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](https://doi.org/10.1016/j.jpaa.2014.05.004)," JPAA 219 (2015), 449–463.



Semi-projective complexes of modules over a finite group
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
$$\cdots\xrightarrow{x}A\xrightarrow{x}A\xrightarrow{x}A\xrightarrow{x}A\xrightarrow{x}\dots$$


