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








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    @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 $mathbb{C}$.
    – Anonymous
    Nov 19 at 16:53










  • Of course it is reduced
    – Francesco Polizzi
    Nov 19 at 17:02















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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







share|cite|improve this question




















  • 4




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








  • 1




    @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 $mathbb{C}$.
    – Anonymous
    Nov 19 at 16:53










  • Of course it is reduced
    – Francesco Polizzi
    Nov 19 at 17:02













up vote
2
down vote

favorite
1









up vote
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1





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















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






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edited Nov 19 at 16:56

























asked Nov 19 at 15:48









Anonymous

1286




1286








  • 4




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








  • 1




    @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 $mathbb{C}$.
    – Anonymous
    Nov 19 at 16:53










  • Of course it is reduced
    – Francesco Polizzi
    Nov 19 at 17:02














  • 4




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








  • 1




    @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 $mathbb{C}$.
    – Anonymous
    Nov 19 at 16:53










  • Of course it is reduced
    – Francesco Polizzi
    Nov 19 at 17:02








4




4




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






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






1




1




@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 $mathbb{C}$.
– Anonymous
Nov 19 at 16:53




@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 $mathbb{C}$.
– Anonymous
Nov 19 at 16:53












Of course it is reduced
– Francesco Polizzi
Nov 19 at 17:02




Of course it is reduced
– Francesco Polizzi
Nov 19 at 17:02










2 Answers
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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 $$pi_1(U) = pi_1(E) = mathbb{Z} oplus mathbb{Z}.$$






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    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.






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











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    2 Answers
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    2 Answers
    2






    active

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    active

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    active

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    up vote
    4
    down vote



    accepted










    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

























      up vote
      4
      down vote



      accepted










      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























        up vote
        4
        down vote



        accepted







        up vote
        4
        down vote



        accepted






        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












        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 at 17:24









        Francesco Polizzi

        47.2k3125202




        47.2k3125202






















            up vote
            3
            down vote













            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























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















            up vote
            3
            down vote













            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























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













            up vote
            3
            down vote










            up vote
            3
            down vote









            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














            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 at 17:16

























            answered Nov 19 at 17:02









            Piotr Achinger

            8,16312752




            8,16312752












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


















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
















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




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


















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