The Littlewood–Richardson rule for products of Schur functions

Fix $\lambda \supseteq \mu$ and put $d=\mathrm{\mid }\lambda \mathrm{\mid }-\mathrm{\mid }\mu \mathrm{\mid }$. If $d=0$, then $\lambda =\mu$, and [F2]–[F4] give the skew expansion $s_{\lambda \mathrm{/}\lambda }=1$ with its unique empty tableau. Suppose henceforth that $d>0$. For a nonnegative tuple $\alpha$ of total $d$, write $h_{\alpha }={\prod }_i{h}_{{\alpha }_i}$. The coefficient of $x^{\alpha }$ in [F4] counts the skew tableaux of content $\alpha$. Symmetry makes this coefficient equal to that of the sorted exponent partition $\pi$; [F10] and Hall duality [F5]–[F6] therefore give $⟨s_{\lambda \mathrm{/}\mu },h_{\alpha }{⟩}_H=\mathrm{\#}\mathrm{Tab}(\lambda \mathrm{/}\mu ,\alpha )$. A tuple with a negative entry contributes zero by [F7].

For a fixed $i$, filter a reading word to the letters $i,i+1$ and match each $i+1$ with the last still-unmatched preceding $i$, when available. After deleting matched pairs the unmatched letters are $(i+1{)}^a{i}^b$. Define $e_i$ by changing the last unmatched $i+1$ to $i$ when $a>0$, and $f_i$ by changing the first unmatched $i$ to $i+1$ when $b>0$. Matching parentheses shows that these are inverse partial operations: $e_i$ replaces $(a,b)$ by $(a-1,b+1)$ and $f_i$ does the reverse, leaving matched positions unchanged. No $e_i$ is available precisely when every prefix has at least as many $i$'s as $i+1$'s. Thus all $e_i$ are unavailable precisely for lattice words.

These operations preserve semistandard skew tableaux. To check this, retain only cells labeled $i,i+1$; they form a skew diagram, since adjoining to $\mu$ all cells with entry at most $j$ gives a partition for each $j$ by the row and column inequalities. Every two-cell column has $i$ above $i+1$. Each maximal rectangle of such columns has reading subword $i^k(i+1{)}^k$, which is neutral for matching; delete these rectangles successively. The remaining columns have one cell each, read from right to left. On them $e_i$ cannot have an $i+1$ immediately to its left, and $f_i$ cannot have an $i$ immediately to its right, by their definitions. A neighbor deleted in a two-row rectangle cannot cause either violation: the skew shape and inequalities would then force the variable cell itself to have a second cell in its column and to belong to that rectangle. Hence weak rows are preserved also before deletion. The variable cell has no other $i$ or $i+1$ in its column, so changing it by one preserves strict columns; other labels cannot violate an inequality. This proves the required tableau closure, including skew and disconnected shapes.

Fix $\nu ⊢d$, pad it to $r=d$, and set $\delta =(r-1,r-2,\dots ,0)$. Expanding the transpose of the Jacobi–Trudi matrix [F7] and applying step 1.1 gives $⟨s_{\lambda \mathrm{/}\mu },s_{\nu }{⟩}_H={\sum }_{\sigma \in S_r}\mathrm{sgn}(\sigma )\mathrm{\#}\mathrm{Tab}(\lambda \mathrm{/}\mu ,\alpha (\sigma ))$, where ${\alpha }_i(\sigma )={\nu }_{\sigma (i)}-\sigma (i)+i$. Thus we count signed pairs $(\sigma ,T)$ with $\mathrm{wt}(T)+\delta =({\nu }_{\sigma (i)}+r-\sigma (i){)}_{i=1}^r$; negative content gives no pairs. Each set is finite, and all entries of these tableaux lie in $\{1,\dots ,r\}$.

Cancel pairs for which $T$ is not lattice. Choose the earliest failing prefix; its final letter is $i+1$ and it is the first unmatched $i+1$ for this $i$, with all earlier prefixes lattice. In its $i$-signature $(i+1{)}^a{i}^b$ we have $a\ge 1$. If $a>b+1$, apply $e_i$ exactly $a-b-1$ times; if $a<b+1$, apply $f_i$ exactly $b+1-a$ times. This changes the signature to $(i+1{)}^{b+1}{i}^{a-1}$, keeping its first unmatched $i+1$ and every letter up to that position fixed. The equality $a=b+1$ cannot occur: it would give ${\alpha }_{i+1}={\alpha }_i+1$, hence equality of entries $i,i+1$ in $\alpha +\delta$, although that vector permutes the distinct numbers ${\nu }_j+r-j$. The new tableau $T^{\mathrm{\prime }}$ exists by steps 1.2 and 2.1 and has content ${\alpha }_i^{\mathrm{\prime }}={\alpha }_{i+1}-1$, ${\alpha }_{i+1}^{\mathrm{\prime }}={\alpha }_i+1$, with other entries unchanged. Replace $\sigma$ by ${\sigma }^{\mathrm{\prime }}=\sigma \circ (ii+1)$; then ${\alpha }^{\mathrm{\prime }}+\delta$ has exactly the permuted entries required in step 2.2. The first failing prefix and its index $i$ are unchanged, and repeating the operation restores $T$ and $\sigma$. This is a sign-reversing involution on all nonlattice pairs.

The uncancelled tableaux are lattice, so their content $\alpha$ is weakly decreasing. Thus $\alpha +\delta$ is strictly decreasing. The only strictly decreasing permutation of the strictly decreasing vector $\nu +\delta$ is itself, so step 2.2 forces $\sigma =\mathrm{id}$ and $\alpha =\nu$. These surviving pairs have positive sign and are exactly the LR tableaux in [F2]. Therefore $⟨s_{\lambda \mathrm{/}\mu },s_{\nu }{⟩}_H=c_{\mu \nu }^{\lambda }$ for every $\nu ⊢d$. The basis [F6] now gives $s_{\lambda \mathrm{/}\mu }={\sum }_{\nu ⊢d}{c}_{\mu \nu }^{\lambda }{s}_{\nu }$.

The coefficient of $s_{\lambda }$ in $s_{\mu }{s}_{\nu }$ is $⟨s_{\lambda },s_{\mu }{s}_{\nu }{⟩}_H$ by [F6], which is $c_{\mu \nu }^{\lambda }$ by [F8] and the skew expansion of steps 1.1 and 4.1. Noncontainment gives zero by [F4], and unequal degrees give zero by [F5], [F9]; these agree with the support rule [F2]. There are finitely many partitions of the product degree by [F1], proving the product formula.

Conversely, the product formula and [F8] give $⟨s_{\lambda \mathrm{/}\mu },s_{\nu }{⟩}_H=c_{\mu \nu }^{\lambda }$, so [F6] recovers the skew expansion. This proves the stated equivalence.

Step 1.1 treats empty skew shapes, including the empty partition. Empty factors are covered by $s_{\mathrm{\varnothing }}=1$ and the general coefficient calculation, while impossible containment, size or tableau conditions give zero by [F2] and step 5.1. For $d=1$ the determinant has size one and the cancellation has no nonlattice pairs. The involution uses the uniquely determined earliest failing prefix and finite signature operations; no representatives, rectifications or choice principle are required.