Plücker embedding - Two definitions
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While reading several papers on the topic of the Grassmannian, I cam about two definitions of the Plücker embedding.
One given as
$$ varphi: mathbbA^n cdot d rightarrow mathbbP^binomnd-1, A mapsto textdet(A^(I))$$
where $ A^(I)$ denotes the submatrix of $A$ obtained by choosing its $I$-th columns. (I know that I can identify the affine space of this dimension with the space of matrices of size $d times n$. Moreover, the map is only well-defined if we look at matrices of rank $d$ which gives us the connection to elements of the Grassmannian). The other one was given as
$$psi: Gr(d,n) rightarrow mathbbP^binomnd-1, U mapsto [u_1 wedge dotsc wedge u_d]$$
So my question concerns the connection of these two maps in the projective space respectively whether there is a connection at all. I assume there must be something as in the papers I've read different authors used them both for talking about the Plücker embedding.
Thank you for your help.
algebraic-geometry projective-space grassmannian
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up vote
3
down vote
favorite
While reading several papers on the topic of the Grassmannian, I cam about two definitions of the Plücker embedding.
One given as
$$ varphi: mathbbA^n cdot d rightarrow mathbbP^binomnd-1, A mapsto textdet(A^(I))$$
where $ A^(I)$ denotes the submatrix of $A$ obtained by choosing its $I$-th columns. (I know that I can identify the affine space of this dimension with the space of matrices of size $d times n$. Moreover, the map is only well-defined if we look at matrices of rank $d$ which gives us the connection to elements of the Grassmannian). The other one was given as
$$psi: Gr(d,n) rightarrow mathbbP^binomnd-1, U mapsto [u_1 wedge dotsc wedge u_d]$$
So my question concerns the connection of these two maps in the projective space respectively whether there is a connection at all. I assume there must be something as in the papers I've read different authors used them both for talking about the Plücker embedding.
Thank you for your help.
algebraic-geometry projective-space grassmannian
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
While reading several papers on the topic of the Grassmannian, I cam about two definitions of the Plücker embedding.
One given as
$$ varphi: mathbbA^n cdot d rightarrow mathbbP^binomnd-1, A mapsto textdet(A^(I))$$
where $ A^(I)$ denotes the submatrix of $A$ obtained by choosing its $I$-th columns. (I know that I can identify the affine space of this dimension with the space of matrices of size $d times n$. Moreover, the map is only well-defined if we look at matrices of rank $d$ which gives us the connection to elements of the Grassmannian). The other one was given as
$$psi: Gr(d,n) rightarrow mathbbP^binomnd-1, U mapsto [u_1 wedge dotsc wedge u_d]$$
So my question concerns the connection of these two maps in the projective space respectively whether there is a connection at all. I assume there must be something as in the papers I've read different authors used them both for talking about the Plücker embedding.
Thank you for your help.
algebraic-geometry projective-space grassmannian
While reading several papers on the topic of the Grassmannian, I cam about two definitions of the Plücker embedding.
One given as
$$ varphi: mathbbA^n cdot d rightarrow mathbbP^binomnd-1, A mapsto textdet(A^(I))$$
where $ A^(I)$ denotes the submatrix of $A$ obtained by choosing its $I$-th columns. (I know that I can identify the affine space of this dimension with the space of matrices of size $d times n$. Moreover, the map is only well-defined if we look at matrices of rank $d$ which gives us the connection to elements of the Grassmannian). The other one was given as
$$psi: Gr(d,n) rightarrow mathbbP^binomnd-1, U mapsto [u_1 wedge dotsc wedge u_d]$$
So my question concerns the connection of these two maps in the projective space respectively whether there is a connection at all. I assume there must be something as in the papers I've read different authors used them both for talking about the Plücker embedding.
Thank you for your help.
algebraic-geometry projective-space grassmannian
algebraic-geometry projective-space grassmannian
asked Sep 3 at 13:05
SallyOwens
284210
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1 Answer
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The idea here is that that the vectors $u_i, i=1,2,ldots,d,$ are the rows of the matrix $A$. More precisely, if $u_i=(a_i1,a_i2,ldots,a_in)$, then the coefficient of
$e_i_1wedge e_i_2wedgecdotswedge e_i_d$ in the wedge product $u_1wedge u_2wedgecdotswedge u_d$ is equal to $det(A^(I))$.
Consider the paper & pencil example of $n=4,d=2$ with $u_1=(a_1,a_2,a_3,a_4)$, $u_2=(b_1,b_2,b_3,b_4)$ when
$$
beginaligned
u_1wedge u_2&=(a_1e_1+a_2e_2+a_3e_3+a_4e_4)wedge(b_1e_1+b_2e_2+b_3e_3+b_4e_4)\
&=sum_i=1^4sum_j=1^4a_ib_je_iwedge e_j\
&=(a_1b_2-a_2b_1)e_1wedge e_2+(a_1b_3-a_3b_1)e_1wedge e_3+(a_1b_4-a_4b_1)e_1wedge e_4\
&+(a_2b_3-a_3b_2)e_2wedge e_3+(a_2b_4-a_4b_2)e_2wedge e_4+(a_3b_4-a_4b_3)e_3wedge e_4.
endaligned
$$
See the six $2times2$ minors of the matrix
$$
A=left(beginarraycccca_1&a_2&a_3&a_4\b_1&b_2&b_3&b_4endarrayright)
$$
emerging?!
For the purposes of introducing coordinates to the Grasmannian we associate a $d$-dimensional subspace $V$ with any matrix $dtimes n$ matrix $A$ that has $V$
as its row space. It is essential that:
- Two matrices $A$ and $A'$ are sharing the same row space if and only if there exists an invertible $dtimes d$ matrix $M$ such that $A'=MA$.
- When we calculate the wedge products of the rows, all the coordinates of the product of the rows of $A'$ are gotten by multiplying the corresponding coordinates of the product of rows of $A$ by $det M$. Implying that as homogeneous coordinates in a projective space the two wedge products refer to the same point. This means that the mapping $G(n,d)toBbbP^N$ is well defined.
- The easiest way of proving the result of the previous bullet is to prove it by direct observation for all three types of elementary matrices, and then writing $M$ as a product of elementary matrices.
Also, proving that $phi$ and $psi$ really give the same mapping may be easiest to do by induction on $d$. Then, at the induction step, you get the usual expansions of $(d+1)times (d+1)$-determinants as linear combinations of $dtimes d$-determinants.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The idea here is that that the vectors $u_i, i=1,2,ldots,d,$ are the rows of the matrix $A$. More precisely, if $u_i=(a_i1,a_i2,ldots,a_in)$, then the coefficient of
$e_i_1wedge e_i_2wedgecdotswedge e_i_d$ in the wedge product $u_1wedge u_2wedgecdotswedge u_d$ is equal to $det(A^(I))$.
Consider the paper & pencil example of $n=4,d=2$ with $u_1=(a_1,a_2,a_3,a_4)$, $u_2=(b_1,b_2,b_3,b_4)$ when
$$
beginaligned
u_1wedge u_2&=(a_1e_1+a_2e_2+a_3e_3+a_4e_4)wedge(b_1e_1+b_2e_2+b_3e_3+b_4e_4)\
&=sum_i=1^4sum_j=1^4a_ib_je_iwedge e_j\
&=(a_1b_2-a_2b_1)e_1wedge e_2+(a_1b_3-a_3b_1)e_1wedge e_3+(a_1b_4-a_4b_1)e_1wedge e_4\
&+(a_2b_3-a_3b_2)e_2wedge e_3+(a_2b_4-a_4b_2)e_2wedge e_4+(a_3b_4-a_4b_3)e_3wedge e_4.
endaligned
$$
See the six $2times2$ minors of the matrix
$$
A=left(beginarraycccca_1&a_2&a_3&a_4\b_1&b_2&b_3&b_4endarrayright)
$$
emerging?!
For the purposes of introducing coordinates to the Grasmannian we associate a $d$-dimensional subspace $V$ with any matrix $dtimes n$ matrix $A$ that has $V$
as its row space. It is essential that:
- Two matrices $A$ and $A'$ are sharing the same row space if and only if there exists an invertible $dtimes d$ matrix $M$ such that $A'=MA$.
- When we calculate the wedge products of the rows, all the coordinates of the product of the rows of $A'$ are gotten by multiplying the corresponding coordinates of the product of rows of $A$ by $det M$. Implying that as homogeneous coordinates in a projective space the two wedge products refer to the same point. This means that the mapping $G(n,d)toBbbP^N$ is well defined.
- The easiest way of proving the result of the previous bullet is to prove it by direct observation for all three types of elementary matrices, and then writing $M$ as a product of elementary matrices.
Also, proving that $phi$ and $psi$ really give the same mapping may be easiest to do by induction on $d$. Then, at the induction step, you get the usual expansions of $(d+1)times (d+1)$-determinants as linear combinations of $dtimes d$-determinants.
add a comment |Â
up vote
2
down vote
accepted
The idea here is that that the vectors $u_i, i=1,2,ldots,d,$ are the rows of the matrix $A$. More precisely, if $u_i=(a_i1,a_i2,ldots,a_in)$, then the coefficient of
$e_i_1wedge e_i_2wedgecdotswedge e_i_d$ in the wedge product $u_1wedge u_2wedgecdotswedge u_d$ is equal to $det(A^(I))$.
Consider the paper & pencil example of $n=4,d=2$ with $u_1=(a_1,a_2,a_3,a_4)$, $u_2=(b_1,b_2,b_3,b_4)$ when
$$
beginaligned
u_1wedge u_2&=(a_1e_1+a_2e_2+a_3e_3+a_4e_4)wedge(b_1e_1+b_2e_2+b_3e_3+b_4e_4)\
&=sum_i=1^4sum_j=1^4a_ib_je_iwedge e_j\
&=(a_1b_2-a_2b_1)e_1wedge e_2+(a_1b_3-a_3b_1)e_1wedge e_3+(a_1b_4-a_4b_1)e_1wedge e_4\
&+(a_2b_3-a_3b_2)e_2wedge e_3+(a_2b_4-a_4b_2)e_2wedge e_4+(a_3b_4-a_4b_3)e_3wedge e_4.
endaligned
$$
See the six $2times2$ minors of the matrix
$$
A=left(beginarraycccca_1&a_2&a_3&a_4\b_1&b_2&b_3&b_4endarrayright)
$$
emerging?!
For the purposes of introducing coordinates to the Grasmannian we associate a $d$-dimensional subspace $V$ with any matrix $dtimes n$ matrix $A$ that has $V$
as its row space. It is essential that:
- Two matrices $A$ and $A'$ are sharing the same row space if and only if there exists an invertible $dtimes d$ matrix $M$ such that $A'=MA$.
- When we calculate the wedge products of the rows, all the coordinates of the product of the rows of $A'$ are gotten by multiplying the corresponding coordinates of the product of rows of $A$ by $det M$. Implying that as homogeneous coordinates in a projective space the two wedge products refer to the same point. This means that the mapping $G(n,d)toBbbP^N$ is well defined.
- The easiest way of proving the result of the previous bullet is to prove it by direct observation for all three types of elementary matrices, and then writing $M$ as a product of elementary matrices.
Also, proving that $phi$ and $psi$ really give the same mapping may be easiest to do by induction on $d$. Then, at the induction step, you get the usual expansions of $(d+1)times (d+1)$-determinants as linear combinations of $dtimes d$-determinants.
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The idea here is that that the vectors $u_i, i=1,2,ldots,d,$ are the rows of the matrix $A$. More precisely, if $u_i=(a_i1,a_i2,ldots,a_in)$, then the coefficient of
$e_i_1wedge e_i_2wedgecdotswedge e_i_d$ in the wedge product $u_1wedge u_2wedgecdotswedge u_d$ is equal to $det(A^(I))$.
Consider the paper & pencil example of $n=4,d=2$ with $u_1=(a_1,a_2,a_3,a_4)$, $u_2=(b_1,b_2,b_3,b_4)$ when
$$
beginaligned
u_1wedge u_2&=(a_1e_1+a_2e_2+a_3e_3+a_4e_4)wedge(b_1e_1+b_2e_2+b_3e_3+b_4e_4)\
&=sum_i=1^4sum_j=1^4a_ib_je_iwedge e_j\
&=(a_1b_2-a_2b_1)e_1wedge e_2+(a_1b_3-a_3b_1)e_1wedge e_3+(a_1b_4-a_4b_1)e_1wedge e_4\
&+(a_2b_3-a_3b_2)e_2wedge e_3+(a_2b_4-a_4b_2)e_2wedge e_4+(a_3b_4-a_4b_3)e_3wedge e_4.
endaligned
$$
See the six $2times2$ minors of the matrix
$$
A=left(beginarraycccca_1&a_2&a_3&a_4\b_1&b_2&b_3&b_4endarrayright)
$$
emerging?!
For the purposes of introducing coordinates to the Grasmannian we associate a $d$-dimensional subspace $V$ with any matrix $dtimes n$ matrix $A$ that has $V$
as its row space. It is essential that:
- Two matrices $A$ and $A'$ are sharing the same row space if and only if there exists an invertible $dtimes d$ matrix $M$ such that $A'=MA$.
- When we calculate the wedge products of the rows, all the coordinates of the product of the rows of $A'$ are gotten by multiplying the corresponding coordinates of the product of rows of $A$ by $det M$. Implying that as homogeneous coordinates in a projective space the two wedge products refer to the same point. This means that the mapping $G(n,d)toBbbP^N$ is well defined.
- The easiest way of proving the result of the previous bullet is to prove it by direct observation for all three types of elementary matrices, and then writing $M$ as a product of elementary matrices.
Also, proving that $phi$ and $psi$ really give the same mapping may be easiest to do by induction on $d$. Then, at the induction step, you get the usual expansions of $(d+1)times (d+1)$-determinants as linear combinations of $dtimes d$-determinants.
The idea here is that that the vectors $u_i, i=1,2,ldots,d,$ are the rows of the matrix $A$. More precisely, if $u_i=(a_i1,a_i2,ldots,a_in)$, then the coefficient of
$e_i_1wedge e_i_2wedgecdotswedge e_i_d$ in the wedge product $u_1wedge u_2wedgecdotswedge u_d$ is equal to $det(A^(I))$.
Consider the paper & pencil example of $n=4,d=2$ with $u_1=(a_1,a_2,a_3,a_4)$, $u_2=(b_1,b_2,b_3,b_4)$ when
$$
beginaligned
u_1wedge u_2&=(a_1e_1+a_2e_2+a_3e_3+a_4e_4)wedge(b_1e_1+b_2e_2+b_3e_3+b_4e_4)\
&=sum_i=1^4sum_j=1^4a_ib_je_iwedge e_j\
&=(a_1b_2-a_2b_1)e_1wedge e_2+(a_1b_3-a_3b_1)e_1wedge e_3+(a_1b_4-a_4b_1)e_1wedge e_4\
&+(a_2b_3-a_3b_2)e_2wedge e_3+(a_2b_4-a_4b_2)e_2wedge e_4+(a_3b_4-a_4b_3)e_3wedge e_4.
endaligned
$$
See the six $2times2$ minors of the matrix
$$
A=left(beginarraycccca_1&a_2&a_3&a_4\b_1&b_2&b_3&b_4endarrayright)
$$
emerging?!
For the purposes of introducing coordinates to the Grasmannian we associate a $d$-dimensional subspace $V$ with any matrix $dtimes n$ matrix $A$ that has $V$
as its row space. It is essential that:
- Two matrices $A$ and $A'$ are sharing the same row space if and only if there exists an invertible $dtimes d$ matrix $M$ such that $A'=MA$.
- When we calculate the wedge products of the rows, all the coordinates of the product of the rows of $A'$ are gotten by multiplying the corresponding coordinates of the product of rows of $A$ by $det M$. Implying that as homogeneous coordinates in a projective space the two wedge products refer to the same point. This means that the mapping $G(n,d)toBbbP^N$ is well defined.
- The easiest way of proving the result of the previous bullet is to prove it by direct observation for all three types of elementary matrices, and then writing $M$ as a product of elementary matrices.
Also, proving that $phi$ and $psi$ really give the same mapping may be easiest to do by induction on $d$. Then, at the induction step, you get the usual expansions of $(d+1)times (d+1)$-determinants as linear combinations of $dtimes d$-determinants.
answered Sep 4 at 10:31
Jyrki Lahtonen
105k12163358
105k12163358
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