Deck transformation group in algebraic geometry
Let $f:Xto Y$ be a finite morphism between (irreducible) varieties. We can define $operatorname{Aut}(X/Y)$ to be the automorphism of $X$ commutes with $f$. For the case over $mathbb C$, we can also define the monodromy group $G$ to be the image of the monodromy action.
My question is
Is it true that $Gcong operatorname{Aut}(X/Y)$ ?
I know this is true in the topological setting (edit: should be "in the birational setting" or "on the unramified part") and also the case of dimension one. So the question is, if the monodromy data, say the monodromy action of some loop, determines a morphism $X to X$ in general?
Edit: I tend to believe this is not true. Assume it is, then for any finite morphism of degree $d>1$, there exists some nontrivial automorphism $g:Xto X$ which preserves the fiber. This might be a way to construct counter-examples.
algebraic-geometry reference-request covering-spaces automorphism-group
asked Dec 31 '18 at 15:22
Akatsuki
1,0271725
add a comment |
Let $f:Xto Y$ be a finite morphism between (irreducible) varieties. We can define $operatorname{Aut}(X/Y)$ to be the automorphism of $X$ commutes with $f$. For the case over $mathbb C$, we can also define the monodromy group $G$ to be the image of the monodromy action.
My question is
Is it true that $Gcong operatorname{Aut}(X/Y)$ ?
I know this is true in the topological setting (edit: should be "in the birational setting" or "on the unramified part") and also the case of dimension one. So the question is, if the monodromy data, say the monodromy action of some loop, determines a morphism $X to X$ in general?
Edit: I tend to believe this is not true. Assume it is, then for any finite morphism of degree $d>1$, there exists some nontrivial automorphism $g:Xto X$ which preserves the fiber. This might be a way to construct counter-examples.
algebraic-geometry reference-request covering-spaces automorphism-group
asked Dec 31 '18 at 15:22
Akatsuki
1,0271725
What do you mean by "automorphism of $X$ commutes with $f$"?
– Armando j18eos
Jan 1 at 11:48
@Armandoj18eos I mean some $g: X to X$ such that $fcirc g=f$.
– Akatsuki
Jan 1 at 12:06
I realized that it should be "rational map" instead of "morphism" which is natural to consider. It is always true that the "rational deck transformation group" is isomorphic to the monodromy group, and these birational maps seems have no reason to be morphisms. Therefore, $mathrm{Aut}(X/Y)$ should be a subgroup of $G$ in general. I will edit the question.
– Akatsuki
yesterday
add a comment |
Let $f:Xto Y$ be a finite morphism between (irreducible) varieties. We can define $operatorname{Aut}(X/Y)$ to be the automorphism of $X$ commutes with $f$. For the case over $mathbb C$, we can also define the monodromy group $G$ to be the image of the monodromy action.
My question is
Is it true that $Gcong operatorname{Aut}(X/Y)$ ?
I know this is true in the topological setting (edit: should be "in the birational setting" or "on the unramified part") and also the case of dimension one. So the question is, if the monodromy data, say the monodromy action of some loop, determines a morphism $X to X$ in general?
Edit: I tend to believe this is not true. Assume it is, then for any finite morphism of degree $d>1$, there exists some nontrivial automorphism $g:Xto X$ which preserves the fiber. This might be a way to construct counter-examples.
algebraic-geometry reference-request covering-spaces automorphism-group
asked Dec 31 '18 at 15:22
Akatsuki
1,0271725
Let $f:Xto Y$ be a finite morphism between (irreducible) varieties. We can define $operatorname{Aut}(X/Y)$ to be the automorphism of $X$ commutes with $f$. For the case over $mathbb C$, we can also define the monodromy group $G$ to be the image of the monodromy action.
My question is
Is it true that $Gcong operatorname{Aut}(X/Y)$ ?
I know this is true in the topological setting (edit: should be "in the birational setting" or "on the unramified part") and also the case of dimension one. So the question is, if the monodromy data, say the monodromy action of some loop, determines a morphism $X to X$ in general?
Edit: I tend to believe this is not true. Assume it is, then for any finite morphism of degree $d>1$, there exists some nontrivial automorphism $g:Xto X$ which preserves the fiber. This might be a way to construct counter-examples.
algebraic-geometry reference-request covering-spaces automorphism-group
algebraic-geometry reference-request covering-spaces automorphism-group
asked Dec 31 '18 at 15:22
Akatsuki
1,0271725
asked Dec 31 '18 at 15:22
Akatsuki
1,0271725
edited 15 hours ago
asked Dec 31 '18 at 15:22
Akatsuki
1,0271725
asked Dec 31 '18 at 15:22
Akatsuki
1,0271725
asked Dec 31 '18 at 15:22
Akatsuki
1,0271725
1,0271725
What do you mean by "automorphism of $X$ commutes with $f$"?
– Armando j18eos
Jan 1 at 11:48
@Armandoj18eos I mean some $g: X to X$ such that $fcirc g=f$.
– Akatsuki
Jan 1 at 12:06
I realized that it should be "rational map" instead of "morphism" which is natural to consider. It is always true that the "rational deck transformation group" is isomorphic to the monodromy group, and these birational maps seems have no reason to be morphisms. Therefore, $mathrm{Aut}(X/Y)$ should be a subgroup of $G$ in general. I will edit the question.
– Akatsuki
yesterday
add a comment |
What do you mean by "automorphism of $X$ commutes with $f$"?
– Armando j18eos
Jan 1 at 11:48
@Armandoj18eos I mean some $g: X to X$ such that $fcirc g=f$.
– Akatsuki
Jan 1 at 12:06
I realized that it should be "rational map" instead of "morphism" which is natural to consider. It is always true that the "rational deck transformation group" is isomorphic to the monodromy group, and these birational maps seems have no reason to be morphisms. Therefore, $mathrm{Aut}(X/Y)$ should be a subgroup of $G$ in general. I will edit the question.
– Akatsuki
yesterday
What do you mean by "automorphism of $X$ commutes with $f$"?
– Armando j18eos
Jan 1 at 11:48
What do you mean by "automorphism of $X$ commutes with $f$"?
– Armando j18eos
Jan 1 at 11:48
@Armandoj18eos I mean some $g: X to X$ such that $fcirc g=f$.
– Akatsuki
Jan 1 at 12:06
@Armandoj18eos I mean some $g: X to X$ such that $fcirc g=f$.
– Akatsuki
Jan 1 at 12:06
I realized that it should be "rational map" instead of "morphism" which is natural to consider. It is always true that the "rational deck transformation group" is isomorphic to the monodromy group, and these birational maps seems have no reason to be morphisms. Therefore, $mathrm{Aut}(X/Y)$ should be a subgroup of $G$ in general. I will edit the question.
– Akatsuki
yesterday
I realized that it should be "rational map" instead of "morphism" which is natural to consider. It is always true that the "rational deck transformation group" is isomorphic to the monodromy group, and these birational maps seems have no reason to be morphisms. Therefore, $mathrm{Aut}(X/Y)$ should be a subgroup of $G$ in general. I will edit the question.
– Akatsuki
yesterday
add a comment |
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I am sorry that I made some mistakes. This is not true, even for smooth curves. For example, let $X$ be a general Riemann surface with genus at least $3$. Then it does not have non-trivial automorphism, but there exists meromorphic functions, hence a branched cover $Xto Y$. In particular, ${rm Aut}(X/Y)$ is trivial. However, the monodromy group is clearly not trivial.
In general, let ${rm Gal}(K(Y)/K(X))$ be the automorphism group of the Galois closure, then
$${rm Aut_{rational}}(X/Y)cong{rm Aut}(K(Y)/K(X))$$
and
$$G cong {rm Gal}(K(Y)/K(X)).$$
For smooth curves, ${rm Aut}(X/Y)={rm Aut_{rational}}(X/Y)$. Then a sufficient and necessary condition to ensure $Gcong {rm Aut(X/Y)}$ is that the covering $Xto Y$ is Galois.
answered 13 hours ago
Akatsuki
1,0271725
add a comment |
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I am sorry that I made some mistakes. This is not true, even for smooth curves. For example, let $X$ be a general Riemann surface with genus at least $3$. Then it does not have non-trivial automorphism, but there exists meromorphic functions, hence a branched cover $Xto Y$. In particular, ${rm Aut}(X/Y)$ is trivial. However, the monodromy group is clearly not trivial.
In general, let ${rm Gal}(K(Y)/K(X))$ be the automorphism group of the Galois closure, then
$${rm Aut_{rational}}(X/Y)cong{rm Aut}(K(Y)/K(X))$$
and
$$G cong {rm Gal}(K(Y)/K(X)).$$
For smooth curves, ${rm Aut}(X/Y)={rm Aut_{rational}}(X/Y)$. Then a sufficient and necessary condition to ensure $Gcong {rm Aut(X/Y)}$ is that the covering $Xto Y$ is Galois.
answered 13 hours ago
Akatsuki
1,0271725
add a comment |
I am sorry that I made some mistakes. This is not true, even for smooth curves. For example, let $X$ be a general Riemann surface with genus at least $3$. Then it does not have non-trivial automorphism, but there exists meromorphic functions, hence a branched cover $Xto Y$. In particular, ${rm Aut}(X/Y)$ is trivial. However, the monodromy group is clearly not trivial.
In general, let ${rm Gal}(K(Y)/K(X))$ be the automorphism group of the Galois closure, then
$${rm Aut_{rational}}(X/Y)cong{rm Aut}(K(Y)/K(X))$$
and
$$G cong {rm Gal}(K(Y)/K(X)).$$
For smooth curves, ${rm Aut}(X/Y)={rm Aut_{rational}}(X/Y)$. Then a sufficient and necessary condition to ensure $Gcong {rm Aut(X/Y)}$ is that the covering $Xto Y$ is Galois.
answered 13 hours ago
Akatsuki
1,0271725
add a comment |
I am sorry that I made some mistakes. This is not true, even for smooth curves. For example, let $X$ be a general Riemann surface with genus at least $3$. Then it does not have non-trivial automorphism, but there exists meromorphic functions, hence a branched cover $Xto Y$. In particular, ${rm Aut}(X/Y)$ is trivial. However, the monodromy group is clearly not trivial.
In general, let ${rm Gal}(K(Y)/K(X))$ be the automorphism group of the Galois closure, then
$${rm Aut_{rational}}(X/Y)cong{rm Aut}(K(Y)/K(X))$$
and
$$G cong {rm Gal}(K(Y)/K(X)).$$
For smooth curves, ${rm Aut}(X/Y)={rm Aut_{rational}}(X/Y)$. Then a sufficient and necessary condition to ensure $Gcong {rm Aut(X/Y)}$ is that the covering $Xto Y$ is Galois.
answered 13 hours ago
Akatsuki
1,0271725
I am sorry that I made some mistakes. This is not true, even for smooth curves. For example, let $X$ be a general Riemann surface with genus at least $3$. Then it does not have non-trivial automorphism, but there exists meromorphic functions, hence a branched cover $Xto Y$. In particular, ${rm Aut}(X/Y)$ is trivial. However, the monodromy group is clearly not trivial.
In general, let ${rm Gal}(K(Y)/K(X))$ be the automorphism group of the Galois closure, then
$${rm Aut_{rational}}(X/Y)cong{rm Aut}(K(Y)/K(X))$$
and
$$G cong {rm Gal}(K(Y)/K(X)).$$
For smooth curves, ${rm Aut}(X/Y)={rm Aut_{rational}}(X/Y)$. Then a sufficient and necessary condition to ensure $Gcong {rm Aut(X/Y)}$ is that the covering $Xto Y$ is Galois.
answered 13 hours ago
Akatsuki
1,0271725
edited 13 hours ago
answered 13 hours ago
Akatsuki
1,0271725
answered 13 hours ago
Akatsuki
1,0271725
answered 13 hours ago
Akatsuki
1,0271725
1,0271725
add a comment |
add a comment |
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What do you mean by "automorphism of $X$ commutes with $f$"?
– Armando j18eos
Jan 1 at 11:48
@Armandoj18eos I mean some $g: X to X$ such that $fcirc g=f$.
– Akatsuki
Jan 1 at 12:06
I realized that it should be "rational map" instead of "morphism" which is natural to consider. It is always true that the "rational deck transformation group" is isomorphic to the monodromy group, and these birational maps seems have no reason to be morphisms. Therefore, $mathrm{Aut}(X/Y)$ should be a subgroup of $G$ in general. I will edit the question.
– Akatsuki
yesterday