\chi }^{\lambda }((n))$ equal to the displayed scalar times $\dim V^{\lambda }$.

For a hook partition, a standard Young tableau is uniquely determined by the choice of its $k$ entries below the top cell, selected from $\{2,\dots ,n\}$. The column and the remaining row are then forced to increase. Hence $\dim V^{(n-k,1^k)}=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$, and cancellation of the factorials yieldsThis uses the central character value of a conjugacy-class sum and tableau counting, without a character rule for removing rim hooks.

The inclusion here is inclusion of Young diagrams, not the dominance order on partitions. The skew Young diagram $\lambda /\mu$ is the set difference of their cells. Two cells are adjacent when they share an edge. It is connected when any two cells can be joined by such steps, and is a rim hook when it is connected and contains no $2\times 2$ square. Under the usual edge-adjacency convention a totally disconnected skew Young diagram has only singleton components, equivalently no two cells share an edge. A horizontal strip instead means at most one cell per column; this distinction matters for the last part of this question.

A standard skew Young tableau is a linear extension of a partially ordered set: the cells are ordered by the row and column inequalities, and the tableau lists them in increasing label order. Write the current list as $x_1,\dots ,x_k$ and let $r_j$ be the desired label of $x_j$ in $R$. Then $(r_1,\dots ,r_k)$ is the one-line notation of the unique permutation $\pi$ with $\pi T=R$.

Whenever this list of desired labels is not increasing, there is an adjacent descent $r_j>r_{j+1}$. The cells $x_j,x_{j+1}$ are incomparable in the cell partial order: if they were comparable, both $T$ and $R$ would have to put them in the same order. Interchanging their consecutive current labels is therefore admissible. It removes exactly one inversion of a permutation from the list of desired labels. Repeating ends at $R$ after exactly the original inversion count, which is the Coxeter length $\ell (\pi )$. ThusThis proves the reduced adjacent-swap path between linear extensions and applies equally to ordinary tableaux.

Put $m=n-k$ and take the subgroup $S_m\times S_k$ permuting the first $m$ and last $k$ letters. Define the skew representation of a symmetric group as the multiplicity spaceThe last-letter copy of $S_k$ commutes with $S_m$, so it acts on a map $f$ by $g\cdot f={\rho }_{\lambda }(g)\circ f$. This gives a genuine $\mathbb{C}{S}_k$-module, without claiming that the whole restricted module is itself the skew representation. Equivalently,

Fix a prefix tableau of shape $\mu$ and complete it to shape $\lambda$. Iterated restriction branching rule for a symmetric group identifies the multiplicity-space orthonormal basis with standard skew Young tableaux, using labels $1,\dots ,k$ for the last $k$ cells. The Young orthogonal form restricts to this basis. When $R=s_{i}T$ is standard and $d=c_T(i+1)-c_T(i)$,An admissible interchange has $|d|\ge 2$, so its off-diagonal coefficient is nonzero. Consequently $w_R$ belongs to the group algebra span of $w_T$. The preceding reduced-path argument reaches every standard skew tableau, so this span is the whole module. Every $w_T$ is a cyclic vector for a group representation. The inherited invariant inner product makes this a unitary representation.

Let $P$ be the averaging linear mapFor a unitary representation of a finite group, $P$ is the orthogonal projection onto the invariant subspace $V^G$, and $P\rho (h)=\rho (h)P=P$. Applying this to $\rho (g)v=\lambda v$ gives $Pv=\lambda Pv$. Since $\lambda \ne 1$, $Pv=0$. The vector $v$ is a cyclic vector for a group representation, so its translates span $V$, and every translate also has projection zero. Thus $P=0$ andThis is the cyclic eigenvector obstruction to invariant vectors. Unitarity also gives a direct proof by pairing $v$ with an invariant vector and then pairing every translate; the averaging proof makes the role of cyclicity especially transparent.

With the usual meaning of totally disconnected skew Young diagram, the assertion in the PDF is false. For example, $\lambda =(3)$ and $\mu =(1)$ give two adjacent cells in one row, so the skew shape is connected, but its sole standard skew Young tableau affords the trivial representation of $S_2$. The multiplicity is one, not zero. The correct criterion is a horizontal strip:If the course uses “totally disconnected” specifically for no repeated column, its terminology must be read as this criterion; it cannot mean disconnected into individual cells.

For necessity, a non-horizontal strip has two vertically adjacent cells. They form a cover in the cell partial order, and some linear extension of a partially ordered set places them consecutively. To justify that assertion, put all predecessors of the lower cell other than the upper cell first, then the upper and lower cells, and complete the order; the cover property prevents any missing intermediate predecessor. The corresponding standard skew Young tableau has consecutive entries in one column. The relevant adjacent transposition acts on its vector by $-1$ in the Young orthogonal form. That vector is cyclic, so the cyclic eigenvector obstruction to invariant vectors excludes invariants.

For sufficiency one can construct an invariant vector explicitly. In a horizontal strip, the occupied column intervals of different rows are disjoint, with every upper interval to the right of every lower interval. Thus an upper-row cell $x$ has content $c(x)$ at least two greater than a lower-row cell $y$ with which it is incomparable. Choose row-reading order as a reference, and for each standard skew Young tableau setOnly pairs in different rows occur in this product; its factors are positive. Put $u={\sum }_T{b}_T{w}_T$. For an admissible swap $R=s_{i}T$ with $d=c_T(i+1)-c_T(i)$, changing the one inversion in the product givesThe two-dimensional Young orthogonal form then fixes $b_T{w}_T+b_R{w}_R$. A nonadmissible swap is within one row and acts by $+1$. Every generator therefore fixes $u\ne 0$, proving existence.

Finally an invariant projection of a cyclic vector generates the invariant subspace, because projection identifies all its translates. That subspace has dimension at most one. This proves the exact multiplicity, including the totally disconnected special case, without asserting the incorrect converse in the PDF.

Let $\lambda ,\mu ⊢n$ each be a partition of an integer. Pad the lists of parts with zeros and use the dominance order on partitionsThe total sums agree, so only finitely many inequalities matter. This is a partial order: reflexivity and transitivity hold term by term, and equality of all partial sums forces equality of every part, proving antisymmetry. For example $(2,2)⊴(3,1)$. It compares concentration in the top rows, whereas diagram inclusion compares individual cells and is a different relation.

A single-box up-move from row $j$ to row $i<j$ changes the partial sum through row $h$ byEvery change is nonnegative, so $\lambda ⊴\mu$ in the dominance order on partitions. Transitivity gives the same conclusion along any chain of up-moves. Thus the existence of the indicated chain implies dominance. A chain of length zero handles equality; each positive-length move must still leave a partition of an integer.

Suppose $\lambda ⊴\mu$ in the dominance order on partitions and $\lambda \ne \mu$. Let $i$ be the first unequal row. Then ${\lambda }_i<{\mu }_i$. PutWe have $D_i>0$, and eventually $D_h=0$. Let $j>i$ be its first return to zero. Then $D_h\ge 1$ for $i\le h<j$ and ${\lambda }_j>{\mu }_j$.

Move one box from row $j$ to row $i$. Increasing row $i$ is legal: if $i>1$, then ${\lambda }_{i-1}={\mu }_{i-1}\ge {\mu }_i>{\lambda }_i$. Decreasing row $j$ is also legal. If ${\lambda }_j={\lambda }_{j+1}$, then ${\mu }_{j+1}\le {\mu }_j<{\lambda }_j={\lambda }_{j+1}$, forcing $D_{j+1}<0$, a contradiction. Also ${\lambda }_j>0$. Thus the new list $\nu$ is a partition of an integer.

The new deficits equal $D_h-1$ on $i\le h<j$ and $D_h$ elsewhere, so $\lambda ⊴\nu ⊴\mu$. The quantity ${\sum }_r|{\lambda }_r-{\mu }_r|$ decreases by two, since the changed rows were a deficit and a surplus. Repeating must terminate at $\mu$. HenceThe construction proves both termination and preservation of the intermediate dominance inequalities.

It suffices to handle one single-box up-move, and then use the chain already constructed. Suppose the affected row sizes are $a={\lambda }_i$ and $b={\lambda }_j$, $i<j$. Since $\lambda$ is a partition of an integer, $a\ge b\ge 1$. Put $N=a+b$. Over the complex numbers, the two-row Young permutation module decomposition givesA zero second row is omitted. Therefore

Let $K$ be the product of the symmetric groups of the unaffected rows and the symmetric group $S_N$ on the union of the affected rows. The two relevant Young subgroups lie in $K$. Transitivity of induced representations expresses $M^{\lambda }$ and $M^{\nu }$ by inducing the preceding two-row modules, tensored with the trivial representations of the unaffected factors, from $K$ to $S_n$. Induction preserves this direct sum, sofor an actual group representation $L_{\lambda ,\nu }$, not merely a difference of characters. Iterate along the chain and take the direct sum of the induced complements. The resulting complement can be realized as an invariant subspace of $M^{\lambda }$, either through these isomorphisms or by Maschke's theorem. ThusIf $\lambda =\mu$, use the zero complement. The characteristic-zero hypothesis is essential to this use of the irreducible two-row decomposition.

The single-column Young diagram has exactly one standard Young tableau, so $V^{(1^n)}$ has dimension one. Consecutive labels are in one column, and the Young seminormal form gives $s_{i}w=-w$ for every adjacent transposition $s_i$. These generators determine the representation, and the sign representation sends every one of them to $-1$. ConsequentlyAt $n=1$ there are no adjacent generators and both modules are the trivial representation.

The conjugate partition is ${\lambda }_j^{\prime }=\#\{i:{\lambda }_i\ge j\}$, obtained by transposing the Young diagram. Transposition sends a standard Young tableau $T$ to a standard tableau $T^{\prime }$ and negates every Content of a Young-diagram cell.

Use the Young orthogonal form. For an admissible pair $R=s_{i}T$ and axial distance $d=c_T(i+1)-c_T(i)$, the action on $(w_T,w_R)$ isIn the transposed pair it is $A_{-d}$, whereas in the tensor product of group representations with the sign representation it is $-A_d$. Their diagonal entries agree, and their off-diagonal entries differ by a sign. Fix a reference tableau and let ${ϵ}_T$ be the sign of the unique label permutation from it to $T$. For every admissible swap ${ϵ}_R=-{ϵ}_T$, sointertwines these two actions. In a nonadmissible pair, transposition exchanges row and column and changes the scalar from $+1$ to $-1$ or conversely, also agreeing with the sign twist. Since adjacent transpositions generate $S_n$, this is an isomorphism:The parity choice is globally well defined because the label permutation is unique; no arbitrary edge-by-edge phase choices are required.

There is an index error in the displayed identity of the original PDF. With its definition of the signed Young permutation module, the tensor identity for induced representations givesBoth superscripts must be $\alpha$. In particular ${\tilde{M}}^{{\lambda }^{\prime }}\cong \mathrm{sgn}\otimes M^{{\lambda }^{\prime }}$, rather than the printed sign twist of $M^{\lambda }$. For $\lambda =(n)$, $n\ge 2$, the printed left side has dimension $n!$, while the printed right side has dimension one. Thus that identity cannot hold as written.

The requested character inner product between $M^{\lambda }$ and ${\tilde{M}}^{{\lambda }^{\prime }}$ nevertheless has a well-defined answer. Let $R$ be the row Young subgroup of a tableau of shape $\lambda$, and let $C$ be its column subgroup, conjugate to $S_{{\lambda }^{\prime }}$. By Frobenius reciprocity and the Mackey restriction formula, the inner product is a sum over $R\backslash S_n/C$ ofThese double cosets are encoded by nonnegative integer matrices $A=(a_{ij})$ with row sums ${\lambda }_i$ and column sums ${\lambda }_j^{\prime }$, recording intersection sizes of row and column blocks. The intersection subgroup is a product of $S_{a_{ij}}$. Its sign representation is trivial exactly when every $a_{ij}\le 1$. Each zero-one matrix contributes one, and every other matrix contributes zero.

There is exactly one zero-one matrix with these margins. Its first row has length ${\lambda }_1$, equal to the number of nonzero column sums, so that row must consist entirely of ones. Delete it, subtract one from every column margin and discard zero columns. The remaining margins are those of $({\lambda }_2,{\lambda }_3,\dots )$ and its conjugate partition; induction forces the rest. The unique matrix is $a_{ij}=1$ precisely for $j\le {\lambda }_i$. ThereforeThis calculation uses the actual two modules requested in the PDF and does not depend on its erroneous displayed isomorphism.

The Specht module construction gives a nonzero inclusion $V^{\lambda }\hookrightarrow M^{\lambda }$ over the complex numbers: a polytabloid has coefficient one at its original tabloid, because the row and column stabilizers have trivial intersection, and its span is the corresponding irreducible representation. This establishes $a=M(\lambda ,\lambda )\ge 1$ without using the multiplicity-one conclusion. Similarly $a^{\prime }=M({\lambda }^{\prime },{\lambda }^{\prime })\ge 1$.

By the corrected tensor identity for induced representations and the conjugate-partition sign twist, the multiplicity of $V^{\lambda }$ in ${\tilde{M}}^{{\lambda }^{\prime }}$ is $a^{\prime }$. The character inner product already computed is the sum of products of multiplicities of common irreducible representations. ThusBoth are positive integers, so $a=a^{\prime }=1$. In particularThis deduction avoids assuming the full Young's rule, which would already contain the answer. Alternatively, the permitted one-dimensional R-module homomorphism space gives the same character inner product immediately.