Flag manifolds as incidence correspondences
$begingroup$
Let $G$ be a reductive group, $B$ a Borel and $P_j$ the maximal parabolics, indexed by the vertices $j$ of the Dynkin diagram. Then $B = bigcap_j P_j$, so the flag manifold $G/B$ injects into $prod_j G/P_j$.
In the classical types, this is a useful concrete description of $G/B$: In type $A_{n-1}$, $GL_n/P_j$ is the Grassmannian of $j$-planes in $n$-space, so a point of $prod_j GL_n/P_j$ is a sequence of subspaces $(V_1, V_2, cdots, V_{n-1})$, with $dim V_j = j$. The subspace $G/B$ is the flags $V_1 subset V_2 subset cdots subset V_{n-1}$.
Similarly, in types $B$ and $C$, the $G/P_j$ are isotropic flags, and we get the standard descriptions of the type $B$/$C$ flag manifolds as complete isotropic flags.
In type $D_n$, the $n-2$ vertices on the long leg of the diagram correspond to isotropic subspaces of dimensions $1 leq j leq n-2$, and the two extra vertices correspond to the two types of isotropic $n$-plane: We get that $G/B$ is flags $V_1 subset V_2 subset cdots subset V_{n-2} subset V_n^+, V_n^-$, where $V_j$ is an isotropic $j$ plane and $V_n^+$ and $V_n^-$ are of the $+$ type and the $-$ type. This isn't the standard description of complete isotropic flags, but it is easy to convert to one: Given $(V_n^+, V_n^-)$, we can recover $V_{n-1}$ as $V_n^+ cap V_n^-$ and, given $(V_{n-1}, V_n^+)$, we can recover $V_n^-$ as the other isotropic $n$-plane between $V_{n-1}$ and $V_{n-1}^{perp}$. Note that the condition $V_{n-2} subseteq V_n^+, V_n^-$ together with the types of $V_n^{pm}$ actually forces $dim (V_n^+ cap V_n^-) = n-1$.
I note that, in each of these cases, I only need to impose conditions on pairs $(V_i, V_j)$ where there is an edge $(i,j)$ in the Dynkin diagram. This leads me to my question:
For two Dynkin vertices $a$, $b$, let $pi_{ab}$ be the projection of $prod G/P_j$ onto $G/P_a times G/P_b$. Is it still right in the exceptional types that $G/B = bigcap pi_{ab}^{-1}left( pi_{ab}( G/B) right) subset prod G/P_j$ where the intersection runs over edges $(a,b)$ of the Dynkin diagram? Does this question make sense, and is the answer yes, for Kac-Moody groups?
Motivation: I've committed to teach a course on Bruhat orders and Schuberty things next Fall, and I'm thinking about how to tell combinatorialists about flag manifolds in other types without getting into the whole construction of reductive/Kac-Moody groups.
lie-groups homogeneous-spaces exceptional-groups
$endgroup$
add a comment |
$begingroup$
Let $G$ be a reductive group, $B$ a Borel and $P_j$ the maximal parabolics, indexed by the vertices $j$ of the Dynkin diagram. Then $B = bigcap_j P_j$, so the flag manifold $G/B$ injects into $prod_j G/P_j$.
In the classical types, this is a useful concrete description of $G/B$: In type $A_{n-1}$, $GL_n/P_j$ is the Grassmannian of $j$-planes in $n$-space, so a point of $prod_j GL_n/P_j$ is a sequence of subspaces $(V_1, V_2, cdots, V_{n-1})$, with $dim V_j = j$. The subspace $G/B$ is the flags $V_1 subset V_2 subset cdots subset V_{n-1}$.
Similarly, in types $B$ and $C$, the $G/P_j$ are isotropic flags, and we get the standard descriptions of the type $B$/$C$ flag manifolds as complete isotropic flags.
In type $D_n$, the $n-2$ vertices on the long leg of the diagram correspond to isotropic subspaces of dimensions $1 leq j leq n-2$, and the two extra vertices correspond to the two types of isotropic $n$-plane: We get that $G/B$ is flags $V_1 subset V_2 subset cdots subset V_{n-2} subset V_n^+, V_n^-$, where $V_j$ is an isotropic $j$ plane and $V_n^+$ and $V_n^-$ are of the $+$ type and the $-$ type. This isn't the standard description of complete isotropic flags, but it is easy to convert to one: Given $(V_n^+, V_n^-)$, we can recover $V_{n-1}$ as $V_n^+ cap V_n^-$ and, given $(V_{n-1}, V_n^+)$, we can recover $V_n^-$ as the other isotropic $n$-plane between $V_{n-1}$ and $V_{n-1}^{perp}$. Note that the condition $V_{n-2} subseteq V_n^+, V_n^-$ together with the types of $V_n^{pm}$ actually forces $dim (V_n^+ cap V_n^-) = n-1$.
I note that, in each of these cases, I only need to impose conditions on pairs $(V_i, V_j)$ where there is an edge $(i,j)$ in the Dynkin diagram. This leads me to my question:
For two Dynkin vertices $a$, $b$, let $pi_{ab}$ be the projection of $prod G/P_j$ onto $G/P_a times G/P_b$. Is it still right in the exceptional types that $G/B = bigcap pi_{ab}^{-1}left( pi_{ab}( G/B) right) subset prod G/P_j$ where the intersection runs over edges $(a,b)$ of the Dynkin diagram? Does this question make sense, and is the answer yes, for Kac-Moody groups?
Motivation: I've committed to teach a course on Bruhat orders and Schuberty things next Fall, and I'm thinking about how to tell combinatorialists about flag manifolds in other types without getting into the whole construction of reductive/Kac-Moody groups.
lie-groups homogeneous-spaces exceptional-groups
$endgroup$
add a comment |
$begingroup$
Let $G$ be a reductive group, $B$ a Borel and $P_j$ the maximal parabolics, indexed by the vertices $j$ of the Dynkin diagram. Then $B = bigcap_j P_j$, so the flag manifold $G/B$ injects into $prod_j G/P_j$.
In the classical types, this is a useful concrete description of $G/B$: In type $A_{n-1}$, $GL_n/P_j$ is the Grassmannian of $j$-planes in $n$-space, so a point of $prod_j GL_n/P_j$ is a sequence of subspaces $(V_1, V_2, cdots, V_{n-1})$, with $dim V_j = j$. The subspace $G/B$ is the flags $V_1 subset V_2 subset cdots subset V_{n-1}$.
Similarly, in types $B$ and $C$, the $G/P_j$ are isotropic flags, and we get the standard descriptions of the type $B$/$C$ flag manifolds as complete isotropic flags.
In type $D_n$, the $n-2$ vertices on the long leg of the diagram correspond to isotropic subspaces of dimensions $1 leq j leq n-2$, and the two extra vertices correspond to the two types of isotropic $n$-plane: We get that $G/B$ is flags $V_1 subset V_2 subset cdots subset V_{n-2} subset V_n^+, V_n^-$, where $V_j$ is an isotropic $j$ plane and $V_n^+$ and $V_n^-$ are of the $+$ type and the $-$ type. This isn't the standard description of complete isotropic flags, but it is easy to convert to one: Given $(V_n^+, V_n^-)$, we can recover $V_{n-1}$ as $V_n^+ cap V_n^-$ and, given $(V_{n-1}, V_n^+)$, we can recover $V_n^-$ as the other isotropic $n$-plane between $V_{n-1}$ and $V_{n-1}^{perp}$. Note that the condition $V_{n-2} subseteq V_n^+, V_n^-$ together with the types of $V_n^{pm}$ actually forces $dim (V_n^+ cap V_n^-) = n-1$.
I note that, in each of these cases, I only need to impose conditions on pairs $(V_i, V_j)$ where there is an edge $(i,j)$ in the Dynkin diagram. This leads me to my question:
For two Dynkin vertices $a$, $b$, let $pi_{ab}$ be the projection of $prod G/P_j$ onto $G/P_a times G/P_b$. Is it still right in the exceptional types that $G/B = bigcap pi_{ab}^{-1}left( pi_{ab}( G/B) right) subset prod G/P_j$ where the intersection runs over edges $(a,b)$ of the Dynkin diagram? Does this question make sense, and is the answer yes, for Kac-Moody groups?
Motivation: I've committed to teach a course on Bruhat orders and Schuberty things next Fall, and I'm thinking about how to tell combinatorialists about flag manifolds in other types without getting into the whole construction of reductive/Kac-Moody groups.
lie-groups homogeneous-spaces exceptional-groups
$endgroup$
Let $G$ be a reductive group, $B$ a Borel and $P_j$ the maximal parabolics, indexed by the vertices $j$ of the Dynkin diagram. Then $B = bigcap_j P_j$, so the flag manifold $G/B$ injects into $prod_j G/P_j$.
In the classical types, this is a useful concrete description of $G/B$: In type $A_{n-1}$, $GL_n/P_j$ is the Grassmannian of $j$-planes in $n$-space, so a point of $prod_j GL_n/P_j$ is a sequence of subspaces $(V_1, V_2, cdots, V_{n-1})$, with $dim V_j = j$. The subspace $G/B$ is the flags $V_1 subset V_2 subset cdots subset V_{n-1}$.
Similarly, in types $B$ and $C$, the $G/P_j$ are isotropic flags, and we get the standard descriptions of the type $B$/$C$ flag manifolds as complete isotropic flags.
In type $D_n$, the $n-2$ vertices on the long leg of the diagram correspond to isotropic subspaces of dimensions $1 leq j leq n-2$, and the two extra vertices correspond to the two types of isotropic $n$-plane: We get that $G/B$ is flags $V_1 subset V_2 subset cdots subset V_{n-2} subset V_n^+, V_n^-$, where $V_j$ is an isotropic $j$ plane and $V_n^+$ and $V_n^-$ are of the $+$ type and the $-$ type. This isn't the standard description of complete isotropic flags, but it is easy to convert to one: Given $(V_n^+, V_n^-)$, we can recover $V_{n-1}$ as $V_n^+ cap V_n^-$ and, given $(V_{n-1}, V_n^+)$, we can recover $V_n^-$ as the other isotropic $n$-plane between $V_{n-1}$ and $V_{n-1}^{perp}$. Note that the condition $V_{n-2} subseteq V_n^+, V_n^-$ together with the types of $V_n^{pm}$ actually forces $dim (V_n^+ cap V_n^-) = n-1$.
I note that, in each of these cases, I only need to impose conditions on pairs $(V_i, V_j)$ where there is an edge $(i,j)$ in the Dynkin diagram. This leads me to my question:
For two Dynkin vertices $a$, $b$, let $pi_{ab}$ be the projection of $prod G/P_j$ onto $G/P_a times G/P_b$. Is it still right in the exceptional types that $G/B = bigcap pi_{ab}^{-1}left( pi_{ab}( G/B) right) subset prod G/P_j$ where the intersection runs over edges $(a,b)$ of the Dynkin diagram? Does this question make sense, and is the answer yes, for Kac-Moody groups?
Motivation: I've committed to teach a course on Bruhat orders and Schuberty things next Fall, and I'm thinking about how to tell combinatorialists about flag manifolds in other types without getting into the whole construction of reductive/Kac-Moody groups.
lie-groups homogeneous-spaces exceptional-groups
lie-groups homogeneous-spaces exceptional-groups
asked 1 hour ago
David E SpeyerDavid E Speyer
106k8275536
106k8275536
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1 Answer
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$begingroup$
The "incidence" relation (at least for types $E_6$ and $E_7$) for all pairs of vertices $i$, $j$ is described in S. Garibaldi, M. Carr "Geometries, the principle of duality, and algebraic groups", Expo. Math. 24 (2006), 195-234. While in most cases it is just inclusion of respective subspaces (in some representation), for pairs (2,5) and (2,6) in $E_6$ case it is more complicated (see 7.21 loc. cit.) and doesn't seem to be deduced from the relations for edges only.
$endgroup$
$begingroup$
Interesting! But I don't think it is a counter-example to what I've asked above. The outer automorphism of $E_6$ interchanges $1$ with $6$ and $3$ with $5$ so, if the $(2,3)$ incidence is a consequence of the $(2,4)$ and $(4,3)$ incidences, then the $(2,5)$ incidence must likewise be a consequence of the $(2,4)$ and $(4,5)$ incidences.
$endgroup$
– David E Speyer
11 mins ago
add a comment |
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1 Answer
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1 Answer
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active
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active
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$begingroup$
The "incidence" relation (at least for types $E_6$ and $E_7$) for all pairs of vertices $i$, $j$ is described in S. Garibaldi, M. Carr "Geometries, the principle of duality, and algebraic groups", Expo. Math. 24 (2006), 195-234. While in most cases it is just inclusion of respective subspaces (in some representation), for pairs (2,5) and (2,6) in $E_6$ case it is more complicated (see 7.21 loc. cit.) and doesn't seem to be deduced from the relations for edges only.
$endgroup$
$begingroup$
Interesting! But I don't think it is a counter-example to what I've asked above. The outer automorphism of $E_6$ interchanges $1$ with $6$ and $3$ with $5$ so, if the $(2,3)$ incidence is a consequence of the $(2,4)$ and $(4,3)$ incidences, then the $(2,5)$ incidence must likewise be a consequence of the $(2,4)$ and $(4,5)$ incidences.
$endgroup$
– David E Speyer
11 mins ago
add a comment |
$begingroup$
The "incidence" relation (at least for types $E_6$ and $E_7$) for all pairs of vertices $i$, $j$ is described in S. Garibaldi, M. Carr "Geometries, the principle of duality, and algebraic groups", Expo. Math. 24 (2006), 195-234. While in most cases it is just inclusion of respective subspaces (in some representation), for pairs (2,5) and (2,6) in $E_6$ case it is more complicated (see 7.21 loc. cit.) and doesn't seem to be deduced from the relations for edges only.
$endgroup$
$begingroup$
Interesting! But I don't think it is a counter-example to what I've asked above. The outer automorphism of $E_6$ interchanges $1$ with $6$ and $3$ with $5$ so, if the $(2,3)$ incidence is a consequence of the $(2,4)$ and $(4,3)$ incidences, then the $(2,5)$ incidence must likewise be a consequence of the $(2,4)$ and $(4,5)$ incidences.
$endgroup$
– David E Speyer
11 mins ago
add a comment |
$begingroup$
The "incidence" relation (at least for types $E_6$ and $E_7$) for all pairs of vertices $i$, $j$ is described in S. Garibaldi, M. Carr "Geometries, the principle of duality, and algebraic groups", Expo. Math. 24 (2006), 195-234. While in most cases it is just inclusion of respective subspaces (in some representation), for pairs (2,5) and (2,6) in $E_6$ case it is more complicated (see 7.21 loc. cit.) and doesn't seem to be deduced from the relations for edges only.
$endgroup$
The "incidence" relation (at least for types $E_6$ and $E_7$) for all pairs of vertices $i$, $j$ is described in S. Garibaldi, M. Carr "Geometries, the principle of duality, and algebraic groups", Expo. Math. 24 (2006), 195-234. While in most cases it is just inclusion of respective subspaces (in some representation), for pairs (2,5) and (2,6) in $E_6$ case it is more complicated (see 7.21 loc. cit.) and doesn't seem to be deduced from the relations for edges only.
answered 48 mins ago
Victor PetrovVictor Petrov
1,00458
1,00458
$begingroup$
Interesting! But I don't think it is a counter-example to what I've asked above. The outer automorphism of $E_6$ interchanges $1$ with $6$ and $3$ with $5$ so, if the $(2,3)$ incidence is a consequence of the $(2,4)$ and $(4,3)$ incidences, then the $(2,5)$ incidence must likewise be a consequence of the $(2,4)$ and $(4,5)$ incidences.
$endgroup$
– David E Speyer
11 mins ago
add a comment |
$begingroup$
Interesting! But I don't think it is a counter-example to what I've asked above. The outer automorphism of $E_6$ interchanges $1$ with $6$ and $3$ with $5$ so, if the $(2,3)$ incidence is a consequence of the $(2,4)$ and $(4,3)$ incidences, then the $(2,5)$ incidence must likewise be a consequence of the $(2,4)$ and $(4,5)$ incidences.
$endgroup$
– David E Speyer
11 mins ago
$begingroup$
Interesting! But I don't think it is a counter-example to what I've asked above. The outer automorphism of $E_6$ interchanges $1$ with $6$ and $3$ with $5$ so, if the $(2,3)$ incidence is a consequence of the $(2,4)$ and $(4,3)$ incidences, then the $(2,5)$ incidence must likewise be a consequence of the $(2,4)$ and $(4,5)$ incidences.
$endgroup$
– David E Speyer
11 mins ago
$begingroup$
Interesting! But I don't think it is a counter-example to what I've asked above. The outer automorphism of $E_6$ interchanges $1$ with $6$ and $3$ with $5$ so, if the $(2,3)$ incidence is a consequence of the $(2,4)$ and $(4,3)$ incidences, then the $(2,5)$ incidence must likewise be a consequence of the $(2,4)$ and $(4,5)$ incidences.
$endgroup$
– David E Speyer
11 mins ago
add a comment |
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