[proofplan] We restrict the two $G$-representations to the compact torus $T$ and use the finite weight-space decompositions of $V$ and $W$. The [tensor product](/page/Tensor%20Product) distributes over these finite direct sums, producing summands $V_\alpha\otimes_{\mathbb C}W_\beta$. A direct computation of the tensor product action shows that this summand has $T$-weight $\alpha+\beta$. Grouping the direct-sum summands by the same total weight and taking dimensions gives the formula. [/proofplan] [step:Restrict the representations to $T$ and decompose them into weight spaces] Let \begin{align*} \rho_T:T\to GL(V) \end{align*} denote the restriction of $\rho$ to $T$, and let \begin{align*} \sigma_T:T\to GL(W) \end{align*} denote the restriction of $\sigma$ to $T$. Since $T$ is a compact torus and $V,W$ are finite-dimensional complex vector spaces carrying continuous representations of $T$, the weight-space decomposition theorem for compact tori, [citetheorem:9726], applies to both $\rho_T$ and $\sigma_T$. Hence \begin{align*} V=\bigoplus_{\alpha\in X^*(T)}V_\alpha \end{align*} and \begin{align*} W=\bigoplus_{\beta\in X^*(T)}W_\beta. \end{align*} Because $V$ and $W$ are finite-dimensional, only finitely many of the spaces $V_\alpha$ and $W_\beta$ are non-zero. [guided] We first forget the ambient group $G$ and keep only the action of the torus $T$. Define \begin{align*} \rho_T:T\to GL(V) \end{align*} to be the map $t\mapsto \rho(t)$, and define \begin{align*} \sigma_T:T\to GL(W) \end{align*} to be the map $t\mapsto \sigma(t)$. These are continuous finite-dimensional complex representations because $\rho$ and $\sigma$ are continuous and restriction of a continuous map to a subgroup is continuous. Now we apply the weight-space decomposition theorem for finite-dimensional representations of compact tori, [citetheorem:9726]. Its hypotheses are satisfied: $T$ is a compact torus, $V$ and $W$ are finite-dimensional complex vector spaces, and $\rho_T$ and $\sigma_T$ are continuous complex representations. Therefore the representation spaces split as direct sums of their character eigenspaces: \begin{align*} V=\bigoplus_{\alpha\in X^*(T)}V_\alpha \end{align*} and \begin{align*} W=\bigoplus_{\beta\in X^*(T)}W_\beta. \end{align*} The direct sums are finite in effect, because a finite-dimensional [vector space](/page/Vector%20Space) cannot contain infinitely many non-zero subspaces in a [direct sum](/page/Direct%20Sum). This is why the later multiplicity sum is finite even though it is written over the character group $X^*(T)$. [/guided] [/step] [step:Distribute the tensor product over the finite weight decompositions] Let \begin{align*} A:=\{\alpha\in X^*(T):V_\alpha\neq 0\} \end{align*} and \begin{align*} B:=\{\beta\in X^*(T):W_\beta\neq 0\}. \end{align*} The sets $A$ and $B$ are finite by the preceding step. The bilinear tensor product construction gives a natural vector-space isomorphism \begin{align*} \Theta:\bigoplus_{(\alpha,\beta)\in A\times B}\left(V_\alpha\otimes_{\mathbb C}W_\beta\right)\to V\otimes_{\mathbb C}W \end{align*} defined on each summand by \begin{align*} \Theta(v\otimes w)=v\otimes w \end{align*} for $v\in V_\alpha$ and $w\in W_\beta$. Thus \begin{align*} V\otimes_{\mathbb C}W = \bigoplus_{(\alpha,\beta)\in A\times B} V_\alpha\otimes_{\mathbb C}W_\beta. \end{align*} [guided] The previous step decomposed each factor. We now pass that decomposition through the tensor product. Define \begin{align*} A:=\{\alpha\in X^*(T):V_\alpha\neq 0\} \end{align*} and \begin{align*} B:=\{\beta\in X^*(T):W_\beta\neq 0\}. \end{align*} Both sets are finite, so all sums below are finite direct sums of finite-dimensional vector spaces. The tensor product is bilinear in the two factors, and finite direct sums distribute over tensor products. Concretely, define \begin{align*} \Theta:\bigoplus_{(\alpha,\beta)\in A\times B}\left(V_\alpha\otimes_{\mathbb C}W_\beta\right)\to V\otimes_{\mathbb C}W \end{align*} by sending a pure tensor $v\otimes w$ in the summand $V_\alpha\otimes_{\mathbb C}W_\beta$ to the same tensor $v\otimes w$ in $V\otimes_{\mathbb C}W$. This map is linear on the finite direct sum. The inverse map is obtained by writing each $v\in V$ uniquely as a finite sum of its weight components and each $w\in W$ uniquely as a finite sum of its weight components, then expanding bilinearly. Thus $\Theta$ is an isomorphism, and we may write \begin{align*} V\otimes_{\mathbb C}W = \bigoplus_{(\alpha,\beta)\in A\times B} V_\alpha\otimes_{\mathbb C}W_\beta. \end{align*} This step is the algebraic place where the [multiplicity formula](/theorems/2420) will come from: each pair of weights in the two factors contributes one tensor-product summand. [/guided] [/step] [step:Compute the weight of each tensor-product summand] Fix $\alpha\in A$ and $\beta\in B$. Let $v\in V_\alpha$ and $w\in W_\beta$. For every $t\in T$, the tensor product representation gives \begin{align*} t\cdot(v\otimes w)=\rho(t)v\otimes\sigma(t)w. \end{align*} Since $v\in V_\alpha$ and $w\in W_\beta$, this becomes \begin{align*} t\cdot(v\otimes w)=\alpha(t)v\otimes\beta(t)w. \end{align*} By scalar bilinearity of the tensor product over $\mathbb C$, \begin{align*} \alpha(t)v\otimes\beta(t)w=\alpha(t)\beta(t)(v\otimes w). \end{align*} Using additive notation for characters, $\alpha(t)\beta(t)=(\alpha+\beta)(t)$, so \begin{align*} t\cdot(v\otimes w)=(\alpha+\beta)(t)(v\otimes w). \end{align*} Therefore \begin{align*} V_\alpha\otimes_{\mathbb C}W_\beta\subseteq (V\otimes_{\mathbb C}W)_{\alpha+\beta}. \end{align*} [/step] [step:Group the summands with total weight $\eta$] For each $\gamma\in X^*(T)$, define the subspace \begin{align*} U_\gamma:=\bigoplus_{\substack{(\alpha, \beta)\in A\times B, \alpha+\beta=\gamma}}V_\alpha\otimes_{\mathbb C}W_\beta \end{align*} of $V\otimes_{\mathbb C}W$. The preceding step gives \begin{align*} U_\gamma\subseteq (V\otimes_{\mathbb C}W)_\gamma \end{align*} for every $\gamma\in X^*(T)$. Also, \begin{align*} V\otimes_{\mathbb C}W=\bigoplus_{\gamma\in X^*(T)}U_\gamma. \end{align*} We now prove the reverse inclusion for a fixed $\eta\in X^*(T)$. Let $x\in (V\otimes_{\mathbb C}W)_\eta$. Using the direct sum above, write \begin{align*} x=\sum_{\gamma\in X^*(T)}x_\gamma \end{align*} with $x_\gamma\in U_\gamma$ and with only finitely many non-zero $x_\gamma$. For every $t\in T$, \begin{align*} t\cdot x=\sum_{\gamma\in X^*(T)}\gamma(t)x_\gamma. \end{align*} Since $x$ has weight $\eta$, also \begin{align*} t\cdot x=\eta(t)x=\sum_{\gamma\in X^*(T)}\eta(t)x_\gamma. \end{align*} Subtracting gives \begin{align*} \sum_{\gamma\in X^*(T)}(\gamma(t)-\eta(t))x_\gamma=0 \end{align*} for every $t\in T$. Because distinct characters of a compact torus are linearly independent as complex-valued functions on $T$, every non-zero component $x_\gamma$ must have $\gamma=\eta$. Hence $x=x_\eta\in U_\eta$. Therefore \begin{align*} (V\otimes_{\mathbb C}W)_\eta=U_\eta. \end{align*} [/step] [step:Take dimensions to obtain the multiplicity formula] From the equality just proved, \begin{align*} (V\otimes_{\mathbb C}W)_\eta = \bigoplus_{\substack{(\alpha, \beta)\in A\times B, \alpha+\beta=\eta}}V_\alpha\otimes_{\mathbb C}W_\beta. \end{align*} Taking complex dimensions of this finite direct sum gives \begin{align*} \dim_{\mathbb C}(V\otimes_{\mathbb C}W)_\eta = \sum_{\substack{(\alpha, \beta)\in A\times B, \alpha+\beta=\eta}} \dim_{\mathbb C}(V_\alpha\otimes_{\mathbb C}W_\beta). \end{align*} For finite-dimensional complex vector spaces, tensor-product dimensions multiply, so \begin{align*} \dim_{\mathbb C}(V_\alpha\otimes_{\mathbb C}W_\beta)=\dim_{\mathbb C}V_\alpha\,\dim_{\mathbb C}W_\beta=m_\alpha n_\beta. \end{align*} Since $m_\alpha n_\beta=0$ whenever $\alpha\notin A$ or $\beta\notin B$, the same finite sum may be written over all characters: \begin{align*} \dim_{\mathbb C}(V\otimes_{\mathbb C}W)_\eta = \sum_{\substack{\alpha, \beta\in X^*(T), \alpha+\beta=\eta}}m_\alpha n_\beta. \end{align*} This is the claimed tensor-product weight multiplicity formula. [/step]