Intersection of Maximal Left Ideals with Finite Dimensional Quotient

Question

Let $\Gamma$ be a finitely generated group and let $A=\mathbb{C}[\Gamma]$ be the corresponding group algebra over $\mathbb{C}$. Let $X$ be the set of all maximal left ideals of $A$ and let $X_0=\{I \in X : \,\, A/I \textrm{ is finite dimensional } \mathbb{C} \textrm{-vector space}\}.$

Now consider $J= \bigcap_{I \in X} I$ and $J_0= \bigcap_{I \in X_0} I$. It is easy to see that $J$ and $J_0$ are two-sided ideals. The ideal $J$ is the Jacobson radical of $A$ and it has been proven that it is the zero ideal.

My question is the following: For which groups $\Gamma$ is $J_0$ also the zero ideal?

Is there some literature on this question? I am interested in necessary and sufficient criteria as well as in examples, but even more in non-examples.

What I know so far: $J_0$ is the zero ideal:

-if $\Gamma$ is finite (that's trivial),

-if $\Gamma$ is commutative,

-if $\Gamma$ is the free group in finitely many generators.

Answer 1

This question is the same as asking when does $\mathbb C\Gamma$ have enough finite dimensional irreducible representations to separate points.

A necessary condition is that $\Gamma$ has enough finite dimensional irreps to separate points. Since a fg linear group is residually finite by a theorem of Malcev, having enough finite dim irreps to separate points implies residually finite.

Conversely, if $\Gamma$ is residually finite and $0\neq a\in \mathbb C\Gamma$ then there is a finite image $G$ of $\Gamma$ such that the image of $a$ in $\mathbb CG$ is nonzero (since the support of $a$ involves only finitely many elements of $\Gamma$). Since for finite $G$ we have the regular rep of $\mathbb CG$ is a direct sum of irreps we can find a finite dim irrep in which $a\neq 0$ (by projecting to an appropriate irreducible summand of the regular rep).

Thus the intersection of the cofinite dimensional maximal left ideals is $0$ iff $\Gamma$ is residually finite.