Is $SU(n)\to GL(n,\mathbb C)$ an irreducible representation? I want to compute the centre of $SU(n)$. In order to do so, I have used Schur’s
lemma, which implies that the only operators that commute with every element of an irreducible representation are those proportional to the identity. In particular, $SU(n)\to GL(n,\mathbb C)$, as a complex matrix group, is a representation over the vector space $\mathbb{C}^n$. I suspect that it is in fact an irreducible representation, but I have not found an elegant way to show it.
In the case of $SO(3)$ this can be easily proven in a "geometric" (or perhaps a bit heuristic) way by noting that there exists no plane in $\mathbb{R}^3$ invariant under every rotation.
 A: The proof for $\mathbb R^3$ carries over to $\mathbb C^n$. I would argue as follows. Consider the unit sphere
$$
S=\{ (z_1, \ldots, z_n)\in\mathbb C^n\ :\ \lvert z_1\rvert^2+\ldots+\lvert z_n\rvert^2=1\}.$$
This is a homogeneous space, that is, for every $z,w\in S$ there is a $R\in SU(n)$ such that $Rz=w$. This is the key step, and it needs proof, I'll give a rough sketch here. It is enough to prove it with $w=(1,0,\ldots,0)$. We can start by multiplying $z$ by a diagonal matrix in $SU(n)$, so that $z$ is mapped to $(\lvert z_1\rvert, \ldots, \lvert z_n\rvert)$. Now we multiply by a real rotation matrix to map this last vector to $(1, 0, \ldots, 0)$; we can do this because we already know that the real $n$-sphere is a homogeneous space. $^{[1]}$
Once we have this homogeneity property of $S$, the proof is basically done. Suppose $V\subset \mathbb C^n$ is a subrepresentation. Then either $V=\{0\}$, or it must contain $S$, because of homogeneity. But then $V=\mathbb C^n$. The proof is complete.

[1]. Actually, it is not hard to construct such matrix $R$ explicitly and directly, using the concept of Householder transformation. There is some detail to be taken care of, because a Householder transformation has determinant $-1$, so it is in $U(n)$ but not in $SU(n)$. I remember a nice answer of "Omnomnomnom" (who later changed username), doing exactly these things, but I cannot find it. EDIT. I found it! https://math.stackexchange.com/a/3388120/8157

P.S.: I now see that runway44 suggested a proof in the comments, and my answer is essentially the same thing.
