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The adjoint representation (over the complex numbers) of SO(4) is 6-dimensional. Is this representation self-dual?

Other than the adjoint representation and its dual, are there other irreducible 6-dimensional representations of SO(4) over the complex numbers?

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up vote 5 down vote accepted

The adjoint representation of any semisimple Lie group $G$ is self-dual. This follows from the fact that the Killing form gives a nondegenerate $G$-invariant bilinear map $\mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ for such $G$ (and we can then extend scalars to $\mathbb{C}$ to get a corresponding such map for $\mathfrak{g} \otimes_{\mathbb{R}} \mathbb{C}$).

The adjoint representation of $\text{SO}(4)$ is also not irreducible (and this is the only $n \ge 2$ for which this is true). This follows from the fact that $\mathfrak{so}(4) \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2)$ as a Lie algebra and so the adjoint representation of $\mathfrak{so}(4)$ on itself is a direct sum of two $3$-dimensional representations.

Edit: In fact $\text{SO}(4)$ admits no irreducible $6$-dimensional representations. Its universal cover is $\text{Spin}(4) \cong \text{SU}(2) \times \text{SU}(2)$ which admits two irreducible $6$-dimensional representations given by $V \otimes S^2(V)$ and $S^2(V) \otimes V$ where $V$ is the defining representation of $\text{SU}(2)$. The kernel of the covering map $\text{Spin}(4) \to \text{SO}(4)$ is generated by $(-1, -1)$ and this acts nontrivially in both of the above representations, so neither descends to a representation of $\text{SO}(4)$.

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