How can I get the matrix form for a schubert cell?
$begingroup$
I am trying to learn how to understand the cell-decomposition for the grassmannian and am following these notes: http://www.math.drexel.edu/~jblasiak/grassmannian.pdf. On page 2 the author considers the cell $e(lambda)$ in $G(3,7)$ for $lambda = (3,3,1)$. From the definition of $e(lambda)$ you can find that
begin{align}
text{dim}(W cap F_1) = 0 && text{dim}(W cap F_2) = 1 && text{dim}(W cap F_3) = 2 \
text{dim}(W cap F_5) = 2 && text{dim}(W cap F_6) = 3
end{align}
where $F_bullet$ is a maximal flag in $mathbb{C}^{n+m}$. The author claims that $e(lambda)$ is the set of matrices
$$
begin{bmatrix}
* & 1 & 0 & 0 & 0 & 0 & 0 \
* & 0 & 1 & 0 & 0 & 0 & 0 \
* & 0 & 0 & * & * & 1 & 0
end{bmatrix}
$$
but I am not sure how this argument works. How can I show the dimension conditions imply that every $3$-plane looks like this matrix?
What I've been able to figure out is that if we consider a basis $v_1,ldots, v_7$ defining the standard flag, then we have
begin{align}
v_2 in & Wcap F_2,ldots F_7 \
v_3 in & W cap F_3, ldots F_7 \
v_6 in & W cap F_6, F_7
end{align}
linear-algebra grassmannian schubert-calculus
edited Jan 5 at 12:37
Matt Samuel
38.1k63767
asked Aug 18 '17 at 17:09
54321user54321user
1,268622
$endgroup$
|
show 1 more comment
$begingroup$
I am trying to learn how to understand the cell-decomposition for the grassmannian and am following these notes: http://www.math.drexel.edu/~jblasiak/grassmannian.pdf. On page 2 the author considers the cell $e(lambda)$ in $G(3,7)$ for $lambda = (3,3,1)$. From the definition of $e(lambda)$ you can find that
begin{align}
text{dim}(W cap F_1) = 0 && text{dim}(W cap F_2) = 1 && text{dim}(W cap F_3) = 2 \
text{dim}(W cap F_5) = 2 && text{dim}(W cap F_6) = 3
end{align}
where $F_bullet$ is a maximal flag in $mathbb{C}^{n+m}$. The author claims that $e(lambda)$ is the set of matrices
$$
begin{bmatrix}
* & 1 & 0 & 0 & 0 & 0 & 0 \
* & 0 & 1 & 0 & 0 & 0 & 0 \
* & 0 & 0 & * & * & 1 & 0
end{bmatrix}
$$
but I am not sure how this argument works. How can I show the dimension conditions imply that every $3$-plane looks like this matrix?
What I've been able to figure out is that if we consider a basis $v_1,ldots, v_7$ defining the standard flag, then we have
begin{align}
v_2 in & Wcap F_2,ldots F_7 \
v_3 in & W cap F_3, ldots F_7 \
v_6 in & W cap F_6, F_7
end{align}
linear-algebra grassmannian schubert-calculus
edited Jan 5 at 12:37
Matt Samuel
38.1k63767
asked Aug 18 '17 at 17:09
54321user54321user
1,268622
$endgroup$
$begingroup$
Did you read the explanation the author gives ? What is unclear ? (They also precise that $dim W cap F_1 = 0$ instead of $1$ as you wrote, this is maybe where your confusion comes from).
$endgroup$
– user171326
Aug 21 '17 at 6:00
$begingroup$
Yes, I did read the explanation. but I do not understand why we get this matrix form.
$endgroup$
– 54321user
Aug 21 '17 at 15:26
$begingroup$
For example, why can we not have in the second row $* * 1 0 0 0 0$?
$endgroup$
– 54321user
Aug 21 '17 at 15:29
1
$begingroup$
$v_1$ is say $pmatrix{* & 1 & 0 & 0 & 0 & 0 & 0}$. Then, any $v_2$ is as you said on the form $pmatrix{* & * & 1 & 0 & 0 & 0 & 0}$ but this is not unique, on the other hand you can make it unique by substracting a suitable multiple of $v_1$, so this $v_1, lambda v_1 + v_2$ will have the form $pmatrix{* & 0 & 1 & 0 & 0 & 0 & 0}$. Having this uniqueness is important as the number of $*$ is the dimension of the corresponding Schubert cell.
$endgroup$
– user171326
Aug 21 '17 at 18:33
$begingroup$
Okay, I understand. Thank you
$endgroup$
– 54321user
Aug 22 '17 at 13:46
|
show 1 more comment
$begingroup$
I am trying to learn how to understand the cell-decomposition for the grassmannian and am following these notes: http://www.math.drexel.edu/~jblasiak/grassmannian.pdf. On page 2 the author considers the cell $e(lambda)$ in $G(3,7)$ for $lambda = (3,3,1)$. From the definition of $e(lambda)$ you can find that
begin{align}
text{dim}(W cap F_1) = 0 && text{dim}(W cap F_2) = 1 && text{dim}(W cap F_3) = 2 \
text{dim}(W cap F_5) = 2 && text{dim}(W cap F_6) = 3
end{align}
where $F_bullet$ is a maximal flag in $mathbb{C}^{n+m}$. The author claims that $e(lambda)$ is the set of matrices
$$
begin{bmatrix}
* & 1 & 0 & 0 & 0 & 0 & 0 \
* & 0 & 1 & 0 & 0 & 0 & 0 \
* & 0 & 0 & * & * & 1 & 0
end{bmatrix}
$$
but I am not sure how this argument works. How can I show the dimension conditions imply that every $3$-plane looks like this matrix?
What I've been able to figure out is that if we consider a basis $v_1,ldots, v_7$ defining the standard flag, then we have
begin{align}
v_2 in & Wcap F_2,ldots F_7 \
v_3 in & W cap F_3, ldots F_7 \
v_6 in & W cap F_6, F_7
end{align}
linear-algebra grassmannian schubert-calculus
edited Jan 5 at 12:37
Matt Samuel
38.1k63767
asked Aug 18 '17 at 17:09
54321user54321user
1,268622
$endgroup$
I am trying to learn how to understand the cell-decomposition for the grassmannian and am following these notes: http://www.math.drexel.edu/~jblasiak/grassmannian.pdf. On page 2 the author considers the cell $e(lambda)$ in $G(3,7)$ for $lambda = (3,3,1)$. From the definition of $e(lambda)$ you can find that
begin{align}
text{dim}(W cap F_1) = 0 && text{dim}(W cap F_2) = 1 && text{dim}(W cap F_3) = 2 \
text{dim}(W cap F_5) = 2 && text{dim}(W cap F_6) = 3
end{align}
where $F_bullet$ is a maximal flag in $mathbb{C}^{n+m}$. The author claims that $e(lambda)$ is the set of matrices
$$
begin{bmatrix}
* & 1 & 0 & 0 & 0 & 0 & 0 \
* & 0 & 1 & 0 & 0 & 0 & 0 \
* & 0 & 0 & * & * & 1 & 0
end{bmatrix}
$$
but I am not sure how this argument works. How can I show the dimension conditions imply that every $3$-plane looks like this matrix?
What I've been able to figure out is that if we consider a basis $v_1,ldots, v_7$ defining the standard flag, then we have
begin{align}
v_2 in & Wcap F_2,ldots F_7 \
v_3 in & W cap F_3, ldots F_7 \
v_6 in & W cap F_6, F_7
end{align}
linear-algebra grassmannian schubert-calculus
linear-algebra grassmannian schubert-calculus
edited Jan 5 at 12:37
Matt Samuel
38.1k63767
asked Aug 18 '17 at 17:09
54321user54321user
1,268622
edited Jan 5 at 12:37
Matt Samuel
38.1k63767
asked Aug 18 '17 at 17:09
54321user54321user
1,268622
edited Jan 5 at 12:37
Matt Samuel
38.1k63767
edited Jan 5 at 12:37
Matt Samuel
38.1k63767
edited Jan 5 at 12:37
Matt Samuel
38.1k63767
38.1k63767
asked Aug 18 '17 at 17:09
54321user54321user
1,268622
asked Aug 18 '17 at 17:09
54321user54321user
1,268622
asked Aug 18 '17 at 17:09
54321user54321user
1,268622
1,268622
$begingroup$
Did you read the explanation the author gives ? What is unclear ? (They also precise that $dim W cap F_1 = 0$ instead of $1$ as you wrote, this is maybe where your confusion comes from).
$endgroup$
– user171326
Aug 21 '17 at 6:00
$begingroup$
Yes, I did read the explanation. but I do not understand why we get this matrix form.
$endgroup$
– 54321user
Aug 21 '17 at 15:26
$begingroup$
For example, why can we not have in the second row $* * 1 0 0 0 0$?
$endgroup$
– 54321user
Aug 21 '17 at 15:29
1
$begingroup$
$v_1$ is say $pmatrix{* & 1 & 0 & 0 & 0 & 0 & 0}$. Then, any $v_2$ is as you said on the form $pmatrix{* & * & 1 & 0 & 0 & 0 & 0}$ but this is not unique, on the other hand you can make it unique by substracting a suitable multiple of $v_1$, so this $v_1, lambda v_1 + v_2$ will have the form $pmatrix{* & 0 & 1 & 0 & 0 & 0 & 0}$. Having this uniqueness is important as the number of $*$ is the dimension of the corresponding Schubert cell.
$endgroup$
– user171326
Aug 21 '17 at 18:33
$begingroup$
Okay, I understand. Thank you
$endgroup$
– 54321user
Aug 22 '17 at 13:46
|
show 1 more comment
$begingroup$
Did you read the explanation the author gives ? What is unclear ? (They also precise that $dim W cap F_1 = 0$ instead of $1$ as you wrote, this is maybe where your confusion comes from).
$endgroup$
– user171326
Aug 21 '17 at 6:00
$begingroup$
Yes, I did read the explanation. but I do not understand why we get this matrix form.
$endgroup$
– 54321user
Aug 21 '17 at 15:26
$begingroup$
For example, why can we not have in the second row $* * 1 0 0 0 0$?
$endgroup$
– 54321user
Aug 21 '17 at 15:29
1
$begingroup$
$v_1$ is say $pmatrix{* & 1 & 0 & 0 & 0 & 0 & 0}$. Then, any $v_2$ is as you said on the form $pmatrix{* & * & 1 & 0 & 0 & 0 & 0}$ but this is not unique, on the other hand you can make it unique by substracting a suitable multiple of $v_1$, so this $v_1, lambda v_1 + v_2$ will have the form $pmatrix{* & 0 & 1 & 0 & 0 & 0 & 0}$. Having this uniqueness is important as the number of $*$ is the dimension of the corresponding Schubert cell.
$endgroup$
– user171326
Aug 21 '17 at 18:33
$begingroup$
Okay, I understand. Thank you
$endgroup$
– 54321user
Aug 22 '17 at 13:46
$begingroup$
Did you read the explanation the author gives ? What is unclear ? (They also precise that $dim W cap F_1 = 0$ instead of $1$ as you wrote, this is maybe where your confusion comes from).
$endgroup$
– user171326
Aug 21 '17 at 6:00
$begingroup$
Did you read the explanation the author gives ? What is unclear ? (They also precise that $dim W cap F_1 = 0$ instead of $1$ as you wrote, this is maybe where your confusion comes from).
$endgroup$
– user171326
Aug 21 '17 at 6:00
$begingroup$
Yes, I did read the explanation. but I do not understand why we get this matrix form.
$endgroup$
– 54321user
Aug 21 '17 at 15:26
$begingroup$
Yes, I did read the explanation. but I do not understand why we get this matrix form.
$endgroup$
– 54321user
Aug 21 '17 at 15:26
$begingroup$
For example, why can we not have in the second row $* * 1 0 0 0 0$?
$endgroup$
– 54321user
Aug 21 '17 at 15:29
$begingroup$
For example, why can we not have in the second row $* * 1 0 0 0 0$?
$endgroup$
– 54321user
Aug 21 '17 at 15:29
1
1
$begingroup$
$v_1$ is say $pmatrix{* & 1 & 0 & 0 & 0 & 0 & 0}$. Then, any $v_2$ is as you said on the form $pmatrix{* & * & 1 & 0 & 0 & 0 & 0}$ but this is not unique, on the other hand you can make it unique by substracting a suitable multiple of $v_1$, so this $v_1, lambda v_1 + v_2$ will have the form $pmatrix{* & 0 & 1 & 0 & 0 & 0 & 0}$. Having this uniqueness is important as the number of $*$ is the dimension of the corresponding Schubert cell.
$endgroup$
– user171326
Aug 21 '17 at 18:33
$begingroup$
$v_1$ is say $pmatrix{* & 1 & 0 & 0 & 0 & 0 & 0}$. Then, any $v_2$ is as you said on the form $pmatrix{* & * & 1 & 0 & 0 & 0 & 0}$ but this is not unique, on the other hand you can make it unique by substracting a suitable multiple of $v_1$, so this $v_1, lambda v_1 + v_2$ will have the form $pmatrix{* & 0 & 1 & 0 & 0 & 0 & 0}$. Having this uniqueness is important as the number of $*$ is the dimension of the corresponding Schubert cell.
$endgroup$
– user171326
Aug 21 '17 at 18:33
$begingroup$
Okay, I understand. Thank you
$endgroup$
– 54321user
Aug 22 '17 at 13:46
$begingroup$
Okay, I understand. Thank you
$endgroup$
– 54321user
Aug 22 '17 at 13:46
|
show 1 more comment
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$begingroup$
Did you read the explanation the author gives ? What is unclear ? (They also precise that $dim W cap F_1 = 0$ instead of $1$ as you wrote, this is maybe where your confusion comes from).
$endgroup$
– user171326
Aug 21 '17 at 6:00
$begingroup$
Yes, I did read the explanation. but I do not understand why we get this matrix form.
$endgroup$
– 54321user
Aug 21 '17 at 15:26
$begingroup$
For example, why can we not have in the second row $* * 1 0 0 0 0$?
$endgroup$
– 54321user
Aug 21 '17 at 15:29
1
$begingroup$
$v_1$ is say $pmatrix{* & 1 & 0 & 0 & 0 & 0 & 0}$. Then, any $v_2$ is as you said on the form $pmatrix{* & * & 1 & 0 & 0 & 0 & 0}$ but this is not unique, on the other hand you can make it unique by substracting a suitable multiple of $v_1$, so this $v_1, lambda v_1 + v_2$ will have the form $pmatrix{* & 0 & 1 & 0 & 0 & 0 & 0}$. Having this uniqueness is important as the number of $*$ is the dimension of the corresponding Schubert cell.
$endgroup$
– user171326
Aug 21 '17 at 18:33
$begingroup$
Okay, I understand. Thank you
$endgroup$
– 54321user
Aug 22 '17 at 13:46