Fixed points of a linear abelian p-group in characteristic p

Question

Let $V=\bigoplus_{n\geq1}\mathbb F_{p}\cdot e_{n}$ be an $\mathbb F_{p}$-vector space of countable dimension, and write $V_{n}=\operatorname{Vect}(e_{1},\dotsc,e_{n})$. Let $G$ be a (possibly infinite) elementary abelian $p$-group which acts linearly on $V$, leaving each $V_{n}$ invariant. I want to prove that

$$\sup_{n\geq m \geq 0}\mathrm{dim}(V_{n}/V_{m})^{G}=+\infty$$

where $(V_{n}/V_{m})^{G}$ denotes the space of $G$-fixed points in $V_{n}/V_{m}$.

My reasoning is as follows. In many cases, the stronger result $\sup_{n}\dim V_{n}^{G}=+\infty$ holds. Let $G_{n}$ be the image of $G$ on $\operatorname{GL}(V_{n})$, and assume that $G_{n}$ is generated by one element, say $g$. The sequence $$\forall i\geq0 , \quad k_{i}=\dim\ker(g-\mathrm{id})^{i}$$ has decreasing differences, and satisfies $k_{p}=n$ since $(g-\mathrm{id})^{p}=g^{p}-\mathrm{id}=0$. We conclude that $k_{1}=\mathrm{dim}V_{n}^{G}\geq n/p$. More generally, if $\operatorname{rk}(G_{n})=o(\log n)$ (for instance, if $G$ is finite), then $\dim V_{n}^{G}\geq n/p^{o(\log n)}$. On the other end of the spectrum, the maximal possible value of $\operatorname{rk}(G_{n})$ is $\lfloor n^{2}/4 \rfloor$, in which case $\dim V_{n}^{G}=\lfloor n/2\rfloor$ (see J. T. Goozeff, "Abelian $p$-subgroups of the general linear group"). My intuition tells me that $\sup_{n}\dim V_{n}^{G}=+\infty$ whenever $\operatorname{rk}(G_{n})$ is of the order of $n^{2}$. Of course, $\operatorname{rk}(G_{n})$ can grow at an intermediate speed. For instance, the elements $$\forall n\geq 1, \quad g_{n}\in \operatorname{GL}(V)$$ defined by $g_{n}(e_{k})=e_{k}+e_{1}$ if $k=n$, $e_{k}$ otherwise, generate an elementary abelian $p$-group such that $\operatorname{rk}(G_{n})=n$ and $\dim V_{n}^{G}=1$. In this case, the action of $G$ on $V_{n}/V_{1}$ is trivial, so $\sup_{n}\mathrm{dim}(V_{n}/V_{1})^{G}=+\infty$.

Maybe there are ways to modify my last example in order to refute my conjecture, but I haven't been able to see them. Such counter-examples are also appreciated.

Answer 1

This post has been extensively edited to expand the second step and thereafter shorten the argument.

Suppose that the sup is finite. We wish to show that $V$ is finite dimensional, contradicting the given setup.

Since $V_n$ is finite dimensional, we have $V_n = \bigcup_i \operatorname{\rm Soc}^iV_n\subseteq \bigcup_i\operatorname{\rm Soc}^i V$. So $$V=\bigcup_nV_n=\bigcup_i\operatorname{\rm Soc}^iV.$$ If $(V_n/V_m)^G$ is bounded in dimension then $(V/V_m)^G$ is contained in some $(V_n/V_m)^G$. So by induction on $i$ each $\operatorname{\rm Soc}^iV$ is contained in some $V_n$.

In particular, $\operatorname{\rm Soc}^2V$ is contained in some $V_n$, so the action of $G$ on $\operatorname{\rm Soc}^2V$ factors through some finite quotient. We may therefore write $G=G_0\times G_1$ with $G_0$ finite and $G_1$ acting trivially on $\operatorname{\rm Soc}^2V$.

Let $\operatorname{\rm Soc}_0^iV$ be the $i$th socle of $V$ under the action of $G_0$ and similarly for $\operatorname{\rm Soc}^i_1$ and $G_1$. Then $\operatorname{\rm Soc}_1^2\operatorname{\rm Soc}_0V$ is contained in $\operatorname{\rm Soc}^2V$, so $G_1$ acts trivially on it. Thus $$\operatorname{\rm Soc}^2_1\operatorname{\rm Soc}_0V=\operatorname{\rm Soc}_1\operatorname{\rm Soc}_0V=\operatorname{\rm Soc}V,$$ and then by induction on $i$, $$\operatorname{\rm Soc}^i_1\operatorname{\rm Soc}_0V=\operatorname{\rm Soc}_1\operatorname{\rm Soc}_0V=\operatorname{\rm Soc}V$$ for all $i\geqslant 1$. Since $$\bigcup_i\operatorname{\rm Soc}^i_1\operatorname{\rm Soc}_0V=\bigcup_i\operatorname{\rm Soc}^i\operatorname{\rm Soc}_0V=\operatorname{\rm Soc}_0V,$$ this implies that $G_1$ acts trivially on $\operatorname{\rm Soc}_0V$.

Thus $\operatorname{\rm Soc}_0V=\operatorname{\rm Soc}V$ is finite dimensional, and since $G_0$ is finite, this implies that $V$ is finite dimensional, because it is a submodule of the injective hull of $\operatorname{\rm Soc}_0V$. This completes the argument by contradiction.

Answer 2

I believe this answer was posted previously but was deleted by its author. Maybe there is an obvious gap in the argument.

Let $J$ be the Jacobson radical of $\mathbb{F}_{p}[G]$, which is generated by elements of the form $g-e$, $g\in G$. Then $V^{G}$ is the submodule annihilated by $J$. More generally, we can define the socle filtration by $\mathrm{soc}_{0}(V)=\{0\}$ and for all $i\geq0$, $\mathrm{soc}_{i+1}(V)$ is the preimage of $(V/\mathrm{soc}_{i}(V))^{G}$. Then $\mathrm{soc}_{i}(V)$ is the submodule annihilated by $J^{i}$. Since $G$ is abelian of exponent $p$, $J^{p}=0$ so $\mathrm{soc}_{p}(V)=V$. If $i>0$ is the smallest integer such that $\mathrm{soc}_{i}(V)$ is infinite-dimensional, then there is some $n$ such that $\mathrm{soc}_{i-1}(V)\subset V_{n}$ and hence $(V/V_{n})^{G}$ is infinite-dimensional.