Mixing
Definitions of “linearly normal” variety
$begingroup$
I wish I understood the explanation of linear normality on Wikipedia a bit better, and it seems there's actually a mistake at one point.
The variety $V$ in its projective embedding is projectively normal if its homogeneous coordinate ring $R$ is integrally closed. This condition implies that $V$ is a normal variety, but not conversely: the property of projective normality is not independent of the projective embedding, as is shown by the example of a rational quartic curve in three dimensions. Another equivalent condition is in terms of the linear system of divisors on $V$ cut out by the tautological line bundle $L$ on projective space, and its dth powers for d = 1, 2, 3, ...; when $V$ is non-singular, it is projectively normal if and only if each such linear system is a complete linear system.
I have no intuition for what it means for the homogenous coordinate ring to be integrally closed.
I also don't have any intuition for this "complete linear system" condition.
That would be okay if I were sure I understood this:
In a more geometric way one can think of $L$ as the Serre twist sheaf $O(1)$ on projective space, and use it to twist the structure sheaf $O_V$ k times, for any k. Then $V$ is called k-normal if the global sections of $O(k)$ map surjectively to those of $O_V(k)$ for a given k.
This is great except for one issue. Isn't $O(1)$ the sheaf of sections of the dual of the tautological line bundle? Yet here Wikipedia seems to be claiming it's the sheaf of sections of $L$, which in the previous passage it claimed was the tautological line bundle!
I think this is just a mistake on their part. I think they should say sections of $L^*$ give the Serre twist sheaf $O(1)$. I'd like to correct this if it's wrong... but I don't understand the previous passage well enough to know if they wanted the tautological line bundle or its dual back there.
Anyway, going on:
If $V$ is 1-normal it is called linearly normal, and projective normality is the condition that $V$ is k-normal for all k ≥ 1.
Is this way of stating projective normality only true for normal varieties, or for all varieties? This Math Stackexchange answer says it's only true for normal varieties.
Linear normality may be said geometrically: $V$ as projective variety cannot be obtained by an isomorphic linear projection from a projective space of higher dimension, except in the trivial way of lying in a proper linear subspace.
This sounds nice and geometrical, but I don't quite understand it. What's a "linear projection from a projective space of a higher dimension"? A linear projection from a vector space of higher dimension to one of lower dimension has a nontrivial kernel so it doesn't give a regular map between their projective spaces: I've been meaning to ask what sort of map we call this partially defined map. (A rational map I guess?) But I guess sometimes it maps a subvariety of the higher-dimensional projective space isomorphically to a subvariety of the lower-dimensional one? Is that the idea?
algebraic-geometry
asked Jan 1 at 21:16
John BaezJohn Baez
30218
$endgroup$
|
show 2 more comments
$begingroup$
I wish I understood the explanation of linear normality on Wikipedia a bit better, and it seems there's actually a mistake at one point.
The variety $V$ in its projective embedding is projectively normal if its homogeneous coordinate ring $R$ is integrally closed. This condition implies that $V$ is a normal variety, but not conversely: the property of projective normality is not independent of the projective embedding, as is shown by the example of a rational quartic curve in three dimensions. Another equivalent condition is in terms of the linear system of divisors on $V$ cut out by the tautological line bundle $L$ on projective space, and its dth powers for d = 1, 2, 3, ...; when $V$ is non-singular, it is projectively normal if and only if each such linear system is a complete linear system.
I have no intuition for what it means for the homogenous coordinate ring to be integrally closed.
I also don't have any intuition for this "complete linear system" condition.
That would be okay if I were sure I understood this:
In a more geometric way one can think of $L$ as the Serre twist sheaf $O(1)$ on projective space, and use it to twist the structure sheaf $O_V$ k times, for any k. Then $V$ is called k-normal if the global sections of $O(k)$ map surjectively to those of $O_V(k)$ for a given k.
This is great except for one issue. Isn't $O(1)$ the sheaf of sections of the dual of the tautological line bundle? Yet here Wikipedia seems to be claiming it's the sheaf of sections of $L$, which in the previous passage it claimed was the tautological line bundle!
I think this is just a mistake on their part. I think they should say sections of $L^*$ give the Serre twist sheaf $O(1)$. I'd like to correct this if it's wrong... but I don't understand the previous passage well enough to know if they wanted the tautological line bundle or its dual back there.
Anyway, going on:
If $V$ is 1-normal it is called linearly normal, and projective normality is the condition that $V$ is k-normal for all k ≥ 1.
Is this way of stating projective normality only true for normal varieties, or for all varieties? This Math Stackexchange answer says it's only true for normal varieties.
Linear normality may be said geometrically: $V$ as projective variety cannot be obtained by an isomorphic linear projection from a projective space of higher dimension, except in the trivial way of lying in a proper linear subspace.
This sounds nice and geometrical, but I don't quite understand it. What's a "linear projection from a projective space of a higher dimension"? A linear projection from a vector space of higher dimension to one of lower dimension has a nontrivial kernel so it doesn't give a regular map between their projective spaces: I've been meaning to ask what sort of map we call this partially defined map. (A rational map I guess?) But I guess sometimes it maps a subvariety of the higher-dimensional projective space isomorphically to a subvariety of the lower-dimensional one? Is that the idea?
algebraic-geometry
asked Jan 1 at 21:16
John BaezJohn Baez
30218
$endgroup$
2
$begingroup$
The reference on Wikipedia to the "tautological line bundle" should indeed refer to the dual of the tautological line bundle.
$endgroup$
– hunter
Jan 1 at 21:22
1
$begingroup$
(The tautological bundle doesn't cut out divisors at all since it has no global sections.)
$endgroup$
– hunter
Jan 1 at 21:25
$begingroup$
So, just to be 200% clear, you're saying every reference to the tautological line bundle in the quoted passages should really be talking about its dual?
$endgroup$
– John Baez
Jan 1 at 21:28
1
$begingroup$
Linear projection is indeed a rational map and not a morphism. But its restriction to a subvariety that misses the kernel is indeed a morphism, as you say. Unfortunately I don't know the answer to your other questions.
$endgroup$
– hunter
Jan 1 at 21:28
1
$begingroup$
I was only referring to the quoted passages, now the whole Wikipedia article. But you seem to be telling me that every time the quoted passages refer to $L$, they should be talking about the dual of the tautological line bundle. I'll change the Wikipedia article.
$endgroup$
– John Baez
Jan 1 at 21:32
|
show 2 more comments
$begingroup$
I wish I understood the explanation of linear normality on Wikipedia a bit better, and it seems there's actually a mistake at one point.
The variety $V$ in its projective embedding is projectively normal if its homogeneous coordinate ring $R$ is integrally closed. This condition implies that $V$ is a normal variety, but not conversely: the property of projective normality is not independent of the projective embedding, as is shown by the example of a rational quartic curve in three dimensions. Another equivalent condition is in terms of the linear system of divisors on $V$ cut out by the tautological line bundle $L$ on projective space, and its dth powers for d = 1, 2, 3, ...; when $V$ is non-singular, it is projectively normal if and only if each such linear system is a complete linear system.
I have no intuition for what it means for the homogenous coordinate ring to be integrally closed.
I also don't have any intuition for this "complete linear system" condition.
That would be okay if I were sure I understood this:
In a more geometric way one can think of $L$ as the Serre twist sheaf $O(1)$ on projective space, and use it to twist the structure sheaf $O_V$ k times, for any k. Then $V$ is called k-normal if the global sections of $O(k)$ map surjectively to those of $O_V(k)$ for a given k.
This is great except for one issue. Isn't $O(1)$ the sheaf of sections of the dual of the tautological line bundle? Yet here Wikipedia seems to be claiming it's the sheaf of sections of $L$, which in the previous passage it claimed was the tautological line bundle!
I think this is just a mistake on their part. I think they should say sections of $L^*$ give the Serre twist sheaf $O(1)$. I'd like to correct this if it's wrong... but I don't understand the previous passage well enough to know if they wanted the tautological line bundle or its dual back there.
Anyway, going on:
If $V$ is 1-normal it is called linearly normal, and projective normality is the condition that $V$ is k-normal for all k ≥ 1.
Is this way of stating projective normality only true for normal varieties, or for all varieties? This Math Stackexchange answer says it's only true for normal varieties.
Linear normality may be said geometrically: $V$ as projective variety cannot be obtained by an isomorphic linear projection from a projective space of higher dimension, except in the trivial way of lying in a proper linear subspace.
This sounds nice and geometrical, but I don't quite understand it. What's a "linear projection from a projective space of a higher dimension"? A linear projection from a vector space of higher dimension to one of lower dimension has a nontrivial kernel so it doesn't give a regular map between their projective spaces: I've been meaning to ask what sort of map we call this partially defined map. (A rational map I guess?) But I guess sometimes it maps a subvariety of the higher-dimensional projective space isomorphically to a subvariety of the lower-dimensional one? Is that the idea?
algebraic-geometry
asked Jan 1 at 21:16
John BaezJohn Baez
30218
$endgroup$
I wish I understood the explanation of linear normality on Wikipedia a bit better, and it seems there's actually a mistake at one point.
The variety $V$ in its projective embedding is projectively normal if its homogeneous coordinate ring $R$ is integrally closed. This condition implies that $V$ is a normal variety, but not conversely: the property of projective normality is not independent of the projective embedding, as is shown by the example of a rational quartic curve in three dimensions. Another equivalent condition is in terms of the linear system of divisors on $V$ cut out by the tautological line bundle $L$ on projective space, and its dth powers for d = 1, 2, 3, ...; when $V$ is non-singular, it is projectively normal if and only if each such linear system is a complete linear system.
I have no intuition for what it means for the homogenous coordinate ring to be integrally closed.
I also don't have any intuition for this "complete linear system" condition.
That would be okay if I were sure I understood this:
In a more geometric way one can think of $L$ as the Serre twist sheaf $O(1)$ on projective space, and use it to twist the structure sheaf $O_V$ k times, for any k. Then $V$ is called k-normal if the global sections of $O(k)$ map surjectively to those of $O_V(k)$ for a given k.
This is great except for one issue. Isn't $O(1)$ the sheaf of sections of the dual of the tautological line bundle? Yet here Wikipedia seems to be claiming it's the sheaf of sections of $L$, which in the previous passage it claimed was the tautological line bundle!
I think this is just a mistake on their part. I think they should say sections of $L^*$ give the Serre twist sheaf $O(1)$. I'd like to correct this if it's wrong... but I don't understand the previous passage well enough to know if they wanted the tautological line bundle or its dual back there.
Anyway, going on:
If $V$ is 1-normal it is called linearly normal, and projective normality is the condition that $V$ is k-normal for all k ≥ 1.
Is this way of stating projective normality only true for normal varieties, or for all varieties? This Math Stackexchange answer says it's only true for normal varieties.
Linear normality may be said geometrically: $V$ as projective variety cannot be obtained by an isomorphic linear projection from a projective space of higher dimension, except in the trivial way of lying in a proper linear subspace.
This sounds nice and geometrical, but I don't quite understand it. What's a "linear projection from a projective space of a higher dimension"? A linear projection from a vector space of higher dimension to one of lower dimension has a nontrivial kernel so it doesn't give a regular map between their projective spaces: I've been meaning to ask what sort of map we call this partially defined map. (A rational map I guess?) But I guess sometimes it maps a subvariety of the higher-dimensional projective space isomorphically to a subvariety of the lower-dimensional one? Is that the idea?
algebraic-geometry
algebraic-geometry
asked Jan 1 at 21:16
John BaezJohn Baez
30218
asked Jan 1 at 21:16
John BaezJohn Baez
30218
asked Jan 1 at 21:16
John BaezJohn Baez
30218
asked Jan 1 at 21:16
John BaezJohn Baez
30218
asked Jan 1 at 21:16
John BaezJohn Baez
30218
30218
2
$begingroup$
The reference on Wikipedia to the "tautological line bundle" should indeed refer to the dual of the tautological line bundle.
$endgroup$
– hunter
Jan 1 at 21:22
1
$begingroup$
(The tautological bundle doesn't cut out divisors at all since it has no global sections.)
$endgroup$
– hunter
Jan 1 at 21:25
$begingroup$
So, just to be 200% clear, you're saying every reference to the tautological line bundle in the quoted passages should really be talking about its dual?
$endgroup$
– John Baez
Jan 1 at 21:28
1
$begingroup$
Linear projection is indeed a rational map and not a morphism. But its restriction to a subvariety that misses the kernel is indeed a morphism, as you say. Unfortunately I don't know the answer to your other questions.
$endgroup$
– hunter
Jan 1 at 21:28
1
$begingroup$
I was only referring to the quoted passages, now the whole Wikipedia article. But you seem to be telling me that every time the quoted passages refer to $L$, they should be talking about the dual of the tautological line bundle. I'll change the Wikipedia article.
$endgroup$
– John Baez
Jan 1 at 21:32
|
show 2 more comments
2
$begingroup$
The reference on Wikipedia to the "tautological line bundle" should indeed refer to the dual of the tautological line bundle.
$endgroup$
– hunter
Jan 1 at 21:22
1
$begingroup$
(The tautological bundle doesn't cut out divisors at all since it has no global sections.)
$endgroup$
– hunter
Jan 1 at 21:25
$begingroup$
So, just to be 200% clear, you're saying every reference to the tautological line bundle in the quoted passages should really be talking about its dual?
$endgroup$
– John Baez
Jan 1 at 21:28
1
$begingroup$
Linear projection is indeed a rational map and not a morphism. But its restriction to a subvariety that misses the kernel is indeed a morphism, as you say. Unfortunately I don't know the answer to your other questions.
$endgroup$
– hunter
Jan 1 at 21:28
1
$begingroup$
I was only referring to the quoted passages, now the whole Wikipedia article. But you seem to be telling me that every time the quoted passages refer to $L$, they should be talking about the dual of the tautological line bundle. I'll change the Wikipedia article.
$endgroup$
– John Baez
Jan 1 at 21:32
2
2
$begingroup$
The reference on Wikipedia to the "tautological line bundle" should indeed refer to the dual of the tautological line bundle.
$endgroup$
– hunter
Jan 1 at 21:22
$begingroup$
The reference on Wikipedia to the "tautological line bundle" should indeed refer to the dual of the tautological line bundle.
$endgroup$
– hunter
Jan 1 at 21:22
1
1
$begingroup$
(The tautological bundle doesn't cut out divisors at all since it has no global sections.)
$endgroup$
– hunter
Jan 1 at 21:25
$begingroup$
(The tautological bundle doesn't cut out divisors at all since it has no global sections.)
$endgroup$
– hunter
Jan 1 at 21:25
$begingroup$
So, just to be 200% clear, you're saying every reference to the tautological line bundle in the quoted passages should really be talking about its dual?
$endgroup$
– John Baez
Jan 1 at 21:28
$begingroup$
So, just to be 200% clear, you're saying every reference to the tautological line bundle in the quoted passages should really be talking about its dual?
$endgroup$
– John Baez
Jan 1 at 21:28
1
1
$begingroup$
Linear projection is indeed a rational map and not a morphism. But its restriction to a subvariety that misses the kernel is indeed a morphism, as you say. Unfortunately I don't know the answer to your other questions.
$endgroup$
– hunter
Jan 1 at 21:28
$begingroup$
Linear projection is indeed a rational map and not a morphism. But its restriction to a subvariety that misses the kernel is indeed a morphism, as you say. Unfortunately I don't know the answer to your other questions.
$endgroup$
– hunter
Jan 1 at 21:28
1
1
$begingroup$
I was only referring to the quoted passages, now the whole Wikipedia article. But you seem to be telling me that every time the quoted passages refer to $L$, they should be talking about the dual of the tautological line bundle. I'll change the Wikipedia article.
$endgroup$
– John Baez
Jan 1 at 21:32
$begingroup$
I was only referring to the quoted passages, now the whole Wikipedia article. But you seem to be telling me that every time the quoted passages refer to $L$, they should be talking about the dual of the tautological line bundle. I'll change the Wikipedia article.
$endgroup$
– John Baez
Jan 1 at 21:32
|
show 2 more comments
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$begingroup$
The reference on Wikipedia to the "tautological line bundle" should indeed refer to the dual of the tautological line bundle.
$endgroup$
– hunter
Jan 1 at 21:22
1
$begingroup$
(The tautological bundle doesn't cut out divisors at all since it has no global sections.)
$endgroup$
– hunter
Jan 1 at 21:25
$begingroup$
So, just to be 200% clear, you're saying every reference to the tautological line bundle in the quoted passages should really be talking about its dual?
$endgroup$
– John Baez
Jan 1 at 21:28
1
$begingroup$
Linear projection is indeed a rational map and not a morphism. But its restriction to a subvariety that misses the kernel is indeed a morphism, as you say. Unfortunately I don't know the answer to your other questions.
$endgroup$
– hunter
Jan 1 at 21:28
1
$begingroup$
I was only referring to the quoted passages, now the whole Wikipedia article. But you seem to be telling me that every time the quoted passages refer to $L$, they should be talking about the dual of the tautological line bundle. I'll change the Wikipedia article.
$endgroup$
– John Baez
Jan 1 at 21:32