Representations of $\mathfrak{sl_3}\mathbb{C}$

lie-algebraslinear algebrarepresentation-theorysemisimple-lie-algebras

I am reading Representation Theory by Fulton & Harris and in the Lecture 13, while constructing irreducible representations of $\mathfrak{sl_3}\mathbb{C}$, to show that $V\otimes V^*$ with $V\cong \mathbb{C}^3$ is not irreducible, he considers the map
$$
V\otimes V^* \to \mathbb C
$$
given by the contraction
$$
v\otimes u^* \mapsto \langle v,u^*\rangle = u^*(v).
$$
Then, in terms of the identification $V\otimes V^*\cong\text{Hom}(V,V)$, they argue that the kernel of this map is the subspace of $V\otimes V^*$ of traceless matrices which is the adjoint representation of $\mathfrak{sl_3}\mathbb C$ which is irreducible. This is clear to me.

To use a similar argument to show that $\text{Sym}^2V\otimes V^*$ is not irreducible, we consider the map
$$
\iota:\text{Sym}^2V\otimes V^*\to V
$$
given by the contraction
$$
vw\otimes u^* \mapsto \langle v,u^*\rangle \cdot w\,+\,\langle w,u^*\rangle\cdot v
$$
The weight diagram of $\text{Sym}^2V\otimes V^*$ looks like

and after introducing the map, they immediately conclude that the weight diagram of the kernel should look like

enter image description here

I have tried to figure out the kernel of the map to no end. Is there an easier way of seeing that the weight diagram of the kernel should look as above?

Best Answer

If $\varphi \colon V \to W$ is a surjective map of $\mathfrak{sl_3}$ representations, then the weights of $\ker \varphi$ are precisely the weights of $V$ minus the weights of $W$.

  • Because $\varphi$ is a homomorphism, for any $h \in \mathfrak{h}$ we have $\varphi(h v) = h \varphi(v)$, and so if $v$ is a vector of weight $\lambda$ we have $h \varphi(v) = \varphi(h v) = \lambda \varphi(v)$, and hence $\varphi$ maps the $V_\lambda$ weight space into $W_\lambda$. Let $\varphi_\lambda \colon V_\lambda \to W_\lambda$ be the resulting map.
  • Since $\varphi$ is surjective, so is each $\varphi_\lambda$. Applying the first isomorphism theorem to $\varphi_\lambda$ gives $V_\lambda / \ker \varphi_\lambda \cong W_\lambda$ as vector spaces, and hence $\dim V_\lambda - \dim \ker \varphi_\lambda = \dim W_\lambda$.
  • Since $\ker \varphi_\lambda = (\ker \varphi)_\lambda$ (the kernel of $\varphi_\lambda$ is the $\lambda$-weight space of $\ker \varphi$) we are done.
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