## Section 2.14. Tensor products and duals of representations of Lie algebras

Problem 2.14.3 Let \(k\) denote the ground field of \(\mathfrak{g}\). Let \(f \in \Hom_{\mathfrak{g}}(V, U \otimes W^*)\). Then we can define \(\varphi(f) \in \Hom_{\mathfrak{g}}(V \otimes W, U)\) by setting \(\varphi(f)(v \otimes w) = f(v)w\) and extending by linearity, where \(f(v) \in U \otimes W^*\) is taken to act on \(w \in W\) the natural way as in Problem 2.11.3, that is, \((u \otimes \lambda)(w) = \lambda(w) u\) where \(\lambda \in W^*\). The \(\varphi(f)\) defined in this manner is well-defined since it's bilinear in \(v\) and \(w\). It's also a linear function of \(f\). Note that for \(\varphi(f)\) to be the zero map, it has to take all \(v \otimes w\) to zero, which is only possible if every \(f(v)\) is the zero tensor in \(U \otimes W^*\); therefore \(f\) is the zero map. Since \(\ker \varphi\) is trivial, \(\varphi\) is one-to-one, and since the domain and codomain have equal (finite) dimension, \(\varphi\) is onto.

Furthermore, fix \(a \in \mathfrak{g}\); then we have \begin{align*} \varphi(f)(a(v \otimes w)) &= \varphi(f)(av \otimes w + v \otimes aw) \\ &= f(av) w + f(v)(aw) \\ &= (a f(v))w + f(v)(aw) \end{align*} and using the definitions of the tensor product and dual of representations, if \(f(v) = u \otimes \lambda\), then: \begin{align*} (a(f(v)))w &= (a(u \otimes \lambda))w \\ &= (au \otimes \lambda)w + (u \otimes a \lambda)w \\ &= \lambda(w)(au) + (u \otimes (w \mapsto -\lambda(aw)))w \\ &= a(f(v)w) - f(v)(aw) \end{align*} which can be extended by linearity to conclude that this equality is true regardless of the form of \(f(v)\); and so, continuing from before, we have that \(\varphi(f)(a(v \otimes w)) = a(f(v)w) = a(\varphi(f)(v \otimes w))\). Extending this by linearity, we conclude that \(\varphi(f)\) is \(\mathfrak{g}\)-linear when \(f\) is, so the isomorphism also takes \(\Hom_\mathfrak{g}(V, U \otimes W)\) to \(\Hom_\mathfrak{g}(V \otimes W, U)\).