Mixing
Projecting a projective variety away from a linear subspace
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I read in Harris' book at page 148, proposition 11.37 and got slightly confused regarding the argument.
Harris mentions a projective space $mathbb{P}^{2n+1}$ and a linear subspace $mathbb{P}^{n}$ denoted $L$, disjoint from a projective variety $J$, which is embedded in $mathbb{P}^{2n+1}$. He then denotes by $pi_L:mathbb{P}^{2n+1}longrightarrow L$, what he refers to as "the projection from $L$", which is a term I both never heard of, nor ever encountered in my reading. He then proceeds to project $J$, claiming that as $Jcap L=emptyset$, the projection is regular map. He finally adds that it is a general fact that the projection of a projective variety from a linear subspace disjoint from it is finite.
My questions are:
1) How is the projection from a linear subspace defined? I am familiar with the notion of projection from a point.
2) Is this projection a regular morphism?
3) Why does a regular morphism in this context has finite fibers?
4) Why is the projection of a projective variety from a linear subspace disjoint from it is finite in general?
I believe all my questions stem from not understanding the answer to 1). Any help is appreciated.
Ciao!
algebraic-geometry grassmannian projective-varieties
asked Jan 22 at 1:22
kindasortakindasorta
9810
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add a comment |
$begingroup$
I read in Harris' book at page 148, proposition 11.37 and got slightly confused regarding the argument.
Harris mentions a projective space $mathbb{P}^{2n+1}$ and a linear subspace $mathbb{P}^{n}$ denoted $L$, disjoint from a projective variety $J$, which is embedded in $mathbb{P}^{2n+1}$. He then denotes by $pi_L:mathbb{P}^{2n+1}longrightarrow L$, what he refers to as "the projection from $L$", which is a term I both never heard of, nor ever encountered in my reading. He then proceeds to project $J$, claiming that as $Jcap L=emptyset$, the projection is regular map. He finally adds that it is a general fact that the projection of a projective variety from a linear subspace disjoint from it is finite.
My questions are:
1) How is the projection from a linear subspace defined? I am familiar with the notion of projection from a point.
2) Is this projection a regular morphism?
3) Why does a regular morphism in this context has finite fibers?
4) Why is the projection of a projective variety from a linear subspace disjoint from it is finite in general?
I believe all my questions stem from not understanding the answer to 1). Any help is appreciated.
Ciao!
algebraic-geometry grassmannian projective-varieties
asked Jan 22 at 1:22
kindasortakindasorta
9810
$endgroup$
add a comment |
$begingroup$
I read in Harris' book at page 148, proposition 11.37 and got slightly confused regarding the argument.
Harris mentions a projective space $mathbb{P}^{2n+1}$ and a linear subspace $mathbb{P}^{n}$ denoted $L$, disjoint from a projective variety $J$, which is embedded in $mathbb{P}^{2n+1}$. He then denotes by $pi_L:mathbb{P}^{2n+1}longrightarrow L$, what he refers to as "the projection from $L$", which is a term I both never heard of, nor ever encountered in my reading. He then proceeds to project $J$, claiming that as $Jcap L=emptyset$, the projection is regular map. He finally adds that it is a general fact that the projection of a projective variety from a linear subspace disjoint from it is finite.
My questions are:
1) How is the projection from a linear subspace defined? I am familiar with the notion of projection from a point.
2) Is this projection a regular morphism?
3) Why does a regular morphism in this context has finite fibers?
4) Why is the projection of a projective variety from a linear subspace disjoint from it is finite in general?
I believe all my questions stem from not understanding the answer to 1). Any help is appreciated.
Ciao!
algebraic-geometry grassmannian projective-varieties
asked Jan 22 at 1:22
kindasortakindasorta
9810
$endgroup$
I read in Harris' book at page 148, proposition 11.37 and got slightly confused regarding the argument.
Harris mentions a projective space $mathbb{P}^{2n+1}$ and a linear subspace $mathbb{P}^{n}$ denoted $L$, disjoint from a projective variety $J$, which is embedded in $mathbb{P}^{2n+1}$. He then denotes by $pi_L:mathbb{P}^{2n+1}longrightarrow L$, what he refers to as "the projection from $L$", which is a term I both never heard of, nor ever encountered in my reading. He then proceeds to project $J$, claiming that as $Jcap L=emptyset$, the projection is regular map. He finally adds that it is a general fact that the projection of a projective variety from a linear subspace disjoint from it is finite.
My questions are:
1) How is the projection from a linear subspace defined? I am familiar with the notion of projection from a point.
2) Is this projection a regular morphism?
3) Why does a regular morphism in this context has finite fibers?
4) Why is the projection of a projective variety from a linear subspace disjoint from it is finite in general?
I believe all my questions stem from not understanding the answer to 1). Any help is appreciated.
Ciao!
algebraic-geometry grassmannian projective-varieties
algebraic-geometry grassmannian projective-varieties
asked Jan 22 at 1:22
kindasortakindasorta
9810
asked Jan 22 at 1:22
kindasortakindasorta
9810
asked Jan 22 at 1:22
kindasortakindasorta
9810
asked Jan 22 at 1:22
kindasortakindasorta
9810
asked Jan 22 at 1:22
kindasortakindasorta
9810
9810
add a comment |
add a comment |
1 Answer
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Any linear subspace $L$ of dimension, after change of variables in $mathbb{P}^{2n+1}$ can be assumed to be given by $x_0=x_1=ldots=x_n=0$. Then the projection from $L$, $mathbb{P}^{2n+1}to mathbb{P}^n$ is given by the map $(x_0:ldots:x_{2n+1})mapsto (x_0:ldots: x_n)$. This is a morphism outside $L$ and thus gives a morphism from any closed subset of $mathbb{P}^{2n+1}$ which is disjoint from $L$. The rest (3), 4)) should be clear.
answered Jan 22 at 1:37
MohanMohan
11.9k1817
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$begingroup$
Lovely, thanks!
$endgroup$
– kindasorta
Jan 22 at 9:38
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Any linear subspace $L$ of dimension, after change of variables in $mathbb{P}^{2n+1}$ can be assumed to be given by $x_0=x_1=ldots=x_n=0$. Then the projection from $L$, $mathbb{P}^{2n+1}to mathbb{P}^n$ is given by the map $(x_0:ldots:x_{2n+1})mapsto (x_0:ldots: x_n)$. This is a morphism outside $L$ and thus gives a morphism from any closed subset of $mathbb{P}^{2n+1}$ which is disjoint from $L$. The rest (3), 4)) should be clear.
answered Jan 22 at 1:37
MohanMohan
11.9k1817
$endgroup$
$begingroup$
Lovely, thanks!
$endgroup$
– kindasorta
Jan 22 at 9:38
add a comment |
$begingroup$
Any linear subspace $L$ of dimension, after change of variables in $mathbb{P}^{2n+1}$ can be assumed to be given by $x_0=x_1=ldots=x_n=0$. Then the projection from $L$, $mathbb{P}^{2n+1}to mathbb{P}^n$ is given by the map $(x_0:ldots:x_{2n+1})mapsto (x_0:ldots: x_n)$. This is a morphism outside $L$ and thus gives a morphism from any closed subset of $mathbb{P}^{2n+1}$ which is disjoint from $L$. The rest (3), 4)) should be clear.
answered Jan 22 at 1:37
MohanMohan
11.9k1817
$endgroup$
$begingroup$
Lovely, thanks!
$endgroup$
– kindasorta
Jan 22 at 9:38
add a comment |
$begingroup$
Any linear subspace $L$ of dimension, after change of variables in $mathbb{P}^{2n+1}$ can be assumed to be given by $x_0=x_1=ldots=x_n=0$. Then the projection from $L$, $mathbb{P}^{2n+1}to mathbb{P}^n$ is given by the map $(x_0:ldots:x_{2n+1})mapsto (x_0:ldots: x_n)$. This is a morphism outside $L$ and thus gives a morphism from any closed subset of $mathbb{P}^{2n+1}$ which is disjoint from $L$. The rest (3), 4)) should be clear.
answered Jan 22 at 1:37
MohanMohan
11.9k1817
$endgroup$
Any linear subspace $L$ of dimension, after change of variables in $mathbb{P}^{2n+1}$ can be assumed to be given by $x_0=x_1=ldots=x_n=0$. Then the projection from $L$, $mathbb{P}^{2n+1}to mathbb{P}^n$ is given by the map $(x_0:ldots:x_{2n+1})mapsto (x_0:ldots: x_n)$. This is a morphism outside $L$ and thus gives a morphism from any closed subset of $mathbb{P}^{2n+1}$ which is disjoint from $L$. The rest (3), 4)) should be clear.
answered Jan 22 at 1:37
MohanMohan
11.9k1817
answered Jan 22 at 1:37
MohanMohan
11.9k1817
answered Jan 22 at 1:37
MohanMohan
11.9k1817
answered Jan 22 at 1:37
MohanMohan
11.9k1817
11.9k1817
$begingroup$
Lovely, thanks!
$endgroup$
– kindasorta
Jan 22 at 9:38
add a comment |
$begingroup$
Lovely, thanks!
$endgroup$
– kindasorta
Jan 22 at 9:38
$begingroup$
Lovely, thanks!
$endgroup$
– kindasorta
Jan 22 at 9:38
$begingroup$
Lovely, thanks!
$endgroup$
– kindasorta
Jan 22 at 9:38
add a comment |
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