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 $1$.

  • $\begingroup$ 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. $\endgroup$ – David Treumann Oct 24 '10 at 23:35
  • $\begingroup$ 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. $\endgroup$ – David Treumann Oct 24 '10 at 23:36
  • $\begingroup$ 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. $\endgroup$ – Ben Webster Oct 25 '10 at 0:01
  • 1
    $\begingroup$ 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... $\endgroup$ – BCnrd Oct 25 '10 at 0:09
  • 2
    $\begingroup$ 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. $\endgroup$ – Angelo Oct 25 '10 at 8:58

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.

[Edit] 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 $f(x,y) = xy + zw$.

  • 2
    $\begingroup$ 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." $\endgroup$ – Allen Knutson Oct 25 '10 at 14:30
  • $\begingroup$ To Allen: nice statement, though! $\endgroup$ – Angelo Oct 25 '10 at 16:09
  • $\begingroup$ 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. $\endgroup$ – Ben Webster Oct 25 '10 at 19:34
  • $\begingroup$ Dear Ben, I believe that the modified version above should satisfy your conditions. $\endgroup$ – Angelo Oct 25 '10 at 21:19
  • $\begingroup$ I prefer to think of the map in the variant as $\det: M_{2\times 2} \to {\mathbb A}^1$. $\endgroup$ – Allen Knutson Oct 26 '10 at 0:31

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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