Consider a (countable) group $G$, a subgroup $H\leq G$ of finite index, and a unitary representation $\pi:G\to \mathcal{U}(\mathcal{H})$.



> 
> If the center of the von Neumann algebra $\pi(H)''$ is finite dimensional, does this imply that the center of $\pi(G)''$ is finite dimensional as well?
> Is it at least true assuming that the representation $\pi$ is induced from a representation of $H$?
> 
> 
> 


Remarks:


* As far as I'm concerned, $H$ can be assumed to be normal if it makes things simpler.
* If $\pi$ is of type I, then the question is whether $\pi$ decomposes to finitely many irreducible representations provided that the restriction $\mathrm{Res}\_H^G\pi$ does. This is obviously true.



The center of a representation von Neumann algebra, and finite index subgroups
The answer is YES and it follows from the general theory of finite-index inclusions of von Neumann subalgebras (a la Jones, Pimsner, Popa...), which says that finite-dimensionality of the center is preserved under taking a finite-index subalgebra/extension. This is perhaps an overkill and so I will try a layman's proof.


**Definition.** An inclusion $M\subset N$ of von Neumann algebras is of finite-index if there is a (necessarily faithful normal) conditional expectation $E$ from $N$ onto $M$ such that $E\geq\lambda\,\mathrm{id}\_N$ for some $\lambda>0$.


Note that $E$ maps $Z(N)$ into $Z(M)$ and that a von Neumann algebra $A$ is finite-dimensional iff $\mathbb{C}1\subset A$ is of finite-index. Thus if $M\subset N$ is of finite-index, then
$$\dim Z(M)<\infty\Rightarrow\dim Z(N)<\infty.$$


**Example.** If $H\le G$ is of finite-index, then the inclusion $\pi(G)'\subset\pi(H)'$ is of finite-index with the conditional expectation given by
$$E(x)=\frac{1}{[G:H]}\sum\_{g\in G/H}\pi(g)x\pi(g)^\*,$$
which satisfies $E\geq\frac{1}{[G:H]}$. Hence $\dim Z(\pi(G)'') <\infty\Rightarrow\dim Z(\pi(H)'') <\infty$.


We want to prove the converse. As I said in the beginning, one way is to use von Neumann algebra machinery (the standard form and the basic construction) to "flip" the inclusion. Here is an alternative way. By passing to a finite-index subgroup, we may assume that $H$ is normal. Take a natural realization (which I'm too lazy to write) of the induced representation $\mathrm{Ind}\_H^G\,\pi$ on $\ell\_2(G/H)\otimes\mathcal{H}$
that satisfies
$$(\mathrm{Ind}\_H^G\,\pi)(H)''\subset(\mathrm{Ind}\_H^G\,\pi)(G)''\subset B(\ell\_2(G/H)) \otimes\pi(H)''.$$
Then one can show the commutant inclusion ${\mathbb C}1\otimes\pi(H)'\subset (\mathrm{Ind}\_H^G\,\pi)(H)'$ is of finite-index (via the conditional expectation coming from the trace on $B(\ell\_2(G/H))$) and so the intermediate inclusion ${\mathbb C}1\otimes\pi(H)'\subset (\mathrm{Ind}\_H^G\,\pi)(G)'$ is also of finite-index. From this and the fact that $\pi\subset\mathrm{Ind}\_H^G\,\pi$,
one obtains
$$\dim Z(\pi(H)'') <\infty\Rightarrow\dim Z(\pi(G)'') <\infty.$$


