Fundamental group of an open subscheme of a normal scheme












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Let $X$ be an irreducible normal projective scheme over $mathbb{C}$. Let $U$ be the open subscheme of smooth points of $X$. Consider the closed subscheme $Z = X setminus U$. Suppose that the codimension of $Z$ in $X$ is at least $2$. Is it true that the fundamental group of $U$ and $X$ are isomorphic?



Edit: Is it true for $X$ an integral normal projective scheme over $mathbb{C}$?






ag.algebraic-geometry







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  • 4




    $begingroup$
    In codimension $2$, is it not a cone over an elliptic curve a counterexample?
    $endgroup$
    – Francesco Polizzi
    Nov 19 '18 at 16:05








  • 1




    $begingroup$
    @FrancescoPolizzi Is cone over an elliptic curve reduced? (sorry for asking a stupid question). Actually, I want to know the result for X normal projective integral scheme over $mathbb{C}$.
    $endgroup$
    – Anonymous
    Nov 19 '18 at 16:53










  • $begingroup$
    Of course it is reduced
    $endgroup$
    – Francesco Polizzi
    Nov 19 '18 at 17:02
















2












$begingroup$


Let $X$ be an irreducible normal projective scheme over $mathbb{C}$. Let $U$ be the open subscheme of smooth points of $X$. Consider the closed subscheme $Z = X setminus U$. Suppose that the codimension of $Z$ in $X$ is at least $2$. Is it true that the fundamental group of $U$ and $X$ are isomorphic?



Edit: Is it true for $X$ an integral normal projective scheme over $mathbb{C}$?






ag.algebraic-geometry







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    In codimension $2$, is it not a cone over an elliptic curve a counterexample?
    $endgroup$
    – Francesco Polizzi
    Nov 19 '18 at 16:05








  • 1




    $begingroup$
    @FrancescoPolizzi Is cone over an elliptic curve reduced? (sorry for asking a stupid question). Actually, I want to know the result for X normal projective integral scheme over $mathbb{C}$.
    $endgroup$
    – Anonymous
    Nov 19 '18 at 16:53










  • $begingroup$
    Of course it is reduced
    $endgroup$
    – Francesco Polizzi
    Nov 19 '18 at 17:02














2












2








2


1



$begingroup$


Let $X$ be an irreducible normal projective scheme over $mathbb{C}$. Let $U$ be the open subscheme of smooth points of $X$. Consider the closed subscheme $Z = X setminus U$. Suppose that the codimension of $Z$ in $X$ is at least $2$. Is it true that the fundamental group of $U$ and $X$ are isomorphic?



Edit: Is it true for $X$ an integral normal projective scheme over $mathbb{C}$?






ag.algebraic-geometry







share|cite|improve this question











$endgroup$




Let $X$ be an irreducible normal projective scheme over $mathbb{C}$. Let $U$ be the open subscheme of smooth points of $X$. Consider the closed subscheme $Z = X setminus U$. Suppose that the codimension of $Z$ in $X$ is at least $2$. Is it true that the fundamental group of $U$ and $X$ are isomorphic?



Edit: Is it true for $X$ an integral normal projective scheme over $mathbb{C}$?






ag.algebraic-geometry




ag.algebraic-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 '18 at 16:56







Anonymous

















asked Nov 19 '18 at 15:48









AnonymousAnonymous

1286




1286








  • 4




    $begingroup$
    In codimension $2$, is it not a cone over an elliptic curve a counterexample?
    $endgroup$
    – Francesco Polizzi
    Nov 19 '18 at 16:05








  • 1




    $begingroup$
    @FrancescoPolizzi Is cone over an elliptic curve reduced? (sorry for asking a stupid question). Actually, I want to know the result for X normal projective integral scheme over $mathbb{C}$.
    $endgroup$
    – Anonymous
    Nov 19 '18 at 16:53










  • $begingroup$
    Of course it is reduced
    $endgroup$
    – Francesco Polizzi
    Nov 19 '18 at 17:02














  • 4




    $begingroup$
    In codimension $2$, is it not a cone over an elliptic curve a counterexample?
    $endgroup$
    – Francesco Polizzi
    Nov 19 '18 at 16:05








  • 1




    $begingroup$
    @FrancescoPolizzi Is cone over an elliptic curve reduced? (sorry for asking a stupid question). Actually, I want to know the result for X normal projective integral scheme over $mathbb{C}$.
    $endgroup$
    – Anonymous
    Nov 19 '18 at 16:53










  • $begingroup$
    Of course it is reduced
    $endgroup$
    – Francesco Polizzi
    Nov 19 '18 at 17:02








4




4




$begingroup$
In codimension $2$, is it not a cone over an elliptic curve a counterexample?
$endgroup$
– Francesco Polizzi
Nov 19 '18 at 16:05






$begingroup$
In codimension $2$, is it not a cone over an elliptic curve a counterexample?
$endgroup$
– Francesco Polizzi
Nov 19 '18 at 16:05






1




1




$begingroup$
@FrancescoPolizzi Is cone over an elliptic curve reduced? (sorry for asking a stupid question). Actually, I want to know the result for X normal projective integral scheme over $mathbb{C}$.
$endgroup$
– Anonymous
Nov 19 '18 at 16:53




$begingroup$
@FrancescoPolizzi Is cone over an elliptic curve reduced? (sorry for asking a stupid question). Actually, I want to know the result for X normal projective integral scheme over $mathbb{C}$.
$endgroup$
– Anonymous
Nov 19 '18 at 16:53












$begingroup$
Of course it is reduced
$endgroup$
– Francesco Polizzi
Nov 19 '18 at 17:02




$begingroup$
Of course it is reduced
$endgroup$
– Francesco Polizzi
Nov 19 '18 at 17:02










2 Answers
2






active

oldest

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4












$begingroup$

Let me expand my comment into an answer.



Take as $X$ the cone of vertex $v$ over an elliptic curve $E$. Then $X$ is simply connected (this is a general property of projective cones). However, $U = X-{v}$ is not simply connected: in fact, the projection $pi colon U to E$ onto the basis gives $X$ the structure of a topological fibration with fiber homeomorphic to $mathbb{R}$, so the corresponding long exact sequence of homotopy groups yields $$pi_1(U) = pi_1(E) = mathbb{Z} oplus mathbb{Z}.$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    In fact, quite the opposite tends to be true. Mumford [1] showed that for $(X,0)$ the germ of a normal surface singularity (over $mathbf{C}$), $U=Xsetminus 0$, one has $pi_1(U)={1}$ if and only if $X$ is smooth. At the same time, $pi_1(X) = {1}$ since $0to X$ is a homotopy equivalence.



    If $X$ is smooth, this is true (the etale variant is called "Zariski-Nagata purity").



    EDIT. To address Francesco's comment: of course the example is not projective. The easiest projective example was given by Francesco in his comment: $X$ is the (projective) cone over an elliptic curve $E$ and $U$ the complement of the vertex. Then $pi_1(U)= pi_1(E) = mathbb{Z}^2$ and $pi_1(X) = {1}$.



    [1] Mumford, D., The topology of normal singularities of an algebraic surface and a criterion for simplicity, Publ. Math., Inst. Hautes Étud. Sci. 9, 5-22 (1961). ZBL0108.16801.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Well, strictly speaking, $X$ is not projective in your example.
      $endgroup$
      – Francesco Polizzi
      Nov 19 '18 at 17:04











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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    4












    $begingroup$

    Let me expand my comment into an answer.



    Take as $X$ the cone of vertex $v$ over an elliptic curve $E$. Then $X$ is simply connected (this is a general property of projective cones). However, $U = X-{v}$ is not simply connected: in fact, the projection $pi colon U to E$ onto the basis gives $X$ the structure of a topological fibration with fiber homeomorphic to $mathbb{R}$, so the corresponding long exact sequence of homotopy groups yields $$pi_1(U) = pi_1(E) = mathbb{Z} oplus mathbb{Z}.$$






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Let me expand my comment into an answer.



      Take as $X$ the cone of vertex $v$ over an elliptic curve $E$. Then $X$ is simply connected (this is a general property of projective cones). However, $U = X-{v}$ is not simply connected: in fact, the projection $pi colon U to E$ onto the basis gives $X$ the structure of a topological fibration with fiber homeomorphic to $mathbb{R}$, so the corresponding long exact sequence of homotopy groups yields $$pi_1(U) = pi_1(E) = mathbb{Z} oplus mathbb{Z}.$$






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Let me expand my comment into an answer.



        Take as $X$ the cone of vertex $v$ over an elliptic curve $E$. Then $X$ is simply connected (this is a general property of projective cones). However, $U = X-{v}$ is not simply connected: in fact, the projection $pi colon U to E$ onto the basis gives $X$ the structure of a topological fibration with fiber homeomorphic to $mathbb{R}$, so the corresponding long exact sequence of homotopy groups yields $$pi_1(U) = pi_1(E) = mathbb{Z} oplus mathbb{Z}.$$






        share|cite|improve this answer









        $endgroup$



        Let me expand my comment into an answer.



        Take as $X$ the cone of vertex $v$ over an elliptic curve $E$. Then $X$ is simply connected (this is a general property of projective cones). However, $U = X-{v}$ is not simply connected: in fact, the projection $pi colon U to E$ onto the basis gives $X$ the structure of a topological fibration with fiber homeomorphic to $mathbb{R}$, so the corresponding long exact sequence of homotopy groups yields $$pi_1(U) = pi_1(E) = mathbb{Z} oplus mathbb{Z}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 '18 at 17:24









        Francesco PolizziFrancesco Polizzi

        47.6k3127207




        47.6k3127207























            3












            $begingroup$

            In fact, quite the opposite tends to be true. Mumford [1] showed that for $(X,0)$ the germ of a normal surface singularity (over $mathbf{C}$), $U=Xsetminus 0$, one has $pi_1(U)={1}$ if and only if $X$ is smooth. At the same time, $pi_1(X) = {1}$ since $0to X$ is a homotopy equivalence.



            If $X$ is smooth, this is true (the etale variant is called "Zariski-Nagata purity").



            EDIT. To address Francesco's comment: of course the example is not projective. The easiest projective example was given by Francesco in his comment: $X$ is the (projective) cone over an elliptic curve $E$ and $U$ the complement of the vertex. Then $pi_1(U)= pi_1(E) = mathbb{Z}^2$ and $pi_1(X) = {1}$.



            [1] Mumford, D., The topology of normal singularities of an algebraic surface and a criterion for simplicity, Publ. Math., Inst. Hautes Étud. Sci. 9, 5-22 (1961). ZBL0108.16801.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Well, strictly speaking, $X$ is not projective in your example.
              $endgroup$
              – Francesco Polizzi
              Nov 19 '18 at 17:04
















            3












            $begingroup$

            In fact, quite the opposite tends to be true. Mumford [1] showed that for $(X,0)$ the germ of a normal surface singularity (over $mathbf{C}$), $U=Xsetminus 0$, one has $pi_1(U)={1}$ if and only if $X$ is smooth. At the same time, $pi_1(X) = {1}$ since $0to X$ is a homotopy equivalence.



            If $X$ is smooth, this is true (the etale variant is called "Zariski-Nagata purity").



            EDIT. To address Francesco's comment: of course the example is not projective. The easiest projective example was given by Francesco in his comment: $X$ is the (projective) cone over an elliptic curve $E$ and $U$ the complement of the vertex. Then $pi_1(U)= pi_1(E) = mathbb{Z}^2$ and $pi_1(X) = {1}$.



            [1] Mumford, D., The topology of normal singularities of an algebraic surface and a criterion for simplicity, Publ. Math., Inst. Hautes Étud. Sci. 9, 5-22 (1961). ZBL0108.16801.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Well, strictly speaking, $X$ is not projective in your example.
              $endgroup$
              – Francesco Polizzi
              Nov 19 '18 at 17:04














            3












            3








            3





            $begingroup$

            In fact, quite the opposite tends to be true. Mumford [1] showed that for $(X,0)$ the germ of a normal surface singularity (over $mathbf{C}$), $U=Xsetminus 0$, one has $pi_1(U)={1}$ if and only if $X$ is smooth. At the same time, $pi_1(X) = {1}$ since $0to X$ is a homotopy equivalence.



            If $X$ is smooth, this is true (the etale variant is called "Zariski-Nagata purity").



            EDIT. To address Francesco's comment: of course the example is not projective. The easiest projective example was given by Francesco in his comment: $X$ is the (projective) cone over an elliptic curve $E$ and $U$ the complement of the vertex. Then $pi_1(U)= pi_1(E) = mathbb{Z}^2$ and $pi_1(X) = {1}$.



            [1] Mumford, D., The topology of normal singularities of an algebraic surface and a criterion for simplicity, Publ. Math., Inst. Hautes Étud. Sci. 9, 5-22 (1961). ZBL0108.16801.






            share|cite|improve this answer











            $endgroup$



            In fact, quite the opposite tends to be true. Mumford [1] showed that for $(X,0)$ the germ of a normal surface singularity (over $mathbf{C}$), $U=Xsetminus 0$, one has $pi_1(U)={1}$ if and only if $X$ is smooth. At the same time, $pi_1(X) = {1}$ since $0to X$ is a homotopy equivalence.



            If $X$ is smooth, this is true (the etale variant is called "Zariski-Nagata purity").



            EDIT. To address Francesco's comment: of course the example is not projective. The easiest projective example was given by Francesco in his comment: $X$ is the (projective) cone over an elliptic curve $E$ and $U$ the complement of the vertex. Then $pi_1(U)= pi_1(E) = mathbb{Z}^2$ and $pi_1(X) = {1}$.



            [1] Mumford, D., The topology of normal singularities of an algebraic surface and a criterion for simplicity, Publ. Math., Inst. Hautes Étud. Sci. 9, 5-22 (1961). ZBL0108.16801.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 19 '18 at 17:16

























            answered Nov 19 '18 at 17:02









            Piotr AchingerPiotr Achinger

            8,27812852




            8,27812852












            • $begingroup$
              Well, strictly speaking, $X$ is not projective in your example.
              $endgroup$
              – Francesco Polizzi
              Nov 19 '18 at 17:04


















            • $begingroup$
              Well, strictly speaking, $X$ is not projective in your example.
              $endgroup$
              – Francesco Polizzi
              Nov 19 '18 at 17:04
















            $begingroup$
            Well, strictly speaking, $X$ is not projective in your example.
            $endgroup$
            – Francesco Polizzi
            Nov 19 '18 at 17:04




            $begingroup$
            Well, strictly speaking, $X$ is not projective in your example.
            $endgroup$
            – Francesco Polizzi
            Nov 19 '18 at 17:04


















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