Brian Bi
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## 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 $$\Hom_\mathfrak{g}(V \otimes W, U)$$.