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
asked Nov 19 '18 at 15:48
AnonymousAnonymous
1286
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add a comment |
$begingroup$
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
asked Nov 19 '18 at 15:48
AnonymousAnonymous
1286
$endgroup$
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
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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 $mathbb{C}$.
$endgroup$
– Anonymous
Nov 19 '18 at 16:53
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Of course it is reduced
$endgroup$
– Francesco Polizzi
Nov 19 '18 at 17:02
add a comment |
$begingroup$
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
asked Nov 19 '18 at 15:48
AnonymousAnonymous
1286
$endgroup$
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
asked Nov 19 '18 at 15:48
AnonymousAnonymous
1286
asked Nov 19 '18 at 15:48
AnonymousAnonymous
1286
edited Nov 19 '18 at 16:56
Anonymous
asked Nov 19 '18 at 15:48
AnonymousAnonymous
1286
asked Nov 19 '18 at 15:48
AnonymousAnonymous
1286
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $$pi_1(U) = pi_1(E) = mathbb{Z} oplus mathbb{Z}.$$
answered Nov 19 '18 at 17:24
Francesco PolizziFrancesco Polizzi
47.6k3127207
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add a comment |
$begingroup$
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.
answered Nov 19 '18 at 17:02
Piotr AchingerPiotr Achinger
8,27812852
$endgroup$
$begingroup$
Well, strictly speaking, $X$ is not projective in your example.
$endgroup$
– Francesco Polizzi
Nov 19 '18 at 17:04
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $$pi_1(U) = pi_1(E) = mathbb{Z} oplus mathbb{Z}.$$
answered Nov 19 '18 at 17:24
Francesco PolizziFrancesco Polizzi
47.6k3127207
$endgroup$
add a comment |
$begingroup$
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 $$pi_1(U) = pi_1(E) = mathbb{Z} oplus mathbb{Z}.$$
answered Nov 19 '18 at 17:24
Francesco PolizziFrancesco Polizzi
47.6k3127207
$endgroup$
add a comment |
$begingroup$
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 $$pi_1(U) = pi_1(E) = mathbb{Z} oplus mathbb{Z}.$$
answered Nov 19 '18 at 17:24
Francesco PolizziFrancesco Polizzi
47.6k3127207
$endgroup$
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 $$pi_1(U) = pi_1(E) = mathbb{Z} oplus mathbb{Z}.$$
answered Nov 19 '18 at 17:24
Francesco PolizziFrancesco Polizzi
47.6k3127207
answered Nov 19 '18 at 17:24
Francesco PolizziFrancesco Polizzi
47.6k3127207
answered Nov 19 '18 at 17:24
Francesco PolizziFrancesco Polizzi
47.6k3127207
answered Nov 19 '18 at 17:24
Francesco PolizziFrancesco Polizzi
47.6k3127207
47.6k3127207
add a comment |
add a comment |
$begingroup$
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.
answered Nov 19 '18 at 17:02
Piotr AchingerPiotr Achinger
8,27812852
$endgroup$
$begingroup$
Well, strictly speaking, $X$ is not projective in your example.
$endgroup$
– Francesco Polizzi
Nov 19 '18 at 17:04
add a comment |
$begingroup$
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.
answered Nov 19 '18 at 17:02
Piotr AchingerPiotr Achinger
8,27812852
$endgroup$
$begingroup$
Well, strictly speaking, $X$ is not projective in your example.
$endgroup$
– Francesco Polizzi
Nov 19 '18 at 17:04
add a comment |
$begingroup$
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.
answered Nov 19 '18 at 17:02
Piotr AchingerPiotr Achinger
8,27812852
$endgroup$
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.
answered Nov 19 '18 at 17:02
Piotr AchingerPiotr Achinger
8,27812852
edited Nov 19 '18 at 17:16
answered Nov 19 '18 at 17:02
Piotr AchingerPiotr Achinger
8,27812852
answered Nov 19 '18 at 17:02
Piotr AchingerPiotr Achinger
8,27812852
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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