Outer Products, Skew Specht Modules, and Littlewood–Richardson Coefficients

Let ${\pi }_G:G\times H\to G$ and ${\pi }_H:G\times H\to H$ be the coordinate projections. Pulling $V$ and $W$ back along these homomorphisms and taking their tensor product gives the stated $G\times H$ action, because the two factor actions multiply componentwise. The tensor-product construction is well defined on $V{\otimes }_{\mathbb{C}}W$.

Let $X$ be an irreducible $G\times H$-module and restrict it to $N:=G\times \{1_H\}$. By [F5] this nonzero restriction has an irreducible constituent $S$ with character $\chi \in \mathrm{Irr}(G)$. Conjugation by $(g,h)$ acts on $N$ by conjugation by $g$, so $\chi$ is fixed because characters are constant on conjugacy classes. Hence the inertia group of $\chi$ in $G\times H$ is all of $G\times H$.

For irreducible characters $\chi ,{\chi }^{\mathrm{\prime }}$ of $G$ and $\psi ,{\psi }^{\mathrm{\prime }}$ of $H$, the finite sum over $G\times H$ factors as $⟨\chi ⊠\psi ,{\chi }^{\mathrm{\prime }}⊠{\psi }^{\mathrm{\prime }}{⟩}_{G\times H}=\frac{1}{\mathrm{\mid }G\mathrm{\mid }\mathrm{\mid }H\mathrm{\mid }}{\sum }_{g\in G,h\in H}\chi (g)\psi (h)\overline{{\chi }^{\mathrm{\prime }}(g)}\overline{{\psi }^{\mathrm{\prime }}(h)}=⟨\chi ,{\chi }^{\mathrm{\prime }}{⟩}_G⟨\psi ,{\psi }^{\mathrm{\prime }}{⟩}_H$. Thus the external products of irreducible characters are orthonormal by [F4].

By [F3], the character of this representation at $(g,h)$ is ${\chi }_V(g){\chi }_W(h)$. Maschke decomposes representations affording honest characters into irreducibles; distributivity of tensor products over direct sums then shows that this formula agrees with the bilinear extension from the irreducible-character bases.

The representation $\tilde{S}(g,h):=S(g)$ extends $S$ to $G\times H$, and the quotient by $N$ is canonically $H$. Gallagher's theorem therefore says that the irreducibles above $\chi$ are exactly $S⊠W$ for $W$ irreducible over $H$, without repetition. Since $X$ lies above its chosen constituent $\chi$, this proves that every irreducible of $G\times H$ occurs exactly once in the family of external products. The argument includes a trivial factor, for which the relevant irreducible-character set is a singleton.

The trivial character is the unit in each character ring, and its external product is the trivial character of $G\times H$. Thus $\mathrm{\Phi }(1\otimes 1)=1$.

Steps 1.3 and 2.2 show that the external products are precisely the irreducible characters of $G\times H$, each once, and are orthonormal. They are therefore an orthonormal $\mathbb{Z}$-basis of $R(G\times H)$ by [F1, F4].

For irreducible basis elements, the pointwise product formula in step 2.1 gives $\mathrm{\Phi }((\chi \otimes \psi )({\chi }^{\mathrm{\prime }}\otimes {\psi }^{\mathrm{\prime }}))=\mathrm{\Phi }(\chi {\chi }^{\mathrm{\prime }}\otimes \psi {\psi }^{\mathrm{\prime }})=(\chi ⊠\psi )({\chi }^{\mathrm{\prime }}⊠{\psi }^{\mathrm{\prime }})$. Bilinearity extends this identity to all virtual characters, including zero and negative combinations. Thus $\mathrm{\Phi }$ is multiplicative under the tensor-product algebra structure [F8].

By Maschke's theorem, any honest character of $G\times H$ is a nonnegative integral sum of its irreducible characters. Step 3.1 identifies each such character as one external product, proving the positivity assertion.

The irreducible characters form free $\mathbb{Z}$-bases of $R(G)$ and $R(H)$, so [F7] gives the basis $(\chi \otimes \psi )$ of $R(G){\otimes }_{\mathbb{Z}}R(H)$. Step 3.1 shows that the bilinear map sends this basis bijectively to the basis of the target; hence it is a $\mathbb{Z}$-module isomorphism.

Steps 2.3, 3.2, and 4.2 prove the unital ring isomorphism; steps 3.1 and 4.1 prove the orthonormal-basis and positivity claims.

For $k=sh{s}^{-1}\in K$, covariance gives $\mathrm{\Phi }(f)(xk)=f(xks)=f(xsh)=h^{-1}\cdot f(xs)=k^{-1}\cdot \mathrm{\Phi }(f)(x)$, where the last action is that of ${}^{s}W$ by [F2]. Thus $\mathrm{\Phi }(f)$ is $K$-covariant and belongs to ${\mathrm{Ind}}_K^G({}^{s}W)$.

Define $\mathrm{\Psi }(\phi )(x):=\phi (x{s}^{-1})$. For $h\in H$, $\mathrm{\Psi }(\phi )(xh)=\phi (xh{s}^{-1})=\phi ((x{s}^{-1})(sh{s}^{-1}))=h^{-1}\cdot \phi (x{s}^{-1})=h^{-1}\cdot \mathrm{\Psi }(\phi )(x)$ by $K$-covariance and [F2]. Hence $\mathrm{\Psi }(\phi )\in {\mathrm{Ind}}_H^{G}W$.

For $x_0,x\in G$, $\mathrm{\Phi }(x_0\cdot f)(x)=f(x_0^{-1}xs)=\mathrm{\Phi }(f)(x_0^{-1}x)=(x_0\cdot \mathrm{\Phi }(f))(x)$, so $\mathrm{\Phi }$ is $G$-equivariant. It is complex-linear by the pointwise module operations.

For $x_0,x\in G$, $\mathrm{\Psi }(x_0\cdot \phi )(x)=\phi (x_0^{-1}x{s}^{-1})=\mathrm{\Psi }(\phi )(x_0^{-1}x)=(x_0\cdot \mathrm{\Psi }(\phi ))(x)$, so $\mathrm{\Psi }$ is $G$-equivariant. It is complex-linear by the pointwise operations.

Substitution gives $\mathrm{\Psi }(\mathrm{\Phi }(f))(x)=f(x{s}^{-1}s)=f(x)$ and $\mathrm{\Phi }(\mathrm{\Psi }(\phi ))(x)=\phi (xs{s}^{-1})=\phi (x)$ for every $x\in G$. Thus $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are inverse $G$-module isomorphisms.

The induced modules in step 3.1 are isomorphic, so their characters are equal by [F3]. Both honest characters lie in $R(G)$ by [F5], giving ${\mathrm{Ind}}_K^G({}^s\chi )={\mathrm{Ind}}_H^G\chi$ there. The formulas use no choice principle.

Since $C\le P$ and $D\le Q$, the componentwise product $C\times D$ contains the identity and is closed under products and inverses in $P\times Q$, so it is a subgroup. The basis inclusion $\mathbb{C}[C\times D]\hookrightarrow \mathbb{C}[P\times Q]$ respects multiplication, giving the right subgroup-ring action used below. All three group indices are finite.

On group-basis and module generators define $\mathrm{\Phi }([(p,q)]\otimes (x\otimes y)):=([p]\otimes x)\otimes ([q]\otimes y)$. For fixed $(p,q)$, the formula is complex-bilinear in $x,y$, so it descends to $X{\otimes }_{\mathbb{C}}Y$. For $(c,d)\in C\times D$, the images of $[(pc,qd)]\otimes (x\otimes y)$ and $[(p,q)]\otimes (cx\otimes dy)$ both equal $([p]\otimes cx)\otimes ([q]\otimes dy)$. Thus the pairing is balanced over $\mathbb{C}[C\times D]$, and [F4] gives a well-defined map on the source tensor product.

Define $\mathrm{\Psi }(([p]\otimes x)\otimes ([q]\otimes y)):=[(p,q)]\otimes (x\otimes y)$. Moving $c\in C$ across the first factor preserves its value because $[(pc,q)]=[(p,q)(c,1)]$ and the balanced relation moves $(c,1)$ to the action $cx$; the same check with $d\in D$ uses $[(p,qd)]=[(p,q)(1,d)]$. The formula is complex-bilinear in the two outer factors, so [F4] gives a well-defined reverse map.

The finite group rings in [F3] have finite bases, so their tensor models on the finite-dimensional inputs [F8] are finite-dimensional. Apply [F2] with $R=\mathbb{C}$ to the subgroup inclusions $C\times D\le P\times Q$, $C\le P$, and $D\le Q$; it identifies the induction terms with the group-ring tensor models used above.

The composites $\mathrm{\Psi }\mathrm{\Phi }$ and $\mathrm{\Phi }\mathrm{\Psi }$ fix every displayed group-basis and pure-tensor generator. These generators span their respective tensor products by [F3, F4], so the composites are identity maps and $\mathrm{\Phi }$ is an isomorphism.

Left multiplication by $(p_0,q_0)$ sends $[(p,q)]$ to $[(p_{0}p,q_{0}q)]$, which $\mathrm{\Phi }$ sends to the corresponding left actions on both target factors. The same formula commutes with $C$- and $D$-module homomorphisms on $X,Y$, so the isomorphism is $(P\times Q)$-equivariant and natural.

The tensor-model identifications in step 2.1 turn $\mathrm{\Phi }$ into the module isomorphism in (i). If $D=Q$, the canonical map $\mathbb{C}[Q]{\otimes }_{\mathbb{C}[Q]}Y\to Y$, $[q]\otimes y\mapsto qy$, is an isomorphism with inverse $y\mapsto [e_Q]\otimes y$; if also $C=P$, the same identity gives ${\mathrm{Ind}}_P^{P}X\cong X$. Thus (ii) holds.

Since $X,Y$ are finite-dimensional [F8] and the finite group rings in [F3] have finite bases, the tensor models of step 2.1 are finite-dimensional and have characters. Taking characters of the isomorphism in step 3.1 and applying [F6] to the two representations pulled back along the coordinate projections of $P\times Q$ gives the stated identity in $R(P\times Q)$ by [F7] and [F9]. All maps were defined explicitly, so no choice principle is used.

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.

For each $t$, define ${\beta }_t(i)=i+1$ from $\{0,\dots ,t-1\}$ to $\{1,\dots ,t\}$, with the empty bijection if $t=0$. The group isomorphism $c_t(\sigma )={\beta }_t\sigma {\beta }_t^{-1}$ preserves products and carries the zero-based ordered blocks to the one-based blocks in [F3]. Pullback along $c_t$ identifies representations and character groups. Explicitly, for a one-based subgroup $H^1$ and module $W$, set $H^0=c_{m+n}^{-1}(H^1)$ and $h{\cdot }_{0}w=c_{m+n}(h){\cdot }_{1}w$. The map $F^1\mapsto F^0=F^1\circ c_{m+n}$ has inverse composition with $c_{m+n}^{-1}$, and $F^0(gh)=c_{m+n}(h{)}^{-1}{\cdot }_1{F}^0(g)=h^{-1}{\cdot }_0{F}^0(g)$. Since $c_{m+n}(x^{-1}g)=c_{m+n}(x{)}^{-1}{c}_{m+n}(g)$, it also intertwines the transported left actions in [F5]. Thus the one-based calculation represents the same outer product in $R_S$.

First take honest characters $\chi ,\psi ,\eta$ of $S_m,S_n,S_r$. In the one-based realization let $B_1=\{1,\dots ,m\}$, $B_2=\{m+1,\dots ,m+n\}$, and $B_3=\{m+n+1,\dots ,m+n+r\}$, allowing empty blocks, and let $J$ be the permutations preserving each $B_i$. The identity, products, and inverses preserve each block, so $J\le S_{m+n+r}$ by [F12]. Restriction to the three blocks identifies $J$ with the componentwise product $S_m\times S_n\times S_r$ by [F10, F11]. Both parenthesized two-block embeddings have image $J$, and on a triple $(a,b,c)$ both parenthesized external product characters have value $\chi (a)\psi (b)\eta (c)$ by [F3]. Hence the two parenthesized external product characters on $J$ agree.

In the one-based realization define $s_{m,n}\in S_{m+n}$ by $s_{m,n}(i)=n+i$ for $1\le i\le m$ and $s_{m,n}(m+j)=j$ for $1\le j\le n$. Its two ranges are disjoint and cover $\{1,\dots ,m+n\}$, so it is a permutation, also when one block is empty. If ${\iota }_{m,n}$ denotes the block embedding in [F3], its action on the two blocks gives $s_{m,n}{\iota }_{m,n}(a,b)s_{m,n}^{-1}={\iota }_{n,m}(b,a)$. Thus $s_{m,n}$ conjugates the first block subgroup to the second.