Intersecting a projectively normal variety and a hyperplane
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If $M$ is a projectively normal projective variety in $mathbb{C}mathrm{P}^n$ and we intersect it with a hyperplane $mathrm{P}V subseteq mathbb{C}mathrm{P}^n$, is the result $M cap mathrm{P}V$ a projectively normal variety in the projective space $mathrm{P}V$?
(This intersection is called a hyperplane section.)
algebraic-geometry
asked Jan 17 at 7:40
John BaezJohn Baez
33619
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add a comment |
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If $M$ is a projectively normal projective variety in $mathbb{C}mathrm{P}^n$ and we intersect it with a hyperplane $mathrm{P}V subseteq mathbb{C}mathrm{P}^n$, is the result $M cap mathrm{P}V$ a projectively normal variety in the projective space $mathrm{P}V$?
(This intersection is called a hyperplane section.)
algebraic-geometry
asked Jan 17 at 7:40
John BaezJohn Baez
33619
$endgroup$
add a comment |
$begingroup$
If $M$ is a projectively normal projective variety in $mathbb{C}mathrm{P}^n$ and we intersect it with a hyperplane $mathrm{P}V subseteq mathbb{C}mathrm{P}^n$, is the result $M cap mathrm{P}V$ a projectively normal variety in the projective space $mathrm{P}V$?
(This intersection is called a hyperplane section.)
algebraic-geometry
asked Jan 17 at 7:40
John BaezJohn Baez
33619
$endgroup$
If $M$ is a projectively normal projective variety in $mathbb{C}mathrm{P}^n$ and we intersect it with a hyperplane $mathrm{P}V subseteq mathbb{C}mathrm{P}^n$, is the result $M cap mathrm{P}V$ a projectively normal variety in the projective space $mathrm{P}V$?
(This intersection is called a hyperplane section.)
algebraic-geometry
algebraic-geometry
asked Jan 17 at 7:40
John BaezJohn Baez
33619
asked Jan 17 at 7:40
John BaezJohn Baez
33619
asked Jan 17 at 7:40
John BaezJohn Baez
33619
asked Jan 17 at 7:40
John BaezJohn Baez
33619
asked Jan 17 at 7:40
John BaezJohn Baez
33619
33619
add a comment |
add a comment |
1 Answer
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Not necessary. For instance, if $M$ is a surface then any its singular hyperplane section is not normal (because it is singular in codimension 1).
answered Jan 17 at 7:47
SashaSasha
5,208139
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Thanks! Is it true that if $M$ is linearly normal than any of its hyperplane sections is linearly normal?
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– John Baez
Jan 19 at 1:20
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No, this is still not true in general --- if $h^1(M,O_M) > h^1(M,O_M(1))$ and $M' subset M$ is a hyperplane section then the map $H^0(M,O_M(1)) to H_0(M',O_{M'}(1))$ is not surjective, hence $M'$ is not linearly normal.
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– Sasha
Jan 19 at 7:58
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Not necessary. For instance, if $M$ is a surface then any its singular hyperplane section is not normal (because it is singular in codimension 1).
answered Jan 17 at 7:47
SashaSasha
5,208139
$endgroup$
$begingroup$
Thanks! Is it true that if $M$ is linearly normal than any of its hyperplane sections is linearly normal?
$endgroup$
– John Baez
Jan 19 at 1:20
$begingroup$
No, this is still not true in general --- if $h^1(M,O_M) > h^1(M,O_M(1))$ and $M' subset M$ is a hyperplane section then the map $H^0(M,O_M(1)) to H_0(M',O_{M'}(1))$ is not surjective, hence $M'$ is not linearly normal.
$endgroup$
– Sasha
Jan 19 at 7:58
add a comment |
$begingroup$
Not necessary. For instance, if $M$ is a surface then any its singular hyperplane section is not normal (because it is singular in codimension 1).
answered Jan 17 at 7:47
SashaSasha
5,208139
$endgroup$
$begingroup$
Thanks! Is it true that if $M$ is linearly normal than any of its hyperplane sections is linearly normal?
$endgroup$
– John Baez
Jan 19 at 1:20
$begingroup$
No, this is still not true in general --- if $h^1(M,O_M) > h^1(M,O_M(1))$ and $M' subset M$ is a hyperplane section then the map $H^0(M,O_M(1)) to H_0(M',O_{M'}(1))$ is not surjective, hence $M'$ is not linearly normal.
$endgroup$
– Sasha
Jan 19 at 7:58
add a comment |
$begingroup$
Not necessary. For instance, if $M$ is a surface then any its singular hyperplane section is not normal (because it is singular in codimension 1).
answered Jan 17 at 7:47
SashaSasha
5,208139
$endgroup$
Not necessary. For instance, if $M$ is a surface then any its singular hyperplane section is not normal (because it is singular in codimension 1).
answered Jan 17 at 7:47
SashaSasha
5,208139
answered Jan 17 at 7:47
SashaSasha
5,208139
answered Jan 17 at 7:47
SashaSasha
5,208139
answered Jan 17 at 7:47
SashaSasha
5,208139
5,208139
$begingroup$
Thanks! Is it true that if $M$ is linearly normal than any of its hyperplane sections is linearly normal?
$endgroup$
– John Baez
Jan 19 at 1:20
$begingroup$
No, this is still not true in general --- if $h^1(M,O_M) > h^1(M,O_M(1))$ and $M' subset M$ is a hyperplane section then the map $H^0(M,O_M(1)) to H_0(M',O_{M'}(1))$ is not surjective, hence $M'$ is not linearly normal.
$endgroup$
– Sasha
Jan 19 at 7:58
add a comment |
$begingroup$
Thanks! Is it true that if $M$ is linearly normal than any of its hyperplane sections is linearly normal?
$endgroup$
– John Baez
Jan 19 at 1:20
$begingroup$
No, this is still not true in general --- if $h^1(M,O_M) > h^1(M,O_M(1))$ and $M' subset M$ is a hyperplane section then the map $H^0(M,O_M(1)) to H_0(M',O_{M'}(1))$ is not surjective, hence $M'$ is not linearly normal.
$endgroup$
– Sasha
Jan 19 at 7:58
$begingroup$
Thanks! Is it true that if $M$ is linearly normal than any of its hyperplane sections is linearly normal?
$endgroup$
– John Baez
Jan 19 at 1:20
$begingroup$
Thanks! Is it true that if $M$ is linearly normal than any of its hyperplane sections is linearly normal?
$endgroup$
– John Baez
Jan 19 at 1:20
$begingroup$
No, this is still not true in general --- if $h^1(M,O_M) > h^1(M,O_M(1))$ and $M' subset M$ is a hyperplane section then the map $H^0(M,O_M(1)) to H_0(M',O_{M'}(1))$ is not surjective, hence $M'$ is not linearly normal.
$endgroup$
– Sasha
Jan 19 at 7:58
$begingroup$
No, this is still not true in general --- if $h^1(M,O_M) > h^1(M,O_M(1))$ and $M' subset M$ is a hyperplane section then the map $H^0(M,O_M(1)) to H_0(M',O_{M'}(1))$ is not surjective, hence $M'$ is not linearly normal.
$endgroup$
– Sasha
Jan 19 at 7:58
add a comment |
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