Bruhat decomposition of flag variety.
Let $GL_{n}/B_{n}$ be the complete flag variety (where $B_{n}$ is the group of upper-triangular nonsingular matrices). Let $W$ be the Weyl group of $U(n)$ (that in this contest is a permutation group). We have $GL_{n}/B_{n} simeq U(n)/T_{n}$ (where $U(n)$ is the unitary group and $T_{n}$ its maximal torus). I have to build an explicit Bruhat decomposition of $U(n)/T_{n}$. Could you help me to find the cells end prove that there are $|W|$ cells of even dimension? (Also in a particular situation, for example $U(3)/(S^{1})^{3}$.)
abstract-algebra algebraic-topology schubert-calculus
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Let $GL_{n}/B_{n}$ be the complete flag variety (where $B_{n}$ is the group of upper-triangular nonsingular matrices). Let $W$ be the Weyl group of $U(n)$ (that in this contest is a permutation group). We have $GL_{n}/B_{n} simeq U(n)/T_{n}$ (where $U(n)$ is the unitary group and $T_{n}$ its maximal torus). I have to build an explicit Bruhat decomposition of $U(n)/T_{n}$. Could you help me to find the cells end prove that there are $|W|$ cells of even dimension? (Also in a particular situation, for example $U(3)/(S^{1})^{3}$.)
abstract-algebra algebraic-topology schubert-calculus
2
How is this different from your previous question: math.stackexchange.com/questions/279270/…? If not substantially different, you may consider deleting this one and editing the previous one.
– Jason DeVito
Jan 16 '13 at 12:58
Aren't these cells essentially subsets of the ${n-1choose 2}$ matrix entries below the diagonal? We do require that the lower triangular matrices with non-zero entries limited to the subset form a subgroup, or equivalently, the subset must meet the condition that position $(i,j), i>j,$ is included if any position on row $i$ AND any position on column $j$ is included. When $n=3$ the "legal" sets are: the empty set, ${(2,1)}$, ${3,2}$, ${(3,1),(3,2)}$, ${(2,1),(3,1)}$ and ${(2,1),(3,1),(3,2)}$. Each position can house a complex number so a cell with $j$ positions has dimension $2j$.
– Jyrki Lahtonen
Jan 16 '13 at 13:19
add a comment |
Let $GL_{n}/B_{n}$ be the complete flag variety (where $B_{n}$ is the group of upper-triangular nonsingular matrices). Let $W$ be the Weyl group of $U(n)$ (that in this contest is a permutation group). We have $GL_{n}/B_{n} simeq U(n)/T_{n}$ (where $U(n)$ is the unitary group and $T_{n}$ its maximal torus). I have to build an explicit Bruhat decomposition of $U(n)/T_{n}$. Could you help me to find the cells end prove that there are $|W|$ cells of even dimension? (Also in a particular situation, for example $U(3)/(S^{1})^{3}$.)
abstract-algebra algebraic-topology schubert-calculus
Let $GL_{n}/B_{n}$ be the complete flag variety (where $B_{n}$ is the group of upper-triangular nonsingular matrices). Let $W$ be the Weyl group of $U(n)$ (that in this contest is a permutation group). We have $GL_{n}/B_{n} simeq U(n)/T_{n}$ (where $U(n)$ is the unitary group and $T_{n}$ its maximal torus). I have to build an explicit Bruhat decomposition of $U(n)/T_{n}$. Could you help me to find the cells end prove that there are $|W|$ cells of even dimension? (Also in a particular situation, for example $U(3)/(S^{1})^{3}$.)
abstract-algebra algebraic-topology schubert-calculus
abstract-algebra algebraic-topology schubert-calculus
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Matt Samuel
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asked Jan 16 '13 at 11:54
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How is this different from your previous question: math.stackexchange.com/questions/279270/…? If not substantially different, you may consider deleting this one and editing the previous one.
– Jason DeVito
Jan 16 '13 at 12:58
Aren't these cells essentially subsets of the ${n-1choose 2}$ matrix entries below the diagonal? We do require that the lower triangular matrices with non-zero entries limited to the subset form a subgroup, or equivalently, the subset must meet the condition that position $(i,j), i>j,$ is included if any position on row $i$ AND any position on column $j$ is included. When $n=3$ the "legal" sets are: the empty set, ${(2,1)}$, ${3,2}$, ${(3,1),(3,2)}$, ${(2,1),(3,1)}$ and ${(2,1),(3,1),(3,2)}$. Each position can house a complex number so a cell with $j$ positions has dimension $2j$.
– Jyrki Lahtonen
Jan 16 '13 at 13:19
add a comment |
2
How is this different from your previous question: math.stackexchange.com/questions/279270/…? If not substantially different, you may consider deleting this one and editing the previous one.
– Jason DeVito
Jan 16 '13 at 12:58
Aren't these cells essentially subsets of the ${n-1choose 2}$ matrix entries below the diagonal? We do require that the lower triangular matrices with non-zero entries limited to the subset form a subgroup, or equivalently, the subset must meet the condition that position $(i,j), i>j,$ is included if any position on row $i$ AND any position on column $j$ is included. When $n=3$ the "legal" sets are: the empty set, ${(2,1)}$, ${3,2}$, ${(3,1),(3,2)}$, ${(2,1),(3,1)}$ and ${(2,1),(3,1),(3,2)}$. Each position can house a complex number so a cell with $j$ positions has dimension $2j$.
– Jyrki Lahtonen
Jan 16 '13 at 13:19
2
2
How is this different from your previous question: math.stackexchange.com/questions/279270/…? If not substantially different, you may consider deleting this one and editing the previous one.
– Jason DeVito
Jan 16 '13 at 12:58
How is this different from your previous question: math.stackexchange.com/questions/279270/…? If not substantially different, you may consider deleting this one and editing the previous one.
– Jason DeVito
Jan 16 '13 at 12:58
Aren't these cells essentially subsets of the ${n-1choose 2}$ matrix entries below the diagonal? We do require that the lower triangular matrices with non-zero entries limited to the subset form a subgroup, or equivalently, the subset must meet the condition that position $(i,j), i>j,$ is included if any position on row $i$ AND any position on column $j$ is included. When $n=3$ the "legal" sets are: the empty set, ${(2,1)}$, ${3,2}$, ${(3,1),(3,2)}$, ${(2,1),(3,1)}$ and ${(2,1),(3,1),(3,2)}$. Each position can house a complex number so a cell with $j$ positions has dimension $2j$.
– Jyrki Lahtonen
Jan 16 '13 at 13:19
Aren't these cells essentially subsets of the ${n-1choose 2}$ matrix entries below the diagonal? We do require that the lower triangular matrices with non-zero entries limited to the subset form a subgroup, or equivalently, the subset must meet the condition that position $(i,j), i>j,$ is included if any position on row $i$ AND any position on column $j$ is included. When $n=3$ the "legal" sets are: the empty set, ${(2,1)}$, ${3,2}$, ${(3,1),(3,2)}$, ${(2,1),(3,1)}$ and ${(2,1),(3,1),(3,2)}$. Each position can house a complex number so a cell with $j$ positions has dimension $2j$.
– Jyrki Lahtonen
Jan 16 '13 at 13:19
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How is this different from your previous question: math.stackexchange.com/questions/279270/…? If not substantially different, you may consider deleting this one and editing the previous one.
– Jason DeVito
Jan 16 '13 at 12:58
Aren't these cells essentially subsets of the ${n-1choose 2}$ matrix entries below the diagonal? We do require that the lower triangular matrices with non-zero entries limited to the subset form a subgroup, or equivalently, the subset must meet the condition that position $(i,j), i>j,$ is included if any position on row $i$ AND any position on column $j$ is included. When $n=3$ the "legal" sets are: the empty set, ${(2,1)}$, ${3,2}$, ${(3,1),(3,2)}$, ${(2,1),(3,1)}$ and ${(2,1),(3,1),(3,2)}$. Each position can house a complex number so a cell with $j$ positions has dimension $2j$.
– Jyrki Lahtonen
Jan 16 '13 at 13:19