# Under what hypotheses are schematic fixed points of a flat deformation themselves flat?

This is something of a follow-up question to this one; I hope people won't think this is a duplicate. At least in my head, it seems like a distinct enough question to merit a fresh start.

All my schemes will be finite type over an algebraically closed field $k$. Let $X\to S$ be a flat family of affine schemes over smooth affine base. Let's say for now that each fiber and the whole family have rational singularities, and thus are Cohen-Macaulay. Assume, furthermore, that $X$ has an action of the group scheme $T=(\mathbb{G}_m)_S$; this is the same data as a grading on $k[X]$ such that $k[S]$ has degree 0.

Now, we can take the schematic fixed points $X^T$ of this family, which is a subscheme of $X$ whose points over any ring are invariant points of $X$. Concretely, this is the vanishing set of the ideal generated by all functions of non-zero degree.

Must the morphism $X^T\to S$ be flat? If not, are there stricter hypotheses than I gave above which would assure it is?

For example, consider the family $$X=\mathrm{Spec}[x,y,z,a_0,\dots, a_{n-1}]/(xy=z^n+a_{n-1}z^{n-1}+\cdots + a_0)$$ where $S=\mathrm{Spec}[a_0,\dots, a_{n-1}]$ with $x$ having degree 1, $y$ degree $-1$ and $z,a_i$ having degree 0. In this case $$X^T=\mathrm{Spec}[z,a_0,\dots, a_{n-1}]/(z^n+a_{n-1}z^{n-1}+\cdots + a_0=0),$$ which is, of course, flat over $S$, even though the number of closed points in a fiber (the number of roots of $z^n+a_{n-1}z^{n-1}+\cdots + a_0$) varies from $n$ to $1$.

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This is false for general group actions. Think of the additive group acting on P^1 x C with x sending (z,t) to (z + tx, t); the dimension of the fixed points at t = 0 jumps. I haven't been able come up with an example like this for torus actions, so I'm starting to believe it. –  David Treumann Oct 24 '10 at 23:35
Another thought: maybe one can reduce to the case of a finite group, thinking of a (char zero) torus as a limit of its finite subgroups. –  David Treumann Oct 24 '10 at 23:36
The behavior of reductive groups is quite different that of unipotent ones, so the counterexample doesn't worry me so much. After all, the degree 0 part of the ring has to vary flatly, and that's the coordinate ring of the categorical quotient, so nothing too horrible can happen. –  Ben Webster Oct 25 '10 at 0:01
Dear David: Good thought. Consider noetherian scheme $S$ and flat $S$-scheme $X$ of f. type equipped with action by $S$-torus $T$. Then $X^T$ exists as closed subscheme of $X$, & formation commutes with base change. To check flatness, WLOG $S$ is local. Pick prime $\ell$ invertible on $S$, and observe that the collection of finite etale $S$-subgroups $G_n=T[\ell^n]$ is relatively sch. dense in $T$. Thus, $X^T = X^{G_n}$ for large $n$ by noetherianness. By finite etale base change on $S$, $G_n$ constant. So problem reduces to analogue for action by finite gp of order invertible on $S$. Hmm... –  BCnrd Oct 25 '10 at 0:09
This is false for finite groups, though. Consider two copies of the affine line glued at the origin, with an action of a cyclic group of order 2 switching the two copies, mapping to the affine line. –  Angelo Oct 25 '10 at 8:58

## 1 Answer

Here is a counterexample. Let $\mathbb G_{\rm m}$ act on $\mathbb A^2$ by $t\cdot(x,y) = (tx,t^{-1}y)$, and let $f\colon \mathbb A^2 \to \mathbb A^1$ be defined by $f(x,y) = xy$.

I am positive that when $X$ is smooth over $Y$, the fixed point scheme is also smooth; but I doubt that one can say much more, in general.

 Here is a variant. Let $\mathbb G_{\rm m}$ act on $\mathbb A^4$ by $t\cdot(x,y, z, w) = (tx,t^{-1}y,tz,t^{-1}w)$, and let $f\colon \mathbb A^4 \to \mathbb A^1$ be defined by $f(x,y) = xy + zw$.

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I had the exact same counterexample typed up and was about to submit! The only remaining value added was the statement "Degeneration can produce new fixed points." –  Allen Knutson Oct 25 '10 at 14:30
To Allen: nice statement, though! –  Angelo Oct 25 '10 at 16:09
While this is an interesting example, it doesn't actually answer my question; the fiber at the origin doesn't have rational singularities, since its normalization is disconnected. –  Ben Webster Oct 25 '10 at 19:34
Dear Ben, I believe that the modified version above should satisfy your conditions. –  Angelo Oct 25 '10 at 21:19
I prefer to think of the map in the variant as $\det: M_{2\times 2} \to {\mathbb A}^1$. –  Allen Knutson Oct 26 '10 at 0:31