Conormal-node composition multiplicities under residue induction

For $d>0$ take $N=d+1$, pad $\mu$ by zeros, set $a_j={\mu }_j-{\mu }_{j+1}$, let $L=L_N(\mu )$ be the actual Schur simple, $V=F^N$, and $M=V\otimes L$. Tensor coefficient operators are the actual polynomial Schur operators of degree $d+1$, obtained by multiplying the two polynomial matrix actions. This works over finite fields without replacing coefficient actions by evaluated group elements. The case $d=0$ is treated below.

Here is the local component identity used below. If a binary root deficit has separated connected components (a zero simple-root coefficient between consecutive components), its negative words split into products supported on those components. All cross-component positive and negative root operators commute, and their Cartan characters on the other component are unchanged: their simple-root supports are nonadjacent. Move the positive words past the other components and evaluate their highest coefficients one component at a time. The positive/negative coefficient matrix is consequently the tensor product of the component matrices. Negative-word spanning and positive detection identify its quotient by the common positive-test kernel with the actual simple weight space. Hence this weight space is the tensor product of the component quotients, including when one is zero. This proves the assertion directly from coefficient words; it does not assume a tensor-product decomposition of the whole simple module.

In particular fix one component $\delta$. Choose a negative basis $f_{1}v,\dots ,f_{t}v$ of its quotient and positive tests $e_1,\dots ,e_t$ dual to it. They are finite linear combinations of actual coefficient words, with coefficients in $F$; their existence follows from positive detection and ordinary finite-dimensional linear algebra. Then $P_{\delta }={\sum }_{h=1}^t{f}_h{e}_h$ is the identity on every binary weight space having this component and any separated other components. The same tests apply because the other components change none of the Cartan data of $\delta$. If $t=0$, these weight spaces are zero. This is an identity of coefficient operators on those spaces, so it descends to subquotients.

Write ${\xi }_i=\mu +{ϵ}_i$. Suppose a simple composition factor with highest weight $\eta$ contributes at ${\xi }_i$. All weights of $M$ are at most $\mu +{ϵ}_1$, and weights of that factor are at most $\eta$. Thus ${\xi }_i\le \eta \le \mu +{ϵ}_1\mathrm{.}$

The deficit $\mathrm{\Gamma }=\mu +{ϵ}_1-\eta$ has simple-root coefficients $0$ or $1$ in positions $1,\dots ,i-1$ and zero elsewhere. On a tensor summand $v_j\otimes L_{\eta -{ϵ}_j}$ the deficit in $L$ is $\mathrm{\Gamma }-({\alpha }_1+\cdots +{\alpha }_{j-1})\mathrm{.}$

It must be nonnegative. Consequently every possible label $j$ lies before the first missing coefficient in $\mathrm{\Gamma }$. If $\mathrm{\Gamma }$ has a component $\delta$ after that first gap, every positive or negative coefficient word supported on $\delta$ fixes the $V$ factor (its nonconstant terms annihilate each of these labels). Its action on this tensor weight is therefore purely its action on $L$. The preceding $P_{\delta }$ is the identity on the entire weight $M_{\eta }$, not just on individual tensor summands.

Choose a composition series and a lift $x\in M_{\eta }$ of the nonzero highest vector of the proposed factor in its series quotient. Each $e_{h}x$ lies in the preceding submodule because it has strictly higher weight, and each $f_h$ preserves that submodule. But $P_{\delta }x=x$ would then place $x$ in it, a contradiction. If the component quotient is zero there was no such tensor weight to start with. Thus $\mathrm{\Gamma }$ has no later component. Its only possible nonzero coefficients form one initial interval, and $\eta =\mu +{ϵ}_u\text{for some }u\le i\mathrm{.}$

This excludes only highest weights of factors contributing to ${\xi }_i$. It asserts nothing about distant tensor factors.

Since every weight of $L$ is at most $\mu$, $M_{{\xi }_i}={⨁}_{j\le i}{v}_j\otimes L_{\mu +{ϵ}_i-{ϵ}_j},d_i:=\dim M_{{\xi }_i}=1+{\sum }_{j<i}{E}_a(j,i-1)\mathrm{.}$

Let $b_u$ be the actual composition multiplicity of $L_N(\mu +{ϵ}_u)$ in $M$ when that weight is dominant; set $b_u=0$ otherwise. No restrictedness condition is imposed on these auxiliary factors. The support lemma and the one-dimensional highest spaces give $d_i=b_i+{\sum }_{u<i}{b}_u{E}_{a+{\omega }_u}(u,i-1)\mathrm{.}$ (equation C).

Indeed $\mu +{ϵ}_u-{\xi }_i={\alpha }_u+\cdots +{\alpha }_{i-1}$. On this interval the fundamental coordinates of the dominant weight $\mu +{ϵ}_u$ are those of $a$ with its first coordinate $u$ increased by one; the coordinate $u-1$ changed by minus one lies outside the interval. Nondominant rows have multiplicity zero, so their formal summands cause no alteration of (C).

For completeness the weight-dimension symmetry needed here is local. Permutation matrices permute the integer weight spaces of $L$. Hence its lowest weight is $w_0\mu$, its space is one dimensional, and every weight lies between $w_0\mu$ and $\mu$. The contragredient dual twisted by ${\det }^d$ is a polynomial simple of highest weight ${\mu }^{\ast }=(d-{\mu }_N,\dots ,d-{\mu }_1)\mathrm{.}$

One can verify polynomiality without a rational-module classification theorem. The tensorword Schur algebra acts faithfully on $V^{\otimes d}$; its regular module embeds in a finite direct sum of copies of this tensor space by its action on a basis. Thus every simple is a subquotient of copies of this tensor space. Duality and the twist make it a subquotient of $(V^{\ast }\otimes \det {)}^{\otimes d}$; the explicit exterior-complement pairing gives $V^{\ast }\otimes \det \cong {\bigwedge }^{N-1}V$ over every field. It is therefore polynomial. Positive and negative coefficients interchange under duality, so the highest weight just identified determines the same actual Schur simple as in the accepted foundation. These statements are coefficient identities; over a finite field they can equivalently be checked over $F(t)$ and descended as the identical finite matrices.

Put ${\bar{a}}_t=a_{N-t}$. Duality reverses roots and preserves weight dimensions, yielding $E_a(j,i-1)=E_{\bar{a}}(N-i+1,N-j)\mathrm{.}$

This argument applies also to any dominant shifted weight. In particular the first-coordinate shift at $u$ becomes the final-coordinate shift at $N-u$ on the reversed interval. No restriction on its fundamental coordinates is needed.

Reverse the indices: $r=N-i+1$, ${\bar{b}}_r=b_i$, ${\bar{d}}_r=d_i$. On indices $1,\dots ,N$, let the strictly upper triangular matrices be $A[r,v]={\nu }_{\bar{a}}^{\mathrm{\prime }}(r,v-1),B[r,v]=E_{\bar{a}}^{\mathrm{\prime }}(r,v-1)(r<v)\mathrm{.}$

For $r<v$, apply the accepted recurrence to $[r,v-1]$ after increasing coordinate $v-1$. Every earlier interval $[r,u-1]$ with $u<v$ is unaffected, while the tail normal count is shifted at $v-1$. Thus $B[r,v]=A[r,v]+{\sum }_{r<u<v}B[r,u]A[u,v]$, or $B=A+BA$. Hence $(I+B)(I-A)=I$. Since these are finite integer matrices, the inverse is two-sided and $(I+B{)}^{-1}=I-A$. Equivalently, the supplier's first-cut recurrence gives $B=A+AB$ and $I+B=I+A+\cdots +A^{N-1}$.

Let $H_r={\sum }_{s=r}^{N-1}{\nu }_{\bar{a}}(r,s)$, with $H_N=0$. Applying the accepted recurrence separately for every endpoint and summing gives $\bar{d}=\mathbf{1}+(I+B)H\mathrm{.}$

Opposition turns (C) into $\bar{d}=(I+B)\bar{b}$. Multiplication by $I-A$ gives, as ordinary integers, $b_i=1+{\sum }_{s=r}^{N-1}({\nu }_{\bar{a}}(r,s)-{\nu }_{\bar{a}}^{\mathrm{\prime }}(r,s)),r=N-i+1.$ (equation D).

This is finite nilpotent matrix arithmetic. It neither divides in $F$ nor replaces a composition multiplicity by a residue in $F$.

Use the accepted normal-stack interpretation; every coordinate and $B,C$ equality tested in this paragraph and the endpoint analysis is modulo $p$, whereas stack heights are ordinary integers. Write $\sigma (r,t)$ for its alive indicator and put $\sigma (r,r-1)=1$. For every dominant coordinate array $\lambda$, ${\nu }_{\lambda }(r,s)-{\nu }_{\lambda }^{\mathrm{\prime }}(r,s)=\sigma (r,s-1)({\mathbf{1}}_{B(r,s)=-1}-{\mathbf{1}}_{B(r,s)=0})\mathrm{.}$ (equation E).

If killed before the final step, both counts are zero. Otherwise let its current height be $h\ge 1$. The endpoint shift keeps $C(r,s)$ fixed and replaces $B$ and ${\lambda }_s$ by $B+1$ and ${\lambda }_s+1$. If $B=-1$ it creates a pop and lowers the count by one; when ${\lambda }_s=-1$, the unshifted step instead pushes and the shifted zero-coordinate step is neutral, with the same difference. If $B=0$ it removes a pop and increases the count by one; when ${\lambda }_s=0$, the old step is neutral and the new step pushes, again with that difference. For other $B$ neither change alters the count. If a pop removes the initial root, its new count is zero, still $h-1=0$. At $s=r$ the two counts are ${\mathbf{1}}_{{\lambda }_r\mathrm{\ne }0}$ and ${\mathbf{1}}_{{\lambda }_r+1\mathrm{\ne }0}$, giving the same formula. These alternatives are exhaustive and valid for $p=2$.

Fix an actual addable row $i$ and put $R_j={\mu }_j+1-j$, $\rho =R_i(\mathrm{m}\mathrm{o}\mathrm{d}p)$. If $i>1$, addability means $a_{i-1}>0$. Since $\mu$ is restricted, $0<a_{i-1}<p$; the reversed root therefore starts alive. For $r=N-i+1$, $s=N-j$, direct substitution gives $B_{\bar{a}}(r,s)=R_j-R_i-1,C_{\bar{a}}(r,s)=R_{j+1}-R_i\mathrm{.}$

Thus the two indicators in (E) respectively detect an addable-position residue $\rho$ in row $j$ and a removable-position residue $\rho$ in row $j$. The normal stack processes the boundary in reverse order, starting with the chosen $+$ in row $i$. At step $j$ its $C=0$ push corresponds to the plus in row $j+1$, and its $B=0$ pop to the minus in row $j$. The top-row plus is a pending final event. This reversal is exactly the stack evaluation of $-+$ cancellation: in a reverse scan, a minus cancels the most recently encountered plus.

Here is the precise check at flat row blocks and at the endpoints. Let $D_s=1+{\sum }_{t=r}^s\sigma (r,t-1)({\mathbf{1}}_{B(r,t)=-1}-{\mathbf{1}}_{B(r,t)=0})\mathrm{.}$

While the chosen root remains alive, $D_s$ is its current normal-stack height plus the pending indicator ${\mathbf{1}}_{R_j=R_i}$, where $j=N-s$. At the initial step the removable event $B(r,r)=0$ cannot occur since $0<a_{i-1}<p$, and this identity follows immediately. At each subsequent live step the change in height is ${\mathbf{1}}_{C(r,s)=0}-{\mathbf{1}}_{B(r,s)=0}\mathrm{.}$

For $a_j\mathrm{\ne }0$ these are disjoint push/pop events. For $a_j=0$ the indicators are equal and the step is neutral, as required by the actual stack. Subtracting this height change from the change in $D_s$ gives ${\mathbf{1}}_{R_j=R_i}-{\mathbf{1}}_{R_{j+1}=R_i},$

which replaces the previous pending indicator by the next. This proves the invariant by induction, including flat blocks. Equivalently, at each flat boundary $a_j=0$, the virtual removable sign in row $j$ and virtual addable sign in row $j+1$ have the same residue and form an adjacent $-+$ pair in the selected-residue row-boundary word. Deleting these pairs restores exactly the actual boundary word. By signature confluence their insertion or deletion has no effect on the surviving actual nodes. The chosen node is an actual addable, so it is never among the virtual pairs.

If the initial root is killed, its preceding height was one, the current event is $B=0$, and $B=-1$ is false. The pending indicator after the step is zero. Thus $D_s=0$, and every later term is zero because the alive indicator remains zero. In the reversed boundary scan, killing the initial plus likewise means that the chosen node is canceled; later pluses above it cannot revive it. At the last step the pending indicator is the actual addable top-row plus. We have therefore evaluated exactly the reverse scan of the prefix ending at the chosen node. A minus below this prefix cannot cancel a plus in it under top-to-bottom $-+$ cancellation; the conormal decisions for this prefix agree with those in the entire signature.

Consequently (D) is zero precisely when the chosen plus is canceled. If it survives, (D) is the number of surviving pluses of residue $\rho$ in the prefix, namely the number of conormal nodes weakly above $B$. For $i=1$, (D) is $b_1=1$: the top plus survives and it is the only node weakly above itself. For completeness, a nondominant row $i>1$ has $a_{i-1}=0$; the first endpoint difference is $-1$, killing the virtual initial unit and all later differences, so (D) gives zero as required by (C). These rows were not silently discarded during the integer inversion.

Let $\omega =(1^N)$ be the regular degree-$N$ weight. Then $M_{\omega }={⨁}_{j=1}^N{v}_j\otimes L_{\omega -{ϵ}_j}\mathrm{.}$

Let $G={\mathfrak{S}}_N$, $H={\mathfrak{S}}_d$ fixing label $N$, and $W=L_{\omega -{ϵ}_N}$. The accepted local $q=1$ corner/form calculation applies to degree $d$ and rank $N$ with regular weight $(1^d,0)$: its surviving semistandard indices are precisely standard tableaux, and evaluating the corner at $1$ gives the same Murphy basis and scalar-$1$ form, regardless of the unused last label. Its radical-corner equality therefore makes $W$ the left group module $D(\mu )$. To fix the action convention, the supplier's right corner element ${\rho }_h$ permutes tensorword labels by $h^{-1}$; star conversion makes left $g$ act by right $g^{-1}$, hence by label permutation $g$. The same conversion in the degree-$N$ corner gives the output action. Thus the output regular-corner action is ordinary permutation of labels: on tensorwords the corresponding orbit matrix is exactly that label permutation. Since these are identities in the actual tensorword Schur algebra, the same equality holds on $M$; it does not depend on evaluated permutation matrices spanning the Schur algebra.

The explicit map $F[G]{\otimes }_{F[H]}W⟶M_{\omega },g\otimes w⟼g(v_N\otimes w)$.

is well defined and $G$-equivariant. Left cosets of $H$ correspond to the image label $gN$, and their summands map bijectively to the displayed summands of $M_{\omega }$. It is therefore an isomorphism. This proves $M_{\omega }\cong {\mathrm{Ind}}_{{\mathfrak{S}}_d}^{{\mathfrak{S}}_{d+1}}D(\mu )$, without Frobenius reciprocity or an adjunction multiplicity assertion.

Corner projection is exact. Each Schur composition factor $L_N(\eta )$ projects to the distinct restricted group simple $D(\eta )$ if $\eta$ is restricted, and to zero otherwise, by the accepted local form/radical and ordinary-group survival proof. Hence for a restricted $\mu +B=\mu +{ϵ}_i$, its total induction multiplicity is exactly $b_i$. Nonrestricted near factors were kept until this exact projection. Distant factors were never excluded; distinct surviving labels ensure that they cannot change this particular count. When $d=0$, the single degree-one tensor/corner gives the unique plus and multiplicity one directly.

The generalized Jucys--Murphy induction summands give a direct decomposition of total induction. For the ordinary group elements $L_k={\sum }_{j<k}(j,k)$, induction by inserting the letter $n$ into a cycle gives the integral identity ${\prod }_{k=1}^n(t+L_k)={\sum }_{g\in {\mathfrak{S}}_n}{t}^{\mathrm{\#}\mathrm{cycles}(g)}g\mathrm{.}$

The factor $t$ makes $n$ a new fixed point; each transposition $(j,n)$ inserts $n$ at the uniquely specified place in the cycle containing $j$. These alternatives are disjoint and exhaust the permutations. Coefficients on the right are class sums, so all coefficients of $Q_n(t)={\prod }_{k=1}^n(t-L_k)$ are central over the integers and every field.

The published The joint spectrum of the Jucys-Murphy elements is the set of tableau content vectors says that $Q_n(t)$ acts on the complex ordinary Specht module of shape $\lambda$ as ${\prod }_{A\in \lambda }(t-{\mathrm{res}}_{\mathbb{Z}}(A))$. The ordinary integral Specht lattice has the free polytabloid basis provided by the accepted integral bridge. The operator equality has integer matrices on that lattice; its validity after scalar extension to $\mathbb{C}$ therefore implies its validity over $\mathbb{Z}$. Reducing and taking its radical quotient proves the same content scalar on ordinary $D^{\lambda }$ over every field. The accepted signed-conjugate dictionary sends each $L_k$ to $-L_k$ under the sign twist, and transpose negates contents. Hence $Q_n(t)$ acts on restricted $D(\lambda )=D^{{\lambda }^{\mathrm{\prime }}}\otimes \mathrm{sgn}$ by the content polynomial of $\lambda$. This use of the published complex spectrum is an integral operator-identity descent; none of the multiplicity inversion uses characteristic-zero composition counts or semisimplicity.

Let $\alpha$ be the multiset of residues of $\mu$. The exact central content polynomial on $D(\mu )$ and the central CRT description of the content sums give $e_{\alpha }^{(d)}D(\mu )=D(\mu )$; this requires no primitive-block assertion. For clarity the functor idempotent $e=e_{\alpha }^{(d)}{P}_{d+1,h}$ implies the following test on $F_{h}D(\mu )=F{\mathfrak{S}}_{d+1}e{\otimes }_{F{\mathfrak{S}}_d}D(\mu )$. On its generating subspace $e\otimes D(\mu )$, the first-$d$ content polynomial has exactly the scalar for $\mu$, and $L_{d+1}-h$ is nilpotent. These operators commute and $(L_{d+1}-h{)}^{m}e=0$ for a fixed primary exponent $m$. A coefficient $z$ of $Q_{d+1}(t)-(t-h)Q_{\mu }(t)$ is central by the integral identity. On $e\otimes w$ it is the corresponding scalar coefficient of $Q_{\mu }(t)$ times $h-L_{d+1}$: balance the first-$d$ central coefficient across $e$ into $w$. Centrality lets this same calculation iterate, so $z^m(e\otimes w)=0$. Since the whole module is generated by these vectors and $z$ commutes with every group translate, $z^m$ is zero on the whole module. On a simple factor a nilpotent central endomorphism is zero, since every nonzero simple-module endomorphism is invertible. Thus every factor in this summand has content polynomial $(t-h)Q_{\mu }(t)$.

Distinct residues give distinct polynomials: $(t-h)Q_{\mu }(t)-(t-k)Q_{\mu }(t)=(k-h)Q_{\mu }(t)\mathrm{\ne }0$ for $h\mathrm{\ne }k$. The content polynomial of $\mu +B$ appends exactly the residue of $B$. Its factors therefore occur only in the corresponding residue summand. The total multiplicity computed above is exactly the multiplicity in $F_{i}D(\mu )$ required by the target. This argument does not use primitivity or exhaustiveness of any content-block idempotent.

Transpose preserves boundary types. Literally $(\mu \cup B{)}^{\mathrm{\prime }}={\mu }^{\mathrm{\prime }}\cup B^{\mathsf{t}},(\mu \setminus A{)}^{\mathrm{\prime }}={\mu }^{\mathrm{\prime }}\setminus A^{\mathsf{t}}\mathrm{.}$

It sends an addable node to an addable node and a removable node to a removable node; it negates the residue. The appropriate boundary reading order is reversed when transporting the signature convention. It does not exchange addable and removable sets. The proof above needs no claimed equality of induction and restriction multiplicities and no transpose transfer of a restriction count.

The finite character inversion, boundary-stack calculation, exact induction corner and residue allocation just proved identify the stated multiplicity with zero off conormality and with the conormal weak-above count on conormality. Both cases hold over every field of characteristic $p$, with nonrestricted auxiliary factors retained until projection and no assertion about other tensor factors.