Define $V_m$ as the space of all homogeneous polynomials in two complex variables of degree $n$.
Then we can define a representation of $SU(2)$ on the space $V_m$ by the formula
$[\Pi_m(U)f](z) = f(U^{-1}z).$
We can then define the Lie algebra representaion of $\text{su}(2)$ by the formula
$\pi_m(X) = \left. \frac{d}{dt}\Pi_m(e^{tX}) \right |_{t=0}$
which works out to be
$\pi_m(X)f = -\frac{\partial f}{\partial z_1}(X_{11} z_1+ X_{12} z_2) - \frac{\partial f}{\partial z_2}(X_{21}z_2 + X_{22}z_2).$
It turns out that this representation extends to a representation of $\text{sl}(2;\mathbb{C})$
Now we have two ways of interpreting the tensor product representation $\pi_m \otimes \pi_m$.
1) We can define $\pi_m \otimes \pi_m$ as a representation of $\text{sl}(2;\mathbb{C}) \oplus \text{sl}(2;\mathbb{C})$ acting on $V_m \otimes V_m$ by $\pi_m \otimes \pi_m (X,Y) = \pi_m(X) \otimes I + I \otimes \pi_m(Y)$
2) We also have $\pi_m \otimes \pi_m$ as a representation of $\text{sl}(2;\mathbb{C})$ acting on $V_m \otimes V_m$ defined by
$\pi_m \otimes \pi_m (X) = \pi_m(X) \otimes I + I \otimes \pi_m(X)$
The exercise is to show that $V_1 \otimes V_1$ as a representaion of $\text{sl}(2;\mathbb{C})$ (our second interpretation) is reducible, whilst as a representation of $\text{sl}(2;\mathbb{C}) \oplus \text{sl}(2;\mathbb{C})$(the first interpretation) it is irreducible.
I think, in the first interpretation, that $\pi_1(X) \otimes \pi_2(Y)$ is irreducible if and only if $\pi_1(X)$ and $\pi_2(Y)$ are irreducible. Is this true?
I can't immediately proof that the second is reducible. I presume it is because we have $\pi_1(X) \otimes I$ and $I \otimes \pi_1(X)$ (as opposed to $X$ and $Y$). Any hints?