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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.

7 added 1 characters in body

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.

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 cordchord.

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.

6 Fixed up the description of the fixed-point scheme.

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.

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 an algebraically closed a field $k$ is defined by looking at the sheaf functor on $k$-algebras defined abstractly by the $\mathbb{G}_m(A)$ invariant (\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}$. Now, the examples I'm looking at have special features which may or may not be revelant, but I mention them in case they strike a cord. 1. I'm looking at examples where$\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.

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