Lie algebra module structure on Hom(U,V) Let $\mathcal{g}$ be a Lie algebra. Let $U$ and $V$ be two representations of $\mathcal{g}$. I know that the dual $U^*$ is also a representation and therefore $U^* \otimes V$ is also a representation
Question: There is a natural isomorphism of vector space $$ \phi: U^* \otimes V \xrightarrow{\sim}Hom_{\mathbb{}C}(U,V).$$ Use this isomorphism to put a $\mathcal{g}$-module structure on $Hom_{\mathbb{}C}(U,V)$ so that $\phi$ becomes a homomorphism of $\mathcal{g}$-modules. Please how do I go ahead solving this problem.
I started by trying to see how the map $\phi$ is defined. I was able to come up with this map $$\phi(\tau \otimes v)(u) = \tau(u)v $$ for $\tau \in U^*, v \in V, u \in U$ but I dont know how to use this iso to put a $\mathcal{g}$-module structure on $Hom_{\mathbb{}C}(U,V)$.
I spoke to my friend about this problem and he advised that if $U=V = \mathbb{C^n}$, then the representation that I must obtain on $Hom_{\mathbb{}C}(\mathbb{C^n},\mathbb{C^n})$ should be the adjoint representation.
 A: Qiaochu's comment is excellent, as always. [Indeed, the natural action of a Lie group element $g$ on any $\tau \otimes v \in U^\star \otimes V$ is $g(\tau \otimes v) = (g^T)^{-1}\tau \otimes gv$, which, via your isomorphism $\phi : U^\star \otimes V \to {\rm Hom}(U,V)$, corresponds to the action $g(\psi) = g\psi g^{-1}$ for any $\psi \in {\rm Hom}(U,V)$. (The two actions "correspond" in the sense that $\phi(g(\tau \otimes v)) = g(\phi(\tau \otimes v))$.) Writing $g = \exp(tX)$ for some Lie algebra element $X$ and expanding to linear order in $t$, you see that $X(\tau \otimes v) = - X^T\tau \otimes v + \tau \otimes Xv$ and $X(\psi) = X\psi - \psi X$, which also "correspond" via $\phi$.]
I just want to comment on your remark that the action $X(\psi) = X\psi - \psi X$ is the adjoint representation when $U = V = \mathbb C^n$. If $\mathfrak g = \mathfrak{gl}_n\mathbb C$, then ${\rm Hom}(\mathbb C^n, \mathbb C^n) = \mathfrak g$, and since $X(\psi) = [X,\psi]$, we see that $\mathfrak g$ indeed acts on ${\rm Hom}(\mathbb C^n, \mathbb C^n)$ in the adjoint representation.
But in general, $\mathfrak g$ is not $\mathfrak{gl}_n\mathbb C$! For example, if $\mathfrak g = \mathfrak{sl}_n\mathbb C$, the Lie algebra of traceless $n\times n$ complex matrices, then decomposing any $\psi \in {\rm Hom}(\mathbb C^n, \mathbb C^n)$ as $\psi = \psi^{\rm Trace} + \psi^{\rm Trace-free}$ where $\psi^{\rm Trace} = \frac 1 n ({\rm Tr}\psi) {\rm  id}$ and $\psi^{\rm Trace-free} = \psi - \psi^{\rm Trace}$, you can identify $\psi^{\rm Trace}$ as a scalar and $\psi^{\rm Trace-free} $ as an $\mathfrak{sl}_n\mathbb C$ matrix. Since $X(\psi^{\rm Trace}) = 0 $ and $X(\psi^{\rm Trace-free}) = [X,\psi^{\rm Trace-free}]$, you see that ${\rm Hom}(\mathbb C^n, \mathbb C^n) $ decomposes into a direct sum of the trivial representation and the adjoint representation.
This decomposition has a really cool application in particle physics! The theory of elementary particles has an approximate $\mathfrak{sl}_3\mathbb C$ symmetry called "flavour symmetry". (Really the Lie algebra of the symmetry group is the real Lie algebra of $SU(3)$, whose complexification is $\mathfrak{sl}_3\mathbb C$.) Particles transforming in the ${\rm Hom}(\mathbb C^3, \mathbb C^3)$ representation of this $\mathfrak{sl}_3\mathbb C$ are called "mesons". The decomposition $\mathbf{3}^\star \otimes \mathbf{3} = \mathbf{1} \oplus \mathbf{8}$ of ${\rm Hom}(\mathbb C^3, \mathbb C^3) $ splits the nine mesons into a $\pi^0$ meson, transforming in the trivial rep, and a further eight mesons, transforming in the adjoint rep (see here). The eight mesons are known as the "Eightfold way" in a reference to Buddhism...
