12
$\begingroup$

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

$\endgroup$
4
  • 2
    $\begingroup$ I went down a rabbit hole looking for a counter-example with $X = \mathbb{C}^n$, and I'll report my failure. According to imsc.uni-graz.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 Cohen-Macaulay. I haven't traced the references in these slides, but this makes me suspect the problem is hard. $\endgroup$ Sep 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'. $\endgroup$
    – Jef
    Sep 2, 2020 at 15:17
  • $\begingroup$ Thanks for the reference! And, in the positive direction, I think I have a counter-example now. $\endgroup$ Sep 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). $\endgroup$ Sep 2, 2020 at 15:20

1 Answer 1

13
$\begingroup$

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 Cohen-Macaulay.

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 scheme-theoretically) 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 non-Cohen-Macaulay ring.

$\endgroup$
2
  • $\begingroup$ Thanks for the great answer! Small typo: $\mathbb{C}^2$ should be $\mathbb{C}^6$. $\endgroup$
    – Jef
    Sep 2, 2020 at 15:41
  • $\begingroup$ Fixed, thank you! $\endgroup$ Sep 2, 2020 at 15:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.