Is the adjoint representation of the unitary group irreducible?

Let $U_n$ be the unitary group of $n \times n$ unitary matrices, and consider its adjoint representation on its Lie algebra $\mathfrak{u}_n$: $$\mathrm{Ad} \, \colon \, U_n \to \mathrm{Aut}(\mathfrak{u}_n) \quad , \quad \mathrm{Ad}_g (x) := g x g^{-1} $$ for any $g \in U_n$ and $x \in \mathfrak{u}_n$.

Is this linear representation irreducible?

Solution 1:

A colleague has already explained to me the situation:

• The adjoint representation of $U_n$ on $\mathfrak{u}_n$ is not irreducible: for the normal subgroup $SU_n \subset U_n$ (of matrices with determinant equal to one) induces a sub-representation $\mathfrak{su}_n \subset \mathfrak{u}_n$.

• The adjoint representation of $U_n$ on $\mathfrak{su}_n$ is irreducible: to see this, let us argue in the algebraic realm. This allows us to change basis to the complex numbers; hence, $U_n$ becomes the group $M_n(\mathbb{C})$ of square matrices over the complex numbers, and $\mathfrak{su}_n$ becomes the complex matrices of trace zero. Now, we finish with a general argument: the decomposition of the group of matrices under the action of the general linear group always has two irreducible addends, one of them being that of trace-free matrices.