Mixing
Number of fixed points of torus action over partial flag variety
$begingroup$
Consider $gin U(n)$ and $tin T$, where $T$ is the diagonal maximal torus in $U(n)$.
Some common manifolds may be obtained as quotients of the $U(n)$ like the complex grassmannian, $Gr(k,n)=U(n)/U(k)times U(n-k)$ and the manifold of complete flags in $mathbb{C}^n$, $F_n=U(n)/T$, and there is an induced action of $T$ coming from the action of $U(n)$ in them. Counting the fixed points of this action yields ${nchoose k}$ and $n!$, respectively. These numbers are precisely $|w_n|/(,|w_{n-k}|cdot|w_{k}|,)$ and $|w_n|$, respectively, where $|w_k|$ denotes the order of the Weyl group of $U(k)$, i.e., $w_k=N_{U(k)}(T)/T$.
From this there is a natural candidate to generalize the result for partial flag varieties $U(n)/U(k_1)timescdotstimes U(k_d)$, where $k_1+cdots+k_d=n$: the number of fixed points of this action will be
$$
frac{|w_n|}{|w_{k_1}|cdots|w_{k_d}|}=frac{n!}{k_1!cdots k_d!}=frac{|N_{U(n)}(T)/T|}{|N_{U(k_1)}(T_1)/T_1|cdots|N_{U(k_d)}(T_d)/T_d|},
$$
where $T_i$ is the diagonal maximal torus in $U(k_i)$.
I would like some help on how to show this, in other words, how to show that there is a bijection between
$$
text{$gin U(n)$ such that $g^*tgin U(k_1)timescdots U(k_d)$ for all $tin T$ (fixed points)}
$$
and
$$
frac{N_{U(n)}(T)/T}{N_{U(k_1)}(T_1)/T_1timescdotstimes N_{U(k_d)}/T_{k_d}}=frac{N_{U(n)}(T)}{N_{U(k_1)}(T_1)timescdotstimes N_{U(k_d)}}.
$$
I'v tried brute forcing some conditions on the matrices $g$, yet I don't see anything that is of help, so I think a conceptual proof may be the way to go.
differential-geometry lie-groups group-actions schubert-calculus
edited Jan 5 at 3:12
Matt Samuel
37.8k63665
asked Aug 2 '16 at 8:43
hjhjhj57hjhjhj57
3,11111233
$endgroup$
|
show 3 more comments
$begingroup$
Consider $gin U(n)$ and $tin T$, where $T$ is the diagonal maximal torus in $U(n)$.
Some common manifolds may be obtained as quotients of the $U(n)$ like the complex grassmannian, $Gr(k,n)=U(n)/U(k)times U(n-k)$ and the manifold of complete flags in $mathbb{C}^n$, $F_n=U(n)/T$, and there is an induced action of $T$ coming from the action of $U(n)$ in them. Counting the fixed points of this action yields ${nchoose k}$ and $n!$, respectively. These numbers are precisely $|w_n|/(,|w_{n-k}|cdot|w_{k}|,)$ and $|w_n|$, respectively, where $|w_k|$ denotes the order of the Weyl group of $U(k)$, i.e., $w_k=N_{U(k)}(T)/T$.
From this there is a natural candidate to generalize the result for partial flag varieties $U(n)/U(k_1)timescdotstimes U(k_d)$, where $k_1+cdots+k_d=n$: the number of fixed points of this action will be
$$
frac{|w_n|}{|w_{k_1}|cdots|w_{k_d}|}=frac{n!}{k_1!cdots k_d!}=frac{|N_{U(n)}(T)/T|}{|N_{U(k_1)}(T_1)/T_1|cdots|N_{U(k_d)}(T_d)/T_d|},
$$
where $T_i$ is the diagonal maximal torus in $U(k_i)$.
I would like some help on how to show this, in other words, how to show that there is a bijection between
$$
text{$gin U(n)$ such that $g^*tgin U(k_1)timescdots U(k_d)$ for all $tin T$ (fixed points)}
$$
and
$$
frac{N_{U(n)}(T)/T}{N_{U(k_1)}(T_1)/T_1timescdotstimes N_{U(k_d)}/T_{k_d}}=frac{N_{U(n)}(T)}{N_{U(k_1)}(T_1)timescdotstimes N_{U(k_d)}}.
$$
I'v tried brute forcing some conditions on the matrices $g$, yet I don't see anything that is of help, so I think a conceptual proof may be the way to go.
differential-geometry lie-groups group-actions schubert-calculus
edited Jan 5 at 3:12
Matt Samuel
37.8k63665
asked Aug 2 '16 at 8:43
hjhjhj57hjhjhj57
3,11111233
$endgroup$
$begingroup$
I can tell you that you are correct, but the proof I know involves explicit parameterization of the partial flag variety and is probably not what you are looking for.
$endgroup$
– Matt Samuel
Aug 7 '16 at 1:11
$begingroup$
Hi @MattSamuel , could you tell me a little bit about how is this parametrization done? I'm mainly interested in counting the points, not in the particular path chosen to do so, as long as it's accessible to me (I know this is very subjective) it's good. I think the parametrization approach could be interesting if it goes deeper into the geometry of the spaces.
$endgroup$
– hjhjhj57
Aug 7 '16 at 1:55
$begingroup$
The textbook Young Tableaux by Fulton gives the parameterization of the Grassmannian and the complete flag variety, and you'd have to sort of interpolate between them to get what you want. It's not a topic I've often seen in something expository. I'm retiring for the night, but it may help to look through lecture notes on Schubert calculus. I found something with that title by Brion.
$endgroup$
– Matt Samuel
Aug 7 '16 at 2:43
$begingroup$
This article should do it for you, it has examples: uni-math.gwdg.de/tschinkel/SS05/school/kresch.pdf The torus acts by multiplying each column by a real number, and everything is renormalized so the $1$'s remain $1$. The fixed points are the matrices with only $1$'s and $0$'s.
$endgroup$
– Matt Samuel
Aug 7 '16 at 21:20
$begingroup$
Multiplying the columns by complex numbers is what I meant.
$endgroup$
– Matt Samuel
Aug 7 '16 at 21:27
|
show 3 more comments
$begingroup$
Consider $gin U(n)$ and $tin T$, where $T$ is the diagonal maximal torus in $U(n)$.
Some common manifolds may be obtained as quotients of the $U(n)$ like the complex grassmannian, $Gr(k,n)=U(n)/U(k)times U(n-k)$ and the manifold of complete flags in $mathbb{C}^n$, $F_n=U(n)/T$, and there is an induced action of $T$ coming from the action of $U(n)$ in them. Counting the fixed points of this action yields ${nchoose k}$ and $n!$, respectively. These numbers are precisely $|w_n|/(,|w_{n-k}|cdot|w_{k}|,)$ and $|w_n|$, respectively, where $|w_k|$ denotes the order of the Weyl group of $U(k)$, i.e., $w_k=N_{U(k)}(T)/T$.
From this there is a natural candidate to generalize the result for partial flag varieties $U(n)/U(k_1)timescdotstimes U(k_d)$, where $k_1+cdots+k_d=n$: the number of fixed points of this action will be
$$
frac{|w_n|}{|w_{k_1}|cdots|w_{k_d}|}=frac{n!}{k_1!cdots k_d!}=frac{|N_{U(n)}(T)/T|}{|N_{U(k_1)}(T_1)/T_1|cdots|N_{U(k_d)}(T_d)/T_d|},
$$
where $T_i$ is the diagonal maximal torus in $U(k_i)$.
I would like some help on how to show this, in other words, how to show that there is a bijection between
$$
text{$gin U(n)$ such that $g^*tgin U(k_1)timescdots U(k_d)$ for all $tin T$ (fixed points)}
$$
and
$$
frac{N_{U(n)}(T)/T}{N_{U(k_1)}(T_1)/T_1timescdotstimes N_{U(k_d)}/T_{k_d}}=frac{N_{U(n)}(T)}{N_{U(k_1)}(T_1)timescdotstimes N_{U(k_d)}}.
$$
I'v tried brute forcing some conditions on the matrices $g$, yet I don't see anything that is of help, so I think a conceptual proof may be the way to go.
differential-geometry lie-groups group-actions schubert-calculus
edited Jan 5 at 3:12
Matt Samuel
37.8k63665
asked Aug 2 '16 at 8:43
hjhjhj57hjhjhj57
3,11111233
$endgroup$
Consider $gin U(n)$ and $tin T$, where $T$ is the diagonal maximal torus in $U(n)$.
Some common manifolds may be obtained as quotients of the $U(n)$ like the complex grassmannian, $Gr(k,n)=U(n)/U(k)times U(n-k)$ and the manifold of complete flags in $mathbb{C}^n$, $F_n=U(n)/T$, and there is an induced action of $T$ coming from the action of $U(n)$ in them. Counting the fixed points of this action yields ${nchoose k}$ and $n!$, respectively. These numbers are precisely $|w_n|/(,|w_{n-k}|cdot|w_{k}|,)$ and $|w_n|$, respectively, where $|w_k|$ denotes the order of the Weyl group of $U(k)$, i.e., $w_k=N_{U(k)}(T)/T$.
From this there is a natural candidate to generalize the result for partial flag varieties $U(n)/U(k_1)timescdotstimes U(k_d)$, where $k_1+cdots+k_d=n$: the number of fixed points of this action will be
$$
frac{|w_n|}{|w_{k_1}|cdots|w_{k_d}|}=frac{n!}{k_1!cdots k_d!}=frac{|N_{U(n)}(T)/T|}{|N_{U(k_1)}(T_1)/T_1|cdots|N_{U(k_d)}(T_d)/T_d|},
$$
where $T_i$ is the diagonal maximal torus in $U(k_i)$.
I would like some help on how to show this, in other words, how to show that there is a bijection between
$$
text{$gin U(n)$ such that $g^*tgin U(k_1)timescdots U(k_d)$ for all $tin T$ (fixed points)}
$$
and
$$
frac{N_{U(n)}(T)/T}{N_{U(k_1)}(T_1)/T_1timescdotstimes N_{U(k_d)}/T_{k_d}}=frac{N_{U(n)}(T)}{N_{U(k_1)}(T_1)timescdotstimes N_{U(k_d)}}.
$$
I'v tried brute forcing some conditions on the matrices $g$, yet I don't see anything that is of help, so I think a conceptual proof may be the way to go.
differential-geometry lie-groups group-actions schubert-calculus
differential-geometry lie-groups group-actions schubert-calculus
edited Jan 5 at 3:12
Matt Samuel
37.8k63665
asked Aug 2 '16 at 8:43
hjhjhj57hjhjhj57
3,11111233
edited Jan 5 at 3:12
Matt Samuel
37.8k63665
asked Aug 2 '16 at 8:43
hjhjhj57hjhjhj57
3,11111233
edited Jan 5 at 3:12
Matt Samuel
37.8k63665
edited Jan 5 at 3:12
Matt Samuel
37.8k63665
edited Jan 5 at 3:12
Matt Samuel
37.8k63665
37.8k63665
asked Aug 2 '16 at 8:43
hjhjhj57hjhjhj57
3,11111233
asked Aug 2 '16 at 8:43
hjhjhj57hjhjhj57
3,11111233
asked Aug 2 '16 at 8:43
hjhjhj57hjhjhj57
3,11111233
3,11111233
$begingroup$
I can tell you that you are correct, but the proof I know involves explicit parameterization of the partial flag variety and is probably not what you are looking for.
$endgroup$
– Matt Samuel
Aug 7 '16 at 1:11
$begingroup$
Hi @MattSamuel , could you tell me a little bit about how is this parametrization done? I'm mainly interested in counting the points, not in the particular path chosen to do so, as long as it's accessible to me (I know this is very subjective) it's good. I think the parametrization approach could be interesting if it goes deeper into the geometry of the spaces.
$endgroup$
– hjhjhj57
Aug 7 '16 at 1:55
$begingroup$
The textbook Young Tableaux by Fulton gives the parameterization of the Grassmannian and the complete flag variety, and you'd have to sort of interpolate between them to get what you want. It's not a topic I've often seen in something expository. I'm retiring for the night, but it may help to look through lecture notes on Schubert calculus. I found something with that title by Brion.
$endgroup$
– Matt Samuel
Aug 7 '16 at 2:43
$begingroup$
This article should do it for you, it has examples: uni-math.gwdg.de/tschinkel/SS05/school/kresch.pdf The torus acts by multiplying each column by a real number, and everything is renormalized so the $1$'s remain $1$. The fixed points are the matrices with only $1$'s and $0$'s.
$endgroup$
– Matt Samuel
Aug 7 '16 at 21:20
$begingroup$
Multiplying the columns by complex numbers is what I meant.
$endgroup$
– Matt Samuel
Aug 7 '16 at 21:27
|
show 3 more comments
$begingroup$
I can tell you that you are correct, but the proof I know involves explicit parameterization of the partial flag variety and is probably not what you are looking for.
$endgroup$
– Matt Samuel
Aug 7 '16 at 1:11
$begingroup$
Hi @MattSamuel , could you tell me a little bit about how is this parametrization done? I'm mainly interested in counting the points, not in the particular path chosen to do so, as long as it's accessible to me (I know this is very subjective) it's good. I think the parametrization approach could be interesting if it goes deeper into the geometry of the spaces.
$endgroup$
– hjhjhj57
Aug 7 '16 at 1:55
$begingroup$
The textbook Young Tableaux by Fulton gives the parameterization of the Grassmannian and the complete flag variety, and you'd have to sort of interpolate between them to get what you want. It's not a topic I've often seen in something expository. I'm retiring for the night, but it may help to look through lecture notes on Schubert calculus. I found something with that title by Brion.
$endgroup$
– Matt Samuel
Aug 7 '16 at 2:43
$begingroup$
This article should do it for you, it has examples: uni-math.gwdg.de/tschinkel/SS05/school/kresch.pdf The torus acts by multiplying each column by a real number, and everything is renormalized so the $1$'s remain $1$. The fixed points are the matrices with only $1$'s and $0$'s.
$endgroup$
– Matt Samuel
Aug 7 '16 at 21:20
$begingroup$
Multiplying the columns by complex numbers is what I meant.
$endgroup$
– Matt Samuel
Aug 7 '16 at 21:27
$begingroup$
I can tell you that you are correct, but the proof I know involves explicit parameterization of the partial flag variety and is probably not what you are looking for.
$endgroup$
– Matt Samuel
Aug 7 '16 at 1:11
$begingroup$
I can tell you that you are correct, but the proof I know involves explicit parameterization of the partial flag variety and is probably not what you are looking for.
$endgroup$
– Matt Samuel
Aug 7 '16 at 1:11
$begingroup$
Hi @MattSamuel , could you tell me a little bit about how is this parametrization done? I'm mainly interested in counting the points, not in the particular path chosen to do so, as long as it's accessible to me (I know this is very subjective) it's good. I think the parametrization approach could be interesting if it goes deeper into the geometry of the spaces.
$endgroup$
– hjhjhj57
Aug 7 '16 at 1:55
$begingroup$
Hi @MattSamuel , could you tell me a little bit about how is this parametrization done? I'm mainly interested in counting the points, not in the particular path chosen to do so, as long as it's accessible to me (I know this is very subjective) it's good. I think the parametrization approach could be interesting if it goes deeper into the geometry of the spaces.
$endgroup$
– hjhjhj57
Aug 7 '16 at 1:55
$begingroup$
The textbook Young Tableaux by Fulton gives the parameterization of the Grassmannian and the complete flag variety, and you'd have to sort of interpolate between them to get what you want. It's not a topic I've often seen in something expository. I'm retiring for the night, but it may help to look through lecture notes on Schubert calculus. I found something with that title by Brion.
$endgroup$
– Matt Samuel
Aug 7 '16 at 2:43
$begingroup$
The textbook Young Tableaux by Fulton gives the parameterization of the Grassmannian and the complete flag variety, and you'd have to sort of interpolate between them to get what you want. It's not a topic I've often seen in something expository. I'm retiring for the night, but it may help to look through lecture notes on Schubert calculus. I found something with that title by Brion.
$endgroup$
– Matt Samuel
Aug 7 '16 at 2:43
$begingroup$
This article should do it for you, it has examples: uni-math.gwdg.de/tschinkel/SS05/school/kresch.pdf The torus acts by multiplying each column by a real number, and everything is renormalized so the $1$'s remain $1$. The fixed points are the matrices with only $1$'s and $0$'s.
$endgroup$
– Matt Samuel
Aug 7 '16 at 21:20
$begingroup$
This article should do it for you, it has examples: uni-math.gwdg.de/tschinkel/SS05/school/kresch.pdf The torus acts by multiplying each column by a real number, and everything is renormalized so the $1$'s remain $1$. The fixed points are the matrices with only $1$'s and $0$'s.
$endgroup$
– Matt Samuel
Aug 7 '16 at 21:20
$begingroup$
Multiplying the columns by complex numbers is what I meant.
$endgroup$
– Matt Samuel
Aug 7 '16 at 21:27
$begingroup$
Multiplying the columns by complex numbers is what I meant.
$endgroup$
– Matt Samuel
Aug 7 '16 at 21:27
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let me propose an absolutely different attempt to this problem.
The idea is to use Lefschetz fixed point theorem.
Recall that it states if $f colon X to X$ is an automorphism of a smooth manifold with all fixed points isolated, then
$$
sum_{f(x) =x} i_x = sum_{k=0}^{n} (-1)^koperatorname{Tr} f^*|_{H^k(X, mathbb{R})}
$$
Here $i_x$ are the indices of fixed points.
In our case the automorphism is homotopic to identity (since $T$ is connected), so the RHS of the Lefschetz formula is just the Euler characteristic of your partial flag variety.
One can check up that for any $t in T$ the map induced by left multiplication by $t$ in fact has only isolated fixed points with all the indices equal to $1$.
Therefore, LHS of the Lefschetz formula is precisely the number of the fixed points.
Thus it is rest to compute the (topological) Euler characteristic of a partial flag variety. Recall, that any partial flag variety can be seen as $G/P$, where $G$ is a complex algebraic group and $P subset G$ is a parabolic subgroup. It has cellular Schubert decomposition, that is $G/P = bigsqcup B_w$ , where all $B_w$ are contractible submanifolds of even dimension numerated by $w in W/W_P$. Here $W$ is the Weyl group of $G$ and $W_P$ is the Weyl group of $P$. These are the so-called Schubert cells.
This implies that
$$b_k(G/P) = begin{cases}
0, text{ if $k$ is odd,}\
text{number of Schubert cells of real dimension $k$ otherwise}
end{cases} $$
and $chi(G/P) = sum_{w in W/W_P} 1 = |W/W_P|$, just as predicted by your hypothesis.
answered Nov 1 '18 at 8:58
V. Rogov V. Rogov
599211
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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votes
$begingroup$
Let me propose an absolutely different attempt to this problem.
The idea is to use Lefschetz fixed point theorem.
Recall that it states if $f colon X to X$ is an automorphism of a smooth manifold with all fixed points isolated, then
$$
sum_{f(x) =x} i_x = sum_{k=0}^{n} (-1)^koperatorname{Tr} f^*|_{H^k(X, mathbb{R})}
$$
Here $i_x$ are the indices of fixed points.
In our case the automorphism is homotopic to identity (since $T$ is connected), so the RHS of the Lefschetz formula is just the Euler characteristic of your partial flag variety.
One can check up that for any $t in T$ the map induced by left multiplication by $t$ in fact has only isolated fixed points with all the indices equal to $1$.
Therefore, LHS of the Lefschetz formula is precisely the number of the fixed points.
Thus it is rest to compute the (topological) Euler characteristic of a partial flag variety. Recall, that any partial flag variety can be seen as $G/P$, where $G$ is a complex algebraic group and $P subset G$ is a parabolic subgroup. It has cellular Schubert decomposition, that is $G/P = bigsqcup B_w$ , where all $B_w$ are contractible submanifolds of even dimension numerated by $w in W/W_P$. Here $W$ is the Weyl group of $G$ and $W_P$ is the Weyl group of $P$. These are the so-called Schubert cells.
This implies that
$$b_k(G/P) = begin{cases}
0, text{ if $k$ is odd,}\
text{number of Schubert cells of real dimension $k$ otherwise}
end{cases} $$
and $chi(G/P) = sum_{w in W/W_P} 1 = |W/W_P|$, just as predicted by your hypothesis.
answered Nov 1 '18 at 8:58
V. Rogov V. Rogov
599211
$endgroup$
add a comment |
$begingroup$
Let me propose an absolutely different attempt to this problem.
The idea is to use Lefschetz fixed point theorem.
Recall that it states if $f colon X to X$ is an automorphism of a smooth manifold with all fixed points isolated, then
$$
sum_{f(x) =x} i_x = sum_{k=0}^{n} (-1)^koperatorname{Tr} f^*|_{H^k(X, mathbb{R})}
$$
Here $i_x$ are the indices of fixed points.
In our case the automorphism is homotopic to identity (since $T$ is connected), so the RHS of the Lefschetz formula is just the Euler characteristic of your partial flag variety.
One can check up that for any $t in T$ the map induced by left multiplication by $t$ in fact has only isolated fixed points with all the indices equal to $1$.
Therefore, LHS of the Lefschetz formula is precisely the number of the fixed points.
Thus it is rest to compute the (topological) Euler characteristic of a partial flag variety. Recall, that any partial flag variety can be seen as $G/P$, where $G$ is a complex algebraic group and $P subset G$ is a parabolic subgroup. It has cellular Schubert decomposition, that is $G/P = bigsqcup B_w$ , where all $B_w$ are contractible submanifolds of even dimension numerated by $w in W/W_P$. Here $W$ is the Weyl group of $G$ and $W_P$ is the Weyl group of $P$. These are the so-called Schubert cells.
This implies that
$$b_k(G/P) = begin{cases}
0, text{ if $k$ is odd,}\
text{number of Schubert cells of real dimension $k$ otherwise}
end{cases} $$
and $chi(G/P) = sum_{w in W/W_P} 1 = |W/W_P|$, just as predicted by your hypothesis.
answered Nov 1 '18 at 8:58
V. Rogov V. Rogov
599211
$endgroup$
add a comment |
$begingroup$
Let me propose an absolutely different attempt to this problem.
The idea is to use Lefschetz fixed point theorem.
Recall that it states if $f colon X to X$ is an automorphism of a smooth manifold with all fixed points isolated, then
$$
sum_{f(x) =x} i_x = sum_{k=0}^{n} (-1)^koperatorname{Tr} f^*|_{H^k(X, mathbb{R})}
$$
Here $i_x$ are the indices of fixed points.
In our case the automorphism is homotopic to identity (since $T$ is connected), so the RHS of the Lefschetz formula is just the Euler characteristic of your partial flag variety.
One can check up that for any $t in T$ the map induced by left multiplication by $t$ in fact has only isolated fixed points with all the indices equal to $1$.
Therefore, LHS of the Lefschetz formula is precisely the number of the fixed points.
Thus it is rest to compute the (topological) Euler characteristic of a partial flag variety. Recall, that any partial flag variety can be seen as $G/P$, where $G$ is a complex algebraic group and $P subset G$ is a parabolic subgroup. It has cellular Schubert decomposition, that is $G/P = bigsqcup B_w$ , where all $B_w$ are contractible submanifolds of even dimension numerated by $w in W/W_P$. Here $W$ is the Weyl group of $G$ and $W_P$ is the Weyl group of $P$. These are the so-called Schubert cells.
This implies that
$$b_k(G/P) = begin{cases}
0, text{ if $k$ is odd,}\
text{number of Schubert cells of real dimension $k$ otherwise}
end{cases} $$
and $chi(G/P) = sum_{w in W/W_P} 1 = |W/W_P|$, just as predicted by your hypothesis.
answered Nov 1 '18 at 8:58
V. Rogov V. Rogov
599211
$endgroup$
Let me propose an absolutely different attempt to this problem.
The idea is to use Lefschetz fixed point theorem.
Recall that it states if $f colon X to X$ is an automorphism of a smooth manifold with all fixed points isolated, then
$$
sum_{f(x) =x} i_x = sum_{k=0}^{n} (-1)^koperatorname{Tr} f^*|_{H^k(X, mathbb{R})}
$$
Here $i_x$ are the indices of fixed points.
In our case the automorphism is homotopic to identity (since $T$ is connected), so the RHS of the Lefschetz formula is just the Euler characteristic of your partial flag variety.
One can check up that for any $t in T$ the map induced by left multiplication by $t$ in fact has only isolated fixed points with all the indices equal to $1$.
Therefore, LHS of the Lefschetz formula is precisely the number of the fixed points.
Thus it is rest to compute the (topological) Euler characteristic of a partial flag variety. Recall, that any partial flag variety can be seen as $G/P$, where $G$ is a complex algebraic group and $P subset G$ is a parabolic subgroup. It has cellular Schubert decomposition, that is $G/P = bigsqcup B_w$ , where all $B_w$ are contractible submanifolds of even dimension numerated by $w in W/W_P$. Here $W$ is the Weyl group of $G$ and $W_P$ is the Weyl group of $P$. These are the so-called Schubert cells.
This implies that
$$b_k(G/P) = begin{cases}
0, text{ if $k$ is odd,}\
text{number of Schubert cells of real dimension $k$ otherwise}
end{cases} $$
and $chi(G/P) = sum_{w in W/W_P} 1 = |W/W_P|$, just as predicted by your hypothesis.
answered Nov 1 '18 at 8:58
V. Rogov V. Rogov
599211
answered Nov 1 '18 at 8:58
V. Rogov V. Rogov
599211
answered Nov 1 '18 at 8:58
V. Rogov V. Rogov
599211
answered Nov 1 '18 at 8:58
V. Rogov V. Rogov
599211
599211
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$begingroup$
I can tell you that you are correct, but the proof I know involves explicit parameterization of the partial flag variety and is probably not what you are looking for.
$endgroup$
– Matt Samuel
Aug 7 '16 at 1:11
$begingroup$
Hi @MattSamuel , could you tell me a little bit about how is this parametrization done? I'm mainly interested in counting the points, not in the particular path chosen to do so, as long as it's accessible to me (I know this is very subjective) it's good. I think the parametrization approach could be interesting if it goes deeper into the geometry of the spaces.
$endgroup$
– hjhjhj57
Aug 7 '16 at 1:55
$begingroup$
The textbook Young Tableaux by Fulton gives the parameterization of the Grassmannian and the complete flag variety, and you'd have to sort of interpolate between them to get what you want. It's not a topic I've often seen in something expository. I'm retiring for the night, but it may help to look through lecture notes on Schubert calculus. I found something with that title by Brion.
$endgroup$
– Matt Samuel
Aug 7 '16 at 2:43
$begingroup$
This article should do it for you, it has examples: uni-math.gwdg.de/tschinkel/SS05/school/kresch.pdf The torus acts by multiplying each column by a real number, and everything is renormalized so the $1$'s remain $1$. The fixed points are the matrices with only $1$'s and $0$'s.
$endgroup$
– Matt Samuel
Aug 7 '16 at 21:20
$begingroup$
Multiplying the columns by complex numbers is what I meant.
$endgroup$
– Matt Samuel
Aug 7 '16 at 21:27