Semi-projective complexes of modules over a finite group

Question

Let $G$ be a finite group and $k$ a field of characteristic $p$ dividing $|G|$. A perfect complex of $kG$-modules is by definition a finite complex of finitely generated projective ($=$ injective $=$ flat) $kG$-modules. A semi-projective complex of $kG$-modules is a filtered colimit of perfect complexes.

My question is this. If a semi-projective complex $X$ has no $G$-fixed points (i.e., there are no homomorphisms from the trivial module to any $X^i$, or equivalently no $X^i$ has the projective cover of the trivial module as a summand), is $X$ then homotopy equivalent to a filtered colimit of perfect complexes with no $G$-fixed points?

I've tried quite hard to prove this, and I've tried quite hard to write down a counterexample, both without success.

A paper relevant to this question is Lars Christensen and Henrik Holm, "The direct limit closure of perfect complexes," JPAA 219 (2015), 449–463.

Answer 1

I think I have a counterexample.

Let $\operatorname{char}(k)=3$ and let $G$ be the symmetric group $S_{3}$.

Then $kG$ has two simple modules: the trivial module $k$ and another one-dimensional module $S$.

The projective covers $P$ and $Q$ of these simples are uniserial with composition series $P= \begin{matrix} k\\S\\k \end{matrix}$ and $Q= \begin{matrix} S\\k\\S \end{matrix}$.

For each $n\geq0$ define a perfect cochain complex $$C(n):=\cdots\to0\to Q\to Q\to\cdots\to Q\to P\to0\to\cdots$$ with $Q$ in degrees $0$ to $n$ and with all differentials between nonzero modules being nonzero nonisomorphisms.

Then for each $n$ there is a map $C(n)\to C(n+2)$ which is the identity in degrees $0$ to $n$. [Note that we need to use $C(n+2)$ and not $C(n+1)$, as the fact that the obvious composition $P\to Q\to P$ is nonzero is an obstruction to contructing a similar map $C(n)\to C(n+1)$.]

Taking the colimit of $C(0)\to C(2)\to C(4)\to\cdots$ we get a complex $$C(\infty)=\cdots\to0\to Q\to Q\to Q\to\cdots$$ with $Q$ in all nonnegative degrees, that has no $G$-fixed points.

The category of projective $kG$-modules that contain no copies of $P$ is equivalent to the category of free modules for $A=k[x]/(x^{2})$, so if $C(\infty)$ were a filtered colimit of perfect complexes of such modules, then the complex $$\cdots\xrightarrow{} 0\xrightarrow{}A\xrightarrow{x}A\xrightarrow{x}A\xrightarrow{}\cdots$$ would be a filtered colimit of perfect complexes of $A$-modules. But it is not, since it has a map to an acyclic complex that is not null-homotopic - namely, the inclusion into $$\cdots\xrightarrow{x}A\xrightarrow{x}A\xrightarrow{x}A\xrightarrow{x}A\xrightarrow{x}\dots$$