Irreducible representation of $\mathfrak{so}(4)$

Question

I am supposed to give a 9-dimensional irreducible representation of $\mathfrak{so}(4)$.

I know that $\mathfrak{so}(4)\cong\mathfrak{so}(3)\oplus\mathfrak{so}(3)$ and hence I have a 6-dimensional reducible representation of $\mathfrak{so}(4)$. But how do I get a irreducible 9-dimensional representation? This may be a stupid question but I couldn't find anything.

Answer 1

Hint: The space of $4\times 4$ symmetric traceless matrices is of dimension $9$.

Answer 2

Almost every treatment of representations of real semisimple Lie algebras explicitly treats the basic case $\mathfrak{su}_2$; the standard result is that for each positive integer $n$, there is up to equivalence exactly one irreducible $n$-dimensional $\mathfrak{su}_2$-representation, call it $\rho_n$.

Now if you already know $\mathfrak{so}_4 \simeq \mathfrak{so}_3 \oplus \mathfrak{so}_3$, and further know $\mathfrak{so}_3 \simeq \mathfrak{su}_2$, then let

$pr_1: \mathfrak{so}_4 \twoheadrightarrow \mathfrak{su}_2, \qquad pr_2: \mathfrak{so}_4 \twoheadrightarrow \mathfrak{su}_2$

be the projections on the respective factors. Then $\rho_9 \circ pr_i$ give two inequivalent (why?), irreducible (why?), $9$-dimensional representations of $\mathfrak{so}_4$.