Can distinct morphisms between curves induce the same morphism on singular cohomology?
$begingroup$
Suppose $f,g:X rightarrow Y$ are finite morphisms between connected smooth curves over $mathbb{C}$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,mathbb{C}) rightarrow H^*(X,mathbb{C})$, does $f=g$?
ag.algebraic-geometry
asked Apr 9 at 22:35
rj7k8rj7k8
195117
$endgroup$
add a comment |
$begingroup$
Suppose $f,g:X rightarrow Y$ are finite morphisms between connected smooth curves over $mathbb{C}$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,mathbb{C}) rightarrow H^*(X,mathbb{C})$, does $f=g$?
ag.algebraic-geometry
asked Apr 9 at 22:35
rj7k8rj7k8
195117
$endgroup$
add a comment |
$begingroup$
Suppose $f,g:X rightarrow Y$ are finite morphisms between connected smooth curves over $mathbb{C}$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,mathbb{C}) rightarrow H^*(X,mathbb{C})$, does $f=g$?
ag.algebraic-geometry
asked Apr 9 at 22:35
rj7k8rj7k8
195117
$endgroup$
Suppose $f,g:X rightarrow Y$ are finite morphisms between connected smooth curves over $mathbb{C}$, with $Y$ of genus at least $2$.
If $f$ and $g$ induce the same morphism $H^*(Y,mathbb{C}) rightarrow H^*(X,mathbb{C})$, does $f=g$?
ag.algebraic-geometry
ag.algebraic-geometry
asked Apr 9 at 22:35
rj7k8rj7k8
195117
asked Apr 9 at 22:35
rj7k8rj7k8
195117
asked Apr 9 at 22:35
rj7k8rj7k8
195117
asked Apr 9 at 22:35
rj7k8rj7k8
195117
asked Apr 9 at 22:35
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 $H^1(-, mathbf{C})$.
answered Apr 9 at 23:01
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
2 days ago
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
2 days ago
$begingroup$
@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
2 days ago
add a comment |
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1 Answer
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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 $H^1(-, mathbf{C})$.
answered Apr 9 at 23:01
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
2 days ago
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
2 days ago
$begingroup$
@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
2 days ago
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 $H^1(-, mathbf{C})$.
answered Apr 9 at 23:01
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
2 days ago
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
2 days ago
$begingroup$
@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
2 days ago
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 $H^1(-, mathbf{C})$.
answered Apr 9 at 23:01
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 $H^1(-, mathbf{C})$.
answered Apr 9 at 23:01
Piotr AchingerPiotr Achinger
8,62212854
answered Apr 9 at 23:01
Piotr AchingerPiotr Achinger
8,62212854
answered Apr 9 at 23:01
Piotr AchingerPiotr Achinger
8,62212854
answered Apr 9 at 23:01
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
2 days ago
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
2 days ago
$begingroup$
@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
2 days ago
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
2 days ago
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
2 days ago
$begingroup$
@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
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
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
2 days ago
$begingroup$
@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
2 days ago
$begingroup$
@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
2 days ago
$begingroup$
@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
2 days ago
add a comment |
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