Schubert Cells of Flags












2












$begingroup$


I have been reading on these notes Undergraduate Lectures on Flag Varieties and I need some explanations on two things:




  1. In page 3, how he modefied the matrices in the "Second Attempt"


  2. In the same manner, how can I get the flag $Fl(1,2,3)(mathbb C^3)$?



Thanks for your help!






combinatorics algebraic-geometry manifolds schubert-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For 1, note that if the star in the second column and second row in the first attempt for the ${1,3}$ matrix is nonzero, then we can divide the row by it and obtain a matrix of the form for ${1,2}$. This causes overlap, and if we insist that instead that spot have a zero in it then there is no overlap. Similar arguments apply for the other matrices.
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 0:44






  • 1




    $begingroup$
    For 2, the book "Young Tableaux" by Fulton has a detailed account for the complete flag variety. Perhaps someone will write an answer but a full explanation would be lengthy.
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 0:47










  • $begingroup$
    I see this is related to a question you asked before: math.stackexchange.com/questions/1508147/…
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 1:44






  • 1




    $begingroup$
    For a quick and dirty explanantion that offers no insight, take a $3times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix.
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 2:05










  • $begingroup$
    @Matt, For 1, I still don't see the case {1,4}, {2,3} can you please explain it like what you did for the case {1,3} earlier?
    $endgroup$
    – Ronald
    Nov 24 '15 at 19:16
















2












$begingroup$


I have been reading on these notes Undergraduate Lectures on Flag Varieties and I need some explanations on two things:




  1. In page 3, how he modefied the matrices in the "Second Attempt"


  2. In the same manner, how can I get the flag $Fl(1,2,3)(mathbb C^3)$?



Thanks for your help!






combinatorics algebraic-geometry manifolds schubert-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For 1, note that if the star in the second column and second row in the first attempt for the ${1,3}$ matrix is nonzero, then we can divide the row by it and obtain a matrix of the form for ${1,2}$. This causes overlap, and if we insist that instead that spot have a zero in it then there is no overlap. Similar arguments apply for the other matrices.
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 0:44






  • 1




    $begingroup$
    For 2, the book "Young Tableaux" by Fulton has a detailed account for the complete flag variety. Perhaps someone will write an answer but a full explanation would be lengthy.
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 0:47










  • $begingroup$
    I see this is related to a question you asked before: math.stackexchange.com/questions/1508147/…
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 1:44






  • 1




    $begingroup$
    For a quick and dirty explanantion that offers no insight, take a $3times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix.
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 2:05










  • $begingroup$
    @Matt, For 1, I still don't see the case {1,4}, {2,3} can you please explain it like what you did for the case {1,3} earlier?
    $endgroup$
    – Ronald
    Nov 24 '15 at 19:16














2












2








2





$begingroup$


I have been reading on these notes Undergraduate Lectures on Flag Varieties and I need some explanations on two things:




  1. In page 3, how he modefied the matrices in the "Second Attempt"


  2. In the same manner, how can I get the flag $Fl(1,2,3)(mathbb C^3)$?



Thanks for your help!






combinatorics algebraic-geometry manifolds schubert-calculus







share|cite|improve this question











$endgroup$




I have been reading on these notes Undergraduate Lectures on Flag Varieties and I need some explanations on two things:




  1. In page 3, how he modefied the matrices in the "Second Attempt"


  2. In the same manner, how can I get the flag $Fl(1,2,3)(mathbb C^3)$?



Thanks for your help!






combinatorics algebraic-geometry manifolds schubert-calculus




combinatorics algebraic-geometry manifolds schubert-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 3:13









Matt Samuel

38.1k63767




38.1k63767










asked Nov 24 '15 at 0:30









RonaldRonald

1,7531921




1,7531921








  • 1




    $begingroup$
    For 1, note that if the star in the second column and second row in the first attempt for the ${1,3}$ matrix is nonzero, then we can divide the row by it and obtain a matrix of the form for ${1,2}$. This causes overlap, and if we insist that instead that spot have a zero in it then there is no overlap. Similar arguments apply for the other matrices.
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 0:44






  • 1




    $begingroup$
    For 2, the book "Young Tableaux" by Fulton has a detailed account for the complete flag variety. Perhaps someone will write an answer but a full explanation would be lengthy.
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 0:47










  • $begingroup$
    I see this is related to a question you asked before: math.stackexchange.com/questions/1508147/…
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 1:44






  • 1




    $begingroup$
    For a quick and dirty explanantion that offers no insight, take a $3times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix.
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 2:05










  • $begingroup$
    @Matt, For 1, I still don't see the case {1,4}, {2,3} can you please explain it like what you did for the case {1,3} earlier?
    $endgroup$
    – Ronald
    Nov 24 '15 at 19:16














  • 1




    $begingroup$
    For 1, note that if the star in the second column and second row in the first attempt for the ${1,3}$ matrix is nonzero, then we can divide the row by it and obtain a matrix of the form for ${1,2}$. This causes overlap, and if we insist that instead that spot have a zero in it then there is no overlap. Similar arguments apply for the other matrices.
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 0:44






  • 1




    $begingroup$
    For 2, the book "Young Tableaux" by Fulton has a detailed account for the complete flag variety. Perhaps someone will write an answer but a full explanation would be lengthy.
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 0:47










  • $begingroup$
    I see this is related to a question you asked before: math.stackexchange.com/questions/1508147/…
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 1:44






  • 1




    $begingroup$
    For a quick and dirty explanantion that offers no insight, take a $3times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix.
    $endgroup$
    – Matt Samuel
    Nov 24 '15 at 2:05










  • $begingroup$
    @Matt, For 1, I still don't see the case {1,4}, {2,3} can you please explain it like what you did for the case {1,3} earlier?
    $endgroup$
    – Ronald
    Nov 24 '15 at 19:16








1




1




$begingroup$
For 1, note that if the star in the second column and second row in the first attempt for the ${1,3}$ matrix is nonzero, then we can divide the row by it and obtain a matrix of the form for ${1,2}$. This causes overlap, and if we insist that instead that spot have a zero in it then there is no overlap. Similar arguments apply for the other matrices.
$endgroup$
– Matt Samuel
Nov 24 '15 at 0:44




$begingroup$
For 1, note that if the star in the second column and second row in the first attempt for the ${1,3}$ matrix is nonzero, then we can divide the row by it and obtain a matrix of the form for ${1,2}$. This causes overlap, and if we insist that instead that spot have a zero in it then there is no overlap. Similar arguments apply for the other matrices.
$endgroup$
– Matt Samuel
Nov 24 '15 at 0:44




1




1




$begingroup$
For 2, the book "Young Tableaux" by Fulton has a detailed account for the complete flag variety. Perhaps someone will write an answer but a full explanation would be lengthy.
$endgroup$
– Matt Samuel
Nov 24 '15 at 0:47




$begingroup$
For 2, the book "Young Tableaux" by Fulton has a detailed account for the complete flag variety. Perhaps someone will write an answer but a full explanation would be lengthy.
$endgroup$
– Matt Samuel
Nov 24 '15 at 0:47












$begingroup$
I see this is related to a question you asked before: math.stackexchange.com/questions/1508147/…
$endgroup$
– Matt Samuel
Nov 24 '15 at 1:44




$begingroup$
I see this is related to a question you asked before: math.stackexchange.com/questions/1508147/…
$endgroup$
– Matt Samuel
Nov 24 '15 at 1:44




1




1




$begingroup$
For a quick and dirty explanantion that offers no insight, take a $3times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix.
$endgroup$
– Matt Samuel
Nov 24 '15 at 2:05




$begingroup$
For a quick and dirty explanantion that offers no insight, take a $3times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix.
$endgroup$
– Matt Samuel
Nov 24 '15 at 2:05












$begingroup$
@Matt, For 1, I still don't see the case {1,4}, {2,3} can you please explain it like what you did for the case {1,3} earlier?
$endgroup$
– Ronald
Nov 24 '15 at 19:16




$begingroup$
@Matt, For 1, I still don't see the case {1,4}, {2,3} can you please explain it like what you did for the case {1,3} earlier?
$endgroup$
– Ronald
Nov 24 '15 at 19:16










1 Answer
1






active

oldest

votes


















2












$begingroup$

For a quick and dirty explanantion that offers no insight (as requested), take a $3times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix.



For example, for the permutation 312 (which has length $2$, so the cell should have codimension $2$, dimension $1$) we have
$$left[begin{array}{ccc}0&0&1\1&0&0\ast&1&0end{array}right]$$
and for 213 (length $1$, codimension $1$, dimension $2$) we have
$$left[begin{array}{ccc}0&1&0\1&0&0\ast&ast&1end{array}right]$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot Matt. However if I have more questions I will ask you :)
    $endgroup$
    – Ronald
    Nov 24 '15 at 2:46










  • $begingroup$
    I have a question here: where we draw stars in this perm matrix $left[begin{array}{ccc}0&1&0\0&0&1\1&0&0end{array}right]$ and why?
    $endgroup$
    – Ronald
    Nov 24 '15 at 23:10













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1 Answer
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$begingroup$

For a quick and dirty explanantion that offers no insight (as requested), take a $3times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix.



For example, for the permutation 312 (which has length $2$, so the cell should have codimension $2$, dimension $1$) we have
$$left[begin{array}{ccc}0&0&1\1&0&0\ast&1&0end{array}right]$$
and for 213 (length $1$, codimension $1$, dimension $2$) we have
$$left[begin{array}{ccc}0&1&0\1&0&0\ast&ast&1end{array}right]$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot Matt. However if I have more questions I will ask you :)
    $endgroup$
    – Ronald
    Nov 24 '15 at 2:46










  • $begingroup$
    I have a question here: where we draw stars in this perm matrix $left[begin{array}{ccc}0&1&0\0&0&1\1&0&0end{array}right]$ and why?
    $endgroup$
    – Ronald
    Nov 24 '15 at 23:10


















2












$begingroup$

For a quick and dirty explanantion that offers no insight (as requested), take a $3times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix.



For example, for the permutation 312 (which has length $2$, so the cell should have codimension $2$, dimension $1$) we have
$$left[begin{array}{ccc}0&0&1\1&0&0\ast&1&0end{array}right]$$
and for 213 (length $1$, codimension $1$, dimension $2$) we have
$$left[begin{array}{ccc}0&1&0\1&0&0\ast&ast&1end{array}right]$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot Matt. However if I have more questions I will ask you :)
    $endgroup$
    – Ronald
    Nov 24 '15 at 2:46










  • $begingroup$
    I have a question here: where we draw stars in this perm matrix $left[begin{array}{ccc}0&1&0\0&0&1\1&0&0end{array}right]$ and why?
    $endgroup$
    – Ronald
    Nov 24 '15 at 23:10
















2












2








2





$begingroup$

For a quick and dirty explanantion that offers no insight (as requested), take a $3times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix.



For example, for the permutation 312 (which has length $2$, so the cell should have codimension $2$, dimension $1$) we have
$$left[begin{array}{ccc}0&0&1\1&0&0\ast&1&0end{array}right]$$
and for 213 (length $1$, codimension $1$, dimension $2$) we have
$$left[begin{array}{ccc}0&1&0\1&0&0\ast&ast&1end{array}right]$$






share|cite|improve this answer









$endgroup$



For a quick and dirty explanantion that offers no insight (as requested), take a $3times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix.



For example, for the permutation 312 (which has length $2$, so the cell should have codimension $2$, dimension $1$) we have
$$left[begin{array}{ccc}0&0&1\1&0&0\ast&1&0end{array}right]$$
and for 213 (length $1$, codimension $1$, dimension $2$) we have
$$left[begin{array}{ccc}0&1&0\1&0&0\ast&ast&1end{array}right]$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 '15 at 2:44









Matt SamuelMatt Samuel

38.1k63767




38.1k63767












  • $begingroup$
    Thanks a lot Matt. However if I have more questions I will ask you :)
    $endgroup$
    – Ronald
    Nov 24 '15 at 2:46










  • $begingroup$
    I have a question here: where we draw stars in this perm matrix $left[begin{array}{ccc}0&1&0\0&0&1\1&0&0end{array}right]$ and why?
    $endgroup$
    – Ronald
    Nov 24 '15 at 23:10




















  • $begingroup$
    Thanks a lot Matt. However if I have more questions I will ask you :)
    $endgroup$
    – Ronald
    Nov 24 '15 at 2:46










  • $begingroup$
    I have a question here: where we draw stars in this perm matrix $left[begin{array}{ccc}0&1&0\0&0&1\1&0&0end{array}right]$ and why?
    $endgroup$
    – Ronald
    Nov 24 '15 at 23:10


















$begingroup$
Thanks a lot Matt. However if I have more questions I will ask you :)
$endgroup$
– Ronald
Nov 24 '15 at 2:46




$begingroup$
Thanks a lot Matt. However if I have more questions I will ask you :)
$endgroup$
– Ronald
Nov 24 '15 at 2:46












$begingroup$
I have a question here: where we draw stars in this perm matrix $left[begin{array}{ccc}0&1&0\0&0&1\1&0&0end{array}right]$ and why?
$endgroup$
– Ronald
Nov 24 '15 at 23:10






$begingroup$
I have a question here: where we draw stars in this perm matrix $left[begin{array}{ccc}0&1&0\0&0&1\1&0&0end{array}right]$ and why?
$endgroup$
– Ronald
Nov 24 '15 at 23:10




















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