TITLE: Projective resolutions of finite-dimensional representations of infinite groups
QUESTION [9 upvotes]: Let $G$ be a group and let $V$ be a finite-dimensional complex representation of $G$.  Question: Under what circumstances can I find a projective resolution
$$ \cdots \longrightarrow P_3 \longrightarrow P_2 \longrightarrow P_1 \longrightarrow V \longrightarrow 0$$
of $\mathbb{C}[G]$-modules such that each $P_i$ is finitely generated?  I believe that this condition is usually expressed by saying that $V$ is of type $FP_{\infty}(\mathbb{C})$.
One obvious first case is where $V$ is the trivial representation $V = \mathbb{C}$.  If a projective resolution as above exists for this $V$, then $G$ is said to be of type $FP_{\infty}(\mathbb{C})$.  The groups I am interested in are all of type $FP_{\infty}(\mathbb{C})$.  It would be really wonderful if this condition was sufficient for these resolutions to exist for all finite-dimensional representations.
Here is a specific example that I don't know how to do and that is typical among the ones I care about:
$$G = SL(n,\mathbb{Z}) \quad \text{and} \quad V = \mathbb{C}^n.$$

REPLY [3 votes]: The answer is yes.   In the answer below I use $M$ instead of $V$ and $k$ is any field. 
Thm 2 of https://www.tandfonline.com/doi/abs/10.1080/00927870600796110 shows that if G is $FP_\infty$ over $k$, then $kG$ has a free resolution as a bimodule by finitely generated free bimodules in each dimension.
If you tensor this resolution with $M$ over $kG$ you get a free resolution of $M$ with the finiteness properties you want.  Tensoring with $M$ gives a resolution because its homology is $Tor^{kG}(M,kG)$.  
It is easy to check that $(kG\otimes_k kG)\otimes_{kG} M\cong kG^{\dim M}$ as a left $kG$-module so that the free resolution is finitely generated in each degree.  The basis as a $kG$-module is the tensors $1\otimes 1\otimes b$ with $b$ running over a basis of $M$.