Can distinct morphisms between curves induce the same morphism on singular cohomology? The 2019 Stack Overflow Developer Survey Results Are InGeneric fiber of morphism between non-singular curvesWhat are unramified morphisms like?Does the normalization of a projective morphism determine the line bundle?Field of definition of canonical morphism between (congruence) modular curvesEtale covers of a hyperelliptic curvepushing out families of curvesWhat happens to the gonality under a finite morphism of curvesCan a birational morphism between two smooth varieties of the same betti numbers exist?Examples of endomorphisms of complex curvesDoes there exist trace maps between $ell$-adic cohomology groups for finite flat morphisms?
Can distinct morphisms between curves induce the same morphism on singular cohomology?
The 2019 Stack Overflow Developer Survey Results Are InGeneric fiber of morphism between non-singular curvesWhat are unramified morphisms like?Does the normalization of a projective morphism determine the line bundle?Field of definition of canonical morphism between (congruence) modular curvesEtale covers of a hyperelliptic curvepushing out families of curvesWhat happens to the gonality under a finite morphism of curvesCan a birational morphism between two smooth varieties of the same betti numbers exist?Examples of endomorphisms of complex curvesDoes there exist trace maps between $ell$-adic cohomology groups for finite flat morphisms?
$begingroup$
Suppose $f,g:X rightarrow Y$ are finite morphisms between connected smooth curves over $mathbbC$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,mathbbC) rightarrow H^*(X,mathbbC)$, does $f=g$?
ag.algebraic-geometry
asked 2 days ago
rj7k8rj7k8
195117
$endgroup$
add a comment |
$begingroup$
Suppose $f,g:X rightarrow Y$ are finite morphisms between connected smooth curves over $mathbbC$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,mathbbC) rightarrow H^*(X,mathbbC)$, does $f=g$?
ag.algebraic-geometry
asked 2 days ago
rj7k8rj7k8
195117
$endgroup$
add a comment |
$begingroup$
Suppose $f,g:X rightarrow Y$ are finite morphisms between connected smooth curves over $mathbbC$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,mathbbC) rightarrow H^*(X,mathbbC)$, does $f=g$?
ag.algebraic-geometry
asked 2 days ago
rj7k8rj7k8
195117
$endgroup$
Suppose $f,g:X rightarrow Y$ are finite morphisms between connected smooth curves over $mathbbC$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,mathbbC) rightarrow H^*(X,mathbbC)$, does $f=g$?
ag.algebraic-geometry
ag.algebraic-geometry
asked 2 days ago
rj7k8rj7k8
195117
asked 2 days ago
rj7k8rj7k8
195117
asked 2 days ago
rj7k8rj7k8
195117
asked 2 days ago
rj7k8rj7k8
195117
asked 2 days ago
rj7k8rj7k8
195117
195117
add a comment |
add a comment |
1 Answer
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Yes. Since $Y$ embeds into its Jacobian $B$, it is enough to prove the statement for pairs of maps to an abelian variety $f, gcolon Xto B$ sending a base point $xin X$ to $0in B$. Every such map factors uniquely through the Albanese variety $A$ of $X$, so we reduce further to the case of pairs of maps $f, gcolon Ato B$ between abelian varieties (sending $0$ to $0$). Every such map is necessarily a group homomorphism, and is uniquely determined by what it does on $pi_1 = H_1$, or on $H^1(-, mathbfC)$.
answered 2 days ago
Piotr AchingerPiotr Achinger
8,62212854
$endgroup$
3
$begingroup$
It seems to me that you have not used that the genus of $Y$ is at least $2$ (which is certainly a necessary condition).
$endgroup$
– ulrich
yesterday
1
$begingroup$
@ulrich I think this condition is used implicitly in the first step. The argument after that point provides the equality $f=alpha circ g$ where $alpha:B rightarrow B$ is a translation, but genus $>1$ is needed to then conclude that $f=g$. I did this using that $alpha$ induces both the identity on cohomology and an automorphism of $Y$, and using the Lefschetz fixed point formula (maybe Piotr had something else in mind).
$endgroup$
– rj7k8
yesterday
$begingroup$
@ulrich You are right, the argument works literally if $f$ and $g$ satisfy $f(x)=g(x)$. Without this assumption, it shows that $f$ and $g$ differ by a translation in $B$. So it remains to show that if $Y$ is a curve of genus $>1$ embedded in its Jacobian $B$, then for every $bin B$, $Ycap (b+Y)$ is finite. Otherwise, $Y = b+Y$ and you can argue as rj7k8 above, but I guess there should be a direct argument using Riemann-Roch.
$endgroup$
– Piotr Achinger
yesterday
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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$begingroup$
Yes. Since $Y$ embeds into its Jacobian $B$, it is enough to prove the statement for pairs of maps to an abelian variety $f, gcolon Xto B$ sending a base point $xin X$ to $0in B$. Every such map factors uniquely through the Albanese variety $A$ of $X$, so we reduce further to the case of pairs of maps $f, gcolon Ato B$ between abelian varieties (sending $0$ to $0$). Every such map is necessarily a group homomorphism, and is uniquely determined by what it does on $pi_1 = H_1$, or on $H^1(-, mathbfC)$.
answered 2 days ago
Piotr AchingerPiotr Achinger
8,62212854
$endgroup$
3
$begingroup$
It seems to me that you have not used that the genus of $Y$ is at least $2$ (which is certainly a necessary condition).
$endgroup$
– ulrich
yesterday
1
$begingroup$
@ulrich I think this condition is used implicitly in the first step. The argument after that point provides the equality $f=alpha circ g$ where $alpha:B rightarrow B$ is a translation, but genus $>1$ is needed to then conclude that $f=g$. I did this using that $alpha$ induces both the identity on cohomology and an automorphism of $Y$, and using the Lefschetz fixed point formula (maybe Piotr had something else in mind).
$endgroup$
– rj7k8
yesterday
$begingroup$
@ulrich You are right, the argument works literally if $f$ and $g$ satisfy $f(x)=g(x)$. Without this assumption, it shows that $f$ and $g$ differ by a translation in $B$. So it remains to show that if $Y$ is a curve of genus $>1$ embedded in its Jacobian $B$, then for every $bin B$, $Ycap (b+Y)$ is finite. Otherwise, $Y = b+Y$ and you can argue as rj7k8 above, but I guess there should be a direct argument using Riemann-Roch.
$endgroup$
– Piotr Achinger
yesterday
add a comment |
$begingroup$
Yes. Since $Y$ embeds into its Jacobian $B$, it is enough to prove the statement for pairs of maps to an abelian variety $f, gcolon Xto B$ sending a base point $xin X$ to $0in B$. Every such map factors uniquely through the Albanese variety $A$ of $X$, so we reduce further to the case of pairs of maps $f, gcolon Ato B$ between abelian varieties (sending $0$ to $0$). Every such map is necessarily a group homomorphism, and is uniquely determined by what it does on $pi_1 = H_1$, or on $H^1(-, mathbfC)$.
answered 2 days ago
Piotr AchingerPiotr Achinger
8,62212854
$endgroup$
3
$begingroup$
It seems to me that you have not used that the genus of $Y$ is at least $2$ (which is certainly a necessary condition).
$endgroup$
– ulrich
yesterday
1
$begingroup$
@ulrich I think this condition is used implicitly in the first step. The argument after that point provides the equality $f=alpha circ g$ where $alpha:B rightarrow B$ is a translation, but genus $>1$ is needed to then conclude that $f=g$. I did this using that $alpha$ induces both the identity on cohomology and an automorphism of $Y$, and using the Lefschetz fixed point formula (maybe Piotr had something else in mind).
$endgroup$
– rj7k8
yesterday
$begingroup$
@ulrich You are right, the argument works literally if $f$ and $g$ satisfy $f(x)=g(x)$. Without this assumption, it shows that $f$ and $g$ differ by a translation in $B$. So it remains to show that if $Y$ is a curve of genus $>1$ embedded in its Jacobian $B$, then for every $bin B$, $Ycap (b+Y)$ is finite. Otherwise, $Y = b+Y$ and you can argue as rj7k8 above, but I guess there should be a direct argument using Riemann-Roch.
$endgroup$
– Piotr Achinger
yesterday
add a comment |
$begingroup$
Yes. Since $Y$ embeds into its Jacobian $B$, it is enough to prove the statement for pairs of maps to an abelian variety $f, gcolon Xto B$ sending a base point $xin X$ to $0in B$. Every such map factors uniquely through the Albanese variety $A$ of $X$, so we reduce further to the case of pairs of maps $f, gcolon Ato B$ between abelian varieties (sending $0$ to $0$). Every such map is necessarily a group homomorphism, and is uniquely determined by what it does on $pi_1 = H_1$, or on $H^1(-, mathbfC)$.
answered 2 days ago
Piotr AchingerPiotr Achinger
8,62212854
$endgroup$
Yes. Since $Y$ embeds into its Jacobian $B$, it is enough to prove the statement for pairs of maps to an abelian variety $f, gcolon Xto B$ sending a base point $xin X$ to $0in B$. Every such map factors uniquely through the Albanese variety $A$ of $X$, so we reduce further to the case of pairs of maps $f, gcolon Ato B$ between abelian varieties (sending $0$ to $0$). Every such map is necessarily a group homomorphism, and is uniquely determined by what it does on $pi_1 = H_1$, or on $H^1(-, mathbfC)$.
answered 2 days ago
Piotr AchingerPiotr Achinger
8,62212854
answered 2 days ago
Piotr AchingerPiotr Achinger
8,62212854
answered 2 days ago
Piotr AchingerPiotr Achinger
8,62212854
answered 2 days ago
Piotr AchingerPiotr Achinger
8,62212854
8,62212854
3
$begingroup$
It seems to me that you have not used that the genus of $Y$ is at least $2$ (which is certainly a necessary condition).
$endgroup$
– ulrich
yesterday
1
$begingroup$
@ulrich I think this condition is used implicitly in the first step. The argument after that point provides the equality $f=alpha circ g$ where $alpha:B rightarrow B$ is a translation, but genus $>1$ is needed to then conclude that $f=g$. I did this using that $alpha$ induces both the identity on cohomology and an automorphism of $Y$, and using the Lefschetz fixed point formula (maybe Piotr had something else in mind).
$endgroup$
– rj7k8
yesterday
$begingroup$
@ulrich You are right, the argument works literally if $f$ and $g$ satisfy $f(x)=g(x)$. Without this assumption, it shows that $f$ and $g$ differ by a translation in $B$. So it remains to show that if $Y$ is a curve of genus $>1$ embedded in its Jacobian $B$, then for every $bin B$, $Ycap (b+Y)$ is finite. Otherwise, $Y = b+Y$ and you can argue as rj7k8 above, but I guess there should be a direct argument using Riemann-Roch.
$endgroup$
– Piotr Achinger
yesterday
add a comment |
3
$begingroup$
It seems to me that you have not used that the genus of $Y$ is at least $2$ (which is certainly a necessary condition).
$endgroup$
– ulrich
yesterday
1
$begingroup$
@ulrich I think this condition is used implicitly in the first step. The argument after that point provides the equality $f=alpha circ g$ where $alpha:B rightarrow B$ is a translation, but genus $>1$ is needed to then conclude that $f=g$. I did this using that $alpha$ induces both the identity on cohomology and an automorphism of $Y$, and using the Lefschetz fixed point formula (maybe Piotr had something else in mind).
$endgroup$
– rj7k8
yesterday
$begingroup$
@ulrich You are right, the argument works literally if $f$ and $g$ satisfy $f(x)=g(x)$. Without this assumption, it shows that $f$ and $g$ differ by a translation in $B$. So it remains to show that if $Y$ is a curve of genus $>1$ embedded in its Jacobian $B$, then for every $bin B$, $Ycap (b+Y)$ is finite. Otherwise, $Y = b+Y$ and you can argue as rj7k8 above, but I guess there should be a direct argument using Riemann-Roch.
$endgroup$
– Piotr Achinger
yesterday
3
3
$begingroup$
It seems to me that you have not used that the genus of $Y$ is at least $2$ (which is certainly a necessary condition).
$endgroup$
– ulrich
yesterday
$begingroup$
It seems to me that you have not used that the genus of $Y$ is at least $2$ (which is certainly a necessary condition).
$endgroup$
– ulrich
yesterday
1
1
$begingroup$
@ulrich I think this condition is used implicitly in the first step. The argument after that point provides the equality $f=alpha circ g$ where $alpha:B rightarrow B$ is a translation, but genus $>1$ is needed to then conclude that $f=g$. I did this using that $alpha$ induces both the identity on cohomology and an automorphism of $Y$, and using the Lefschetz fixed point formula (maybe Piotr had something else in mind).
$endgroup$
– rj7k8
yesterday
$begingroup$
@ulrich I think this condition is used implicitly in the first step. The argument after that point provides the equality $f=alpha circ g$ where $alpha:B rightarrow B$ is a translation, but genus $>1$ is needed to then conclude that $f=g$. I did this using that $alpha$ induces both the identity on cohomology and an automorphism of $Y$, and using the Lefschetz fixed point formula (maybe Piotr had something else in mind).
$endgroup$
– rj7k8
yesterday
$begingroup$
@ulrich You are right, the argument works literally if $f$ and $g$ satisfy $f(x)=g(x)$. Without this assumption, it shows that $f$ and $g$ differ by a translation in $B$. So it remains to show that if $Y$ is a curve of genus $>1$ embedded in its Jacobian $B$, then for every $bin B$, $Ycap (b+Y)$ is finite. Otherwise, $Y = b+Y$ and you can argue as rj7k8 above, but I guess there should be a direct argument using Riemann-Roch.
$endgroup$
– Piotr Achinger
yesterday
$begingroup$
@ulrich You are right, the argument works literally if $f$ and $g$ satisfy $f(x)=g(x)$. Without this assumption, it shows that $f$ and $g$ differ by a translation in $B$. So it remains to show that if $Y$ is a curve of genus $>1$ embedded in its Jacobian $B$, then for every $bin B$, $Ycap (b+Y)$ is finite. Otherwise, $Y = b+Y$ and you can argue as rj7k8 above, but I guess there should be a direct argument using Riemann-Roch.
$endgroup$
– Piotr Achinger
yesterday
add a comment |
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