Solution verification tensor product of representations of $\mathbf{sl}(2;\mathbb{C})$ I am not very familiar with representation theory, but I have an assignment in a Lie groups course which I am the only student taking, and my lecturer is overseas currently so I don’t have anyone else I can ask about this homework assignment. The question is from Hall’s “Lie groups, Lie algebras, and representations” and it is exercise 12 from chapter 4 of the latest edition. I would like it if someone could please tell me if I am on track with the first part of the question, and also some hints about how to get started with the second part would be very much appreciated. Thanks.
Recall the spaces $V_m$ introduced in section 4.2, viewed as representations of the Lie algebra $\mathbf{sl}(2;\mathbb{C})$. In particular, consider the space $V_1$ (which has dimension 2).

Question 1:
Regard $V_1\otimes V_1$ as a representation of $\mathbf{sl}(2;\mathbb{C})$, as in definition 4.20. Show that this representation is not irreducible.

Solution:
Definition 4.20 states that if we have two representations $\pi_1,\pi_2$ of $\mathbf{sl}(2;\mathbb{C})$ both acting on the space $$V_1 := \{f(z_1,z_2) =a_0z_1+a_1z_2~:~\text{$a_0,a_1\in \mathbb{C}$, $z_1, z_2$ complex variables}\}$$ then for any $X\in \mathbf{sl}(2;\mathbb{C})$ we find the tensor product of representations $\pi_1\otimes \pi_2$ is a representation of $\mathbf{sl}(2;\mathbb{C})$ acting on $V_1\otimes V_1$ given by $$(\pi_1\otimes\pi_2)(X) = \pi_1(X)\otimes I + I\otimes \pi_2(X) \,.$$
We will show that this representation is not irreducible by finding a nontrivial invariant subspace. Without loss of generality we may assume that $\pi_1, \pi_2$ are each irreducible representations of $\textbf{sl}(2;\mathbb{C})$ over $V_1$ otherwise the result is obvious.
If we consider the action of some $X\in \textbf{sl}(2;\mathbb{C})$ on an element $f\otimes g\in V_1\otimes V_1$, this representation looks like
\begin{align*}
  (\pi_1\otimes\pi_2)(X)(f\otimes g)
  &= \pi_1(X)\otimes I (f\otimes g) + I\otimes \pi_2(X) (f\otimes g)\\
  &= \pi_1(X)(f)\otimes I(g) +I(f)\otimes \pi_2(X)(g) \,.
\end{align*}
Consider the nontrivial subspace of $V_1\otimes V_1$ given by $V_1\otimes \{0\}$. It is clear that this is an invariant subrepresentation of the above representation, since we have for any $X\in \textbf{sl}(2;\mathbb{C})$, given some $f\otimes 0 \in V_1\otimes \{0\}$:
$$(\pi_1\otimes \pi_2)(X)(f\otimes 0) = \pi_1(X)(f)\otimes 0+f\otimes 0 \in V_1\otimes \{0\} \,.$$
Therefore, we have found a nontrivial invariant subspace, and so this representation is not irreducible.

Question 2:
Now, view $V_1\otimes V_1$ as a representation of $\mathbf{sl}(2;\mathbb{C})\oplus \mathbf{sl}(2;\mathbb{C})$, as in definition 4.19. Show that this representation is irreducible.

Defintion 4.19 states that a tensor product representation of $\mathbf{sl}(2;\mathbb{C})\oplus \mathbf{sl}(2;\mathbb{C})$ takes the form $$(\pi_1\otimes\pi_2)(X,Y) = \pi_1(X)\otimes I + I\otimes \pi_2(Y) \,.$$
 A: As I noted in the comments, $V_1 \otimes \{0\}$ is in fact a trivial subspace (by which I mean it is $\{0\}$ rather than a trivial representation). So we instead want to find that $V_1 \otimes V_1 = \Lambda^2 V_1 \oplus S^2 V_1$ as representations. Note you do not need to show that $\pi_1 = \pi_2$, nor that $V_1$ is irreducible as that is in the definition of $V_1$. It is defined not just as a vector space but one carrying a particular representation.
We can prove a much more general fact here.
Let $\mathfrak{g}$ be a Lie algebra and $\pi:\mathfrak{g} \to \mathfrak{gl}(V)$ a representation. Then $V \otimes V = \Lambda^2 V \oplus S^2 V$ is a decomposition into subrepresentations.
Let's denote
$$ v \odot w = v\otimes w + w \otimes v$$
and
$$ v \wedge w = v\otimes w - w \otimes v$$
We call the span of all $v \odot w$, $S^2V$ and the span of $v \wedge w$, $\Lambda^2V$. It is not too hard to see that $V \otimes V = \Lambda^2 V \oplus S^2 V$ as a vector space.
Then $$\begin{aligned}(\pi \otimes \pi)(X)(v\odot w) &= (\pi(X)v) \otimes w + v\otimes(\pi(X)w) + (\pi(X)w) \otimes v + w\otimes(\pi(X)v) \\&= \pi(X)v \odot w + v \odot \pi(X)w.\end{aligned}$$
This (by linearity) shows $S^2 V$ is closed under the action of $\mathfrak{g}$ and so is a subrepresentation and we can do the same with $\Lambda^2 V$. So $V \otimes V$ is never irreducible. Note I have made no assumptions about the irreducibility of $V$ or even semisimplicity,etc. of $\mathfrak{g}$ and I don't show that $S^2 V, \Lambda^2 V$ are irreducible. In general, they do not have to be even if $V$ is (although they are in your example).
The ideas here generalise to a very important part of the representation theory of Lie algebras. For example, you can show that the action of $\mathfrak{sl}_n$ on $V^{\otimes k} = V\otimes \cdots \otimes V$ commutes with the action of the symmetric group $S_k$, permuting the order of the tensor products. The resulting decomposition into simultaneous irreducible representations of $\mathfrak{sl}_n$ and $S_k$ is called Schur-Weyl duality.
As to question 2, you can show, by considering the action of elements like $(X,0)$ and $(0,Y)$, that any invariant subspace of $V_1 \otimes V_1$ must be a tensor product of invariant subspaces in each copy and since they are each irreducible there are no proper invariant subspaces.
