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

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.

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 $$\mathfrak{S}_n$$?

• It seems that after your original question was answered, you added a somewhat different question at the end. I think if you want answers to that one, you should ask it as a new question. Commented Aug 22, 2022 at 17:04
• @TorstenSchoeneberg , for $\mathfrak{sl}_n\mathbb{C}$, $\wedge ^{n-1}V \cong V^*$. I just mention it here.
– Jino
Commented Aug 23, 2022 at 7:38

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.

• Funny, I never thought about that: Not only is $tr(X.t) = X.tr(t)$ but indeed $tr(X.t) =0$ for all $X \in \mathfrak g$ and $t \in V \otimes V^*$. And that's saying the same here, because the codomain of $tr$ is the ground field with trivial $\mathfrak g$-action. Commented Aug 12, 2022 at 16:53