how to compute the cohomology ring of grassmannian G(4,2)
$begingroup$
I need to compute the ring of cohomologies over the integers of the complex grassmannian G(4,2).
As I understand, one can use the Schubert cells and cellular homology to show that the homology groups of G(4,2) are free abelian with bases corresponding to the appropriate Schubert cells.
And the cohomology groups have the same structure, am I right?
Now - I want to understand how the cup product looks like in this case without using some general formulas (like Pieri's or Giambelli's) which seem to be hard to prove. How can I do that? I don't understand how one can compute cup products using cellular cohomology.
Thank you.
algebraic-topology schubert-calculus
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
asked May 4 '13 at 16:27
zakharzakhar
562
$endgroup$
add a comment |
$begingroup$
I need to compute the ring of cohomologies over the integers of the complex grassmannian G(4,2).
As I understand, one can use the Schubert cells and cellular homology to show that the homology groups of G(4,2) are free abelian with bases corresponding to the appropriate Schubert cells.
And the cohomology groups have the same structure, am I right?
Now - I want to understand how the cup product looks like in this case without using some general formulas (like Pieri's or Giambelli's) which seem to be hard to prove. How can I do that? I don't understand how one can compute cup products using cellular cohomology.
Thank you.
algebraic-topology schubert-calculus
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
asked May 4 '13 at 16:27
zakharzakhar
562
$endgroup$
$begingroup$
Only the Euler characteristic is discussed here, but the answers (other than mine) may have useful references?
$endgroup$
– Jyrki Lahtonen
Aug 15 '15 at 20:41
add a comment |
$begingroup$
I need to compute the ring of cohomologies over the integers of the complex grassmannian G(4,2).
As I understand, one can use the Schubert cells and cellular homology to show that the homology groups of G(4,2) are free abelian with bases corresponding to the appropriate Schubert cells.
And the cohomology groups have the same structure, am I right?
Now - I want to understand how the cup product looks like in this case without using some general formulas (like Pieri's or Giambelli's) which seem to be hard to prove. How can I do that? I don't understand how one can compute cup products using cellular cohomology.
Thank you.
algebraic-topology schubert-calculus
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
asked May 4 '13 at 16:27
zakharzakhar
562
$endgroup$
I need to compute the ring of cohomologies over the integers of the complex grassmannian G(4,2).
As I understand, one can use the Schubert cells and cellular homology to show that the homology groups of G(4,2) are free abelian with bases corresponding to the appropriate Schubert cells.
And the cohomology groups have the same structure, am I right?
Now - I want to understand how the cup product looks like in this case without using some general formulas (like Pieri's or Giambelli's) which seem to be hard to prove. How can I do that? I don't understand how one can compute cup products using cellular cohomology.
Thank you.
algebraic-topology schubert-calculus
algebraic-topology schubert-calculus
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
asked May 4 '13 at 16:27
zakharzakhar
562
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
asked May 4 '13 at 16:27
zakharzakhar
562
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
edited Jan 5 at 11:19
Matt Samuel
39.2k63770
39.2k63770
asked May 4 '13 at 16:27
zakharzakhar
562
asked May 4 '13 at 16:27
zakharzakhar
562
asked May 4 '13 at 16:27
zakharzakhar
562
562
$begingroup$
Only the Euler characteristic is discussed here, but the answers (other than mine) may have useful references?
$endgroup$
– Jyrki Lahtonen
Aug 15 '15 at 20:41
add a comment |
$begingroup$
Only the Euler characteristic is discussed here, but the answers (other than mine) may have useful references?
$endgroup$
– Jyrki Lahtonen
Aug 15 '15 at 20:41
$begingroup$
Only the Euler characteristic is discussed here, but the answers (other than mine) may have useful references?
$endgroup$
– Jyrki Lahtonen
Aug 15 '15 at 20:41
$begingroup$
Only the Euler characteristic is discussed here, but the answers (other than mine) may have useful references?
$endgroup$
– Jyrki Lahtonen
Aug 15 '15 at 20:41
add a comment |
1 Answer
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I don't know, how can one compute product for cellular cohomology, maybe using some tricks.
In case complex grassmannian $G(4,2)$ Schubert cells give you that $H^0=H^2=H^6=H^8=mathbb Z$ and $H^4=mathbb Z^2$ (other are zero). Define their generators by $a_2,a_4,a_4',a_6$ and $a_8$. Some theorem about cohomology of a manifold (I don't remember the name) gives us $a_2cdot a_6=a_8$. And considering cell subspace $mathbb CP^2=G(3,2)subset G(4,2)$ shows that $a_2^2=a'_4$.
[Instead this you may write down a spectral sequence for $U(2)$-fibration $V(4,2)to G(4,2)$]
To show that $a'_4cdot a_4=0$ consider two Schubert $4$-cells of $G(4,2)$: for basis $v_1,v_2,v_3,v_4inmathbb C^4$ they defined as sets of hyperplanes $langle sv_1+v_2,,tv_1+v_3rangle$ and $langle sv_2+tv_3+v_4,,v_1rangle$ for complex parameters $s,t$. Therefore cocycles $a_4$ and $a'_4$ are Poincare-dual for submanifolds $M':={Win G(4,2):Wsubsetlangle v_2,v_3,v_4rangle}$ and $M:={Win G(4,2):Wni v_1}$ (because intersection cell with corresponding submanifold is transversal). $Mcap M'=emptyset$, thus $a'_4cdot a_4=0$, $a_2cdot a_4=0$. Also observe that self-intersection indices of $M$ and $M'$ are $1$, therefore $a_4^2=a_4'^2=a_8$.
So, $H^*(G(4,2))=mathbb Z[a_2,a_4]/(a_2^5,,,a_2cdot a_4,,,a_4^2-a_2^4)$.
answered Aug 15 '15 at 18:59
Andrey RyabichevAndrey Ryabichev
2,182720
$endgroup$
$begingroup$
Very nice explanation.
$endgroup$
– King Khan
Feb 26 '17 at 11:09
add a comment |
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$begingroup$
I don't know, how can one compute product for cellular cohomology, maybe using some tricks.
In case complex grassmannian $G(4,2)$ Schubert cells give you that $H^0=H^2=H^6=H^8=mathbb Z$ and $H^4=mathbb Z^2$ (other are zero). Define their generators by $a_2,a_4,a_4',a_6$ and $a_8$. Some theorem about cohomology of a manifold (I don't remember the name) gives us $a_2cdot a_6=a_8$. And considering cell subspace $mathbb CP^2=G(3,2)subset G(4,2)$ shows that $a_2^2=a'_4$.
[Instead this you may write down a spectral sequence for $U(2)$-fibration $V(4,2)to G(4,2)$]
To show that $a'_4cdot a_4=0$ consider two Schubert $4$-cells of $G(4,2)$: for basis $v_1,v_2,v_3,v_4inmathbb C^4$ they defined as sets of hyperplanes $langle sv_1+v_2,,tv_1+v_3rangle$ and $langle sv_2+tv_3+v_4,,v_1rangle$ for complex parameters $s,t$. Therefore cocycles $a_4$ and $a'_4$ are Poincare-dual for submanifolds $M':={Win G(4,2):Wsubsetlangle v_2,v_3,v_4rangle}$ and $M:={Win G(4,2):Wni v_1}$ (because intersection cell with corresponding submanifold is transversal). $Mcap M'=emptyset$, thus $a'_4cdot a_4=0$, $a_2cdot a_4=0$. Also observe that self-intersection indices of $M$ and $M'$ are $1$, therefore $a_4^2=a_4'^2=a_8$.
So, $H^*(G(4,2))=mathbb Z[a_2,a_4]/(a_2^5,,,a_2cdot a_4,,,a_4^2-a_2^4)$.
answered Aug 15 '15 at 18:59
Andrey RyabichevAndrey Ryabichev
2,182720
$endgroup$
$begingroup$
Very nice explanation.
$endgroup$
– King Khan
Feb 26 '17 at 11:09
add a comment |
$begingroup$
I don't know, how can one compute product for cellular cohomology, maybe using some tricks.
In case complex grassmannian $G(4,2)$ Schubert cells give you that $H^0=H^2=H^6=H^8=mathbb Z$ and $H^4=mathbb Z^2$ (other are zero). Define their generators by $a_2,a_4,a_4',a_6$ and $a_8$. Some theorem about cohomology of a manifold (I don't remember the name) gives us $a_2cdot a_6=a_8$. And considering cell subspace $mathbb CP^2=G(3,2)subset G(4,2)$ shows that $a_2^2=a'_4$.
[Instead this you may write down a spectral sequence for $U(2)$-fibration $V(4,2)to G(4,2)$]
To show that $a'_4cdot a_4=0$ consider two Schubert $4$-cells of $G(4,2)$: for basis $v_1,v_2,v_3,v_4inmathbb C^4$ they defined as sets of hyperplanes $langle sv_1+v_2,,tv_1+v_3rangle$ and $langle sv_2+tv_3+v_4,,v_1rangle$ for complex parameters $s,t$. Therefore cocycles $a_4$ and $a'_4$ are Poincare-dual for submanifolds $M':={Win G(4,2):Wsubsetlangle v_2,v_3,v_4rangle}$ and $M:={Win G(4,2):Wni v_1}$ (because intersection cell with corresponding submanifold is transversal). $Mcap M'=emptyset$, thus $a'_4cdot a_4=0$, $a_2cdot a_4=0$. Also observe that self-intersection indices of $M$ and $M'$ are $1$, therefore $a_4^2=a_4'^2=a_8$.
So, $H^*(G(4,2))=mathbb Z[a_2,a_4]/(a_2^5,,,a_2cdot a_4,,,a_4^2-a_2^4)$.
answered Aug 15 '15 at 18:59
Andrey RyabichevAndrey Ryabichev
2,182720
$endgroup$
$begingroup$
Very nice explanation.
$endgroup$
– King Khan
Feb 26 '17 at 11:09
add a comment |
$begingroup$
I don't know, how can one compute product for cellular cohomology, maybe using some tricks.
In case complex grassmannian $G(4,2)$ Schubert cells give you that $H^0=H^2=H^6=H^8=mathbb Z$ and $H^4=mathbb Z^2$ (other are zero). Define their generators by $a_2,a_4,a_4',a_6$ and $a_8$. Some theorem about cohomology of a manifold (I don't remember the name) gives us $a_2cdot a_6=a_8$. And considering cell subspace $mathbb CP^2=G(3,2)subset G(4,2)$ shows that $a_2^2=a'_4$.
[Instead this you may write down a spectral sequence for $U(2)$-fibration $V(4,2)to G(4,2)$]
To show that $a'_4cdot a_4=0$ consider two Schubert $4$-cells of $G(4,2)$: for basis $v_1,v_2,v_3,v_4inmathbb C^4$ they defined as sets of hyperplanes $langle sv_1+v_2,,tv_1+v_3rangle$ and $langle sv_2+tv_3+v_4,,v_1rangle$ for complex parameters $s,t$. Therefore cocycles $a_4$ and $a'_4$ are Poincare-dual for submanifolds $M':={Win G(4,2):Wsubsetlangle v_2,v_3,v_4rangle}$ and $M:={Win G(4,2):Wni v_1}$ (because intersection cell with corresponding submanifold is transversal). $Mcap M'=emptyset$, thus $a'_4cdot a_4=0$, $a_2cdot a_4=0$. Also observe that self-intersection indices of $M$ and $M'$ are $1$, therefore $a_4^2=a_4'^2=a_8$.
So, $H^*(G(4,2))=mathbb Z[a_2,a_4]/(a_2^5,,,a_2cdot a_4,,,a_4^2-a_2^4)$.
answered Aug 15 '15 at 18:59
Andrey RyabichevAndrey Ryabichev
2,182720
$endgroup$
I don't know, how can one compute product for cellular cohomology, maybe using some tricks.
In case complex grassmannian $G(4,2)$ Schubert cells give you that $H^0=H^2=H^6=H^8=mathbb Z$ and $H^4=mathbb Z^2$ (other are zero). Define their generators by $a_2,a_4,a_4',a_6$ and $a_8$. Some theorem about cohomology of a manifold (I don't remember the name) gives us $a_2cdot a_6=a_8$. And considering cell subspace $mathbb CP^2=G(3,2)subset G(4,2)$ shows that $a_2^2=a'_4$.
[Instead this you may write down a spectral sequence for $U(2)$-fibration $V(4,2)to G(4,2)$]
To show that $a'_4cdot a_4=0$ consider two Schubert $4$-cells of $G(4,2)$: for basis $v_1,v_2,v_3,v_4inmathbb C^4$ they defined as sets of hyperplanes $langle sv_1+v_2,,tv_1+v_3rangle$ and $langle sv_2+tv_3+v_4,,v_1rangle$ for complex parameters $s,t$. Therefore cocycles $a_4$ and $a'_4$ are Poincare-dual for submanifolds $M':={Win G(4,2):Wsubsetlangle v_2,v_3,v_4rangle}$ and $M:={Win G(4,2):Wni v_1}$ (because intersection cell with corresponding submanifold is transversal). $Mcap M'=emptyset$, thus $a'_4cdot a_4=0$, $a_2cdot a_4=0$. Also observe that self-intersection indices of $M$ and $M'$ are $1$, therefore $a_4^2=a_4'^2=a_8$.
So, $H^*(G(4,2))=mathbb Z[a_2,a_4]/(a_2^5,,,a_2cdot a_4,,,a_4^2-a_2^4)$.
answered Aug 15 '15 at 18:59
Andrey RyabichevAndrey Ryabichev
2,182720
edited Aug 15 '15 at 20:32
answered Aug 15 '15 at 18:59
Andrey RyabichevAndrey Ryabichev
2,182720
answered Aug 15 '15 at 18:59
Andrey RyabichevAndrey Ryabichev
2,182720
answered Aug 15 '15 at 18:59
Andrey RyabichevAndrey Ryabichev
2,182720
2,182720
$begingroup$
Very nice explanation.
$endgroup$
– King Khan
Feb 26 '17 at 11:09
add a comment |
$begingroup$
Very nice explanation.
$endgroup$
– King Khan
Feb 26 '17 at 11:09
$begingroup$
Very nice explanation.
$endgroup$
– King Khan
Feb 26 '17 at 11:09
$begingroup$
Very nice explanation.
$endgroup$
– King Khan
Feb 26 '17 at 11:09
add a comment |
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$begingroup$
Only the Euler characteristic is discussed here, but the answers (other than mine) may have useful references?
$endgroup$
– Jyrki Lahtonen
Aug 15 '15 at 20:41