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Intersection of lines in $ mathbbP^3 $ can be given as zero locus of homogeneous linear polynomial in $ mathbbP^5 $
Every curve in $mathbbA^3$ is the zero-locus of $3$ polynomialsWorking out an example of a Chern classGeneric fibers of the Gauss map are linear spacesOpen cover of a grassmannianFinding ideal generators of the ideal of relations among a certain set of polynomials$mathbb P^1(mathbb C)$ inside $Gr(2,4)$Linear polynomial whose multiplication with a given quadratic polynomial vanishes in the Jacobi ringDimension of irreducible components of single homogeneous polynomialWhich subspaces of exterior power have decomposable bases?Projection of singular del Pezzo from the sigunlar point
$begingroup$
I'm currently stuck at Exercise 8.19 b) in the notes to Algebraic Geometry from Gathmann
Let $L subset mathbbP^3$ be an arbitrary line. Show that the set of lines in $mathbbP^3$ that intersect $L$, considered as a subset of $G(2,4) subset mathbbP^5 $, is the zero locus of a homogeneous linear polynomial.
I tried solving the problem by first fixing a line in $mathbbP^3$ and looking at all lines that intersect it and then derive the form of the needed linear homogeneous linear polynomial. But have no idea how.
So i started with fixing $L$ as $Lin(e_1 ,e_2) $ this leads to coordinate $(1:0:0:0:0:0)$ in $mathbbP^5$ with Plücker embedding. Obviously the lines $(e_1,e_3),(e_1,e_4),(e_2,e_3),(e_2,e_4),(e_3,e_4)$ intersect $L$ and have coordinates $(0:1:0:0:0:0),...,(0:0:0:0:1:0)$.
Now i need to also include lines that are linear combinations of my basis vectors for example $(e_1+e_2,e_2+e_3)$. My question is how to find all of these lines and how to build a homogeneous linear polynomial from them.
algebraic-geometry grassmannian
$endgroup$
add a comment |
$begingroup$
I'm currently stuck at Exercise 8.19 b) in the notes to Algebraic Geometry from Gathmann
Let $L subset mathbbP^3$ be an arbitrary line. Show that the set of lines in $mathbbP^3$ that intersect $L$, considered as a subset of $G(2,4) subset mathbbP^5 $, is the zero locus of a homogeneous linear polynomial.
I tried solving the problem by first fixing a line in $mathbbP^3$ and looking at all lines that intersect it and then derive the form of the needed linear homogeneous linear polynomial. But have no idea how.
So i started with fixing $L$ as $Lin(e_1 ,e_2) $ this leads to coordinate $(1:0:0:0:0:0)$ in $mathbbP^5$ with Plücker embedding. Obviously the lines $(e_1,e_3),(e_1,e_4),(e_2,e_3),(e_2,e_4),(e_3,e_4)$ intersect $L$ and have coordinates $(0:1:0:0:0:0),...,(0:0:0:0:1:0)$.
Now i need to also include lines that are linear combinations of my basis vectors for example $(e_1+e_2,e_2+e_3)$. My question is how to find all of these lines and how to build a homogeneous linear polynomial from them.
algebraic-geometry grassmannian
$endgroup$
add a comment |
$begingroup$
I'm currently stuck at Exercise 8.19 b) in the notes to Algebraic Geometry from Gathmann
Let $L subset mathbbP^3$ be an arbitrary line. Show that the set of lines in $mathbbP^3$ that intersect $L$, considered as a subset of $G(2,4) subset mathbbP^5 $, is the zero locus of a homogeneous linear polynomial.
I tried solving the problem by first fixing a line in $mathbbP^3$ and looking at all lines that intersect it and then derive the form of the needed linear homogeneous linear polynomial. But have no idea how.
So i started with fixing $L$ as $Lin(e_1 ,e_2) $ this leads to coordinate $(1:0:0:0:0:0)$ in $mathbbP^5$ with Plücker embedding. Obviously the lines $(e_1,e_3),(e_1,e_4),(e_2,e_3),(e_2,e_4),(e_3,e_4)$ intersect $L$ and have coordinates $(0:1:0:0:0:0),...,(0:0:0:0:1:0)$.
Now i need to also include lines that are linear combinations of my basis vectors for example $(e_1+e_2,e_2+e_3)$. My question is how to find all of these lines and how to build a homogeneous linear polynomial from them.
algebraic-geometry grassmannian
$endgroup$
I'm currently stuck at Exercise 8.19 b) in the notes to Algebraic Geometry from Gathmann
Let $L subset mathbbP^3$ be an arbitrary line. Show that the set of lines in $mathbbP^3$ that intersect $L$, considered as a subset of $G(2,4) subset mathbbP^5 $, is the zero locus of a homogeneous linear polynomial.
I tried solving the problem by first fixing a line in $mathbbP^3$ and looking at all lines that intersect it and then derive the form of the needed linear homogeneous linear polynomial. But have no idea how.
So i started with fixing $L$ as $Lin(e_1 ,e_2) $ this leads to coordinate $(1:0:0:0:0:0)$ in $mathbbP^5$ with Plücker embedding. Obviously the lines $(e_1,e_3),(e_1,e_4),(e_2,e_3),(e_2,e_4),(e_3,e_4)$ intersect $L$ and have coordinates $(0:1:0:0:0:0),...,(0:0:0:0:1:0)$.
Now i need to also include lines that are linear combinations of my basis vectors for example $(e_1+e_2,e_2+e_3)$. My question is how to find all of these lines and how to build a homogeneous linear polynomial from them.
algebraic-geometry grassmannian
algebraic-geometry grassmannian
asked Mar 28 at 21:26
ASPASP
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Let $L$ be the line given by the projectivization of the span of $e_1, e_2$. In other words, $L$ is the vanishing locus of $e_3$ and $e_4$. Let $v, w in (mathbbC^4)^*$ be linearly independent dual vectors (i.e., homogeneous linear forms, whose vanishing loci are planes), and let $L'$ be the line whose image in the Plucker embedding is given by $v wedge w$ (i.e., $L'$ is the vanishing locus of the functionals $v,w$). Then $L cap L'$ is nonempty precisely when there exists $[a : b ] in mathbbP^1$ such that $v(ae_1 + be_2) = w(ae_1 +be_2) = 0$. Let $v = sum_i = 1^4 c_ie_i^*$ and $w = sum_i = 1^4 d_ie_i^*$. Then $v(ae_1 + be_2) = w(ae_1 +be_2) = 0$ is equivalent to saying that $ac_1 + bc_2 = ad_1 + bd_2 = 0$, so $L cap L'$ is nonempty precisely when the matrix $left[beginarraycc c_1 & c_2 \ d_1 & d_2endarrayright]$ has vanishing determinant, which occurs when $c_1d_2 - d_1c_2 = 0$. But the $e_1^* wedge e_2^*$ coordinate of $v wedge w$ is given by $c_1d_2 - d_1c_2$, so $L cap L'$ is nonempty precisely when $L'$ has vanishing $e_1^* wedge e_2^*$ coordinate under the Plucker embedding, and this is obviously a hyperplane condition on the $mathbbP^5$ in which $G(2,4)$ sits.
$endgroup$
$begingroup$
Hi, thank you for answering my question. Now i have two questions with regards two to answer. First im unfamiliar with the notation $e_i^*$ what is the difference to $e_i$. Second you wrote that $L cap L' $ is nonempty when $e_1^* land e_2^*$ vanished under the Plücker embedding. But $L cap L$ is nonempty and $e_1 land e_2 neq 0 $ and would this not also mean that $L cap (e_3 land e_4) $ should have a nonempty intersection.
$endgroup$
– ASP
Mar 29 at 9:19
1
$begingroup$
$e_i^*$ simply means the linear function $mathbbC^4 to mathbbC$ that sends $e_i$ to $1$ and $e_j$ to $0$ for all $j neq i$. The Plucker embedding sends a line in $mathbbP^3$ defined as the vanishing locus of a pair of linear forms $v,w$ to the vector $v wedge w$. I mean that $L cap L'$ is nonempty when the coordinate of the basis vector $e_1^* wedge e_2^*$ in $v wedge w$ is $0$ (i.e., write $v wedge w$ in the basis $e_i^* wedge e_j^*: 1 leq i < j leq 4$, and the coefficient of $e_1^* wedge e_2^*$ has to be $0$). I am not saying that $e_1 wedge e_2 = 0$.
$endgroup$
– Ashvin Swaminathan
Mar 29 at 11:32
$begingroup$
I try understand your explanation by running through it for fixed $ L = Lin(e_1,e_2) $ and $L' = Lin(e_1+e_4, e_2 -e_4) $ correct me if i go wrong somewhere. Then $ Lcap L' $ is nonempty when $v(ae_1+be_2)=w(ae_1+be_2)=0$ , in this case if i take as base for my dual space $e_1,e_2,e_3,e_4$. This should be $ a = b = 0$ , but i should have something wrong here since $(0:0) notin mathbbP^1$.
$endgroup$
– ASP
Mar 31 at 1:03
1
$begingroup$
I think you may have mistakenly taken $v = e_1^* + e_4^*$ and $w = e_2^* - e_4^*$. In this case you ought to take, for example, $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$, because you need $L'$ to be the intersection of the vanishing loci of $v$ and $w$. Note that with my choice of $v$ and $w$, you get $v(e_1 + e_4) = v(e_2 - e_4) = w(e_1 + e_4) = w(e_2 - e_4) = 0$. So $v(ae_1+be_2) = a-b$ and $w(ae_1+be_2) = 0$. So taking $a = b neq 0$ works.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 1:33
1
$begingroup$
(1) Notice that every point on $operatornameLin(e_1, e_2)$ corresponds to a linear combination $ae_1 + be_2$, so for that point to lie on the line $L'$, it must be in the vanishing locus of both $v,w$, because the line $L'$ is defined to be the vanishing locus of $v$ and $w$. (2) You just need to choose any $2$ different homogeneous linear forms vanishing on both $e_1 + e_4$ and $e_2 - e_4$. After some playing around, I found the forms $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$ in my previous comment.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 23:44
|
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$begingroup$
Let $L$ be the line given by the projectivization of the span of $e_1, e_2$. In other words, $L$ is the vanishing locus of $e_3$ and $e_4$. Let $v, w in (mathbbC^4)^*$ be linearly independent dual vectors (i.e., homogeneous linear forms, whose vanishing loci are planes), and let $L'$ be the line whose image in the Plucker embedding is given by $v wedge w$ (i.e., $L'$ is the vanishing locus of the functionals $v,w$). Then $L cap L'$ is nonempty precisely when there exists $[a : b ] in mathbbP^1$ such that $v(ae_1 + be_2) = w(ae_1 +be_2) = 0$. Let $v = sum_i = 1^4 c_ie_i^*$ and $w = sum_i = 1^4 d_ie_i^*$. Then $v(ae_1 + be_2) = w(ae_1 +be_2) = 0$ is equivalent to saying that $ac_1 + bc_2 = ad_1 + bd_2 = 0$, so $L cap L'$ is nonempty precisely when the matrix $left[beginarraycc c_1 & c_2 \ d_1 & d_2endarrayright]$ has vanishing determinant, which occurs when $c_1d_2 - d_1c_2 = 0$. But the $e_1^* wedge e_2^*$ coordinate of $v wedge w$ is given by $c_1d_2 - d_1c_2$, so $L cap L'$ is nonempty precisely when $L'$ has vanishing $e_1^* wedge e_2^*$ coordinate under the Plucker embedding, and this is obviously a hyperplane condition on the $mathbbP^5$ in which $G(2,4)$ sits.
$endgroup$
$begingroup$
Hi, thank you for answering my question. Now i have two questions with regards two to answer. First im unfamiliar with the notation $e_i^*$ what is the difference to $e_i$. Second you wrote that $L cap L' $ is nonempty when $e_1^* land e_2^*$ vanished under the Plücker embedding. But $L cap L$ is nonempty and $e_1 land e_2 neq 0 $ and would this not also mean that $L cap (e_3 land e_4) $ should have a nonempty intersection.
$endgroup$
– ASP
Mar 29 at 9:19
1
$begingroup$
$e_i^*$ simply means the linear function $mathbbC^4 to mathbbC$ that sends $e_i$ to $1$ and $e_j$ to $0$ for all $j neq i$. The Plucker embedding sends a line in $mathbbP^3$ defined as the vanishing locus of a pair of linear forms $v,w$ to the vector $v wedge w$. I mean that $L cap L'$ is nonempty when the coordinate of the basis vector $e_1^* wedge e_2^*$ in $v wedge w$ is $0$ (i.e., write $v wedge w$ in the basis $e_i^* wedge e_j^*: 1 leq i < j leq 4$, and the coefficient of $e_1^* wedge e_2^*$ has to be $0$). I am not saying that $e_1 wedge e_2 = 0$.
$endgroup$
– Ashvin Swaminathan
Mar 29 at 11:32
$begingroup$
I try understand your explanation by running through it for fixed $ L = Lin(e_1,e_2) $ and $L' = Lin(e_1+e_4, e_2 -e_4) $ correct me if i go wrong somewhere. Then $ Lcap L' $ is nonempty when $v(ae_1+be_2)=w(ae_1+be_2)=0$ , in this case if i take as base for my dual space $e_1,e_2,e_3,e_4$. This should be $ a = b = 0$ , but i should have something wrong here since $(0:0) notin mathbbP^1$.
$endgroup$
– ASP
Mar 31 at 1:03
1
$begingroup$
I think you may have mistakenly taken $v = e_1^* + e_4^*$ and $w = e_2^* - e_4^*$. In this case you ought to take, for example, $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$, because you need $L'$ to be the intersection of the vanishing loci of $v$ and $w$. Note that with my choice of $v$ and $w$, you get $v(e_1 + e_4) = v(e_2 - e_4) = w(e_1 + e_4) = w(e_2 - e_4) = 0$. So $v(ae_1+be_2) = a-b$ and $w(ae_1+be_2) = 0$. So taking $a = b neq 0$ works.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 1:33
1
$begingroup$
(1) Notice that every point on $operatornameLin(e_1, e_2)$ corresponds to a linear combination $ae_1 + be_2$, so for that point to lie on the line $L'$, it must be in the vanishing locus of both $v,w$, because the line $L'$ is defined to be the vanishing locus of $v$ and $w$. (2) You just need to choose any $2$ different homogeneous linear forms vanishing on both $e_1 + e_4$ and $e_2 - e_4$. After some playing around, I found the forms $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$ in my previous comment.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 23:44
|
show 1 more comment
$begingroup$
Let $L$ be the line given by the projectivization of the span of $e_1, e_2$. In other words, $L$ is the vanishing locus of $e_3$ and $e_4$. Let $v, w in (mathbbC^4)^*$ be linearly independent dual vectors (i.e., homogeneous linear forms, whose vanishing loci are planes), and let $L'$ be the line whose image in the Plucker embedding is given by $v wedge w$ (i.e., $L'$ is the vanishing locus of the functionals $v,w$). Then $L cap L'$ is nonempty precisely when there exists $[a : b ] in mathbbP^1$ such that $v(ae_1 + be_2) = w(ae_1 +be_2) = 0$. Let $v = sum_i = 1^4 c_ie_i^*$ and $w = sum_i = 1^4 d_ie_i^*$. Then $v(ae_1 + be_2) = w(ae_1 +be_2) = 0$ is equivalent to saying that $ac_1 + bc_2 = ad_1 + bd_2 = 0$, so $L cap L'$ is nonempty precisely when the matrix $left[beginarraycc c_1 & c_2 \ d_1 & d_2endarrayright]$ has vanishing determinant, which occurs when $c_1d_2 - d_1c_2 = 0$. But the $e_1^* wedge e_2^*$ coordinate of $v wedge w$ is given by $c_1d_2 - d_1c_2$, so $L cap L'$ is nonempty precisely when $L'$ has vanishing $e_1^* wedge e_2^*$ coordinate under the Plucker embedding, and this is obviously a hyperplane condition on the $mathbbP^5$ in which $G(2,4)$ sits.
$endgroup$
$begingroup$
Hi, thank you for answering my question. Now i have two questions with regards two to answer. First im unfamiliar with the notation $e_i^*$ what is the difference to $e_i$. Second you wrote that $L cap L' $ is nonempty when $e_1^* land e_2^*$ vanished under the Plücker embedding. But $L cap L$ is nonempty and $e_1 land e_2 neq 0 $ and would this not also mean that $L cap (e_3 land e_4) $ should have a nonempty intersection.
$endgroup$
– ASP
Mar 29 at 9:19
1
$begingroup$
$e_i^*$ simply means the linear function $mathbbC^4 to mathbbC$ that sends $e_i$ to $1$ and $e_j$ to $0$ for all $j neq i$. The Plucker embedding sends a line in $mathbbP^3$ defined as the vanishing locus of a pair of linear forms $v,w$ to the vector $v wedge w$. I mean that $L cap L'$ is nonempty when the coordinate of the basis vector $e_1^* wedge e_2^*$ in $v wedge w$ is $0$ (i.e., write $v wedge w$ in the basis $e_i^* wedge e_j^*: 1 leq i < j leq 4$, and the coefficient of $e_1^* wedge e_2^*$ has to be $0$). I am not saying that $e_1 wedge e_2 = 0$.
$endgroup$
– Ashvin Swaminathan
Mar 29 at 11:32
$begingroup$
I try understand your explanation by running through it for fixed $ L = Lin(e_1,e_2) $ and $L' = Lin(e_1+e_4, e_2 -e_4) $ correct me if i go wrong somewhere. Then $ Lcap L' $ is nonempty when $v(ae_1+be_2)=w(ae_1+be_2)=0$ , in this case if i take as base for my dual space $e_1,e_2,e_3,e_4$. This should be $ a = b = 0$ , but i should have something wrong here since $(0:0) notin mathbbP^1$.
$endgroup$
– ASP
Mar 31 at 1:03
1
$begingroup$
I think you may have mistakenly taken $v = e_1^* + e_4^*$ and $w = e_2^* - e_4^*$. In this case you ought to take, for example, $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$, because you need $L'$ to be the intersection of the vanishing loci of $v$ and $w$. Note that with my choice of $v$ and $w$, you get $v(e_1 + e_4) = v(e_2 - e_4) = w(e_1 + e_4) = w(e_2 - e_4) = 0$. So $v(ae_1+be_2) = a-b$ and $w(ae_1+be_2) = 0$. So taking $a = b neq 0$ works.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 1:33
1
$begingroup$
(1) Notice that every point on $operatornameLin(e_1, e_2)$ corresponds to a linear combination $ae_1 + be_2$, so for that point to lie on the line $L'$, it must be in the vanishing locus of both $v,w$, because the line $L'$ is defined to be the vanishing locus of $v$ and $w$. (2) You just need to choose any $2$ different homogeneous linear forms vanishing on both $e_1 + e_4$ and $e_2 - e_4$. After some playing around, I found the forms $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$ in my previous comment.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 23:44
|
show 1 more comment
$begingroup$
Let $L$ be the line given by the projectivization of the span of $e_1, e_2$. In other words, $L$ is the vanishing locus of $e_3$ and $e_4$. Let $v, w in (mathbbC^4)^*$ be linearly independent dual vectors (i.e., homogeneous linear forms, whose vanishing loci are planes), and let $L'$ be the line whose image in the Plucker embedding is given by $v wedge w$ (i.e., $L'$ is the vanishing locus of the functionals $v,w$). Then $L cap L'$ is nonempty precisely when there exists $[a : b ] in mathbbP^1$ such that $v(ae_1 + be_2) = w(ae_1 +be_2) = 0$. Let $v = sum_i = 1^4 c_ie_i^*$ and $w = sum_i = 1^4 d_ie_i^*$. Then $v(ae_1 + be_2) = w(ae_1 +be_2) = 0$ is equivalent to saying that $ac_1 + bc_2 = ad_1 + bd_2 = 0$, so $L cap L'$ is nonempty precisely when the matrix $left[beginarraycc c_1 & c_2 \ d_1 & d_2endarrayright]$ has vanishing determinant, which occurs when $c_1d_2 - d_1c_2 = 0$. But the $e_1^* wedge e_2^*$ coordinate of $v wedge w$ is given by $c_1d_2 - d_1c_2$, so $L cap L'$ is nonempty precisely when $L'$ has vanishing $e_1^* wedge e_2^*$ coordinate under the Plucker embedding, and this is obviously a hyperplane condition on the $mathbbP^5$ in which $G(2,4)$ sits.
$endgroup$
Let $L$ be the line given by the projectivization of the span of $e_1, e_2$. In other words, $L$ is the vanishing locus of $e_3$ and $e_4$. Let $v, w in (mathbbC^4)^*$ be linearly independent dual vectors (i.e., homogeneous linear forms, whose vanishing loci are planes), and let $L'$ be the line whose image in the Plucker embedding is given by $v wedge w$ (i.e., $L'$ is the vanishing locus of the functionals $v,w$). Then $L cap L'$ is nonempty precisely when there exists $[a : b ] in mathbbP^1$ such that $v(ae_1 + be_2) = w(ae_1 +be_2) = 0$. Let $v = sum_i = 1^4 c_ie_i^*$ and $w = sum_i = 1^4 d_ie_i^*$. Then $v(ae_1 + be_2) = w(ae_1 +be_2) = 0$ is equivalent to saying that $ac_1 + bc_2 = ad_1 + bd_2 = 0$, so $L cap L'$ is nonempty precisely when the matrix $left[beginarraycc c_1 & c_2 \ d_1 & d_2endarrayright]$ has vanishing determinant, which occurs when $c_1d_2 - d_1c_2 = 0$. But the $e_1^* wedge e_2^*$ coordinate of $v wedge w$ is given by $c_1d_2 - d_1c_2$, so $L cap L'$ is nonempty precisely when $L'$ has vanishing $e_1^* wedge e_2^*$ coordinate under the Plucker embedding, and this is obviously a hyperplane condition on the $mathbbP^5$ in which $G(2,4)$ sits.
edited Mar 29 at 21:50
answered Mar 28 at 23:43
Ashvin SwaminathanAshvin Swaminathan
1,695520
1,695520
$begingroup$
Hi, thank you for answering my question. Now i have two questions with regards two to answer. First im unfamiliar with the notation $e_i^*$ what is the difference to $e_i$. Second you wrote that $L cap L' $ is nonempty when $e_1^* land e_2^*$ vanished under the Plücker embedding. But $L cap L$ is nonempty and $e_1 land e_2 neq 0 $ and would this not also mean that $L cap (e_3 land e_4) $ should have a nonempty intersection.
$endgroup$
– ASP
Mar 29 at 9:19
1
$begingroup$
$e_i^*$ simply means the linear function $mathbbC^4 to mathbbC$ that sends $e_i$ to $1$ and $e_j$ to $0$ for all $j neq i$. The Plucker embedding sends a line in $mathbbP^3$ defined as the vanishing locus of a pair of linear forms $v,w$ to the vector $v wedge w$. I mean that $L cap L'$ is nonempty when the coordinate of the basis vector $e_1^* wedge e_2^*$ in $v wedge w$ is $0$ (i.e., write $v wedge w$ in the basis $e_i^* wedge e_j^*: 1 leq i < j leq 4$, and the coefficient of $e_1^* wedge e_2^*$ has to be $0$). I am not saying that $e_1 wedge e_2 = 0$.
$endgroup$
– Ashvin Swaminathan
Mar 29 at 11:32
$begingroup$
I try understand your explanation by running through it for fixed $ L = Lin(e_1,e_2) $ and $L' = Lin(e_1+e_4, e_2 -e_4) $ correct me if i go wrong somewhere. Then $ Lcap L' $ is nonempty when $v(ae_1+be_2)=w(ae_1+be_2)=0$ , in this case if i take as base for my dual space $e_1,e_2,e_3,e_4$. This should be $ a = b = 0$ , but i should have something wrong here since $(0:0) notin mathbbP^1$.
$endgroup$
– ASP
Mar 31 at 1:03
1
$begingroup$
I think you may have mistakenly taken $v = e_1^* + e_4^*$ and $w = e_2^* - e_4^*$. In this case you ought to take, for example, $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$, because you need $L'$ to be the intersection of the vanishing loci of $v$ and $w$. Note that with my choice of $v$ and $w$, you get $v(e_1 + e_4) = v(e_2 - e_4) = w(e_1 + e_4) = w(e_2 - e_4) = 0$. So $v(ae_1+be_2) = a-b$ and $w(ae_1+be_2) = 0$. So taking $a = b neq 0$ works.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 1:33
1
$begingroup$
(1) Notice that every point on $operatornameLin(e_1, e_2)$ corresponds to a linear combination $ae_1 + be_2$, so for that point to lie on the line $L'$, it must be in the vanishing locus of both $v,w$, because the line $L'$ is defined to be the vanishing locus of $v$ and $w$. (2) You just need to choose any $2$ different homogeneous linear forms vanishing on both $e_1 + e_4$ and $e_2 - e_4$. After some playing around, I found the forms $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$ in my previous comment.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 23:44
|
show 1 more comment
$begingroup$
Hi, thank you for answering my question. Now i have two questions with regards two to answer. First im unfamiliar with the notation $e_i^*$ what is the difference to $e_i$. Second you wrote that $L cap L' $ is nonempty when $e_1^* land e_2^*$ vanished under the Plücker embedding. But $L cap L$ is nonempty and $e_1 land e_2 neq 0 $ and would this not also mean that $L cap (e_3 land e_4) $ should have a nonempty intersection.
$endgroup$
– ASP
Mar 29 at 9:19
1
$begingroup$
$e_i^*$ simply means the linear function $mathbbC^4 to mathbbC$ that sends $e_i$ to $1$ and $e_j$ to $0$ for all $j neq i$. The Plucker embedding sends a line in $mathbbP^3$ defined as the vanishing locus of a pair of linear forms $v,w$ to the vector $v wedge w$. I mean that $L cap L'$ is nonempty when the coordinate of the basis vector $e_1^* wedge e_2^*$ in $v wedge w$ is $0$ (i.e., write $v wedge w$ in the basis $e_i^* wedge e_j^*: 1 leq i < j leq 4$, and the coefficient of $e_1^* wedge e_2^*$ has to be $0$). I am not saying that $e_1 wedge e_2 = 0$.
$endgroup$
– Ashvin Swaminathan
Mar 29 at 11:32
$begingroup$
I try understand your explanation by running through it for fixed $ L = Lin(e_1,e_2) $ and $L' = Lin(e_1+e_4, e_2 -e_4) $ correct me if i go wrong somewhere. Then $ Lcap L' $ is nonempty when $v(ae_1+be_2)=w(ae_1+be_2)=0$ , in this case if i take as base for my dual space $e_1,e_2,e_3,e_4$. This should be $ a = b = 0$ , but i should have something wrong here since $(0:0) notin mathbbP^1$.
$endgroup$
– ASP
Mar 31 at 1:03
1
$begingroup$
I think you may have mistakenly taken $v = e_1^* + e_4^*$ and $w = e_2^* - e_4^*$. In this case you ought to take, for example, $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$, because you need $L'$ to be the intersection of the vanishing loci of $v$ and $w$. Note that with my choice of $v$ and $w$, you get $v(e_1 + e_4) = v(e_2 - e_4) = w(e_1 + e_4) = w(e_2 - e_4) = 0$. So $v(ae_1+be_2) = a-b$ and $w(ae_1+be_2) = 0$. So taking $a = b neq 0$ works.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 1:33
1
$begingroup$
(1) Notice that every point on $operatornameLin(e_1, e_2)$ corresponds to a linear combination $ae_1 + be_2$, so for that point to lie on the line $L'$, it must be in the vanishing locus of both $v,w$, because the line $L'$ is defined to be the vanishing locus of $v$ and $w$. (2) You just need to choose any $2$ different homogeneous linear forms vanishing on both $e_1 + e_4$ and $e_2 - e_4$. After some playing around, I found the forms $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$ in my previous comment.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 23:44
$begingroup$
Hi, thank you for answering my question. Now i have two questions with regards two to answer. First im unfamiliar with the notation $e_i^*$ what is the difference to $e_i$. Second you wrote that $L cap L' $ is nonempty when $e_1^* land e_2^*$ vanished under the Plücker embedding. But $L cap L$ is nonempty and $e_1 land e_2 neq 0 $ and would this not also mean that $L cap (e_3 land e_4) $ should have a nonempty intersection.
$endgroup$
– ASP
Mar 29 at 9:19
$begingroup$
Hi, thank you for answering my question. Now i have two questions with regards two to answer. First im unfamiliar with the notation $e_i^*$ what is the difference to $e_i$. Second you wrote that $L cap L' $ is nonempty when $e_1^* land e_2^*$ vanished under the Plücker embedding. But $L cap L$ is nonempty and $e_1 land e_2 neq 0 $ and would this not also mean that $L cap (e_3 land e_4) $ should have a nonempty intersection.
$endgroup$
– ASP
Mar 29 at 9:19
1
1
$begingroup$
$e_i^*$ simply means the linear function $mathbbC^4 to mathbbC$ that sends $e_i$ to $1$ and $e_j$ to $0$ for all $j neq i$. The Plucker embedding sends a line in $mathbbP^3$ defined as the vanishing locus of a pair of linear forms $v,w$ to the vector $v wedge w$. I mean that $L cap L'$ is nonempty when the coordinate of the basis vector $e_1^* wedge e_2^*$ in $v wedge w$ is $0$ (i.e., write $v wedge w$ in the basis $e_i^* wedge e_j^*: 1 leq i < j leq 4$, and the coefficient of $e_1^* wedge e_2^*$ has to be $0$). I am not saying that $e_1 wedge e_2 = 0$.
$endgroup$
– Ashvin Swaminathan
Mar 29 at 11:32
$begingroup$
$e_i^*$ simply means the linear function $mathbbC^4 to mathbbC$ that sends $e_i$ to $1$ and $e_j$ to $0$ for all $j neq i$. The Plucker embedding sends a line in $mathbbP^3$ defined as the vanishing locus of a pair of linear forms $v,w$ to the vector $v wedge w$. I mean that $L cap L'$ is nonempty when the coordinate of the basis vector $e_1^* wedge e_2^*$ in $v wedge w$ is $0$ (i.e., write $v wedge w$ in the basis $e_i^* wedge e_j^*: 1 leq i < j leq 4$, and the coefficient of $e_1^* wedge e_2^*$ has to be $0$). I am not saying that $e_1 wedge e_2 = 0$.
$endgroup$
– Ashvin Swaminathan
Mar 29 at 11:32
$begingroup$
I try understand your explanation by running through it for fixed $ L = Lin(e_1,e_2) $ and $L' = Lin(e_1+e_4, e_2 -e_4) $ correct me if i go wrong somewhere. Then $ Lcap L' $ is nonempty when $v(ae_1+be_2)=w(ae_1+be_2)=0$ , in this case if i take as base for my dual space $e_1,e_2,e_3,e_4$. This should be $ a = b = 0$ , but i should have something wrong here since $(0:0) notin mathbbP^1$.
$endgroup$
– ASP
Mar 31 at 1:03
$begingroup$
I try understand your explanation by running through it for fixed $ L = Lin(e_1,e_2) $ and $L' = Lin(e_1+e_4, e_2 -e_4) $ correct me if i go wrong somewhere. Then $ Lcap L' $ is nonempty when $v(ae_1+be_2)=w(ae_1+be_2)=0$ , in this case if i take as base for my dual space $e_1,e_2,e_3,e_4$. This should be $ a = b = 0$ , but i should have something wrong here since $(0:0) notin mathbbP^1$.
$endgroup$
– ASP
Mar 31 at 1:03
1
1
$begingroup$
I think you may have mistakenly taken $v = e_1^* + e_4^*$ and $w = e_2^* - e_4^*$. In this case you ought to take, for example, $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$, because you need $L'$ to be the intersection of the vanishing loci of $v$ and $w$. Note that with my choice of $v$ and $w$, you get $v(e_1 + e_4) = v(e_2 - e_4) = w(e_1 + e_4) = w(e_2 - e_4) = 0$. So $v(ae_1+be_2) = a-b$ and $w(ae_1+be_2) = 0$. So taking $a = b neq 0$ works.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 1:33
$begingroup$
I think you may have mistakenly taken $v = e_1^* + e_4^*$ and $w = e_2^* - e_4^*$. In this case you ought to take, for example, $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$, because you need $L'$ to be the intersection of the vanishing loci of $v$ and $w$. Note that with my choice of $v$ and $w$, you get $v(e_1 + e_4) = v(e_2 - e_4) = w(e_1 + e_4) = w(e_2 - e_4) = 0$. So $v(ae_1+be_2) = a-b$ and $w(ae_1+be_2) = 0$. So taking $a = b neq 0$ works.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 1:33
1
1
$begingroup$
(1) Notice that every point on $operatornameLin(e_1, e_2)$ corresponds to a linear combination $ae_1 + be_2$, so for that point to lie on the line $L'$, it must be in the vanishing locus of both $v,w$, because the line $L'$ is defined to be the vanishing locus of $v$ and $w$. (2) You just need to choose any $2$ different homogeneous linear forms vanishing on both $e_1 + e_4$ and $e_2 - e_4$. After some playing around, I found the forms $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$ in my previous comment.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 23:44
$begingroup$
(1) Notice that every point on $operatornameLin(e_1, e_2)$ corresponds to a linear combination $ae_1 + be_2$, so for that point to lie on the line $L'$, it must be in the vanishing locus of both $v,w$, because the line $L'$ is defined to be the vanishing locus of $v$ and $w$. (2) You just need to choose any $2$ different homogeneous linear forms vanishing on both $e_1 + e_4$ and $e_2 - e_4$. After some playing around, I found the forms $v = e_1^* - e_2^* - e_4^*$ and $w = e_3^*$ in my previous comment.
$endgroup$
– Ashvin Swaminathan
Mar 31 at 23:44
|
show 1 more comment
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