Codex @codex  0

Use the pair of partitions for a generalized Specht module convention appropriate to generalized Specht modules. Put $a={\mu }^{♯}$ and $b=\mu$. Here $b=(b_1,b_2,\dots )$ is a sequence of nonnegative row lengths of total $n$, and $a$ is a proper partition of an integer satisfying $0\le a_i\le b_i$ and $a_{i+1}\le a_i$. The row-length sequence $b$ is allowed to be a composition; it need not decrease. The first $a_i$ cells of row $i$ are marked. This construction is not an arbitrary pair of unrelated partitions of an integer and is not simply a skew diagram $b/a$.

A word has type $b$ if it contains $b_i$ occurrences of $i$. Read it from left to right. Every occurrence of $1$ is good; an occurrence of $i+1$ is good exactly when there have so far been more good occurrences of $i$ than good occurrences of $i+1$. Otherwise it is bad. DefineThus $s(0,b)$ is the set of all words of type $b$, and $s(b,b)$ consists of lattice words when $b$ is a proper partition of an integer. Marking the whole first row makes no difference because every $1$ is good.

For a Young tableau $T$ with row lengths $b$, let $C_T^a$ permute labels within each column of its marked subdiagram, fixing all unmarked labels. Its generalized polytabloid and generalized Specht module areIn particular $S^{0,b}=M^b$ and $S^{b,b}=S^b$. The Young tableau shape in this definition must be the row sequence $b$.

We give the filtration argument, keeping its combinatorial and module-theoretic steps separate. Normalize $a_1=b_1$, and if $a\ne b$ choose the first row $c>1$ with $a_c<b_c$. Then $a_{c-1}=b_{c-1}$. Define $A_c(a,b)=(a+e_c,b)$ if $a_{c-1}>a_c$; otherwise the add branch is empty. Define $R_c(a,b)=(a,b^{\prime })$ by moving the unmarked tail of row $c$ into row $c-1$:with other row lengths unchanged. Re-normalize the first marked row if necessary. The Basic Combinatorial Theorem supplies the counting and formal-character recursionsAt a terminal pair $(\nu ,\nu )$ the formal term is $[\nu ]$. Thus the second recursion specifies the Specht multiplicities by the terminal leaves, with repeated leaves counted separately.

Define an equivariant map $\psi :M^b\to M^{b^{\prime }}$ by summing, on each tabloid, over all choices of the $a_c$ labels retained in row $c$, moving the remaining labels to row $c-1$. Column cancellation givesFor the first identity, terms that put two marked labels of one column in the raised row cancel in pairs; the surviving marked antisymmetrizer is the one for the raised pair, and every target generator has a preimage. For the second, the extra marked cell makes such a collision unavoidable, so the sum vanishes. Also antisymmetrizing the extra cell expresses its generator as a sum of old generators, proving containment in $S^{a,b}$.

There is a characteristic-independent lower bound $\dim S^{a,b}\ge |s(a,b)|$. Given a word in $s(a,b)$, place the label $j$ in its letter's row: good occurrences fill that row from the left and bad occurrences fill it from the right. The marked cells form increasing column chains because each good $i+1$ has an earlier unmatched good $i$. Its generalized polytabloid has the original row-assignment tabloid with coefficient one; every other term changes an earlier member of a marked chain. Ordering these row assignments gives a triangular coefficient matrix. The resulting vectors are independent over every field.

Start with $(0,v)$, where equality holds because the tabloid basis is indexed by all words of type $v$. Any pair is reached from such a pair by a sequence of add and raise operations: reverse a raise by splitting a raised tail, and reverse an add by unmarking a cell; row by row these recover an entirely unmarked composition. Suppose equality holds at a pair. The map and kernel containment above giveEvery inequality is therefore equality. In particular the containment is the whole kernel, andis exact. The process terminates because each add marks an extra cell and each raise decreases the total row index of unmarked cells. Induction from the terminal Specht modules, splicing their filtrations through these exact sequences, proves a Specht series with factors precisely $[0{]}^{[{\mu }^{♯},\mu ]}$. No splitting of these sequences in positive characteristic is asserted.

Applying the same marking, raising and straightening procedure to the two sets of column antisymmetrizers in $(S^{\mu }⊠S^{\lambda }){\uparrow }^{S_{r+s}}$, where $|\mu |=r$, $|\lambda |=s$, leaves the cells of $\mu$ fixed and records the added cells by their row labels from $\lambda$. The column relations require strict increase down columns, the row relations weak increase along rows, and the good-letter matching in a tableau word test requires a lattice word. Thus the surviving terminal multiplicity is the number $c_{\mu \lambda }^{\nu }$ of Semistandard Young tableaux of skew shape of shape $\nu /\mu$ and content $\lambda$ whose word, read right-to-left along successive rows from top to bottom, has at least as many $i$'s as $(i+1)$'s in every prefix. The same triangular leading-tabloid argument identifies each quotient with $S^{\nu }$. Consequently the Littlewood–Richardson rule isIt describes ordinary irreducible constituents in characteristic zero and Specht filtration factors over arbitrary $F$.

For induction by one letter, content $(1)$ permits exactly one added cell and each multiplicity is one. A Specht series for the requested induced module has successive quotients, ordered by addable nodes from bottom to top,Equivalently there is $0=N_0\subset N_1\subset N_2\subset N_3\subset N_4=S^{(4,2,2,1)}{\uparrow }^{S_{10}}$ with these four quotients in order. Their dimensions $567,300,768,525$ sum to $2160=10\cdot 216$, the dimension of the induced module.