Deck transformation group in algebraic geometry












10














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







share|cite|improve this question
























  • 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


















10














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







share|cite|improve this question
























  • 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
















10












10








10


3





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 15 hours ago

























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




















  • 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












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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.






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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.






    share|cite|improve this answer




























      0














      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.






      share|cite|improve this answer


























        0












        0








        0






        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.






        share|cite|improve this answer














        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 13 hours ago

























        answered 13 hours ago









        Akatsuki

        1,0271725




        1,0271725






























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