Coordinates on a Richardson variety
$begingroup$
I'm looking for convenient coordinates to use to describe the intersection of a Schubert cell $X^circ_lambda$ and an opposite Schubert cell $Omega^circ_mu$ in a Grassmannian $G(k,n)$.
Describing a single Schubert variety is no problem as its maximal cell is isomorphic to an affine space $mathbb{A^r}$ -- explicitly written down as matrices with many entries set to 0 or 1, and the rest free. And there's the corresponding description for opposite Schuberts. For describing the intersection $X^circ_lambda cap Omega^circ_mu$, is there a simple way to 'combine' the forward- and opposite-Schubert descriptions? For example, in $G(2,5)$,
$$begin{pmatrix}
* &* & 1 & 0 & 0 \ * & * & 0 & * & 1
end{pmatrix} cap begin{pmatrix}
1 & * & 0 & * & *\ 0 & 0 & 1 & * & *
end{pmatrix}$$
has a dense open subset consisting of matrices of the form
$$begin{pmatrix}
*' & * & 1 & 0 & 0 \ 0 & 0 & *' & * & 1
end{pmatrix},$$
where $*'$ means the entry can't be zero. (Note that we can't just "overlay" the two types of matrix, since e.g. they disagree in the middle column.)
Ideally I'd like to have exactly $dim(X^circ_lambda cap Omega^circ_mu)$ coordinates, as in this example. (And, while I'm happy to describe, e.g. a dense open subset of $X^circ_lambda cap Omega^circ_mu$, it would be helpful to know what exactly I'm ignoring.)
algebraic-geometry schubert-calculus
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
asked Jan 27 '15 at 3:29
Jake LevinsonJake Levinson
3,4631712
$endgroup$
add a comment |
$begingroup$
I'm looking for convenient coordinates to use to describe the intersection of a Schubert cell $X^circ_lambda$ and an opposite Schubert cell $Omega^circ_mu$ in a Grassmannian $G(k,n)$.
Describing a single Schubert variety is no problem as its maximal cell is isomorphic to an affine space $mathbb{A^r}$ -- explicitly written down as matrices with many entries set to 0 or 1, and the rest free. And there's the corresponding description for opposite Schuberts. For describing the intersection $X^circ_lambda cap Omega^circ_mu$, is there a simple way to 'combine' the forward- and opposite-Schubert descriptions? For example, in $G(2,5)$,
$$begin{pmatrix}
* &* & 1 & 0 & 0 \ * & * & 0 & * & 1
end{pmatrix} cap begin{pmatrix}
1 & * & 0 & * & *\ 0 & 0 & 1 & * & *
end{pmatrix}$$
has a dense open subset consisting of matrices of the form
$$begin{pmatrix}
*' & * & 1 & 0 & 0 \ 0 & 0 & *' & * & 1
end{pmatrix},$$
where $*'$ means the entry can't be zero. (Note that we can't just "overlay" the two types of matrix, since e.g. they disagree in the middle column.)
Ideally I'd like to have exactly $dim(X^circ_lambda cap Omega^circ_mu)$ coordinates, as in this example. (And, while I'm happy to describe, e.g. a dense open subset of $X^circ_lambda cap Omega^circ_mu$, it would be helpful to know what exactly I'm ignoring.)
algebraic-geometry schubert-calculus
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
asked Jan 27 '15 at 3:29
Jake LevinsonJake Levinson
3,4631712
$endgroup$
$begingroup$
(I think the general answer should be similar to the example! But I'm having trouble working it out myself.)
$endgroup$
– Jake Levinson
Jan 27 '15 at 3:31
$begingroup$
I can think of an algorithm for doing this. Judging by the example you may already know this algorithm. Am I right?
$endgroup$
– Matt Samuel
Jan 27 '15 at 22:37
$begingroup$
It was sort of ad hoc - just trying to mimic the placement of the ones and zeros. But it's not obvious to me what's going on in general (especially thinking geometrically).
$endgroup$
– Jake Levinson
Jan 28 '15 at 1:58
$begingroup$
I can't comment on what's going on geometrically, but row operations is the trick. I'm sure there aren't very many cases to cover to get a general algorithm. You get them so that they have 1's in the same places then you take the most restrictive condition on the remaining entries.
$endgroup$
– Matt Samuel
Jan 28 '15 at 2:01
$begingroup$
You can get the whole variety this way by checking intersections of all the cells. This will quickly become intractible, of course, as does anything in Schubert calculus.
$endgroup$
– Matt Samuel
Jan 28 '15 at 2:37
add a comment |
$begingroup$
I'm looking for convenient coordinates to use to describe the intersection of a Schubert cell $X^circ_lambda$ and an opposite Schubert cell $Omega^circ_mu$ in a Grassmannian $G(k,n)$.
Describing a single Schubert variety is no problem as its maximal cell is isomorphic to an affine space $mathbb{A^r}$ -- explicitly written down as matrices with many entries set to 0 or 1, and the rest free. And there's the corresponding description for opposite Schuberts. For describing the intersection $X^circ_lambda cap Omega^circ_mu$, is there a simple way to 'combine' the forward- and opposite-Schubert descriptions? For example, in $G(2,5)$,
$$begin{pmatrix}
* &* & 1 & 0 & 0 \ * & * & 0 & * & 1
end{pmatrix} cap begin{pmatrix}
1 & * & 0 & * & *\ 0 & 0 & 1 & * & *
end{pmatrix}$$
has a dense open subset consisting of matrices of the form
$$begin{pmatrix}
*' & * & 1 & 0 & 0 \ 0 & 0 & *' & * & 1
end{pmatrix},$$
where $*'$ means the entry can't be zero. (Note that we can't just "overlay" the two types of matrix, since e.g. they disagree in the middle column.)
Ideally I'd like to have exactly $dim(X^circ_lambda cap Omega^circ_mu)$ coordinates, as in this example. (And, while I'm happy to describe, e.g. a dense open subset of $X^circ_lambda cap Omega^circ_mu$, it would be helpful to know what exactly I'm ignoring.)
algebraic-geometry schubert-calculus
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
asked Jan 27 '15 at 3:29
Jake LevinsonJake Levinson
3,4631712
$endgroup$
I'm looking for convenient coordinates to use to describe the intersection of a Schubert cell $X^circ_lambda$ and an opposite Schubert cell $Omega^circ_mu$ in a Grassmannian $G(k,n)$.
Describing a single Schubert variety is no problem as its maximal cell is isomorphic to an affine space $mathbb{A^r}$ -- explicitly written down as matrices with many entries set to 0 or 1, and the rest free. And there's the corresponding description for opposite Schuberts. For describing the intersection $X^circ_lambda cap Omega^circ_mu$, is there a simple way to 'combine' the forward- and opposite-Schubert descriptions? For example, in $G(2,5)$,
$$begin{pmatrix}
* &* & 1 & 0 & 0 \ * & * & 0 & * & 1
end{pmatrix} cap begin{pmatrix}
1 & * & 0 & * & *\ 0 & 0 & 1 & * & *
end{pmatrix}$$
has a dense open subset consisting of matrices of the form
$$begin{pmatrix}
*' & * & 1 & 0 & 0 \ 0 & 0 & *' & * & 1
end{pmatrix},$$
where $*'$ means the entry can't be zero. (Note that we can't just "overlay" the two types of matrix, since e.g. they disagree in the middle column.)
Ideally I'd like to have exactly $dim(X^circ_lambda cap Omega^circ_mu)$ coordinates, as in this example. (And, while I'm happy to describe, e.g. a dense open subset of $X^circ_lambda cap Omega^circ_mu$, it would be helpful to know what exactly I'm ignoring.)
algebraic-geometry schubert-calculus
algebraic-geometry schubert-calculus
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
asked Jan 27 '15 at 3:29
Jake LevinsonJake Levinson
3,4631712
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
asked Jan 27 '15 at 3:29
Jake LevinsonJake Levinson
3,4631712
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
39.2k63770
asked Jan 27 '15 at 3:29
Jake LevinsonJake Levinson
3,4631712
asked Jan 27 '15 at 3:29
Jake LevinsonJake Levinson
3,4631712
asked Jan 27 '15 at 3:29
Jake LevinsonJake Levinson
3,4631712
3,4631712
$begingroup$
(I think the general answer should be similar to the example! But I'm having trouble working it out myself.)
$endgroup$
– Jake Levinson
Jan 27 '15 at 3:31
$begingroup$
I can think of an algorithm for doing this. Judging by the example you may already know this algorithm. Am I right?
$endgroup$
– Matt Samuel
Jan 27 '15 at 22:37
$begingroup$
It was sort of ad hoc - just trying to mimic the placement of the ones and zeros. But it's not obvious to me what's going on in general (especially thinking geometrically).
$endgroup$
– Jake Levinson
Jan 28 '15 at 1:58
$begingroup$
I can't comment on what's going on geometrically, but row operations is the trick. I'm sure there aren't very many cases to cover to get a general algorithm. You get them so that they have 1's in the same places then you take the most restrictive condition on the remaining entries.
$endgroup$
– Matt Samuel
Jan 28 '15 at 2:01
$begingroup$
You can get the whole variety this way by checking intersections of all the cells. This will quickly become intractible, of course, as does anything in Schubert calculus.
$endgroup$
– Matt Samuel
Jan 28 '15 at 2:37
add a comment |
$begingroup$
(I think the general answer should be similar to the example! But I'm having trouble working it out myself.)
$endgroup$
– Jake Levinson
Jan 27 '15 at 3:31
$begingroup$
I can think of an algorithm for doing this. Judging by the example you may already know this algorithm. Am I right?
$endgroup$
– Matt Samuel
Jan 27 '15 at 22:37
$begingroup$
It was sort of ad hoc - just trying to mimic the placement of the ones and zeros. But it's not obvious to me what's going on in general (especially thinking geometrically).
$endgroup$
– Jake Levinson
Jan 28 '15 at 1:58
$begingroup$
I can't comment on what's going on geometrically, but row operations is the trick. I'm sure there aren't very many cases to cover to get a general algorithm. You get them so that they have 1's in the same places then you take the most restrictive condition on the remaining entries.
$endgroup$
– Matt Samuel
Jan 28 '15 at 2:01
$begingroup$
You can get the whole variety this way by checking intersections of all the cells. This will quickly become intractible, of course, as does anything in Schubert calculus.
$endgroup$
– Matt Samuel
Jan 28 '15 at 2:37
$begingroup$
(I think the general answer should be similar to the example! But I'm having trouble working it out myself.)
$endgroup$
– Jake Levinson
Jan 27 '15 at 3:31
$begingroup$
(I think the general answer should be similar to the example! But I'm having trouble working it out myself.)
$endgroup$
– Jake Levinson
Jan 27 '15 at 3:31
$begingroup$
I can think of an algorithm for doing this. Judging by the example you may already know this algorithm. Am I right?
$endgroup$
– Matt Samuel
Jan 27 '15 at 22:37
$begingroup$
I can think of an algorithm for doing this. Judging by the example you may already know this algorithm. Am I right?
$endgroup$
– Matt Samuel
Jan 27 '15 at 22:37
$begingroup$
It was sort of ad hoc - just trying to mimic the placement of the ones and zeros. But it's not obvious to me what's going on in general (especially thinking geometrically).
$endgroup$
– Jake Levinson
Jan 28 '15 at 1:58
$begingroup$
It was sort of ad hoc - just trying to mimic the placement of the ones and zeros. But it's not obvious to me what's going on in general (especially thinking geometrically).
$endgroup$
– Jake Levinson
Jan 28 '15 at 1:58
$begingroup$
I can't comment on what's going on geometrically, but row operations is the trick. I'm sure there aren't very many cases to cover to get a general algorithm. You get them so that they have 1's in the same places then you take the most restrictive condition on the remaining entries.
$endgroup$
– Matt Samuel
Jan 28 '15 at 2:01
$begingroup$
I can't comment on what's going on geometrically, but row operations is the trick. I'm sure there aren't very many cases to cover to get a general algorithm. You get them so that they have 1's in the same places then you take the most restrictive condition on the remaining entries.
$endgroup$
– Matt Samuel
Jan 28 '15 at 2:01
$begingroup$
You can get the whole variety this way by checking intersections of all the cells. This will quickly become intractible, of course, as does anything in Schubert calculus.
$endgroup$
– Matt Samuel
Jan 28 '15 at 2:37
$begingroup$
You can get the whole variety this way by checking intersections of all the cells. This will quickly become intractible, of course, as does anything in Schubert calculus.
$endgroup$
– Matt Samuel
Jan 28 '15 at 2:37
add a comment |
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$begingroup$
(I think the general answer should be similar to the example! But I'm having trouble working it out myself.)
$endgroup$
– Jake Levinson
Jan 27 '15 at 3:31
$begingroup$
I can think of an algorithm for doing this. Judging by the example you may already know this algorithm. Am I right?
$endgroup$
– Matt Samuel
Jan 27 '15 at 22:37
$begingroup$
It was sort of ad hoc - just trying to mimic the placement of the ones and zeros. But it's not obvious to me what's going on in general (especially thinking geometrically).
$endgroup$
– Jake Levinson
Jan 28 '15 at 1:58
$begingroup$
I can't comment on what's going on geometrically, but row operations is the trick. I'm sure there aren't very many cases to cover to get a general algorithm. You get them so that they have 1's in the same places then you take the most restrictive condition on the remaining entries.
$endgroup$
– Matt Samuel
Jan 28 '15 at 2:01
$begingroup$
You can get the whole variety this way by checking intersections of all the cells. This will quickly become intractible, of course, as does anything in Schubert calculus.
$endgroup$
– Matt Samuel
Jan 28 '15 at 2:37