4
$\begingroup$

I am reading the paper Space with $\mathbb{G}_{m}$-action, hyperbolic localization and nearby cycles by Timo Richarz and I am having some troubles in understanding the proof of Lemma 1.10.

The setting is as follows. Let $S$ be a scheme and $U \longrightarrow X$ be a morphism of $S$-algebraic spaces endowed with $\mathbb{G}_{m,S}$-actions. Lemma 1.10 in loc. cit. says that if $u \colon U \longrightarrow X$ is étale then $U^0 = U \times_X X^0$ where $(-)^0$ denotes the functor of fixed points (of the given $\mathbb{G}_{m,S}$-actions).

In case $U \longrightarrow X$ is an open immersion, the lemma seems to be obvious but I could not understand the proof of the étale case. Here the proof: let $T/S$ be a scheme with trivial $\mathbb{G}_{m,S}$-action and an equivariant morphism $T \longrightarrow X$. Given a morphism $\widetilde{f} \colon T \longrightarrow U$ such that $u \circ \widetilde{f} = f$, then the problem is to prove that $\widetilde{f}$ is equivariant.

  • The author claims that it suffices to prove that $\widetilde{f}$ is locally étale equivariant.
  • If $T$ is the spectrum of a strictly henselian local ring then $\widetilde{f}$ is equivariant because $\mathbb{G}_{m,S}$ is connected and $u$ is étale.
  • Finally, he finishes the proof by some limit argument.

Since I am not really familiar with neither algebraic spaces nor algebraic groups, I hope you can help me to understand the proof or give me some references for each of my confusions. Thank you in advance.

$\endgroup$
3
  • $\begingroup$ You may be interested in the Appendix to the paper "Algebraic Magnetism" my Arnaud Mayeux, see arxiv.org/abs/2203.08231v3. It contains a generalization as well as more detailed proofs. $\endgroup$ Commented Jun 15 at 13:35
  • $\begingroup$ Hi Matthieu, it is probably too generalized for me. I found an argument by Alper in the case of algebraic spaces over a field. The problem reduces to: if $ U \to X$ is an open immersion of spaces with $G$-actions ($G$: connected), then by connectedness, any $G$-orbit $Gu$ ($u \in U$) is connected, hence $U$ is $G$-invariant. I do not really understand the last conclusion. I think somehow you made a similar argument to this in the appendix. $\endgroup$
    – Alexey Do
    Commented Jun 15 at 16:12
  • $\begingroup$ Well, but saying that $i:U\to X$ is an open immersion and $G$-equivariant means that $U$ (seen as a subspace of $X$) is $G$-invariant, doesn't it? $\endgroup$ Commented Jun 16 at 10:04

0