Mixing
Group Acting on Variety
$begingroup$
My question refers to some comments occured in following thread:
Galois morphism - group acting on the variety
The setting is that we have a finite Galois morphism $f: X to S$, where $X$ and $S$ are non-singular and connected projective varieties over $mathbb{C}$.
Galois means here that if we denote with $G$ the automorphism group of $X$ over $S$ then the quotient $X/G$ exists and the natural morphism $X/G to S $ is an isomorphism.
While the discussion there occured two points making me curious:
Firstly, @user52991 question if one start with exact sequence
$$0rightarrow F'rightarrow Frightarrow F''rightarrow 0$$
of coherent sheaves on $X$ and applying the pushforward $f_*$ gives an short exact, (since $f$ finite !) sequence
$$0 rightarrow f_*F'rightarrow f_*Frightarrow f_*F''rightarrow 0$$
over $S= X/G$.
My first question is how and why does $G$ act on these pushforward sheaves explicitely? The cruical point is how does it act on local sections $f_*F(U)$?
And second question:
If we were able to answer that the $G$-action is welldefined on the sheaves above then we can indeed deride the functor of invariants $E to E^G$. And there ocured the question of this functor is exact.
@Mohan explained that the exactness is based on Hilbert 90 thm. Could anybody explain this argument a bit more how Hilbert 90 is inveolved in the problem?
algebraic-geometry group-actions projective-varieties
asked Jan 12 at 1:30
KarlPeterKarlPeter
5701315
$endgroup$
add a comment |
$begingroup$
My question refers to some comments occured in following thread:
Galois morphism - group acting on the variety
The setting is that we have a finite Galois morphism $f: X to S$, where $X$ and $S$ are non-singular and connected projective varieties over $mathbb{C}$.
Galois means here that if we denote with $G$ the automorphism group of $X$ over $S$ then the quotient $X/G$ exists and the natural morphism $X/G to S $ is an isomorphism.
While the discussion there occured two points making me curious:
Firstly, @user52991 question if one start with exact sequence
$$0rightarrow F'rightarrow Frightarrow F''rightarrow 0$$
of coherent sheaves on $X$ and applying the pushforward $f_*$ gives an short exact, (since $f$ finite !) sequence
$$0 rightarrow f_*F'rightarrow f_*Frightarrow f_*F''rightarrow 0$$
over $S= X/G$.
My first question is how and why does $G$ act on these pushforward sheaves explicitely? The cruical point is how does it act on local sections $f_*F(U)$?
And second question:
If we were able to answer that the $G$-action is welldefined on the sheaves above then we can indeed deride the functor of invariants $E to E^G$. And there ocured the question of this functor is exact.
@Mohan explained that the exactness is based on Hilbert 90 thm. Could anybody explain this argument a bit more how Hilbert 90 is inveolved in the problem?
algebraic-geometry group-actions projective-varieties
asked Jan 12 at 1:30
KarlPeterKarlPeter
5701315
$endgroup$
$begingroup$
Start with $mathbb{C}(S) = mathbb{C}(X)^G$ and $G$ a finite group of birational maps $X to X$ ?
$endgroup$
– reuns
Jan 12 at 4:32
$begingroup$
@reuns: Ah ok the $G$-action make sense on global section, but what about the local sections? Concretely an automorphism $g: X to X$ induces the rings morphisms on local sections in following way: Let $U subset X$, then $$g_U ^{#}:Gamma(U, mathcal{O}_X) to Gamma(U, g_*mathcal{O}_X)= Gamma(g^{-1}(U), mathcal{O}_X)$$ And here occurs to me the question how $G$ can act on the local sections $Gamma(U, mathcal{O}_X)$? This action only make sense if $g^{-1}(U)$ for all $g in G$ and $ U subset X$, right?
$endgroup$
– KarlPeter
Jan 12 at 21:24
$begingroup$
@reuns: Therefore I don't know why it make sense to define $Gamma(U, mathcal{O}_X)^G$. On the other hand since $g^{-1}(X)=X$ it make sense to define the action on global sections $Gamma(X, mathcal{O}_X)$, and this induces well defined $mathbb{C}(S) = mathbb{C}(X)^G$, but since we expect that $X/G$ has a variety structure (especially a scheme structure) it should be possible to talk about local sections of $X/G$. What are they? The similar problem occures when we want to transfer the action of $G$ to pushforards $f_*F$. How does this concretely look like on local sections?
$endgroup$
– KarlPeter
Jan 12 at 21:25
$begingroup$
Let $X = mathbb{A^1_C}$ and $f(x) = 1/x$ and $G = {1,f}$. Then $G$ is regular on $X - {0} cong X'={(x,y),xy=1} subset mathbb{A^2_C}$ and $mathbb{C}(X)=mathbb{C}(X-{0})$. On $X'$, $G$ become a group of polynomial maps and $mathbb{C}(X')^G =mathbb{C}(Y)$ where $Y= { { (x,y),(f(x),f(y))},xy=1}={ { (x,y),(y,x)},xy=1}cong $ (symmetric polynomials) ${ (x+y,xy),xy=1} = h(W)$ where $W={ (x,y,u,v), u=x+y,v=xy,xy=1})$ and $h(x,y,u,v)= (u,v)$ so $h(W) = V( (u-x-y,v-xy,xy-1) cap mathbb{C}[u,v])$ an affine variety. Won't the general case be similar ?
$endgroup$
– reuns
Jan 13 at 10:28
add a comment |
$begingroup$
My question refers to some comments occured in following thread:
Galois morphism - group acting on the variety
The setting is that we have a finite Galois morphism $f: X to S$, where $X$ and $S$ are non-singular and connected projective varieties over $mathbb{C}$.
Galois means here that if we denote with $G$ the automorphism group of $X$ over $S$ then the quotient $X/G$ exists and the natural morphism $X/G to S $ is an isomorphism.
While the discussion there occured two points making me curious:
Firstly, @user52991 question if one start with exact sequence
$$0rightarrow F'rightarrow Frightarrow F''rightarrow 0$$
of coherent sheaves on $X$ and applying the pushforward $f_*$ gives an short exact, (since $f$ finite !) sequence
$$0 rightarrow f_*F'rightarrow f_*Frightarrow f_*F''rightarrow 0$$
over $S= X/G$.
My first question is how and why does $G$ act on these pushforward sheaves explicitely? The cruical point is how does it act on local sections $f_*F(U)$?
And second question:
If we were able to answer that the $G$-action is welldefined on the sheaves above then we can indeed deride the functor of invariants $E to E^G$. And there ocured the question of this functor is exact.
@Mohan explained that the exactness is based on Hilbert 90 thm. Could anybody explain this argument a bit more how Hilbert 90 is inveolved in the problem?
algebraic-geometry group-actions projective-varieties
asked Jan 12 at 1:30
KarlPeterKarlPeter
5701315
$endgroup$
My question refers to some comments occured in following thread:
Galois morphism - group acting on the variety
The setting is that we have a finite Galois morphism $f: X to S$, where $X$ and $S$ are non-singular and connected projective varieties over $mathbb{C}$.
Galois means here that if we denote with $G$ the automorphism group of $X$ over $S$ then the quotient $X/G$ exists and the natural morphism $X/G to S $ is an isomorphism.
While the discussion there occured two points making me curious:
Firstly, @user52991 question if one start with exact sequence
$$0rightarrow F'rightarrow Frightarrow F''rightarrow 0$$
of coherent sheaves on $X$ and applying the pushforward $f_*$ gives an short exact, (since $f$ finite !) sequence
$$0 rightarrow f_*F'rightarrow f_*Frightarrow f_*F''rightarrow 0$$
over $S= X/G$.
My first question is how and why does $G$ act on these pushforward sheaves explicitely? The cruical point is how does it act on local sections $f_*F(U)$?
And second question:
If we were able to answer that the $G$-action is welldefined on the sheaves above then we can indeed deride the functor of invariants $E to E^G$. And there ocured the question of this functor is exact.
@Mohan explained that the exactness is based on Hilbert 90 thm. Could anybody explain this argument a bit more how Hilbert 90 is inveolved in the problem?
algebraic-geometry group-actions projective-varieties
algebraic-geometry group-actions projective-varieties
asked Jan 12 at 1:30
KarlPeterKarlPeter
5701315
asked Jan 12 at 1:30
KarlPeterKarlPeter
5701315
edited Jan 12 at 20:06
KarlPeter
asked Jan 12 at 1:30
KarlPeterKarlPeter
5701315
asked Jan 12 at 1:30
KarlPeterKarlPeter
5701315
asked Jan 12 at 1:30
KarlPeterKarlPeter
5701315
5701315
$begingroup$
Start with $mathbb{C}(S) = mathbb{C}(X)^G$ and $G$ a finite group of birational maps $X to X$ ?
$endgroup$
– reuns
Jan 12 at 4:32
$begingroup$
@reuns: Ah ok the $G$-action make sense on global section, but what about the local sections? Concretely an automorphism $g: X to X$ induces the rings morphisms on local sections in following way: Let $U subset X$, then $$g_U ^{#}:Gamma(U, mathcal{O}_X) to Gamma(U, g_*mathcal{O}_X)= Gamma(g^{-1}(U), mathcal{O}_X)$$ And here occurs to me the question how $G$ can act on the local sections $Gamma(U, mathcal{O}_X)$? This action only make sense if $g^{-1}(U)$ for all $g in G$ and $ U subset X$, right?
$endgroup$
– KarlPeter
Jan 12 at 21:24
$begingroup$
@reuns: Therefore I don't know why it make sense to define $Gamma(U, mathcal{O}_X)^G$. On the other hand since $g^{-1}(X)=X$ it make sense to define the action on global sections $Gamma(X, mathcal{O}_X)$, and this induces well defined $mathbb{C}(S) = mathbb{C}(X)^G$, but since we expect that $X/G$ has a variety structure (especially a scheme structure) it should be possible to talk about local sections of $X/G$. What are they? The similar problem occures when we want to transfer the action of $G$ to pushforards $f_*F$. How does this concretely look like on local sections?
$endgroup$
– KarlPeter
Jan 12 at 21:25
$begingroup$
Let $X = mathbb{A^1_C}$ and $f(x) = 1/x$ and $G = {1,f}$. Then $G$ is regular on $X - {0} cong X'={(x,y),xy=1} subset mathbb{A^2_C}$ and $mathbb{C}(X)=mathbb{C}(X-{0})$. On $X'$, $G$ become a group of polynomial maps and $mathbb{C}(X')^G =mathbb{C}(Y)$ where $Y= { { (x,y),(f(x),f(y))},xy=1}={ { (x,y),(y,x)},xy=1}cong $ (symmetric polynomials) ${ (x+y,xy),xy=1} = h(W)$ where $W={ (x,y,u,v), u=x+y,v=xy,xy=1})$ and $h(x,y,u,v)= (u,v)$ so $h(W) = V( (u-x-y,v-xy,xy-1) cap mathbb{C}[u,v])$ an affine variety. Won't the general case be similar ?
$endgroup$
– reuns
Jan 13 at 10:28
add a comment |
$begingroup$
Start with $mathbb{C}(S) = mathbb{C}(X)^G$ and $G$ a finite group of birational maps $X to X$ ?
$endgroup$
– reuns
Jan 12 at 4:32
$begingroup$
@reuns: Ah ok the $G$-action make sense on global section, but what about the local sections? Concretely an automorphism $g: X to X$ induces the rings morphisms on local sections in following way: Let $U subset X$, then $$g_U ^{#}:Gamma(U, mathcal{O}_X) to Gamma(U, g_*mathcal{O}_X)= Gamma(g^{-1}(U), mathcal{O}_X)$$ And here occurs to me the question how $G$ can act on the local sections $Gamma(U, mathcal{O}_X)$? This action only make sense if $g^{-1}(U)$ for all $g in G$ and $ U subset X$, right?
$endgroup$
– KarlPeter
Jan 12 at 21:24
$begingroup$
@reuns: Therefore I don't know why it make sense to define $Gamma(U, mathcal{O}_X)^G$. On the other hand since $g^{-1}(X)=X$ it make sense to define the action on global sections $Gamma(X, mathcal{O}_X)$, and this induces well defined $mathbb{C}(S) = mathbb{C}(X)^G$, but since we expect that $X/G$ has a variety structure (especially a scheme structure) it should be possible to talk about local sections of $X/G$. What are they? The similar problem occures when we want to transfer the action of $G$ to pushforards $f_*F$. How does this concretely look like on local sections?
$endgroup$
– KarlPeter
Jan 12 at 21:25
$begingroup$
Let $X = mathbb{A^1_C}$ and $f(x) = 1/x$ and $G = {1,f}$. Then $G$ is regular on $X - {0} cong X'={(x,y),xy=1} subset mathbb{A^2_C}$ and $mathbb{C}(X)=mathbb{C}(X-{0})$. On $X'$, $G$ become a group of polynomial maps and $mathbb{C}(X')^G =mathbb{C}(Y)$ where $Y= { { (x,y),(f(x),f(y))},xy=1}={ { (x,y),(y,x)},xy=1}cong $ (symmetric polynomials) ${ (x+y,xy),xy=1} = h(W)$ where $W={ (x,y,u,v), u=x+y,v=xy,xy=1})$ and $h(x,y,u,v)= (u,v)$ so $h(W) = V( (u-x-y,v-xy,xy-1) cap mathbb{C}[u,v])$ an affine variety. Won't the general case be similar ?
$endgroup$
– reuns
Jan 13 at 10:28
$begingroup$
Start with $mathbb{C}(S) = mathbb{C}(X)^G$ and $G$ a finite group of birational maps $X to X$ ?
$endgroup$
– reuns
Jan 12 at 4:32
$begingroup$
Start with $mathbb{C}(S) = mathbb{C}(X)^G$ and $G$ a finite group of birational maps $X to X$ ?
$endgroup$
– reuns
Jan 12 at 4:32
$begingroup$
@reuns: Ah ok the $G$-action make sense on global section, but what about the local sections? Concretely an automorphism $g: X to X$ induces the rings morphisms on local sections in following way: Let $U subset X$, then $$g_U ^{#}:Gamma(U, mathcal{O}_X) to Gamma(U, g_*mathcal{O}_X)= Gamma(g^{-1}(U), mathcal{O}_X)$$ And here occurs to me the question how $G$ can act on the local sections $Gamma(U, mathcal{O}_X)$? This action only make sense if $g^{-1}(U)$ for all $g in G$ and $ U subset X$, right?
$endgroup$
– KarlPeter
Jan 12 at 21:24
$begingroup$
@reuns: Ah ok the $G$-action make sense on global section, but what about the local sections? Concretely an automorphism $g: X to X$ induces the rings morphisms on local sections in following way: Let $U subset X$, then $$g_U ^{#}:Gamma(U, mathcal{O}_X) to Gamma(U, g_*mathcal{O}_X)= Gamma(g^{-1}(U), mathcal{O}_X)$$ And here occurs to me the question how $G$ can act on the local sections $Gamma(U, mathcal{O}_X)$? This action only make sense if $g^{-1}(U)$ for all $g in G$ and $ U subset X$, right?
$endgroup$
– KarlPeter
Jan 12 at 21:24
$begingroup$
@reuns: Therefore I don't know why it make sense to define $Gamma(U, mathcal{O}_X)^G$. On the other hand since $g^{-1}(X)=X$ it make sense to define the action on global sections $Gamma(X, mathcal{O}_X)$, and this induces well defined $mathbb{C}(S) = mathbb{C}(X)^G$, but since we expect that $X/G$ has a variety structure (especially a scheme structure) it should be possible to talk about local sections of $X/G$. What are they? The similar problem occures when we want to transfer the action of $G$ to pushforards $f_*F$. How does this concretely look like on local sections?
$endgroup$
– KarlPeter
Jan 12 at 21:25
$begingroup$
@reuns: Therefore I don't know why it make sense to define $Gamma(U, mathcal{O}_X)^G$. On the other hand since $g^{-1}(X)=X$ it make sense to define the action on global sections $Gamma(X, mathcal{O}_X)$, and this induces well defined $mathbb{C}(S) = mathbb{C}(X)^G$, but since we expect that $X/G$ has a variety structure (especially a scheme structure) it should be possible to talk about local sections of $X/G$. What are they? The similar problem occures when we want to transfer the action of $G$ to pushforards $f_*F$. How does this concretely look like on local sections?
$endgroup$
– KarlPeter
Jan 12 at 21:25
$begingroup$
Let $X = mathbb{A^1_C}$ and $f(x) = 1/x$ and $G = {1,f}$. Then $G$ is regular on $X - {0} cong X'={(x,y),xy=1} subset mathbb{A^2_C}$ and $mathbb{C}(X)=mathbb{C}(X-{0})$. On $X'$, $G$ become a group of polynomial maps and $mathbb{C}(X')^G =mathbb{C}(Y)$ where $Y= { { (x,y),(f(x),f(y))},xy=1}={ { (x,y),(y,x)},xy=1}cong $ (symmetric polynomials) ${ (x+y,xy),xy=1} = h(W)$ where $W={ (x,y,u,v), u=x+y,v=xy,xy=1})$ and $h(x,y,u,v)= (u,v)$ so $h(W) = V( (u-x-y,v-xy,xy-1) cap mathbb{C}[u,v])$ an affine variety. Won't the general case be similar ?
$endgroup$
– reuns
Jan 13 at 10:28
$begingroup$
Let $X = mathbb{A^1_C}$ and $f(x) = 1/x$ and $G = {1,f}$. Then $G$ is regular on $X - {0} cong X'={(x,y),xy=1} subset mathbb{A^2_C}$ and $mathbb{C}(X)=mathbb{C}(X-{0})$. On $X'$, $G$ become a group of polynomial maps and $mathbb{C}(X')^G =mathbb{C}(Y)$ where $Y= { { (x,y),(f(x),f(y))},xy=1}={ { (x,y),(y,x)},xy=1}cong $ (symmetric polynomials) ${ (x+y,xy),xy=1} = h(W)$ where $W={ (x,y,u,v), u=x+y,v=xy,xy=1})$ and $h(x,y,u,v)= (u,v)$ so $h(W) = V( (u-x-y,v-xy,xy-1) cap mathbb{C}[u,v])$ an affine variety. Won't the general case be similar ?
$endgroup$
– reuns
Jan 13 at 10:28
add a comment |
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$begingroup$
Start with $mathbb{C}(S) = mathbb{C}(X)^G$ and $G$ a finite group of birational maps $X to X$ ?
$endgroup$
– reuns
Jan 12 at 4:32
$begingroup$
@reuns: Ah ok the $G$-action make sense on global section, but what about the local sections? Concretely an automorphism $g: X to X$ induces the rings morphisms on local sections in following way: Let $U subset X$, then $$g_U ^{#}:Gamma(U, mathcal{O}_X) to Gamma(U, g_*mathcal{O}_X)= Gamma(g^{-1}(U), mathcal{O}_X)$$ And here occurs to me the question how $G$ can act on the local sections $Gamma(U, mathcal{O}_X)$? This action only make sense if $g^{-1}(U)$ for all $g in G$ and $ U subset X$, right?
$endgroup$
– KarlPeter
Jan 12 at 21:24
$begingroup$
@reuns: Therefore I don't know why it make sense to define $Gamma(U, mathcal{O}_X)^G$. On the other hand since $g^{-1}(X)=X$ it make sense to define the action on global sections $Gamma(X, mathcal{O}_X)$, and this induces well defined $mathbb{C}(S) = mathbb{C}(X)^G$, but since we expect that $X/G$ has a variety structure (especially a scheme structure) it should be possible to talk about local sections of $X/G$. What are they? The similar problem occures when we want to transfer the action of $G$ to pushforards $f_*F$. How does this concretely look like on local sections?
$endgroup$
– KarlPeter
Jan 12 at 21:25
$begingroup$
Let $X = mathbb{A^1_C}$ and $f(x) = 1/x$ and $G = {1,f}$. Then $G$ is regular on $X - {0} cong X'={(x,y),xy=1} subset mathbb{A^2_C}$ and $mathbb{C}(X)=mathbb{C}(X-{0})$. On $X'$, $G$ become a group of polynomial maps and $mathbb{C}(X')^G =mathbb{C}(Y)$ where $Y= { { (x,y),(f(x),f(y))},xy=1}={ { (x,y),(y,x)},xy=1}cong $ (symmetric polynomials) ${ (x+y,xy),xy=1} = h(W)$ where $W={ (x,y,u,v), u=x+y,v=xy,xy=1})$ and $h(x,y,u,v)= (u,v)$ so $h(W) = V( (u-x-y,v-xy,xy-1) cap mathbb{C}[u,v])$ an affine variety. Won't the general case be similar ?
$endgroup$
– reuns
Jan 13 at 10:28