Understanding the unreducedness of a subscheme supported on fixed points

Question

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.

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

      f
  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.