Let $\pi$ be the fundamental group of a closed orientable surface of positive genus (though I don't think this will be particularly important) and $\phi : \pi \to U(n)$ a morphism (in particular, a representation of $\pi$ in $\mathbb{C}^n$). In Goldman's The Symplectic Nature of Fundamental Groups of Surfaces, he claims that the irreducibility of $\phi$ is equivalent to $\dim \zeta(\phi)/\zeta(U(n)) = 0$, where $\zeta(\phi)$ is the centralizer of the image of $\phi$ and $\zeta(U(n))$ is the center of $U(n)$. I'm trying to better understand this fact and some generalizations of it.
First, if $\zeta(\phi)$ has some element not in the center of $U(n)$, then the eigenspaces of that element give a $\pi$-invariant and nontrivial decomposition of $\mathbb{C}^n$, showing that it is reducible. This shows one direction.
For the other direction, I suppose we could argue like: take some nontrivial $\pi$-invariant decomposition of $\mathbb{C}^n$. Any unitary operator whose eigenspaces are exactly the factors in that decomposition will commute with the image of $\phi$. This gives me one degree of freedom for each factor in the decomposition, and since the decomposition is nontrivial, that grants me at least two degrees; since the center of $U(n)$ is one-dimensional, this already shows $\dim \zeta(\phi)/\zeta(U(n)) > 0$.
I think this argument works (or does it not?), but what I'm trying to do is to understand better what the number $k = \dim \zeta(\phi)/\zeta(U(n))$ means. Question: is $k+1$ the number of irreducible factors that $\mathbb{C}^n$ is decomposed into? i.e. are there subspaces $V_0, \ldots, V_k$ of $\mathbb{C}^n$ each of which are irreducible and $\pi$-invariant such that $\mathbb{C}^n = V_0 \oplus V_1 \oplus \cdots \oplus V_k$? If not, how does this number $k$ measure the reducibility of $\phi$?
Note: Here I am interpreting 'reducible' to mean that it can be decomposed as a nontrivial sum of invariant subspaces. I do not know if in this case this is the same as admitting a proper invariant subspace (that is also an interesting question).