Betti numbers for the isotropic grassmannian
$begingroup$
I want to know if there is some type of combinatorial formula for computing the Betti numbers of the isotropic grassmannian $IG(r,2n)$ for $rleq n$.
I'm thinking of this as the homogenous space $G/P$ where $G=Sp(2n)$, and $P$ is the maximal parabolic generated by the subset of simple roots $S-{alpha_r}$, where $S={alpha_1,...,alpha_n}$ are the simple roots for $Sp(2n)$.
I know in the case of the Grassmannian $Gr(r,n)$ the $i^{th}$ Betti number is the number of partitons of the integer $i$ which can for inside a $rtimes (n-r)$ box. These are the coefficients of the Poicare series of the Grassmanian which happens to be the Gaussian polynomial or q-binomial coefficient
http://mathworld.wolfram.com/q-BinomialCoefficient.html
So I wonder if there is a similar combinatorial description for the betti numbers of the isotropic grassmannian or it the Poincare Series has a known form. I can write down the Poincare series for the isotrpic Grassmannian, but its not clear to me if this has some more well-known form.
In particular I'd like to know whether $dim(H^i(IG,r,2n))$ is still bound by the number of partitions of the integer $i$ ?
combinatorics algebraic-geometry algebraic-topology representation-theory schubert-calculus
edited Jan 5 at 11:18
Matt Samuel
39.2k63770
asked Mar 10 '17 at 1:22
user103033user103033
162
$endgroup$
add a comment |
$begingroup$
I want to know if there is some type of combinatorial formula for computing the Betti numbers of the isotropic grassmannian $IG(r,2n)$ for $rleq n$.
I'm thinking of this as the homogenous space $G/P$ where $G=Sp(2n)$, and $P$ is the maximal parabolic generated by the subset of simple roots $S-{alpha_r}$, where $S={alpha_1,...,alpha_n}$ are the simple roots for $Sp(2n)$.
I know in the case of the Grassmannian $Gr(r,n)$ the $i^{th}$ Betti number is the number of partitons of the integer $i$ which can for inside a $rtimes (n-r)$ box. These are the coefficients of the Poicare series of the Grassmanian which happens to be the Gaussian polynomial or q-binomial coefficient
http://mathworld.wolfram.com/q-BinomialCoefficient.html
So I wonder if there is a similar combinatorial description for the betti numbers of the isotropic grassmannian or it the Poincare Series has a known form. I can write down the Poincare series for the isotrpic Grassmannian, but its not clear to me if this has some more well-known form.
In particular I'd like to know whether $dim(H^i(IG,r,2n))$ is still bound by the number of partitions of the integer $i$ ?
combinatorics algebraic-geometry algebraic-topology representation-theory schubert-calculus
edited Jan 5 at 11:18
Matt Samuel
39.2k63770
asked Mar 10 '17 at 1:22
user103033user103033
162
$endgroup$
$begingroup$
The $i$th Betti number is the number of strict partitions of $i$ with all parts $le n$. I'd write a full answer but I'm about to head to bed.
$endgroup$
– Matt Samuel
Mar 10 '17 at 3:46
$begingroup$
Oh, you're allowing $r<n$. That's a bit uglier, but there's still a combinatorial description. You could look at some papers by my advisor, Anders Buch.
$endgroup$
– Matt Samuel
Mar 10 '17 at 3:48
$begingroup$
@Matt Thanks for the suggestion. From this paper math.rutgers.edu/~asbuch/papers/qgig.pdf It seems the $i^{th}$ Betti number of $IG(n-k,2k)$ is the number of k-strict partitions of $i$ which fit into a $n-ktimes n+k$ rectangle
$endgroup$
– user103033
Mar 10 '17 at 7:13
$begingroup$
Sounds right. And no problem.
$endgroup$
– Matt Samuel
Mar 10 '17 at 10:33
$begingroup$
By the way, if you know about Coxeter groups and the relation to the Betti numbers it's a good exercise to prove this yourself. You can decompose the minimal length coset representatives into the product of a type $A$ representative and a maximal type $B$ representative. It can be done with utterly elementary reasoning. I had fun with it at least!
$endgroup$
– Matt Samuel
Mar 10 '17 at 23:49
add a comment |
$begingroup$
I want to know if there is some type of combinatorial formula for computing the Betti numbers of the isotropic grassmannian $IG(r,2n)$ for $rleq n$.
I'm thinking of this as the homogenous space $G/P$ where $G=Sp(2n)$, and $P$ is the maximal parabolic generated by the subset of simple roots $S-{alpha_r}$, where $S={alpha_1,...,alpha_n}$ are the simple roots for $Sp(2n)$.
I know in the case of the Grassmannian $Gr(r,n)$ the $i^{th}$ Betti number is the number of partitons of the integer $i$ which can for inside a $rtimes (n-r)$ box. These are the coefficients of the Poicare series of the Grassmanian which happens to be the Gaussian polynomial or q-binomial coefficient
http://mathworld.wolfram.com/q-BinomialCoefficient.html
So I wonder if there is a similar combinatorial description for the betti numbers of the isotropic grassmannian or it the Poincare Series has a known form. I can write down the Poincare series for the isotrpic Grassmannian, but its not clear to me if this has some more well-known form.
In particular I'd like to know whether $dim(H^i(IG,r,2n))$ is still bound by the number of partitions of the integer $i$ ?
combinatorics algebraic-geometry algebraic-topology representation-theory schubert-calculus
edited Jan 5 at 11:18
Matt Samuel
39.2k63770
asked Mar 10 '17 at 1:22
user103033user103033
162
$endgroup$
I want to know if there is some type of combinatorial formula for computing the Betti numbers of the isotropic grassmannian $IG(r,2n)$ for $rleq n$.
I'm thinking of this as the homogenous space $G/P$ where $G=Sp(2n)$, and $P$ is the maximal parabolic generated by the subset of simple roots $S-{alpha_r}$, where $S={alpha_1,...,alpha_n}$ are the simple roots for $Sp(2n)$.
I know in the case of the Grassmannian $Gr(r,n)$ the $i^{th}$ Betti number is the number of partitons of the integer $i$ which can for inside a $rtimes (n-r)$ box. These are the coefficients of the Poicare series of the Grassmanian which happens to be the Gaussian polynomial or q-binomial coefficient
http://mathworld.wolfram.com/q-BinomialCoefficient.html
So I wonder if there is a similar combinatorial description for the betti numbers of the isotropic grassmannian or it the Poincare Series has a known form. I can write down the Poincare series for the isotrpic Grassmannian, but its not clear to me if this has some more well-known form.
In particular I'd like to know whether $dim(H^i(IG,r,2n))$ is still bound by the number of partitions of the integer $i$ ?
combinatorics algebraic-geometry algebraic-topology representation-theory schubert-calculus
combinatorics algebraic-geometry algebraic-topology representation-theory schubert-calculus
edited Jan 5 at 11:18
Matt Samuel
39.2k63770
asked Mar 10 '17 at 1:22
user103033user103033
162
edited Jan 5 at 11:18
Matt Samuel
39.2k63770
asked Mar 10 '17 at 1:22
user103033user103033
162
edited Jan 5 at 11:18
Matt Samuel
39.2k63770
edited Jan 5 at 11:18
Matt Samuel
39.2k63770
edited Jan 5 at 11:18
Matt Samuel
39.2k63770
39.2k63770
asked Mar 10 '17 at 1:22
user103033user103033
162
asked Mar 10 '17 at 1:22
user103033user103033
162
asked Mar 10 '17 at 1:22
user103033user103033
162
162
$begingroup$
The $i$th Betti number is the number of strict partitions of $i$ with all parts $le n$. I'd write a full answer but I'm about to head to bed.
$endgroup$
– Matt Samuel
Mar 10 '17 at 3:46
$begingroup$
Oh, you're allowing $r<n$. That's a bit uglier, but there's still a combinatorial description. You could look at some papers by my advisor, Anders Buch.
$endgroup$
– Matt Samuel
Mar 10 '17 at 3:48
$begingroup$
@Matt Thanks for the suggestion. From this paper math.rutgers.edu/~asbuch/papers/qgig.pdf It seems the $i^{th}$ Betti number of $IG(n-k,2k)$ is the number of k-strict partitions of $i$ which fit into a $n-ktimes n+k$ rectangle
$endgroup$
– user103033
Mar 10 '17 at 7:13
$begingroup$
Sounds right. And no problem.
$endgroup$
– Matt Samuel
Mar 10 '17 at 10:33
$begingroup$
By the way, if you know about Coxeter groups and the relation to the Betti numbers it's a good exercise to prove this yourself. You can decompose the minimal length coset representatives into the product of a type $A$ representative and a maximal type $B$ representative. It can be done with utterly elementary reasoning. I had fun with it at least!
$endgroup$
– Matt Samuel
Mar 10 '17 at 23:49
add a comment |
$begingroup$
The $i$th Betti number is the number of strict partitions of $i$ with all parts $le n$. I'd write a full answer but I'm about to head to bed.
$endgroup$
– Matt Samuel
Mar 10 '17 at 3:46
$begingroup$
Oh, you're allowing $r<n$. That's a bit uglier, but there's still a combinatorial description. You could look at some papers by my advisor, Anders Buch.
$endgroup$
– Matt Samuel
Mar 10 '17 at 3:48
$begingroup$
@Matt Thanks for the suggestion. From this paper math.rutgers.edu/~asbuch/papers/qgig.pdf It seems the $i^{th}$ Betti number of $IG(n-k,2k)$ is the number of k-strict partitions of $i$ which fit into a $n-ktimes n+k$ rectangle
$endgroup$
– user103033
Mar 10 '17 at 7:13
$begingroup$
Sounds right. And no problem.
$endgroup$
– Matt Samuel
Mar 10 '17 at 10:33
$begingroup$
By the way, if you know about Coxeter groups and the relation to the Betti numbers it's a good exercise to prove this yourself. You can decompose the minimal length coset representatives into the product of a type $A$ representative and a maximal type $B$ representative. It can be done with utterly elementary reasoning. I had fun with it at least!
$endgroup$
– Matt Samuel
Mar 10 '17 at 23:49
$begingroup$
The $i$th Betti number is the number of strict partitions of $i$ with all parts $le n$. I'd write a full answer but I'm about to head to bed.
$endgroup$
– Matt Samuel
Mar 10 '17 at 3:46
$begingroup$
The $i$th Betti number is the number of strict partitions of $i$ with all parts $le n$. I'd write a full answer but I'm about to head to bed.
$endgroup$
– Matt Samuel
Mar 10 '17 at 3:46
$begingroup$
Oh, you're allowing $r<n$. That's a bit uglier, but there's still a combinatorial description. You could look at some papers by my advisor, Anders Buch.
$endgroup$
– Matt Samuel
Mar 10 '17 at 3:48
$begingroup$
Oh, you're allowing $r<n$. That's a bit uglier, but there's still a combinatorial description. You could look at some papers by my advisor, Anders Buch.
$endgroup$
– Matt Samuel
Mar 10 '17 at 3:48
$begingroup$
@Matt Thanks for the suggestion. From this paper math.rutgers.edu/~asbuch/papers/qgig.pdf It seems the $i^{th}$ Betti number of $IG(n-k,2k)$ is the number of k-strict partitions of $i$ which fit into a $n-ktimes n+k$ rectangle
$endgroup$
– user103033
Mar 10 '17 at 7:13
$begingroup$
@Matt Thanks for the suggestion. From this paper math.rutgers.edu/~asbuch/papers/qgig.pdf It seems the $i^{th}$ Betti number of $IG(n-k,2k)$ is the number of k-strict partitions of $i$ which fit into a $n-ktimes n+k$ rectangle
$endgroup$
– user103033
Mar 10 '17 at 7:13
$begingroup$
Sounds right. And no problem.
$endgroup$
– Matt Samuel
Mar 10 '17 at 10:33
$begingroup$
Sounds right. And no problem.
$endgroup$
– Matt Samuel
Mar 10 '17 at 10:33
$begingroup$
By the way, if you know about Coxeter groups and the relation to the Betti numbers it's a good exercise to prove this yourself. You can decompose the minimal length coset representatives into the product of a type $A$ representative and a maximal type $B$ representative. It can be done with utterly elementary reasoning. I had fun with it at least!
$endgroup$
– Matt Samuel
Mar 10 '17 at 23:49
$begingroup$
By the way, if you know about Coxeter groups and the relation to the Betti numbers it's a good exercise to prove this yourself. You can decompose the minimal length coset representatives into the product of a type $A$ representative and a maximal type $B$ representative. It can be done with utterly elementary reasoning. I had fun with it at least!
$endgroup$
– Matt Samuel
Mar 10 '17 at 23:49
add a comment |
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$begingroup$
The $i$th Betti number is the number of strict partitions of $i$ with all parts $le n$. I'd write a full answer but I'm about to head to bed.
$endgroup$
– Matt Samuel
Mar 10 '17 at 3:46
$begingroup$
Oh, you're allowing $r<n$. That's a bit uglier, but there's still a combinatorial description. You could look at some papers by my advisor, Anders Buch.
$endgroup$
– Matt Samuel
Mar 10 '17 at 3:48
$begingroup$
@Matt Thanks for the suggestion. From this paper math.rutgers.edu/~asbuch/papers/qgig.pdf It seems the $i^{th}$ Betti number of $IG(n-k,2k)$ is the number of k-strict partitions of $i$ which fit into a $n-ktimes n+k$ rectangle
$endgroup$
– user103033
Mar 10 '17 at 7:13
$begingroup$
Sounds right. And no problem.
$endgroup$
– Matt Samuel
Mar 10 '17 at 10:33
$begingroup$
By the way, if you know about Coxeter groups and the relation to the Betti numbers it's a good exercise to prove this yourself. You can decompose the minimal length coset representatives into the product of a type $A$ representative and a maximal type $B$ representative. It can be done with utterly elementary reasoning. I had fun with it at least!
$endgroup$
– Matt Samuel
Mar 10 '17 at 23:49