EDIT: Based on comments, I've decided to essentially rewrite this question. My apologies to those who commented below, whose comments seem a bit off topic when you try to match them with the current question. I've also posted a better-thought out follow-up here.

Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. Most people I know just think about fixed points as a set, but they have a canonical scheme structure which carries more information.

As Bcnrd points out below, this subscheme of a scheme $X$ over a field $k$ is defined by looking at the functor on $k$-algebras defined abstractly by the $(\mathbb{G}_m)_A$-invariant points of $X(A)$. As Dave Anderson points out, if $X=\mathrm{Spec}(R)$, then this is simply the subscheme defined by the ideal generated by all elements of non-zero weight $I=R^{>0}R+R^{<0}R$.

For example, consider $R=\mathbb{C}[x,y,z]/(xy=z^n)$ where $x$ has weight 1, $y$ has weight -1 and $z$ has weight 0. In this case $I=(x,y)$, so $X^{\mathbb{G}_m}=\mathrm{Spec} \:\:\:\mathbb{C}[z]/(z^n)$. So, as you can see, the fixed point scheme doesn't have to be reduced, though if $X^{\mathbb{G}_m}$ is 0-dimensional, this can only happen if $X$ is not regular at the corresponding fixed point.

Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities $\tilde X$, with a compatible $\mathbb{G}_m$ action, and $\tilde X^{\mathbb{G}_m}$ also 0 dimensional.

Can I conclude anything about the length of $X^{\mathbb{G}_m}$ from knowing the length of $\tilde X^{\mathbb{G}_m}$ (which is just the number of fixed $k$-points by smoothness of $\tilde X$)? In a number of examples I'm looking at, these turn out to be equal, and I'm wondering how general a phenomenon this is.

For example, the example $R=\mathbb{C}[x,y,z]/(xy=z^n)$ has a resolution with $n$ fixed points, and indeed, that's the length of $X^{\mathbb{G}_m}$.

Now, the examples I'm looking at have special features which may or may not be revelant, but I mention them in case they strike a chord.

  1. I'm looking at examples where $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian.

  2. Also in my examples, $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation $\tilde Y$, where the generic fiber is affine, and the fixed point scheme $\tilde Y^{\mathbb{G}_m}$ is flat and finite over the base.

  • $\begingroup$ Dear Ben: Let $k$ be the (arbitrary) ground field, and $T$ a $k$-torus acting on sep'td $k$-scheme $X$ of finite type. Are you interested in your "ad hoc" construction (for affine $X$), or the closed subscheme $X^T$ that represents the functor associating to any $k$-algebra $A$ the set of points in $X(A)$ that are fixed by the $T_A$-action on the $A$-scheme $X_A$? (Not obvious to me that these agree.) See Prop. A.8.10 of "Pseudo-reductive groups" for existence of $X^T$, and proof that ${\rm{Tan}}_x(X^T) = {\rm{Tan}}_x(X)^T$ for any $x \in X^T(k)$. That may help to tell you about dimensions. $\endgroup$
    – BCnrd
    Oct 23, 2010 at 0:46
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    $\begingroup$ BCrnd & Ben: Isn't this equal to the fixed point scheme? Could be wrong, but I think two things are true: (1) if $X$ is an affine $\Bbb{G}_m$-scheme (=$\Bbb{Z}$-graded affine $k$-algebra $R$), then the scheme $X^T$ is Spec of $R/I$, where $I = R^{>0}R+R^{<0}R$ is the ideal generated by nonzero-degree elements; and (2) the obvious map $R^0/(R^0 \cap R^{<0} R^{>0}) \to R/I$ is an isomorphism. (Please correct me!) $\endgroup$ Oct 23, 2010 at 5:29
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    $\begingroup$ Your definition fixed point subscheme is a little off, because it is missing a base change. See also BCnrd's answer to mathoverflow.net/questions/3190/… $\endgroup$
    – S. Carnahan
    Oct 23, 2010 at 11:02
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    $\begingroup$ This is a very good question. On the (rather feeble) basis of some examples, one could conjecture that the length of the fixed point scheme of the singular variety is at most equal to that of the desingularization, and that you have equality if the resolution is crepant (this would be somehow related to your conditions). I wonder if one can relate this with one of the many McKay type conjectures floating around. $\endgroup$
    – Angelo
    Oct 23, 2010 at 13:39
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    $\begingroup$ Angelo, a simple example of a 3-dimensional affine quadratic cone shows that one can't expect to have equality for crepant resolutions. Indeed, the small resolution of this singularity is crepant, and has 2 torus fixed points on the exceptional locus, while the singular point has length 1. $\endgroup$
    – Sasha
    Oct 23, 2010 at 19:31

1 Answer 1


Since a good picture for a length $n$ scheme set-theoretically supported at a point is some kind of limit of $n$ different reduced points getting closer together, let's try to understand what's going on that way. We can appeal to this

Also in my examples, $\tilde{X}$ has a smooth $G_m$-equivariant deformation $\tilde{Y}$, where the generic fiber is affine, and the fixed point scheme $\tilde{Y}^{G_m}$ is flat and finite over the base.

Probably in your examples this deformation (call it $\tilde{Y}_t$, where $t$ runs through the base space) descends to a deformation of $X$ (call it $Y_t$). ``Descends to'' means in particular that generically we have $Y_t = \tilde{Y}_t$. This is the case with the A_n resolution of C[x,y,z]/xy=z^n for example, and also for other cones associated with Springer theory. Now as $t \to 0$ the fixed points in $Y_t$ will collide at the singular fixed point in $X$, suggesting that the number of them is equal to the length. And as $t \to 0$ the fixed points of $\tilde{Y}_t$ will move bijectively to the fixed points of $\tilde{X}$.

I haven't worked out what general hypotheses make this argument legitimate--at least if $X$ is a complete intersection in a vector space on which the torus acts linearly, and your $\tilde{Y}$ is smooth over the base of the deformation, we seem to be all right.

Edit: Sorry, I think "complete intersection" is a distraction. Here's a formalization of the above argument. The situation is pretty special but there's not much you have to check. Say we have a commutative diagram

  Y~ ---> Y
g~|       |g
  v       v
  S~ ---> S

that's Cartesian over a Zariski open subset of $S$, with $T$-actions on $\tilde{Y}$ and $Y$ compatible with $f$ and preserving the fibers of the vertical maps. Say that $\tilde{Y}$ and $Y$ are smooth, that $\tilde{g}$ is smooth and that $\tilde{g}\vert_{\tilde{Y}^T}$ is smooth and finite, that $g$ is flat and that $g\vert_{Y^T}$ is flat and finite. Finally, suppose that $X^T = Y^T \times_Y X$, $\tilde{X}^T = \tilde{Y}^T \times_{\tilde{Y}} \tilde{X}$. (Is this always true?) Then the length of $X^T$ = the length of $\tilde{X}^T$ because length is constant in flat families.

  • $\begingroup$ In case it's relevant, the X's I have in mind have rational singularities, and in particular, are Cohen-Macaulay. It would be a real beast to try to prove that they are complete intersections, since there's no really natural choice of embedding, but something like Gorenstein or l.c.i. is more conceivable. $\endgroup$
    – Ben Webster
    Oct 24, 2010 at 19:11
  • $\begingroup$ I believe your argument, but it takes an input the point which is the crux of the matter for me (i.e. the barrier between me and the result I ultimately want) which is whether X^T is flat over S. I just posted a question about whether/when this is true, probably literally while you were writing your addendum. $\endgroup$
    – Ben Webster
    Oct 24, 2010 at 22:19

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