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Brill-Noether Theory-ish Question
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I'm not much of an expert on Brill-Noether Theory, so apologies if this is is a silly question. Given $p_1,ldots, p_n$ distinct points in $mathbb P^n-1$, where $ngeq 4$, is it possible to find at least $n+1$ linearly independent quadrics that pass through $p_1,ldots, p_n$. If this is not true in general, are there specific instances where this is true?
For example, if $v_1,ldots, v_n$ are an orthogonal basis for $mathbb K^n$ (where $mathbb K$ need not be algebraically closed), then if $ell_i = v_icdot [x_1,ldots, x_n]^T$, the quadrics $ell_iell_j$ are linearly independent and all pass through $v_1,ldots, v_n$. As $ngeq 4$, there are at least $n+1$ linearly independent quadrics passing through $v_1,ldots, v_n$.
algebraic-geometry commutative-algebra
asked Jul 19 at 2:37
Zach Flores
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up vote
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down vote
favorite
I'm not much of an expert on Brill-Noether Theory, so apologies if this is is a silly question. Given $p_1,ldots, p_n$ distinct points in $mathbb P^n-1$, where $ngeq 4$, is it possible to find at least $n+1$ linearly independent quadrics that pass through $p_1,ldots, p_n$. If this is not true in general, are there specific instances where this is true?
For example, if $v_1,ldots, v_n$ are an orthogonal basis for $mathbb K^n$ (where $mathbb K$ need not be algebraically closed), then if $ell_i = v_icdot [x_1,ldots, x_n]^T$, the quadrics $ell_iell_j$ are linearly independent and all pass through $v_1,ldots, v_n$. As $ngeq 4$, there are at least $n+1$ linearly independent quadrics passing through $v_1,ldots, v_n$.
algebraic-geometry commutative-algebra
asked Jul 19 at 2:37
Zach Flores
666
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm not much of an expert on Brill-Noether Theory, so apologies if this is is a silly question. Given $p_1,ldots, p_n$ distinct points in $mathbb P^n-1$, where $ngeq 4$, is it possible to find at least $n+1$ linearly independent quadrics that pass through $p_1,ldots, p_n$. If this is not true in general, are there specific instances where this is true?
For example, if $v_1,ldots, v_n$ are an orthogonal basis for $mathbb K^n$ (where $mathbb K$ need not be algebraically closed), then if $ell_i = v_icdot [x_1,ldots, x_n]^T$, the quadrics $ell_iell_j$ are linearly independent and all pass through $v_1,ldots, v_n$. As $ngeq 4$, there are at least $n+1$ linearly independent quadrics passing through $v_1,ldots, v_n$.
algebraic-geometry commutative-algebra
asked Jul 19 at 2:37
Zach Flores
666
I'm not much of an expert on Brill-Noether Theory, so apologies if this is is a silly question. Given $p_1,ldots, p_n$ distinct points in $mathbb P^n-1$, where $ngeq 4$, is it possible to find at least $n+1$ linearly independent quadrics that pass through $p_1,ldots, p_n$. If this is not true in general, are there specific instances where this is true?
For example, if $v_1,ldots, v_n$ are an orthogonal basis for $mathbb K^n$ (where $mathbb K$ need not be algebraically closed), then if $ell_i = v_icdot [x_1,ldots, x_n]^T$, the quadrics $ell_iell_j$ are linearly independent and all pass through $v_1,ldots, v_n$. As $ngeq 4$, there are at least $n+1$ linearly independent quadrics passing through $v_1,ldots, v_n$.
algebraic-geometry commutative-algebra
asked Jul 19 at 2:37
Zach Flores
666
asked Jul 19 at 2:37
Zach Flores
666
asked Jul 19 at 2:37
Zach Flores
666
asked Jul 19 at 2:37
Zach Flores
666
666
add a comment |Â
add a comment |Â
1 Answer
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This is true. Here's one way to see this if your points are in general position, i.e. if I let $l_i subset k^n$ to be the line corresponding to the point $p_i$, then I'm assuming that taking $0ne v_iin l_i$, you have that $v_i$ are linearly independent in $k^n$.
1) WLOG you may assume that your points are given by $p_1 = [1:0:ldots:0]$, $p_2=[0:1:ldots:0], ldots, p_n = [0:0:ldots:0:1]$. To see this, because $v_i$ are linearly independent, there exists $Ain GL(n)$ such that $Av_i = langle e_i rangle$, where $e_i$ are your standard basis of $k^n$. Then it follows that $A$ defines an automorphism of $mathbbP^n-1$ mapping $p_1$ to $[1:0:ldots:0]$ etc.
2) Let $X_0,ldots,X_n-1$ denote homogeneous coordinates on $mathbbP^n-1$. A quadric can be written as $sum_i=0^n-1 a_ii X_i^2 + sum_ine j a_ij X_iX_j = 0$, where $a_ij in k$. Imposing the conditions that they vanish on $p_i$ for all $i$, we deduce that $a_ii = 0$ for all $i$. This leaves us that the vector space of quadrics vanishing on $p_i$ for all $i$ has dimension $binomn2 ge n+1$ for $nge 4$.
If your points are not in general position, i.e. if the $v_i$ are not linearly independent, then basically tracing through the same argument will find that the vector space of quadrics is going to have even higher dimension.
Another way of saying this is simply to note that you have an exact sequence
$$0rightarrow mathcalI(2) rightarrow mathcalO_mathbbP^n-1(2) rightarrow mathcalO_bigcup p_i rightarrow 0$$
Applying the long exact sequence in cohomology and noting that $h^0(mathcalO_mathbbP^n-1(2)) = binomn+12$ and $h^0(mathcalO_bigcup p_i) = n$ tells you that $h^0(mathcalI(2)) ge binomn-12 ge n+1$ when $nge 4$, which is the same claim as above.
answered Jul 19 at 3:17
loch
1,403179
If $A$ is an automorphism of $mathbb P^n-1$, then it's not always true that $g(Av_i) = 0$ if $g(v_i) = 0$, so I'm not entirely sure we can assme that $p_1,ldots, p_n$ is given by the standard basis.
– Zach Flores
Aug 4 at 1:05
@ZachFlores If I understood you correctly, your concern is that if $g$ is a quadric containing $v_i$, i.e. $g(v_i) = 0$, it's not necessarily true that $g(Av_i) = 0$, where $A$ is an automorphism of $mathbbP^n-1$. While this is true, what we do get is a different quadric which will contain these new $Av_i$'s.
– loch
Aug 4 at 1:43
ah, I see! Apologies. I'm not a geometer :)
– Zach Flores
Aug 4 at 1:45
@ZachFlores In particular let's say your automorphism maps $X_i mapsto sum a_ij X_j$, with inverse $X_imapsto sum b_ij X_j$ then if the quadric you had in the beginning containing $v_i$ was $g(X_0,X_1,ldots,X_n-1)$, then the new quadric you have is $g(sum b_0jX_j, sum b_1jX_j,ldots )$. The new quadric will contain $Av_i$.
– loch
Aug 4 at 2:08
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This is true. Here's one way to see this if your points are in general position, i.e. if I let $l_i subset k^n$ to be the line corresponding to the point $p_i$, then I'm assuming that taking $0ne v_iin l_i$, you have that $v_i$ are linearly independent in $k^n$.
1) WLOG you may assume that your points are given by $p_1 = [1:0:ldots:0]$, $p_2=[0:1:ldots:0], ldots, p_n = [0:0:ldots:0:1]$. To see this, because $v_i$ are linearly independent, there exists $Ain GL(n)$ such that $Av_i = langle e_i rangle$, where $e_i$ are your standard basis of $k^n$. Then it follows that $A$ defines an automorphism of $mathbbP^n-1$ mapping $p_1$ to $[1:0:ldots:0]$ etc.
2) Let $X_0,ldots,X_n-1$ denote homogeneous coordinates on $mathbbP^n-1$. A quadric can be written as $sum_i=0^n-1 a_ii X_i^2 + sum_ine j a_ij X_iX_j = 0$, where $a_ij in k$. Imposing the conditions that they vanish on $p_i$ for all $i$, we deduce that $a_ii = 0$ for all $i$. This leaves us that the vector space of quadrics vanishing on $p_i$ for all $i$ has dimension $binomn2 ge n+1$ for $nge 4$.
If your points are not in general position, i.e. if the $v_i$ are not linearly independent, then basically tracing through the same argument will find that the vector space of quadrics is going to have even higher dimension.
Another way of saying this is simply to note that you have an exact sequence
$$0rightarrow mathcalI(2) rightarrow mathcalO_mathbbP^n-1(2) rightarrow mathcalO_bigcup p_i rightarrow 0$$
Applying the long exact sequence in cohomology and noting that $h^0(mathcalO_mathbbP^n-1(2)) = binomn+12$ and $h^0(mathcalO_bigcup p_i) = n$ tells you that $h^0(mathcalI(2)) ge binomn-12 ge n+1$ when $nge 4$, which is the same claim as above.
answered Jul 19 at 3:17
loch
1,403179
If $A$ is an automorphism of $mathbb P^n-1$, then it's not always true that $g(Av_i) = 0$ if $g(v_i) = 0$, so I'm not entirely sure we can assme that $p_1,ldots, p_n$ is given by the standard basis.
– Zach Flores
Aug 4 at 1:05
@ZachFlores If I understood you correctly, your concern is that if $g$ is a quadric containing $v_i$, i.e. $g(v_i) = 0$, it's not necessarily true that $g(Av_i) = 0$, where $A$ is an automorphism of $mathbbP^n-1$. While this is true, what we do get is a different quadric which will contain these new $Av_i$'s.
– loch
Aug 4 at 1:43
ah, I see! Apologies. I'm not a geometer :)
– Zach Flores
Aug 4 at 1:45
@ZachFlores In particular let's say your automorphism maps $X_i mapsto sum a_ij X_j$, with inverse $X_imapsto sum b_ij X_j$ then if the quadric you had in the beginning containing $v_i$ was $g(X_0,X_1,ldots,X_n-1)$, then the new quadric you have is $g(sum b_0jX_j, sum b_1jX_j,ldots )$. The new quadric will contain $Av_i$.
– loch
Aug 4 at 2:08
add a comment |Â
up vote
1
down vote
This is true. Here's one way to see this if your points are in general position, i.e. if I let $l_i subset k^n$ to be the line corresponding to the point $p_i$, then I'm assuming that taking $0ne v_iin l_i$, you have that $v_i$ are linearly independent in $k^n$.
1) WLOG you may assume that your points are given by $p_1 = [1:0:ldots:0]$, $p_2=[0:1:ldots:0], ldots, p_n = [0:0:ldots:0:1]$. To see this, because $v_i$ are linearly independent, there exists $Ain GL(n)$ such that $Av_i = langle e_i rangle$, where $e_i$ are your standard basis of $k^n$. Then it follows that $A$ defines an automorphism of $mathbbP^n-1$ mapping $p_1$ to $[1:0:ldots:0]$ etc.
2) Let $X_0,ldots,X_n-1$ denote homogeneous coordinates on $mathbbP^n-1$. A quadric can be written as $sum_i=0^n-1 a_ii X_i^2 + sum_ine j a_ij X_iX_j = 0$, where $a_ij in k$. Imposing the conditions that they vanish on $p_i$ for all $i$, we deduce that $a_ii = 0$ for all $i$. This leaves us that the vector space of quadrics vanishing on $p_i$ for all $i$ has dimension $binomn2 ge n+1$ for $nge 4$.
If your points are not in general position, i.e. if the $v_i$ are not linearly independent, then basically tracing through the same argument will find that the vector space of quadrics is going to have even higher dimension.
Another way of saying this is simply to note that you have an exact sequence
$$0rightarrow mathcalI(2) rightarrow mathcalO_mathbbP^n-1(2) rightarrow mathcalO_bigcup p_i rightarrow 0$$
Applying the long exact sequence in cohomology and noting that $h^0(mathcalO_mathbbP^n-1(2)) = binomn+12$ and $h^0(mathcalO_bigcup p_i) = n$ tells you that $h^0(mathcalI(2)) ge binomn-12 ge n+1$ when $nge 4$, which is the same claim as above.
answered Jul 19 at 3:17
loch
1,403179
If $A$ is an automorphism of $mathbb P^n-1$, then it's not always true that $g(Av_i) = 0$ if $g(v_i) = 0$, so I'm not entirely sure we can assme that $p_1,ldots, p_n$ is given by the standard basis.
– Zach Flores
Aug 4 at 1:05
@ZachFlores If I understood you correctly, your concern is that if $g$ is a quadric containing $v_i$, i.e. $g(v_i) = 0$, it's not necessarily true that $g(Av_i) = 0$, where $A$ is an automorphism of $mathbbP^n-1$. While this is true, what we do get is a different quadric which will contain these new $Av_i$'s.
– loch
Aug 4 at 1:43
ah, I see! Apologies. I'm not a geometer :)
– Zach Flores
Aug 4 at 1:45
@ZachFlores In particular let's say your automorphism maps $X_i mapsto sum a_ij X_j$, with inverse $X_imapsto sum b_ij X_j$ then if the quadric you had in the beginning containing $v_i$ was $g(X_0,X_1,ldots,X_n-1)$, then the new quadric you have is $g(sum b_0jX_j, sum b_1jX_j,ldots )$. The new quadric will contain $Av_i$.
– loch
Aug 4 at 2:08
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is true. Here's one way to see this if your points are in general position, i.e. if I let $l_i subset k^n$ to be the line corresponding to the point $p_i$, then I'm assuming that taking $0ne v_iin l_i$, you have that $v_i$ are linearly independent in $k^n$.
1) WLOG you may assume that your points are given by $p_1 = [1:0:ldots:0]$, $p_2=[0:1:ldots:0], ldots, p_n = [0:0:ldots:0:1]$. To see this, because $v_i$ are linearly independent, there exists $Ain GL(n)$ such that $Av_i = langle e_i rangle$, where $e_i$ are your standard basis of $k^n$. Then it follows that $A$ defines an automorphism of $mathbbP^n-1$ mapping $p_1$ to $[1:0:ldots:0]$ etc.
2) Let $X_0,ldots,X_n-1$ denote homogeneous coordinates on $mathbbP^n-1$. A quadric can be written as $sum_i=0^n-1 a_ii X_i^2 + sum_ine j a_ij X_iX_j = 0$, where $a_ij in k$. Imposing the conditions that they vanish on $p_i$ for all $i$, we deduce that $a_ii = 0$ for all $i$. This leaves us that the vector space of quadrics vanishing on $p_i$ for all $i$ has dimension $binomn2 ge n+1$ for $nge 4$.
If your points are not in general position, i.e. if the $v_i$ are not linearly independent, then basically tracing through the same argument will find that the vector space of quadrics is going to have even higher dimension.
Another way of saying this is simply to note that you have an exact sequence
$$0rightarrow mathcalI(2) rightarrow mathcalO_mathbbP^n-1(2) rightarrow mathcalO_bigcup p_i rightarrow 0$$
Applying the long exact sequence in cohomology and noting that $h^0(mathcalO_mathbbP^n-1(2)) = binomn+12$ and $h^0(mathcalO_bigcup p_i) = n$ tells you that $h^0(mathcalI(2)) ge binomn-12 ge n+1$ when $nge 4$, which is the same claim as above.
answered Jul 19 at 3:17
loch
1,403179
This is true. Here's one way to see this if your points are in general position, i.e. if I let $l_i subset k^n$ to be the line corresponding to the point $p_i$, then I'm assuming that taking $0ne v_iin l_i$, you have that $v_i$ are linearly independent in $k^n$.
1) WLOG you may assume that your points are given by $p_1 = [1:0:ldots:0]$, $p_2=[0:1:ldots:0], ldots, p_n = [0:0:ldots:0:1]$. To see this, because $v_i$ are linearly independent, there exists $Ain GL(n)$ such that $Av_i = langle e_i rangle$, where $e_i$ are your standard basis of $k^n$. Then it follows that $A$ defines an automorphism of $mathbbP^n-1$ mapping $p_1$ to $[1:0:ldots:0]$ etc.
2) Let $X_0,ldots,X_n-1$ denote homogeneous coordinates on $mathbbP^n-1$. A quadric can be written as $sum_i=0^n-1 a_ii X_i^2 + sum_ine j a_ij X_iX_j = 0$, where $a_ij in k$. Imposing the conditions that they vanish on $p_i$ for all $i$, we deduce that $a_ii = 0$ for all $i$. This leaves us that the vector space of quadrics vanishing on $p_i$ for all $i$ has dimension $binomn2 ge n+1$ for $nge 4$.
If your points are not in general position, i.e. if the $v_i$ are not linearly independent, then basically tracing through the same argument will find that the vector space of quadrics is going to have even higher dimension.
Another way of saying this is simply to note that you have an exact sequence
$$0rightarrow mathcalI(2) rightarrow mathcalO_mathbbP^n-1(2) rightarrow mathcalO_bigcup p_i rightarrow 0$$
Applying the long exact sequence in cohomology and noting that $h^0(mathcalO_mathbbP^n-1(2)) = binomn+12$ and $h^0(mathcalO_bigcup p_i) = n$ tells you that $h^0(mathcalI(2)) ge binomn-12 ge n+1$ when $nge 4$, which is the same claim as above.
answered Jul 19 at 3:17
loch
1,403179
edited Jul 19 at 3:56
answered Jul 19 at 3:17
loch
1,403179
answered Jul 19 at 3:17
loch
1,403179
answered Jul 19 at 3:17
loch
1,403179
1,403179
If $A$ is an automorphism of $mathbb P^n-1$, then it's not always true that $g(Av_i) = 0$ if $g(v_i) = 0$, so I'm not entirely sure we can assme that $p_1,ldots, p_n$ is given by the standard basis.
– Zach Flores
Aug 4 at 1:05
@ZachFlores If I understood you correctly, your concern is that if $g$ is a quadric containing $v_i$, i.e. $g(v_i) = 0$, it's not necessarily true that $g(Av_i) = 0$, where $A$ is an automorphism of $mathbbP^n-1$. While this is true, what we do get is a different quadric which will contain these new $Av_i$'s.
– loch
Aug 4 at 1:43
ah, I see! Apologies. I'm not a geometer :)
– Zach Flores
Aug 4 at 1:45
@ZachFlores In particular let's say your automorphism maps $X_i mapsto sum a_ij X_j$, with inverse $X_imapsto sum b_ij X_j$ then if the quadric you had in the beginning containing $v_i$ was $g(X_0,X_1,ldots,X_n-1)$, then the new quadric you have is $g(sum b_0jX_j, sum b_1jX_j,ldots )$. The new quadric will contain $Av_i$.
– loch
Aug 4 at 2:08
add a comment |Â
If $A$ is an automorphism of $mathbb P^n-1$, then it's not always true that $g(Av_i) = 0$ if $g(v_i) = 0$, so I'm not entirely sure we can assme that $p_1,ldots, p_n$ is given by the standard basis.
– Zach Flores
Aug 4 at 1:05
@ZachFlores If I understood you correctly, your concern is that if $g$ is a quadric containing $v_i$, i.e. $g(v_i) = 0$, it's not necessarily true that $g(Av_i) = 0$, where $A$ is an automorphism of $mathbbP^n-1$. While this is true, what we do get is a different quadric which will contain these new $Av_i$'s.
– loch
Aug 4 at 1:43
ah, I see! Apologies. I'm not a geometer :)
– Zach Flores
Aug 4 at 1:45
@ZachFlores In particular let's say your automorphism maps $X_i mapsto sum a_ij X_j$, with inverse $X_imapsto sum b_ij X_j$ then if the quadric you had in the beginning containing $v_i$ was $g(X_0,X_1,ldots,X_n-1)$, then the new quadric you have is $g(sum b_0jX_j, sum b_1jX_j,ldots )$. The new quadric will contain $Av_i$.
– loch
Aug 4 at 2:08
If $A$ is an automorphism of $mathbb P^n-1$, then it's not always true that $g(Av_i) = 0$ if $g(v_i) = 0$, so I'm not entirely sure we can assme that $p_1,ldots, p_n$ is given by the standard basis.
– Zach Flores
Aug 4 at 1:05
If $A$ is an automorphism of $mathbb P^n-1$, then it's not always true that $g(Av_i) = 0$ if $g(v_i) = 0$, so I'm not entirely sure we can assme that $p_1,ldots, p_n$ is given by the standard basis.
– Zach Flores
Aug 4 at 1:05
@ZachFlores If I understood you correctly, your concern is that if $g$ is a quadric containing $v_i$, i.e. $g(v_i) = 0$, it's not necessarily true that $g(Av_i) = 0$, where $A$ is an automorphism of $mathbbP^n-1$. While this is true, what we do get is a different quadric which will contain these new $Av_i$'s.
– loch
Aug 4 at 1:43
@ZachFlores If I understood you correctly, your concern is that if $g$ is a quadric containing $v_i$, i.e. $g(v_i) = 0$, it's not necessarily true that $g(Av_i) = 0$, where $A$ is an automorphism of $mathbbP^n-1$. While this is true, what we do get is a different quadric which will contain these new $Av_i$'s.
– loch
Aug 4 at 1:43
ah, I see! Apologies. I'm not a geometer :)
– Zach Flores
Aug 4 at 1:45
ah, I see! Apologies. I'm not a geometer :)
– Zach Flores
Aug 4 at 1:45
@ZachFlores In particular let's say your automorphism maps $X_i mapsto sum a_ij X_j$, with inverse $X_imapsto sum b_ij X_j$ then if the quadric you had in the beginning containing $v_i$ was $g(X_0,X_1,ldots,X_n-1)$, then the new quadric you have is $g(sum b_0jX_j, sum b_1jX_j,ldots )$. The new quadric will contain $Av_i$.
– loch
Aug 4 at 2:08
@ZachFlores In particular let's say your automorphism maps $X_i mapsto sum a_ij X_j$, with inverse $X_imapsto sum b_ij X_j$ then if the quadric you had in the beginning containing $v_i$ was $g(X_0,X_1,ldots,X_n-1)$, then the new quadric you have is $g(sum b_0jX_j, sum b_1jX_j,ldots )$. The new quadric will contain $Av_i$.
– loch
Aug 4 at 2:08
add a comment |Â
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