Top summand in a tensor product

The sum $\lambda +\mu$ is dominant integral because its simple-coroot values are sums of nonnegative integers (Integral, dominant, and strictly dominant weights). The vector $v_{\lambda }\otimes v_{\mu }$ is nonzero and is a highest weight vector of weight $\lambda +\mu$: ${\mathfrak{n}}^{+}$ acts by $x\cdot (v_{\lambda }\otimes v_{\mu })=x\cdot v_{\lambda }\otimes v_{\mu }+v_{\lambda }\otimes x\cdot v_{\mu }=0$ by [L1], and $H\cdot (v_{\lambda }\otimes v_{\mu })=(\lambda +\mu )(H)v_{\lambda }\otimes v_{\mu }$.

By [L4], negative-root words applied to a highest vector span each factor and are weight vectors or zero. Independence of distinct weight spaces therefore gives a direct weight-space decomposition of each factor; finite dimensionality makes it a finite sum. Tensoring bases of those spaces gives a weight basis of the tensor product under the action in [L1]. Every weight of the tensor product is a sum ${\nu }_1+{\nu }_2$ of a weight of $V(\lambda )$ and a weight of $V(\mu )$ (the direct sum decomposition of the tensor product into weight spaces has components of this form), so by [L1] and [L3] every weight satisfies ${\nu }_1+{\nu }_2\le \lambda +\mu$.

The $(\lambda +\mu )$-weight space of the tensor product is the direct sum of the spaces $V(\lambda {)}_{{\nu }_1}\otimes V(\mu {)}_{{\nu }_2}$ over pairs with ${\nu }_1+{\nu }_2=\lambda +\mu$; by [L3] the only pair with ${\nu }_1\le \lambda$, ${\nu }_2\le \mu$ and ${\nu }_1+{\nu }_2=\lambda +\mu$ is $(\lambda ,\mu )$, so this weight space is $V(\lambda {)}_{\lambda }\otimes V(\mu {)}_{\mu }=\mathbb{C}(v_{\lambda }\otimes v_{\mu })$ and has dimension one by [L1].

Decompose $V(\lambda )\otimes V(\mu )={⨁}_r{W}_r$ into irreducibles as in [L2], with $W_r\cong L({\nu }_r)$; the top weight ${\nu }_r$ of each summand is a weight of the tensor product, so ${\nu }_r\le \lambda +\mu$ by step 1.2, and $(W_r{)}_{\lambda +\mu }=0$ unless ${\nu }_r=\lambda +\mu$ because all weights of $L({\nu }_r)$ are $\le {\nu }_r$ by [L2]; each summand with top weight $\lambda +\mu$ contributes exactly one dimension by [L2], so comparison with step 2.1 shows that exactly one summand has highest weight $\lambda +\mu$.

By step 3.1 the tensor product contains exactly one summand $L(\lambda +\mu )\cong V(\lambda +\mu )$. Every other summand has highest weight ${\nu }_r\le \lambda +\mu$ by step 1.2 and cannot have equality, so its highest weight is strictly below $\lambda +\mu$. If $\mathfrak{g}=0$, then ${\mathfrak{h}}^{\ast }=0$, both highest-weight modules are one-dimensional, and the conclusion is a single trivial summand. More generally zero dominant weights are allowed throughout: their top lines remain nonzero, and no division by a weight occurs. This is the assertion.