Representation of $\mathfrak{sl}_n\mathbb{C}$, the kernel of $V \otimes V^*$ will be adjiont representation

Question

I am reading the book: Representation theory a first course by Fulton & Harris.The book give some examples about representation of $\mathfrak{sl}_3\mathbb{C}$ and $\mathfrak{sl}_4\mathbb{C}$. (p178 & p221)

In the case of $\mathfrak{sl}_3\mathbb{C}$, V is standard representation on $\mathbb{C^3}$, $V^*\cong \wedge ^2V$. We can see the representation $V \otimes V^*$ is not irreducible by the following map：

\begin{align} tr: V\otimes V^* &\to \mathbb{C} \\ v\ \otimes u^* &\mapsto \langle v,u^*\rangle = u^*(v) \end{align} The kernel will be adjiont representation. We can identify $V \otimes V^*$ with Hom$(V, V)$, then the map will be trace of $M(3, \mathbb{C})$, and the kernel will be traceless matrices.

My question is how to see the map is a map of $\mathfrak{sl}_3\mathbb{C}$ module.

I found the answer https://math.stackexchange.com/q/3579163, but I can not add comment because of my reputation.

Thank Callum's answer.

Some thinking after Callum's answer. If we consider the map(for $\mathfrak{sl}_n\mathbb{C},V=\mathbb{C^n})$: \begin{align} \wedge:\quad \quad V\otimes \wedge^{n-1} V &\to \wedge^{n}V\cong\mathbb{C} \\ v_1\ \otimes v_2\wedge...\wedge v_n &\mapsto v_1\wedge v_2\wedge...\wedge v_n \end{align} It seems the map commutes with the action of $\mathfrak{sl}_n\mathbb{C}$ obviously, because $\mathfrak{sl}_n\mathbb{C}$(even $\mathfrak{gl}_n\mathbb{C})$ commutes with symmetric group $\mathfrak{S}_n$?

Answer 1

Firstly, I think it's easier not to identify $V^* \cong\bigwedge^2 V$ here and instead just use the dual (aka contragredient) representation directly. Specifically this is the representation $\rho^*:\mathfrak{g}\to \mathfrak{gl}(V^*)$ where $\rho^{*}(X)(f) := - f \circ \rho(X)$ for $\rho$ the representation on $V$.

So then the representation on the tensor product looks like $$ v \otimes f \mapsto \rho(X)(v) \otimes f - v \otimes f \circ \rho(X) =[\rho(X),v \otimes f].$$

Hopefully, it is clear from here that this action preserves the kernel of $\operatorname{tr}$ (indeed has image there) and its kernel is precisely the span of the identity. From this, we can quickly see that $\operatorname{tr}$ identifies the span of the identity with the trivial representation.

Note that this argument does not depend on whatever $\mathfrak{g}$ is, nor on whichever representation $V$ we are considering so there is always a copy of the trivial representation in $V\otimes V^*$ for any such choices of $\mathfrak{g}$-representation $V$. Of course in our specific example we have found the adjoint representation as the other component.