# Are schematic fixed-points of a Cohen-Macaulay scheme Cohen-Macaulay?

I'm not sure how long this iterative questions can go on, but let me try again. Let's say $X$ is a Cohen-Macaulay scheme with an action of $\mathbb{G}_m$ (i.e. if $X$ is affine, a grading on the coordinate ring). Are the schematic fixed points $X^{\mathbb{G}_m}$ of $X$ Cohen-Macaulay?

• I am not sure what the usual business is, but it is not true that if $\mathbb G_{\rm m}$ acts on a variety $X$ with a fixed point $p$, then this induces an action of $\mathbb G_{\rm m}$ on $\mathop{\rm Spec}\mathcal O_{X,p}$. Oct 25, 2010 at 21:39
• Is this an issue of not being able to find a $\mathbb{G}_m$-invariant affine open containing $p$? Oct 25, 2010 at 22:23
• Those invariant affines are not cofinal in all neighborhoods. Algebraically, the localization of C[x] at the ideal (x) does not admit a grading...does it? Oct 25, 2010 at 22:34
• Ah, right. That was complete nonsense. Removed. I think you probably you can reduce to the graded local case, but let me not worry about that. Oct 25, 2010 at 23:03
• Let $T^1$ act on ${\mathbb A}^3$ with weights $0,1,1$. Then on the ${\mathbb P}^2$, the fixed-point set is not equidimensional. So you can only hope to have a local statement. Oct 26, 2010 at 0:39

Here is a counterexample. Consider the action of $\mathbb G_{\rm m}$ on $\mathbb A^4$ defined by $t \cdot(x,y,z,w) = (x, y, tz, t^{-1}w)$, and let $X$ be the invariant closed subscheme with ideal $(xy, y^2 + zw)$; this is a complete intersection, hence it is Cohen-Macaulay. The fixed point subscheme is obtained by intersecting with the fixed point subscheme in $\mathbb A^4$, which is given by $z = w = 0$; hence it is the subscheme of $\mathbb A^2$ given by $xy = y^2 = 0$, which is of course the canonical example of a non Cohen-Macaulay scheme.

Developing this idea a little, one can show that any kind of horrible singularity can appear in the fixed point subscheme of a $\mathbb G_{\rm m}$-action on a complete intersection variety.

• Murphy's law strikes again! Oct 27, 2010 at 6:50

Edit: the following does not answer Ben's question. It gives an example of the subring fixed by $G_m$ being not CM, while the question asked about the subscheme of fixed points, see the comments for more details.

Let $R$ be the (homogenous) cone of a curve $C$ of genus $g>0$, for example $R=\mathbb C[x,y,z]/(x^3+y^3+z^3)$. Let $S=R[u,v]$, $X=\text{Spec}(S)$ and $G_m$ acts by

$a.(x,y,z,u,v) = (ax,ay,az,a^{-1}u, a^{-1}v)$.

Then $A= S^{G_m}$ would be a homogenous coordinate ring for $Y= C\times \mathbb P^1$, so it is not Cohen-Macaulay (if $A$ is CM, it would mean that $H^1(Y,\mathcal O_Y)=0$, impossible, see here for an explanation).

(I learned this idea from Hochster, let me try to find a reference)

• Hailong- this is the categorical quotient, not schematic fixed points. The schematic fixed points here are a single point with reduced structure and thus Cohen-Macaulay. Oct 25, 2010 at 23:12
• Hmm, sorry, I always thought of this notation as the invariants. What do you mean by schematic fixed pts? Are they just literally the pts of $X$ fixed by the group action? Oct 26, 2010 at 0:10
• @Hailong: you applied the fixed-point functor to the algebra rather than to the space. That is, the notation itself has a consistent meaning. (I'm not surely how to reasonable define schematic fixed points; in characteristic zero, with a connected group, one could look where the generating vector fields vanish.) Oct 26, 2010 at 0:43
• Dear Hailong: No ad hoc constructions/definitions are required. For any scheme $X$ over a ring $k$ and action on $X$ by a $k$-gp scheme $G$, define functor $X^G$ on $k$-algebras as follows: for $k$-algebra $R$, $X^G(R)$ is set of $x \in X(R)$ fixed by the $G_R$-action on $X_R$ (i.e., for any $R$-algebra $R'$, $x$ viewed in $X(R')$ is $G(R')$-invariant). Is this represented by a closed subscheme of $X$? Yes, provided $X$ is locally of finite type and separated over $k$ and $G$ is affine and fppf over $k$ with connected fibers. (See Prop. A.8.10(1)ff. in "Pseudo-reductive groups" for details.) Oct 26, 2010 at 1:28
• @ Allen and BCnrd: thank you! I will leave my answer, so people might benefit from your explanations and avoid my mistakes! Oct 26, 2010 at 2:07