Linear subspaces in quadric hypersurfaces
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Consider $H_1,H_2,H_3subsetmathbb{P}^{2m+1}$ three general linear subspaces of projective dimension $m$.
Then there exists a quadric hypersurface $Q^{2m}subsetmathbb{P}^{2m+1}$ containing $H_1,H_2,H_3$. This is just a dimension count.
Does there exist a smooth quadric hypersurface $Q^{2m}subsetmathbb{P}^{2m+1}$ containing $H_1,H_2,H_3$?
The answer is positive when $m = 1$, since if $Q^2$ is a cone then the three lines must intersect, and if $Q^2$ is the union of two planes or a double plane then the three lines must intersect as well.
ag.algebraic-geometry linear-algebra projective-geometry grassmannians quadrics
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Consider $H_1,H_2,H_3subsetmathbb{P}^{2m+1}$ three general linear subspaces of projective dimension $m$.
Then there exists a quadric hypersurface $Q^{2m}subsetmathbb{P}^{2m+1}$ containing $H_1,H_2,H_3$. This is just a dimension count.
Does there exist a smooth quadric hypersurface $Q^{2m}subsetmathbb{P}^{2m+1}$ containing $H_1,H_2,H_3$?
The answer is positive when $m = 1$, since if $Q^2$ is a cone then the three lines must intersect, and if $Q^2$ is the union of two planes or a double plane then the three lines must intersect as well.
ag.algebraic-geometry linear-algebra projective-geometry grassmannians quadrics
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider $H_1,H_2,H_3subsetmathbb{P}^{2m+1}$ three general linear subspaces of projective dimension $m$.
Then there exists a quadric hypersurface $Q^{2m}subsetmathbb{P}^{2m+1}$ containing $H_1,H_2,H_3$. This is just a dimension count.
Does there exist a smooth quadric hypersurface $Q^{2m}subsetmathbb{P}^{2m+1}$ containing $H_1,H_2,H_3$?
The answer is positive when $m = 1$, since if $Q^2$ is a cone then the three lines must intersect, and if $Q^2$ is the union of two planes or a double plane then the three lines must intersect as well.
ag.algebraic-geometry linear-algebra projective-geometry grassmannians quadrics
Consider $H_1,H_2,H_3subsetmathbb{P}^{2m+1}$ three general linear subspaces of projective dimension $m$.
Then there exists a quadric hypersurface $Q^{2m}subsetmathbb{P}^{2m+1}$ containing $H_1,H_2,H_3$. This is just a dimension count.
Does there exist a smooth quadric hypersurface $Q^{2m}subsetmathbb{P}^{2m+1}$ containing $H_1,H_2,H_3$?
The answer is positive when $m = 1$, since if $Q^2$ is a cone then the three lines must intersect, and if $Q^2$ is the union of two planes or a double plane then the three lines must intersect as well.
ag.algebraic-geometry linear-algebra projective-geometry grassmannians quadrics
ag.algebraic-geometry linear-algebra projective-geometry grassmannians quadrics
edited Nov 18 at 7:04
Kugelblitz
1185
1185
asked Nov 17 at 21:01
SMB
1085
1085
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1 Answer
1
active
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up vote
4
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This is true when $m$ is odd and false otherwise.
Indeed, let $mathbb{P}^{2m+1} = mathbb{P}(V)$. Assuming $H_1 cap H_2 = emptyset$ (by genericity), we have $H_1 = mathbb{P}(V_1)$, $H_2 = mathbb{P}(V_2)$ and
$$
V = V_1 oplus V_2.
$$
Furthermore, assuming $H_3 cap H_1 = H_3 cap H_2 = emptyset$, we see that $H_3 = mathbb{P}(V_3)$, where $V_3 subset V_1 oplus V_2$ is the graph of an isomorphism $V_1 to V_2$. Thus, choosing bases appropriately we may assume
$$
V_1 = langle e_1,dots,e_{m+1} rangle,quad
V_2 = langle e_{m+2},dots,e_{2m+2} rangle,quad
V_3 = langle e_1 + e_{m+2}, dots, e_{m+1} + e_{2m+2} rangle.
$$
Let $A = (a_{ij})$ be the matrix of the quadric $Q$. It follows that
$a_{ij} = 0$ when either $1 le i,j le m+1$ or $m+2 le i,j le 2m+2$, and
$$
b_{ij} = a_{i,m+1+j}
$$
is a skew-symmetric matrix. It remains to note that
$$
det(A) = pm det(B)^2,
$$
and that a skew-symmetric matrix of odd size is always degenerate.
So, if I got correctly you argument, $B$ is an $(m+1)times (m+1)$ matrix. If $m$ is odd then $m+1$ is even, $det(B)neq 0$ and so $det(A)neq 0$ and the quadric is smooth. If $m$ is even then $m+1$ is odd, $det(B)= 0$ and so $det(A)= 0$ and the quadric is singular. Is this right?
– SMB
Nov 17 at 21:35
Yes, precisely.
– Sasha
Nov 17 at 21:36
Thank you very much.
– SMB
Nov 17 at 21:58
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
This is true when $m$ is odd and false otherwise.
Indeed, let $mathbb{P}^{2m+1} = mathbb{P}(V)$. Assuming $H_1 cap H_2 = emptyset$ (by genericity), we have $H_1 = mathbb{P}(V_1)$, $H_2 = mathbb{P}(V_2)$ and
$$
V = V_1 oplus V_2.
$$
Furthermore, assuming $H_3 cap H_1 = H_3 cap H_2 = emptyset$, we see that $H_3 = mathbb{P}(V_3)$, where $V_3 subset V_1 oplus V_2$ is the graph of an isomorphism $V_1 to V_2$. Thus, choosing bases appropriately we may assume
$$
V_1 = langle e_1,dots,e_{m+1} rangle,quad
V_2 = langle e_{m+2},dots,e_{2m+2} rangle,quad
V_3 = langle e_1 + e_{m+2}, dots, e_{m+1} + e_{2m+2} rangle.
$$
Let $A = (a_{ij})$ be the matrix of the quadric $Q$. It follows that
$a_{ij} = 0$ when either $1 le i,j le m+1$ or $m+2 le i,j le 2m+2$, and
$$
b_{ij} = a_{i,m+1+j}
$$
is a skew-symmetric matrix. It remains to note that
$$
det(A) = pm det(B)^2,
$$
and that a skew-symmetric matrix of odd size is always degenerate.
So, if I got correctly you argument, $B$ is an $(m+1)times (m+1)$ matrix. If $m$ is odd then $m+1$ is even, $det(B)neq 0$ and so $det(A)neq 0$ and the quadric is smooth. If $m$ is even then $m+1$ is odd, $det(B)= 0$ and so $det(A)= 0$ and the quadric is singular. Is this right?
– SMB
Nov 17 at 21:35
Yes, precisely.
– Sasha
Nov 17 at 21:36
Thank you very much.
– SMB
Nov 17 at 21:58
add a comment |
up vote
4
down vote
accepted
This is true when $m$ is odd and false otherwise.
Indeed, let $mathbb{P}^{2m+1} = mathbb{P}(V)$. Assuming $H_1 cap H_2 = emptyset$ (by genericity), we have $H_1 = mathbb{P}(V_1)$, $H_2 = mathbb{P}(V_2)$ and
$$
V = V_1 oplus V_2.
$$
Furthermore, assuming $H_3 cap H_1 = H_3 cap H_2 = emptyset$, we see that $H_3 = mathbb{P}(V_3)$, where $V_3 subset V_1 oplus V_2$ is the graph of an isomorphism $V_1 to V_2$. Thus, choosing bases appropriately we may assume
$$
V_1 = langle e_1,dots,e_{m+1} rangle,quad
V_2 = langle e_{m+2},dots,e_{2m+2} rangle,quad
V_3 = langle e_1 + e_{m+2}, dots, e_{m+1} + e_{2m+2} rangle.
$$
Let $A = (a_{ij})$ be the matrix of the quadric $Q$. It follows that
$a_{ij} = 0$ when either $1 le i,j le m+1$ or $m+2 le i,j le 2m+2$, and
$$
b_{ij} = a_{i,m+1+j}
$$
is a skew-symmetric matrix. It remains to note that
$$
det(A) = pm det(B)^2,
$$
and that a skew-symmetric matrix of odd size is always degenerate.
So, if I got correctly you argument, $B$ is an $(m+1)times (m+1)$ matrix. If $m$ is odd then $m+1$ is even, $det(B)neq 0$ and so $det(A)neq 0$ and the quadric is smooth. If $m$ is even then $m+1$ is odd, $det(B)= 0$ and so $det(A)= 0$ and the quadric is singular. Is this right?
– SMB
Nov 17 at 21:35
Yes, precisely.
– Sasha
Nov 17 at 21:36
Thank you very much.
– SMB
Nov 17 at 21:58
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
This is true when $m$ is odd and false otherwise.
Indeed, let $mathbb{P}^{2m+1} = mathbb{P}(V)$. Assuming $H_1 cap H_2 = emptyset$ (by genericity), we have $H_1 = mathbb{P}(V_1)$, $H_2 = mathbb{P}(V_2)$ and
$$
V = V_1 oplus V_2.
$$
Furthermore, assuming $H_3 cap H_1 = H_3 cap H_2 = emptyset$, we see that $H_3 = mathbb{P}(V_3)$, where $V_3 subset V_1 oplus V_2$ is the graph of an isomorphism $V_1 to V_2$. Thus, choosing bases appropriately we may assume
$$
V_1 = langle e_1,dots,e_{m+1} rangle,quad
V_2 = langle e_{m+2},dots,e_{2m+2} rangle,quad
V_3 = langle e_1 + e_{m+2}, dots, e_{m+1} + e_{2m+2} rangle.
$$
Let $A = (a_{ij})$ be the matrix of the quadric $Q$. It follows that
$a_{ij} = 0$ when either $1 le i,j le m+1$ or $m+2 le i,j le 2m+2$, and
$$
b_{ij} = a_{i,m+1+j}
$$
is a skew-symmetric matrix. It remains to note that
$$
det(A) = pm det(B)^2,
$$
and that a skew-symmetric matrix of odd size is always degenerate.
This is true when $m$ is odd and false otherwise.
Indeed, let $mathbb{P}^{2m+1} = mathbb{P}(V)$. Assuming $H_1 cap H_2 = emptyset$ (by genericity), we have $H_1 = mathbb{P}(V_1)$, $H_2 = mathbb{P}(V_2)$ and
$$
V = V_1 oplus V_2.
$$
Furthermore, assuming $H_3 cap H_1 = H_3 cap H_2 = emptyset$, we see that $H_3 = mathbb{P}(V_3)$, where $V_3 subset V_1 oplus V_2$ is the graph of an isomorphism $V_1 to V_2$. Thus, choosing bases appropriately we may assume
$$
V_1 = langle e_1,dots,e_{m+1} rangle,quad
V_2 = langle e_{m+2},dots,e_{2m+2} rangle,quad
V_3 = langle e_1 + e_{m+2}, dots, e_{m+1} + e_{2m+2} rangle.
$$
Let $A = (a_{ij})$ be the matrix of the quadric $Q$. It follows that
$a_{ij} = 0$ when either $1 le i,j le m+1$ or $m+2 le i,j le 2m+2$, and
$$
b_{ij} = a_{i,m+1+j}
$$
is a skew-symmetric matrix. It remains to note that
$$
det(A) = pm det(B)^2,
$$
and that a skew-symmetric matrix of odd size is always degenerate.
edited Nov 17 at 21:37
answered Nov 17 at 21:20
Sasha
19.8k22652
19.8k22652
So, if I got correctly you argument, $B$ is an $(m+1)times (m+1)$ matrix. If $m$ is odd then $m+1$ is even, $det(B)neq 0$ and so $det(A)neq 0$ and the quadric is smooth. If $m$ is even then $m+1$ is odd, $det(B)= 0$ and so $det(A)= 0$ and the quadric is singular. Is this right?
– SMB
Nov 17 at 21:35
Yes, precisely.
– Sasha
Nov 17 at 21:36
Thank you very much.
– SMB
Nov 17 at 21:58
add a comment |
So, if I got correctly you argument, $B$ is an $(m+1)times (m+1)$ matrix. If $m$ is odd then $m+1$ is even, $det(B)neq 0$ and so $det(A)neq 0$ and the quadric is smooth. If $m$ is even then $m+1$ is odd, $det(B)= 0$ and so $det(A)= 0$ and the quadric is singular. Is this right?
– SMB
Nov 17 at 21:35
Yes, precisely.
– Sasha
Nov 17 at 21:36
Thank you very much.
– SMB
Nov 17 at 21:58
So, if I got correctly you argument, $B$ is an $(m+1)times (m+1)$ matrix. If $m$ is odd then $m+1$ is even, $det(B)neq 0$ and so $det(A)neq 0$ and the quadric is smooth. If $m$ is even then $m+1$ is odd, $det(B)= 0$ and so $det(A)= 0$ and the quadric is singular. Is this right?
– SMB
Nov 17 at 21:35
So, if I got correctly you argument, $B$ is an $(m+1)times (m+1)$ matrix. If $m$ is odd then $m+1$ is even, $det(B)neq 0$ and so $det(A)neq 0$ and the quadric is smooth. If $m$ is even then $m+1$ is odd, $det(B)= 0$ and so $det(A)= 0$ and the quadric is singular. Is this right?
– SMB
Nov 17 at 21:35
Yes, precisely.
– Sasha
Nov 17 at 21:36
Yes, precisely.
– Sasha
Nov 17 at 21:36
Thank you very much.
– SMB
Nov 17 at 21:58
Thank you very much.
– SMB
Nov 17 at 21:58
add a comment |
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