Schubert cell decomposition and full flags












0












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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







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    0












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 25 '18 at 13:15









      Matt Samuel

      38.5k63768




      38.5k63768










      asked Nov 25 '15 at 14:34









      RonaldRonald

      1,7531921




      1,7531921






















          1 Answer
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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 $mathbb C^3$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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











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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 $mathbb C^3$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          3












          $begingroup$

          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 $mathbb C^3$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          3












          3








          3





          $begingroup$

          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 $mathbb C^3$.






          share|cite|improve this answer









          $endgroup$



          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 $mathbb C^3$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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