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edited Oct 24 2010 at 22:07 |
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.
I'm looking at examples where $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian. 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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edited Oct 23 2010 at 20:01 |
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.
I'm looking at examples where $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian. 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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edited Oct 23 2010 at 15:05 |
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.
I'm looking at examples where $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian. 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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edited Oct 23 2010 at 7:41 |
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 field $k$ is defined by looking at the sheaf 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 cord.
I'm looking at examples where $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian. 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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edited Oct 23 2010 at 6:53 |
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. Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. Most people I recently in my research came upon know just think about fixed points as a variation on this that I've been stuck on some details ofset, but they have a canonical scheme structure which carries more information. Given an action As Bcnrd points out below, this subscheme of $\mathbb{G}_m$ on an affine a scheme $X=\mathrm{Spec}(R)$ (lets' say X$ over an algebraically closed field), we can consider the set of closed points which are field $\mathbb{G}_m$-invariant, or we can study the scheme k$ is defined by looking at the ring $B=R^0/(R^0\cap R^{>0}R^{<0})$ where sheaf on $R^\star$ denotes k$-algebras defined abstractly by the set of elements $\mathbb{G}_m(A)$ invariant points of $R$ whose weights are in that set. Geometrically, this should correspond to the fixed X(A)$. As Dave Anderson points , but maybe with some funny scheme structure. Indeedout, the closed points of this scheme are just if $X^{\mathbb{G}_m}$, but not necessarily with X=\mathrm{Spec}(R)$, then this is simply the reduced scheme structure. Somehow, subscheme defined by the unreducedness ideal generated by all elements of this measures how singular the variety is at fixed points (for example, if you're smooth at a fixed point, it will be reduced).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 $B=\mathbb{C}[z]/(z^n)$.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 isolated a compatible $\mathbb{G}_m$ fixed pointsaction, and $\tilde X^{\mathbb{G}_m}$ also 0 dimensional. Can I conclude anything about the dimension length of $B=R^0/(R^0\cap R^{>0}R^{<0})$ X^{\mathbb{G}_m}$ from knowing the number length of fixed points on the resolution $\tilde X$? Especially, can I find upper bounds? What I would love would be if X^{\mathbb{G}_m}$ (which is just the number of fixed points in the resolution were the same as the dimension $k$-points by smoothness of $B$. \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 dimension length of $B$.X^{\mathbb{G}_m}$. Now, the examples I'm willing to assume looking at have special features which may or may not be revelant, but I mention them in case they strike a large number of things about this situationcord.For example: I'm willing to assume that looking at examples where $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian. I'm also willing to assume that 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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edited Oct 23 2010 at 5:45 |
Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. I recently in my research came upon a variation on this that I've been stuck on some details of.
Given an action of $\mathbb{G}_m$ on an affine scheme $X=\mathrm{Spec}(R)$ (lets' say over an algebraically closed field), we can consider the set of closed points which are $\mathbb{G}_m$-invariant, or we can study the subscheme scheme defined by the ideal ring $B=R^0/(R^0\cap R^{>0}R^{<0})$ where $R^\star$ denotes the set of elements of $R$ whose weights are in that set. Geometrically, this can be thought of as passing should correspond to the subscheme where positive weights vanish, and then passing tofixed fixed points, but maybe with some funny scheme structure. Indeed, the closed points of this subscheme scheme are just $X^{\mathbb{G}_m}$, but not necessarily with the reduced scheme structure. Somehow, the unreducedness of this measures how singular the variety is at fixed points (for example, if you're smooth at a fixed point, it will be reduced).
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 $B=\mathbb{C}[z]/(z^n)$.
Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities $\tilde X$, with isolated $\mathbb{G}_m$ fixed points.
Can I conclude anything about the dimension of $B=R^0/(R^0\cap R^{>0}R^{<0})$ from knowing the number of fixed points on the resolution $\tilde X$? Especially, can I find upper bounds?
What I would love would be if the number of fixed points in the resolution were the same as the dimension of $B$. 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 dimension of $B$.
I'm willing to assume a large number of things about this situation. For example:
I'm willing to assume that $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian. I'm also willing to assume that $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation, where the generic fiber is affine.
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edited Oct 23 2010 at 4:16 |
Understanding the unreducedness of "scheme-y a subscheme supported on fixed points"
Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. I recently in my research came upon a variation on this that I've been stuck on some details of.
Given an action of $\mathbb{G}_m$ on an affine scheme $X=\mathrm{Spec}(R)$ (lets' say over an algebraically closed field), we can consider the set of closed points which are $\mathbb{G}_m$-invariant, or we can study the subscheme defined by the ideal $R^0/(R^0\cap B=R^0/(R^0\cap R^{>0}R^{<0})$ where $R^\star$ denotes the set of elements of $R$ whose weights are in that set. Geometrically, this can be thought of as passing to the subscheme where positive weights vanish, and then passing tofixed points, but maybe with some funny scheme structure. Indeed, the closed points of this subscheme are just $X^{\mathbb{G}_m}$, but not necessarily with the reduced scheme structure. Somehow, the unreducedness of this measures how singular the variety is at fixed points (for example, if you're smooth at a fixed point, it will be reduced).
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 $B=\mathbb{C}[z]/(z^n)$.
Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities $\tilde X$, with isolated $\mathbb{G}_m$ fixed points.
Can I conclude anything about the dimension of $R^0/(R^0\cap B=R^0/(R^0\cap R^{>0}R^{<0})$ from knowing the number of fixed points on the resolution $\tilde X$? Especially, can I find upper bounds?
What I would love would be if the number of fixed points in the resolution were the same as the dimension of $B$. 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 dimension of $B$.
I'm willing to assume a large number of things about this situation. For example:
I'm willing to assume that $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian. I'm also willing to assume that $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation, where the generic fiber is affine.
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asked Oct 23 2010 at 0:21 |
Understanding the unreducedness of "scheme-y fixed points"
Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. I recently in my research came upon a variation on this that I've been stuck on some details of.
Given an action of $\mathbb{G}_m$ on an affine scheme $X=\mathrm{Spec}(R)$ (lets' say over an algebraically closed field), we can consider the set of closed points which are $\mathbb{G}_m$-invariant, or we can study the subscheme defined by the ideal $R^0/(R^0\cap R^{>0}R^{<0})$ where $R^\star$ denotes the set of elements of $R$ whose weights are in that set. Geometrically, this can be thought of as passing to the subscheme where positive weights vanish, and then passing tofixed points, but maybe with some funny scheme structure. Indeed, the closed points of this subscheme are just $X^{\mathbb{G}_m}$, but not necessarily with the reduced scheme structure. Somehow, the unreducedness of this measures how singular the variety is at fixed points (for example, if you're smooth at a fixed point, it will be reduced).
Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities, with isolated $\mathbb{G}_m$ fixed points.
Can I conclude anything about the dimension of $R^0/(R^0\cap R^{>0}R^{<0})$ from knowing the number of fixed points on the resolution? Especially, can I find upper bounds?
I'm willing to assume a large number of things about this situation. For example:
I'm willing to assume that $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian. I'm also willing to assume that $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation, where the generic fiber is affine.
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