Let $X$ be a quasiprojective scheme over a field $k$. Let $G$ be a finite group acting on $X$ whose order is invertible in $k$. If $X$ is CohenMacaulay, can we conclude that the subscheme of fixed points $X^G$ is CohenMacaulay?

2$\begingroup$ I went down a rabbit hole looking for a counterexample with $X = \mathbb{C}^n$, and I'll report my failure. According to imsc.unigraz.at/baur/AGIT/Talks/Kraft_Ascona.pdf (slide 33, bullet point 3), every known algebraic action of a finite group on $\mathbb{C}^n$ is holomorphically equivalent to a linear action. So the fixed point locus is holomorphically isomorphic to $\mathbb{C}^k$, and hence smooth, and hence CohenMacaulay. I haven't traced the references in these slides, but this makes me suspect the problem is hard. $\endgroup$– David E SpeyerSep 2, 2020 at 15:09

$\begingroup$ That's interesting! I'm not sure if I understand you're suggestion correctly, but $X^G$ is certainly smooth if $X$ is, see for example Proposition 3.5 of Edixhoven 'Neron models and tame ramification'. $\endgroup$– JefSep 2, 2020 at 15:17

$\begingroup$ Thanks for the reference! And, in the positive direction, I think I have a counterexample now. $\endgroup$– David E SpeyerSep 2, 2020 at 15:17

2$\begingroup$ @David E Speyer. Near each fixed point of the (full) group action, the action is linearizable. Smoothness of the fixed point scheme for an action of a tame, linearly reductive group on a smooth scheme is usually attributed to Iversen (who only considered the complex case). $\endgroup$– Jason StarrSep 2, 2020 at 15:20
1 Answer
Here is a simpler example than the one I left before, using the same strategy. Let $$X = \{ x_1 x_3 = x_1 x_4 = x_1 x_5 = x_2 x_4 = x_2 x_5 = x_3 x_5 = 0 \} \subset \mathbb{C}^5.$$ This is the reduced union of four $2$planes. Here is a projective picture, where $j$ represents the point where $x_j$ is the sole nonzero coordinate: $$1  2  3  4  5.$$ The graph above is shellable, so this is CohenMacaulay.
Now, let $C_2$ act on $X$ by $(x_1, x_2, x_3, x_4, x_5) \mapsto (x_1, x_2,  x_3, x_4, x_5)$. Then the fixed locus of $C_2$ (even schemetheoretically) is $$Y = \{ x_1 x_4 = x_1 x_5 = x_2 x_4 = x_2 x_5 = x_3 = 0 \}.$$
This is the reduced union of two $2$planes; we can visualize it as $$1  2 \phantom{ 3 } 4  5.$$ That is a standard example of a nonCohenMacaulay ring.

$\begingroup$ Thanks for the great answer! Small typo: $\mathbb{C}^2$ should be $\mathbb{C}^6$. $\endgroup$– JefSep 2, 2020 at 15:41
