Tensor Product Multiplicities and Littlewood Richardson

Part (i) is the multiplicativity and additivity of the formal character applied to the decomposition: the tensor product distributes over the direct sum, so $\mathrm{ch}(L(\lambda )\otimes L(\mu ))={\sum }_{\nu }{c}_{\lambda \mu }^{\nu }\mathrm{ch}L(\nu )$ in $\mathcal{R}$; the sum is finite by Tensor-product multiplicities for finite-dimensional simple modules.

Part (ii), comparison of coefficients. Let $V$ be finite-dimensional with decomposition $V\cong {⨁}_{\nu }L(\nu {)}^{\oplus a_{\nu }^{\mathrm{\prime }}}$, $a_{\nu }^{\mathrm{\prime }}=[V:L(\nu )]$. Then $\mathrm{ch}V={\sum }_{\nu }{a}_{\nu }^{\mathrm{\prime }}\mathrm{ch}L(\nu )$. Suppose also $\mathrm{ch}V={\sum }_{\nu }{a}_{\nu }\mathrm{ch}L(\nu )$ with integers $a_{\nu }$, both sums finite. Let ${\nu }_0$ be maximal in the root order among the indices with $a_{{\nu }_0}\mathrm{\ne }{a}_{{\nu }_0}^{\mathrm{\prime }}$ (if there is none, the two families are equal). Evaluating both characters in the weight ${\nu }_0$ and using that $m_{\eta }({\nu }_0)=0$ unless ${\nu }_0\le \eta$ with equality only for $\eta ={\nu }_0$, while $m_{{\nu }_0}({\nu }_0)=1$ [F2], gives $0=\dim V_{{\nu }_0}-\dim V_{{\nu }_0}={\sum }_{\eta \ge {\nu }_0}(a_{\eta }-a_{\eta }^{\mathrm{\prime }})m_{\eta }({\nu }_0)=(a_{{\nu }_0}-a_{{\nu }_0}^{\mathrm{\prime }})\mathrm{\ne }0,$ a contradiction. Hence $a_{\nu }=a_{\nu }^{\mathrm{\prime }}=[V:L(\nu )]$ for all $\nu$.

Part (ii), linear independence. Suppose ${\sum }_{\nu \in F}{b}_{\nu }\mathrm{ch}L(\nu )=0$ with a finite nonempty set $F$ and integers $b_{\nu }$, not all zero. Split at the $\nu$ with $b_{\nu }>0$ and $b_{\nu }<0$ and let $V^{+}={⨁}_{\nu \in F,b_{\nu }>0}L(\nu {)}^{\oplus b_{\nu }}$ and $V^{-}={⨁}_{\nu \in F,b_{\nu }<0}L(\nu {)}^{\oplus (-b_{\nu })}$; the vanishing of the alternating sum gives $\mathrm{ch}{V}^{+}=\mathrm{ch}{V}^{-}$ in $\mathcal{R}$. Both are characters of finite-dimensional modules, so by step 1.2 the multiplicity families $(b_{\nu }{)}_{\nu :b_{\nu }>0}$ and $(-b_{\nu }{)}_{\nu :b_{\nu }<0}$ agree on every $\nu$, forcing $b_{\nu }=0$ against the choice of $F$. Hence the characters $\mathrm{ch}L(\nu )$ are linearly independent.

Part (iii): by the tensor-product weight-space formula of [F1], $(L(\lambda )\otimes L(\mu ){)}_{\gamma }={⨁}_{\sigma +\tau =\gamma }L(\lambda {)}_{\sigma }\otimes L(\mu {)}_{\tau }$, and taking dimensions gives $\dim (L(\lambda )\otimes L(\mu ){)}_{\gamma }={\sum }_{\sigma +\tau =\gamma }{m}_{\lambda }(\sigma )m_{\mu }(\tau )$; only the finitely many pairs of weights of $L(\lambda )$ and $L(\mu )$ can contribute, so the sum is finite.

The decomposition of $V$ into simple summands and the additivity of the formal character [F2] give $\mathrm{ch}V={\sum }_{\mu \in {\mathrm{\Lambda }}^{+}}[V:L(\mu )]\mathrm{ch}L(\mu )$, a finite sum. Multiplying by the ring element $A(\rho )$ and using $A(\rho )\mathrm{ch}L(\mu )=A(\mu +\rho )$ from [F1] gives $A(\rho )\mathrm{ch}V={\sum }_{\mu \in {\mathrm{\Lambda }}^{+}}[V:L(\mu )]A(\mu +\rho )\mathrm{.}$

Coefficient of a dominant translate. For $\nu \in {\mathrm{\Lambda }}^{+}$ and any $\mu \in {\mathrm{\Lambda }}^{+}$, $[e^{\nu +\rho }]A(\mu +\rho )=[e^{\nu +\rho }]{\sum }_{w\in W}(-1{)}^{\mathrm{\ell }(w)}{e}^{w(\mu +\rho )}={\sum }_{w\in W}(-1{)}^{\mathrm{\ell }(w)}[w(\mu +\rho )=\nu +\rho ],$ a finite sum. The term $w=1$ contributes $1$ when $\mu =\nu$. If $w\mathrm{\ne }1$ and $w(\mu +\rho )=\nu +\rho$, then $\mu +\rho =w^{-1}(\nu +\rho )$ would be a strictly dominant weight conjugate to the strictly dominant weight $\nu +\rho$, which by [F3] forces $w=1$, a contradiction; hence all such terms are $0$ and $[e^{\nu +\rho }]A(\mu +\rho )={\delta }_{\mu \nu }$.

Extraction. Taking the coefficient of $e^{\nu +\rho }$ in step 1.1 and using step 1.2 gives $[e^{\nu +\rho }]A(\rho )\mathrm{ch}V={\sum }_{\mu \in {\mathrm{\Lambda }}^{+}}[V:L(\mu )]{\delta }_{\mu \nu }=[V:L(\nu )],$ which is the asserted extraction formula; here the sum over $\mu$ is finite because $V$ is finite-dimensional.

Let $V=L(\lambda )\otimes L(\mu )$. By multiplicativity and the character formula [F3], $A(\rho )\mathrm{ch}V=A(\lambda +\rho )\mathrm{ch}L(\mu )$. Expanding gives ${\sum }_{w,\sigma }(-1{)}^{\mathrm{\ell }(w)}{m}_{\mu }(\sigma )e^{w(\lambda +\rho )+\sigma }$. For each fixed $w$, put $\sigma =w\tau$; Weyl invariance [F4] gives $m_{\mu }(w\tau )=m_{\mu }(\tau )$. The finite double sum is therefore ${\sum }_{w,\tau }(-1{)}^{\mathrm{\ell }(w)}{m}_{\mu }(\tau )e^{w(\lambda +\rho +\tau )}={\sum }_{\tau }{m}_{\mu }(\tau )A(\lambda +\rho +\tau )$.

Extract the coefficient of $e^{\nu +\rho }$ using [F2]: $c_{\lambda \mu }^{\nu }=[e^{\nu +\rho }]A(\rho )\mathrm{ch}V={\sum }_{\sigma }{m}_{\mu }(\sigma )[e^{\nu +\rho }]A(\lambda +\rho +\sigma ),$ and $[e^{\nu +\rho }]A(\lambda +\rho +\sigma )={\sum }_{w\in W}(-1{)}^{\mathrm{\ell }(w)}[w(\lambda +\rho +\sigma )=\nu +\rho ]$. Hence $c_{\lambda \mu }^{\nu }={\sum }_{w\in W}(-1{)}^{\mathrm{\ell }(w)}{m}_{\mu }(w^{-1}(\nu +\rho )-(\lambda +\rho )),$ because the condition $w(\lambda +\rho +\sigma )=\nu +\rho$ is equivalent to $\sigma =w^{-1}(\nu +\rho )-(\lambda +\rho )$, and terms with $\sigma$ outside the finite weight set of $L(\mu )$ contribute $m_{\mu }(\sigma )=0$.

Equivalent form. Substituting $w\mapsto w^{-1}$ in step 2.1 and using $(-1{)}^{\mathrm{\ell }(w^{-1})}=(-1{)}^{\mathrm{\ell }(w)}$ gives $c_{\lambda \mu }^{\nu }={\sum }_w(-1{)}^{\mathrm{\ell }(w)}{m}_{\mu }(w(\nu +\rho )-(\lambda +\rho ))$, which is the first displayed formula. Applying to $m_{\mu }$ the Weyl-invariance of weight multiplicities [F4] with the group element $w^{-1}$ gives $m_{\mu }(w(\nu +\rho )-(\lambda +\rho ))=m_{\mu }(\nu +\rho -w^{-1}(\lambda +\rho ));$ relabelling $w^{\mathrm{\prime }}\mapsto w^{-1}$ in the sum yields the equivalent form $c_{\lambda \mu }^{\nu }={\sum }_{w\in W}(-1{)}^{\mathrm{\ell }(w)}{m}_{\mu }(\nu +\rho -w(\lambda +\rho ))\mathrm{.}$ Only finitely many summands are nonzero in either form, since $W$ is finite [F4] and $m_{\mu }$ has finite support.

Rewrite the Steinberg sum by the second argument. For $w\in W$ put ${\phi }_w:=w^{-1}(\nu +\rho )-(\lambda +\rho )$ and ${\psi }_w+\rho :=w^{-1}(\nu +\rho )={\phi }_w+\lambda +\rho$, so that $c_{\lambda \mu }^{\nu }={\sum }_w(-1{)}^{\mathrm{\ell }(w)}{m}_{\mu }({\phi }_w)$ by [F1], the sum being finite. Fix a weight $\phi$ of $L(\mu )$ and let $W(\phi )=\{w\in W:{\phi }_w=\phi \}=\{w:w^{-1}(\nu +\rho )=\psi +\rho \}$ where $\psi =\phi +\lambda$. This set is nonempty exactly when $\psi +\rho$ is $W$-conjugate to $\nu +\rho$.

Regular weights contribute one term each. Suppose $\phi$ is regular, so that $\psi +\rho$ has trivial stabilizer. If $W(\phi )\mathrm{\ne }\mathrm{\varnothing }$, then $\nu +\rho =w(\psi +\rho )$ is strictly dominant for any $w\in W(\phi )$, so by the uniqueness in [F2] there is exactly one such $w$, namely $w=u(\phi )$ where $u(\phi )$ is the unique element with $u(\phi )(\psi +\rho )$ strictly dominant; in that case $\nu (\phi )=u(\phi )(\psi +\rho )-\rho =\nu$, and the contribution of $\phi$ to $c_{\lambda \mu }^{\nu }$ is $(-1{)}^{\mathrm{\ell }(u(\phi ))}{m}_{\mu }(\phi )$. If $W(\phi )=\mathrm{\varnothing }$, or if $\nu (\phi )\mathrm{\ne }\nu$, the weight $\phi$ contributes nothing to $c_{\lambda \mu }^{\nu }$.

Irregular weights cancel. Suppose $\phi$ is not regular: $\psi +\rho$ is fixed by some reflection $s$. Then $W(\phi )$ is stable under right multiplication by $s$, because $(ws{)}^{-1}(\nu +\rho )=s{w}^{-1}(\nu +\rho )=s(\psi +\rho )=\psi +\rho$ for every $w\in W(\phi )$; the map $w\mapsto ws$ is a fixed-point-free involution of $W(\phi )$, and by [F3] the signs of paired terms are opposite. Since the multiplicity $m_{\mu }(\phi )$ is the same for paired terms, the total contribution of the group $W(\phi )$ to the sum of step 1.1 is $0$.

Combining step 1.2 and 2.1, the value of $c_{\lambda \mu }^{\nu }$ is the sum of $(-1{)}^{\mathrm{\ell }(u(\phi ))}{m}_{\mu }(\phi )$ over the regular weights $\phi$ of $L(\mu )$ with $\nu (\phi )=\nu$, which is the displayed Racah--Speiser formula; the sum is finite because $L(\mu )$ has finitely many weights. If $c_{\lambda \mu }^{\nu }\mathrm{\ne }0$, the displayed sum is nonzero, so at least one regular weight $\phi$ satisfies $\nu (\phi )=\nu$; this proves the final assertion.

First suppose that $\omega$ is minuscule and nonzero; by [F1] it is a fundamental weight ${\omega }_i$. We argue by induction on the rank of the irreducible root system. Let $\lambda$ be dominant integral and write $\beta :={\omega }_i-\lambda ={\sum }_j{m}_j{\alpha }_j\in Q_{+}$. If some $k\mathrm{\ne }i$ has $m_k=0$, delete the vertex $k$ from the Dynkin diagram. By [F6], the root subsystem generated by the remaining simple roots is the orthogonal direct sum of the subsystems ${\mathrm{\Phi }}_C$ for the connected components $C$ of the deleted diagram; their spans are mutually orthogonal, and $\beta$ is the sum of its component projections ${\beta }_C$. In each component not containing $i$, the projection of ${\omega }_i$ is zero, so ${\beta }_C=-{\lambda }_C$, where ${\lambda }_C$ is dominant for that component. Writing ${\beta }_C={\sum }_{j\in C}{m}_j{\alpha }_j$, we have $({\beta }_C,{\beta }_C)=-{\sum }_{j\in C}{m}_j({\lambda }_C,{\alpha }_j)\le 0$ because $({\lambda }_C,{\alpha }_j)=\frac{1}{2}({\alpha }_j,{\alpha }_j)⟨{\lambda }_C,{\alpha }_j^{\vee }⟩\ge 0$. Positive definiteness gives ${\beta }_C={\lambda }_C=0$. The component $C_i$ containing $i$ has smaller rank; the projection of ${\omega }_i$ is its fundamental weight, still minuscule, and ${\lambda }_{C_i}$ is dominant integral with ${\omega }_i{\mathrm{\mid }}_{C_i}-{\lambda }_{C_i}={\beta }_{C_i}\in Q_{+}(C_i)$. The induction hypothesis applies to the irreducible lower-rank system ${\mathrm{\Phi }}_{C_i}$ and gives ${\beta }_{C_i}=0$, hence $\beta =0$. Thus, if $\beta \mathrm{\ne }0$, then $m_j>0$ for every $j\mathrm{\ne }i$.

A root-lattice element with all pairings bounded by $1$ vanishes: if $\xi \in Q$ satisfies $\mathrm{\mid }⟨\xi ,{\beta }^{\vee }⟩\mathrm{\mid }\le 1$ for every coroot ${\beta }^{\vee }$, then $\xi =0$. Suppose not, and choose a counterexample $\xi ={\sum }_k{m}_k{\alpha }_k$ with ${\sum }_k\mathrm{\mid }{m}_k\mathrm{\mid }$ minimal. Then $(\xi ,\xi )={\sum }_k{m}_k(\xi ,{\alpha }_k)>0$ by positive definiteness, so some $k$ has $m_k\mathrm{\ne }0$ and $(\xi ,{\alpha }_k)$ of the same sign as $m_k$; replacing $\xi$ by $-\xi$ if necessary, we may assume $m_k>0$ and $(\xi ,{\alpha }_k)>0$, so $⟨\xi ,{\alpha }_k^{\vee }⟩=2(\xi ,{\alpha }_k)\mathrm{/}({\alpha }_k,{\alpha }_k)=1$ by the bound. Then $s_k\xi =\xi -{\alpha }_k$ is again a counterexample, since $\mathrm{\mid }⟨s_k\xi ,{\beta }^{\vee }⟩\mathrm{\mid }=\mathrm{\mid }⟨\xi ,s_k{\beta }^{\vee }⟩\mathrm{\mid }\le 1$ for all coroots (the set of coroots is $W$-stable), and it has coordinate sum ${\sum }_j\mathrm{\mid }{m}_j\mathrm{\mid }-1$, contradicting minimality. Hence $\xi =0$.

The weight set of a finite-dimensional module is $W$-stable. Let $M$ be such a module, let $\mu$ be a weight, and let $\alpha ={\alpha }_k$ be simple with $t=⟨\mu ,{\alpha }^{\vee }⟩$. The subalgebra ${\mathfrak{s}\mathfrak{l}}_2(\alpha )$ acts on $M$. Decompose it into irreducibles and write a nonzero weight vector $v$ as the sum of its components in their weight-$t$ spaces. In each irreducible summand where that component is nonzero, the highest weight is some $m\ge \mathrm{\mid }t\mathrm{\mid }$ with $m\equiv t(\mathrm{m}\mathrm{o}\mathrm{d}2)$. If $t\ge 0$, lowering that component by $t$ steps is nonzero and has weight $-t$; if $t<0$, raising it by $-t$ steps is nonzero and has weight $-t$. Their direct sum is nonzero and has weight $\mu -t\alpha =s_k\mu$. Thus each simple reflection preserves the set of weights, and these reflections generate $W$.

(2)$\Rightarrow$(1): suppose (2) holds and $\omega$ is not minuscule. Then by [F1] there is a positive root $\alpha$ with $⟨\omega ,{\alpha }^{\vee }⟩\ge 2$. Let $v_{\omega }\mathrm{\ne }0$ be a highest weight vector; the root vector $x_{-\alpha }$ of the ${\mathfrak{s}\mathfrak{l}}_2(\alpha )$-triple [F5] satisfies $x_{-\alpha }{v}_{\omega }\mathrm{\ne }0$, because $v_{\omega }$ is annihilated by all positive root vectors and spans the highest weight space of the ${\mathfrak{s}\mathfrak{l}}_2(\alpha )$-module it generates, of highest weight $⟨\omega ,{\alpha }^{\vee }⟩\ge 1$; this vector has weight $\omega -\alpha$ (Root vectors shift weights). By (2) there is $w\in W$ with $\omega -\alpha =w\omega$, and the $W$-invariance of the form [F2] gives $(\omega -\alpha ,\omega -\alpha )=(\omega ,\omega )$, while $(\omega -\alpha ,\omega -\alpha )=(\omega ,\omega )-2(\omega ,\alpha )+(\alpha ,\alpha )<(\omega ,\omega )$ because $2(\omega ,\alpha )=⟨\omega ,{\alpha }^{\vee }⟩(\alpha ,\alpha )\ge 2(\alpha ,\alpha )>(\alpha ,\alpha )$. This contradiction proves (2)$\Rightarrow$(1).

Suppose $⟨{\omega }_i-\lambda ,{\alpha }_i^{\vee }⟩\le 0$. For every $j\mathrm{\ne }i$, $⟨{\omega }_i-\lambda ,{\alpha }_j^{\vee }⟩=-⟨\lambda ,{\alpha }_j^{\vee }⟩\le 0$, since ${\omega }_i$ is the $i$th fundamental weight and $\lambda$ is dominant. Together with the assumed nonpositivity at $i$, all simple-coroot pairings of $\beta ={\omega }_i-\lambda ={\sum }_k{m}_k{\alpha }_k$ are nonpositive. Therefore $(\beta ,\beta )={\sum }_k{m}_k(\beta ,{\alpha }_k)={\sum }_k{m}_k\frac{({\alpha }_k,{\alpha }_k)}{2}⟨\beta ,{\alpha }_k^{\vee }⟩\le 0$, because each $m_k\ge 0$. Positive definiteness gives $\beta =0$ and $\lambda ={\omega }_i$.

It remains to exclude the case $⟨{\omega }_i-\lambda ,{\alpha }_i^{\vee }⟩>0$, which is $1$ because ${\omega }_i$ pairs by $1$ and $\lambda$ is dominant integral. Write $\beta ={\omega }_i-\lambda ={\sum }_k{m}_k{\alpha }_k$. Then $m_k>0$ for $k\mathrm{\ne }i$ by step 1.1, and the Cartan-integer formula gives $1=⟨\beta ,{\alpha }_i^{\vee }⟩=2m_i+{\sum }_{j\mathrm{\ne }i}{m}_j⟨{\alpha }_j,{\alpha }_i^{\vee }⟩$. The off-diagonal Cartan integers are nonpositive, so $2m_i\ge 1$ and the integer $m_i$ is positive; hence every $m_j\ge 1$. Let $\theta$ be the highest root and write ${\theta }^{\vee }={\sum }_j{c}_j{\alpha }_j^{\vee }$ with all $c_j>0$ as in [F1]. By Existence and uniqueness of the highest root, $(\theta ,{\alpha }_j)\ge 0$ for every simple root. Since ${\theta }^{\vee }$ is a positive scalar multiple of $\theta$, this gives $⟨{\alpha }_j,{\theta }^{\vee }⟩=2({\alpha }_j,\theta )\mathrm{/}(\theta ,\theta )\ge 0$; at least one is positive because the simple roots span and $\theta \mathrm{\ne }0$. Therefore $\beta ({\theta }^{\vee })={\sum }_j{m}_j⟨{\alpha }_j,{\theta }^{\vee }⟩>0$. By [F1], $⟨{\omega }_i,{\theta }^{\vee }⟩=c_i=1$, so $\lambda ({\theta }^{\vee })=1-\beta ({\theta }^{\vee })<1$. This is a nonnegative integer because $\lambda$ is dominant integral and every $c_j>0$; hence it is zero, all simple-coroot pairings of $\lambda$ vanish, and $\lambda =0$. Thus $\beta ={\omega }_i$; since $\beta$ was in $Q_{+}$, this proves ${\omega }_i\in Q$ before applying the root-lattice lemma.