Schubert cell decomposition and full flags
$begingroup$
I am looking for a self-contained basic theory of Schubert-cells through finding the decomposition of the full flag $Fl_3(mathbb C^3)$.
algebraic-geometry manifolds lie-algebras homogeneous-spaces schubert-calculus
edited Dec 25 '18 at 13:15
Matt Samuel
38.5k63768
asked Nov 25 '15 at 14:34
RonaldRonald
1,7531921
$endgroup$
add a comment |
$begingroup$
I am looking for a self-contained basic theory of Schubert-cells through finding the decomposition of the full flag $Fl_3(mathbb C^3)$.
algebraic-geometry manifolds lie-algebras homogeneous-spaces schubert-calculus
edited Dec 25 '18 at 13:15
Matt Samuel
38.5k63768
asked Nov 25 '15 at 14:34
RonaldRonald
1,7531921
$endgroup$
add a comment |
$begingroup$
I am looking for a self-contained basic theory of Schubert-cells through finding the decomposition of the full flag $Fl_3(mathbb C^3)$.
algebraic-geometry manifolds lie-algebras homogeneous-spaces schubert-calculus
edited Dec 25 '18 at 13:15
Matt Samuel
38.5k63768
asked Nov 25 '15 at 14:34
RonaldRonald
1,7531921
$endgroup$
I am looking for a self-contained basic theory of Schubert-cells through finding the decomposition of the full flag $Fl_3(mathbb C^3)$.
algebraic-geometry manifolds lie-algebras homogeneous-spaces schubert-calculus
algebraic-geometry manifolds lie-algebras homogeneous-spaces schubert-calculus
edited Dec 25 '18 at 13:15
Matt Samuel
38.5k63768
asked Nov 25 '15 at 14:34
RonaldRonald
1,7531921
edited Dec 25 '18 at 13:15
Matt Samuel
38.5k63768
asked Nov 25 '15 at 14:34
RonaldRonald
1,7531921
edited Dec 25 '18 at 13:15
Matt Samuel
38.5k63768
edited Dec 25 '18 at 13:15
Matt Samuel
38.5k63768
edited Dec 25 '18 at 13:15
Matt Samuel
38.5k63768
38.5k63768
asked Nov 25 '15 at 14:34
RonaldRonald
1,7531921
asked Nov 25 '15 at 14:34
RonaldRonald
1,7531921
asked Nov 25 '15 at 14:34
RonaldRonald
1,7531921
1,7531921
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Usually Schubert cells are constructed using the structure theory of semisimple Lie groups (in this case $SL(3,mathbb C)$), but for this case, one can give an explicit description as follows. Start by fixing one flag, say the standard one $mathbb Csubsetmathbb C^2subsetmathbb C^3$. This will be the unique Schubert-cell of dimension $0$. The further cells can be indexed by the dimensions of the intersections of their constituents with the spaces in the standard flag.
Explicitly, there are two Schubert-cells of dimension $1$. One of those consists of flags $ellsubsetmathbb C^2$ with $ellneqBbb C$. This is the complement of the point defined by $mathbb C$ in $mathbb CP^2$ and hence diffeomorphic to $mathbb C$. The second cell of dimension one consists of flags of the form $mathbb Csubset W$ with $Wneqmathbb C^2$. This again looks like the complement of a point in the projectivization of $mathbb C^3/mathbb C$, and hence like $mathbb C$.
From now on, we only have to look at flags $ellsubset W$, where $ellneqmathbb C$ and $Wneqmathbb C^2$. There are two Schubert-cells of Dimension $2$. One of them is formed by those flags, for which $mathbb Csubset W$. Since $ellneqmathbb C$ and $Wneqmathbb C^2$, we get $W=mathbb Coplusell$. Hence the flag is determined by the line $ell$, which is not contained in $mathbb C^2$. So this looks like $mathbb CP^3setminusmathbb CP^2congmathbb C^2$. The second of these cells consists of those flags, for which $ellsubsetmathbb C^2$. Since $Wneqmathbb C^2$, this implies that $ell=Wcapmathbb C^2$, so this time, the flag is determined by the plane $W$, which should not contain $mathbb C$. Again this shows that the space of such planes is isomorphic to $mathbb C^2$.
Finally, there is one cell of dimension $3$ consisting of the "generic" flags, for which $ell$ is not contained in $mathbb C^2$, and $mathbb C$ is not contained in $W$. Such a flag is determined by $ell$, which lies in $mathbb CP^3setminusmathbb CP^2$ and $Wcapmathbb C^2$, which lies in the complement of a point in $mathbb CP^2$. Hence the space of generic flags looks like $mathbb C^3$.
answered Nov 25 '15 at 15:25
Andreas CapAndreas Cap
11.2k923
$endgroup$
$begingroup$
thanks a lot Andreas .. this is very nice!
$endgroup$
– Ronald
Nov 25 '15 at 17:41
$begingroup$
I have a question here: What do you mean by the projectivization of $mathbb C^3/ mathbb C$?
$endgroup$
– Ronald
Dec 2 '15 at 2:17
$begingroup$
What I mean is take the qoutient space $mathbb C^3/mathbb C$ and consider the space of 1-dimensional complex subspaces in there. Then the canonical projection $pi: mathbb C^3tomathbb C^3/mathbb C$ induces a bijection between the set of planes in $mathbb C^3$ containing the line $mathbb C$ and the space of lines in the qoutient.
$endgroup$
– Andreas Cap
Dec 2 '15 at 12:05
add a comment |
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$begingroup$
Usually Schubert cells are constructed using the structure theory of semisimple Lie groups (in this case $SL(3,mathbb C)$), but for this case, one can give an explicit description as follows. Start by fixing one flag, say the standard one $mathbb Csubsetmathbb C^2subsetmathbb C^3$. This will be the unique Schubert-cell of dimension $0$. The further cells can be indexed by the dimensions of the intersections of their constituents with the spaces in the standard flag.
Explicitly, there are two Schubert-cells of dimension $1$. One of those consists of flags $ellsubsetmathbb C^2$ with $ellneqBbb C$. This is the complement of the point defined by $mathbb C$ in $mathbb CP^2$ and hence diffeomorphic to $mathbb C$. The second cell of dimension one consists of flags of the form $mathbb Csubset W$ with $Wneqmathbb C^2$. This again looks like the complement of a point in the projectivization of $mathbb C^3/mathbb C$, and hence like $mathbb C$.
From now on, we only have to look at flags $ellsubset W$, where $ellneqmathbb C$ and $Wneqmathbb C^2$. There are two Schubert-cells of Dimension $2$. One of them is formed by those flags, for which $mathbb Csubset W$. Since $ellneqmathbb C$ and $Wneqmathbb C^2$, we get $W=mathbb Coplusell$. Hence the flag is determined by the line $ell$, which is not contained in $mathbb C^2$. So this looks like $mathbb CP^3setminusmathbb CP^2congmathbb C^2$. The second of these cells consists of those flags, for which $ellsubsetmathbb C^2$. Since $Wneqmathbb C^2$, this implies that $ell=Wcapmathbb C^2$, so this time, the flag is determined by the plane $W$, which should not contain $mathbb C$. Again this shows that the space of such planes is isomorphic to $mathbb C^2$.
Finally, there is one cell of dimension $3$ consisting of the "generic" flags, for which $ell$ is not contained in $mathbb C^2$, and $mathbb C$ is not contained in $W$. Such a flag is determined by $ell$, which lies in $mathbb CP^3setminusmathbb CP^2$ and $Wcapmathbb C^2$, which lies in the complement of a point in $mathbb CP^2$. Hence the space of generic flags looks like $mathbb C^3$.
answered Nov 25 '15 at 15:25
Andreas CapAndreas Cap
11.2k923
$endgroup$
$begingroup$
thanks a lot Andreas .. this is very nice!
$endgroup$
– Ronald
Nov 25 '15 at 17:41
$begingroup$
I have a question here: What do you mean by the projectivization of $mathbb C^3/ mathbb C$?
$endgroup$
– Ronald
Dec 2 '15 at 2:17
$begingroup$
What I mean is take the qoutient space $mathbb C^3/mathbb C$ and consider the space of 1-dimensional complex subspaces in there. Then the canonical projection $pi: mathbb C^3tomathbb C^3/mathbb C$ induces a bijection between the set of planes in $mathbb C^3$ containing the line $mathbb C$ and the space of lines in the qoutient.
$endgroup$
– Andreas Cap
Dec 2 '15 at 12:05
add a comment |
$begingroup$
Usually Schubert cells are constructed using the structure theory of semisimple Lie groups (in this case $SL(3,mathbb C)$), but for this case, one can give an explicit description as follows. Start by fixing one flag, say the standard one $mathbb Csubsetmathbb C^2subsetmathbb C^3$. This will be the unique Schubert-cell of dimension $0$. The further cells can be indexed by the dimensions of the intersections of their constituents with the spaces in the standard flag.
Explicitly, there are two Schubert-cells of dimension $1$. One of those consists of flags $ellsubsetmathbb C^2$ with $ellneqBbb C$. This is the complement of the point defined by $mathbb C$ in $mathbb CP^2$ and hence diffeomorphic to $mathbb C$. The second cell of dimension one consists of flags of the form $mathbb Csubset W$ with $Wneqmathbb C^2$. This again looks like the complement of a point in the projectivization of $mathbb C^3/mathbb C$, and hence like $mathbb C$.
From now on, we only have to look at flags $ellsubset W$, where $ellneqmathbb C$ and $Wneqmathbb C^2$. There are two Schubert-cells of Dimension $2$. One of them is formed by those flags, for which $mathbb Csubset W$. Since $ellneqmathbb C$ and $Wneqmathbb C^2$, we get $W=mathbb Coplusell$. Hence the flag is determined by the line $ell$, which is not contained in $mathbb C^2$. So this looks like $mathbb CP^3setminusmathbb CP^2congmathbb C^2$. The second of these cells consists of those flags, for which $ellsubsetmathbb C^2$. Since $Wneqmathbb C^2$, this implies that $ell=Wcapmathbb C^2$, so this time, the flag is determined by the plane $W$, which should not contain $mathbb C$. Again this shows that the space of such planes is isomorphic to $mathbb C^2$.
Finally, there is one cell of dimension $3$ consisting of the "generic" flags, for which $ell$ is not contained in $mathbb C^2$, and $mathbb C$ is not contained in $W$. Such a flag is determined by $ell$, which lies in $mathbb CP^3setminusmathbb CP^2$ and $Wcapmathbb C^2$, which lies in the complement of a point in $mathbb CP^2$. Hence the space of generic flags looks like $mathbb C^3$.
answered Nov 25 '15 at 15:25
Andreas CapAndreas Cap
11.2k923
$endgroup$
$begingroup$
thanks a lot Andreas .. this is very nice!
$endgroup$
– Ronald
Nov 25 '15 at 17:41
$begingroup$
I have a question here: What do you mean by the projectivization of $mathbb C^3/ mathbb C$?
$endgroup$
– Ronald
Dec 2 '15 at 2:17
$begingroup$
What I mean is take the qoutient space $mathbb C^3/mathbb C$ and consider the space of 1-dimensional complex subspaces in there. Then the canonical projection $pi: mathbb C^3tomathbb C^3/mathbb C$ induces a bijection between the set of planes in $mathbb C^3$ containing the line $mathbb C$ and the space of lines in the qoutient.
$endgroup$
– Andreas Cap
Dec 2 '15 at 12:05
add a comment |
$begingroup$
Usually Schubert cells are constructed using the structure theory of semisimple Lie groups (in this case $SL(3,mathbb C)$), but for this case, one can give an explicit description as follows. Start by fixing one flag, say the standard one $mathbb Csubsetmathbb C^2subsetmathbb C^3$. This will be the unique Schubert-cell of dimension $0$. The further cells can be indexed by the dimensions of the intersections of their constituents with the spaces in the standard flag.
Explicitly, there are two Schubert-cells of dimension $1$. One of those consists of flags $ellsubsetmathbb C^2$ with $ellneqBbb C$. This is the complement of the point defined by $mathbb C$ in $mathbb CP^2$ and hence diffeomorphic to $mathbb C$. The second cell of dimension one consists of flags of the form $mathbb Csubset W$ with $Wneqmathbb C^2$. This again looks like the complement of a point in the projectivization of $mathbb C^3/mathbb C$, and hence like $mathbb C$.
From now on, we only have to look at flags $ellsubset W$, where $ellneqmathbb C$ and $Wneqmathbb C^2$. There are two Schubert-cells of Dimension $2$. One of them is formed by those flags, for which $mathbb Csubset W$. Since $ellneqmathbb C$ and $Wneqmathbb C^2$, we get $W=mathbb Coplusell$. Hence the flag is determined by the line $ell$, which is not contained in $mathbb C^2$. So this looks like $mathbb CP^3setminusmathbb CP^2congmathbb C^2$. The second of these cells consists of those flags, for which $ellsubsetmathbb C^2$. Since $Wneqmathbb C^2$, this implies that $ell=Wcapmathbb C^2$, so this time, the flag is determined by the plane $W$, which should not contain $mathbb C$. Again this shows that the space of such planes is isomorphic to $mathbb C^2$.
Finally, there is one cell of dimension $3$ consisting of the "generic" flags, for which $ell$ is not contained in $mathbb C^2$, and $mathbb C$ is not contained in $W$. Such a flag is determined by $ell$, which lies in $mathbb CP^3setminusmathbb CP^2$ and $Wcapmathbb C^2$, which lies in the complement of a point in $mathbb CP^2$. Hence the space of generic flags looks like $mathbb C^3$.
answered Nov 25 '15 at 15:25
Andreas CapAndreas Cap
11.2k923
$endgroup$
Usually Schubert cells are constructed using the structure theory of semisimple Lie groups (in this case $SL(3,mathbb C)$), but for this case, one can give an explicit description as follows. Start by fixing one flag, say the standard one $mathbb Csubsetmathbb C^2subsetmathbb C^3$. This will be the unique Schubert-cell of dimension $0$. The further cells can be indexed by the dimensions of the intersections of their constituents with the spaces in the standard flag.
Explicitly, there are two Schubert-cells of dimension $1$. One of those consists of flags $ellsubsetmathbb C^2$ with $ellneqBbb C$. This is the complement of the point defined by $mathbb C$ in $mathbb CP^2$ and hence diffeomorphic to $mathbb C$. The second cell of dimension one consists of flags of the form $mathbb Csubset W$ with $Wneqmathbb C^2$. This again looks like the complement of a point in the projectivization of $mathbb C^3/mathbb C$, and hence like $mathbb C$.
From now on, we only have to look at flags $ellsubset W$, where $ellneqmathbb C$ and $Wneqmathbb C^2$. There are two Schubert-cells of Dimension $2$. One of them is formed by those flags, for which $mathbb Csubset W$. Since $ellneqmathbb C$ and $Wneqmathbb C^2$, we get $W=mathbb Coplusell$. Hence the flag is determined by the line $ell$, which is not contained in $mathbb C^2$. So this looks like $mathbb CP^3setminusmathbb CP^2congmathbb C^2$. The second of these cells consists of those flags, for which $ellsubsetmathbb C^2$. Since $Wneqmathbb C^2$, this implies that $ell=Wcapmathbb C^2$, so this time, the flag is determined by the plane $W$, which should not contain $mathbb C$. Again this shows that the space of such planes is isomorphic to $mathbb C^2$.
Finally, there is one cell of dimension $3$ consisting of the "generic" flags, for which $ell$ is not contained in $mathbb C^2$, and $mathbb C$ is not contained in $W$. Such a flag is determined by $ell$, which lies in $mathbb CP^3setminusmathbb CP^2$ and $Wcapmathbb C^2$, which lies in the complement of a point in $mathbb CP^2$. Hence the space of generic flags looks like $mathbb C^3$.
answered Nov 25 '15 at 15:25
Andreas CapAndreas Cap
11.2k923
answered Nov 25 '15 at 15:25
Andreas CapAndreas Cap
11.2k923
answered Nov 25 '15 at 15:25
Andreas CapAndreas Cap
11.2k923
answered Nov 25 '15 at 15:25
Andreas CapAndreas Cap
11.2k923
11.2k923
$begingroup$
thanks a lot Andreas .. this is very nice!
$endgroup$
– Ronald
Nov 25 '15 at 17:41
$begingroup$
I have a question here: What do you mean by the projectivization of $mathbb C^3/ mathbb C$?
$endgroup$
– Ronald
Dec 2 '15 at 2:17
$begingroup$
What I mean is take the qoutient space $mathbb C^3/mathbb C$ and consider the space of 1-dimensional complex subspaces in there. Then the canonical projection $pi: mathbb C^3tomathbb C^3/mathbb C$ induces a bijection between the set of planes in $mathbb C^3$ containing the line $mathbb C$ and the space of lines in the qoutient.
$endgroup$
– Andreas Cap
Dec 2 '15 at 12:05
add a comment |
$begingroup$
thanks a lot Andreas .. this is very nice!
$endgroup$
– Ronald
Nov 25 '15 at 17:41
$begingroup$
I have a question here: What do you mean by the projectivization of $mathbb C^3/ mathbb C$?
$endgroup$
– Ronald
Dec 2 '15 at 2:17
$begingroup$
What I mean is take the qoutient space $mathbb C^3/mathbb C$ and consider the space of 1-dimensional complex subspaces in there. Then the canonical projection $pi: mathbb C^3tomathbb C^3/mathbb C$ induces a bijection between the set of planes in $mathbb C^3$ containing the line $mathbb C$ and the space of lines in the qoutient.
$endgroup$
– Andreas Cap
Dec 2 '15 at 12:05
$begingroup$
thanks a lot Andreas .. this is very nice!
$endgroup$
– Ronald
Nov 25 '15 at 17:41
$begingroup$
thanks a lot Andreas .. this is very nice!
$endgroup$
– Ronald
Nov 25 '15 at 17:41
$begingroup$
I have a question here: What do you mean by the projectivization of $mathbb C^3/ mathbb C$?
$endgroup$
– Ronald
Dec 2 '15 at 2:17
$begingroup$
I have a question here: What do you mean by the projectivization of $mathbb C^3/ mathbb C$?
$endgroup$
– Ronald
Dec 2 '15 at 2:17
$begingroup$
What I mean is take the qoutient space $mathbb C^3/mathbb C$ and consider the space of 1-dimensional complex subspaces in there. Then the canonical projection $pi: mathbb C^3tomathbb C^3/mathbb C$ induces a bijection between the set of planes in $mathbb C^3$ containing the line $mathbb C$ and the space of lines in the qoutient.
$endgroup$
– Andreas Cap
Dec 2 '15 at 12:05
$begingroup$
What I mean is take the qoutient space $mathbb C^3/mathbb C$ and consider the space of 1-dimensional complex subspaces in there. Then the canonical projection $pi: mathbb C^3tomathbb C^3/mathbb C$ induces a bijection between the set of planes in $mathbb C^3$ containing the line $mathbb C$ and the space of lines in the qoutient.
$endgroup$
– Andreas Cap
Dec 2 '15 at 12:05
add a comment |
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