[proofplan] The highest-weight constraints are exactly the [tensor product highest weight bound](/theorems/9389): the top possible highest weight is $\lambda+\mu$, it occurs once, and no irreducible summand can have highest weight larger than $\lambda+\mu$ in the dominance order. After those multiplicity constraints are known, complete reducibility gives a finite direct-sum decomposition of $V(\lambda)\otimes V(\mu)$. Taking dimensions in that decomposition gives the character dimension identity, because dimension is additive on finite direct sums and multiplicative on tensor products. [/proofplan] [step:Apply the tensor product highest weight bound to get the top multiplicity and vanishing range] The modules $V(\lambda)$ and $V(\mu)$ are irreducible finite-dimensional highest weight $\mathfrak g$-modules with dominant integral highest weights $\lambda$ and $\mu$. Therefore [citetheorem:9389] applies to the [tensor product](/page/Tensor%20Product) $V(\lambda)\otimes V(\mu)$. It gives that $V(\lambda+\mu)$ occurs as an irreducible summand with multiplicity $1$, and that if $V(\nu)$ occurs as an irreducible summand, then \begin{align*} \nu\leq \lambda+\mu \end{align*} in the dominance order. By the definition of $c_{\lambda\mu}^{\nu}$ as the multiplicity of $V(\nu)$ in $V(\lambda)\otimes V(\mu)$, this proves \begin{align*} c_{\lambda\mu}^{\lambda+\mu}=1 \end{align*} and \begin{align*} c_{\lambda\mu}^{\nu}=0 \quad \text{unless } \nu\leq \lambda+\mu. \end{align*} [guided] We first isolate the purely representation-theoretic constraint on which irreducible summands can appear. The hypotheses needed for [citetheorem:9389] are exactly present here: $\mathfrak g$ is a complex semisimple [Lie algebra](/page/Lie%20Algebra), and $\lambda,\mu\in P^+$ are dominant integral weights, so $V(\lambda)$ and $V(\mu)$ are irreducible finite-dimensional highest weight modules. Applying [citetheorem:9389] to $V(\lambda)\otimes V(\mu)$ gives two conclusions. First, the irreducible highest weight module with highest weight $\lambda+\mu$ appears once: \begin{align*} c_{\lambda\mu}^{\lambda+\mu}=1. \end{align*} Second, if an irreducible module $V(\nu)$ appears in the tensor product, then its highest weight cannot exceed the sum of the highest weights: \begin{align*} \nu\leq \lambda+\mu. \end{align*} Since $c_{\lambda\mu}^{\nu}$ is defined to be the multiplicity of $V(\nu)$ in the tensor product decomposition, non-occurrence is exactly the statement $c_{\lambda\mu}^{\nu}=0$. Hence \begin{align*} c_{\lambda\mu}^{\nu}=0 \quad \text{unless } \nu\leq \lambda+\mu. \end{align*} This proves the first two displayed constraints. [/guided] [/step] [step:Take dimensions in the finite semisimple decomposition] By complete reducibility for finite-dimensional modules over a complex semisimple Lie algebra, the finite-dimensional module $V(\lambda)\otimes V(\mu)$ decomposes as \begin{align*} V(\lambda)\otimes V(\mu)\cong \bigoplus_{\nu\in P^+} V(\nu)^{\oplus c_{\lambda\mu}^{\nu}}, \end{align*} with only finitely many nonzero multiplicities $c_{\lambda\mu}^{\nu}$. Taking dimensions of both sides gives \begin{align*} \dim\bigl(V(\lambda)\otimes V(\mu)\bigr)=\sum_{\nu\in P^+}\dim\bigl(V(\nu)^{\oplus c_{\lambda\mu}^{\nu}}\bigr). \end{align*} For finite-dimensional complex vector spaces, dimension is multiplicative on tensor products and additive on finite direct sums, so \begin{align*} \dim\bigl(V(\lambda)\otimes V(\mu)\bigr)=\dim V(\lambda)\dim V(\mu) \end{align*} and, for each $\nu\in P^+$, \begin{align*} \dim\bigl(V(\nu)^{\oplus c_{\lambda\mu}^{\nu}}\bigr)=c_{\lambda\mu}^{\nu}\dim V(\nu). \end{align*} Substituting these two identities into the preceding dimension equality yields \begin{align*} \dim V(\lambda)\dim V(\mu)=\sum_{\nu\in P^+}c_{\lambda\mu}^{\nu}\dim V(\nu). \end{align*} Together with the multiplicity and vanishing statements proved above, this is the desired result. [/step]