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.

  • $\begingroup$ Only the Euler characteristic is discussed here, but the answers (other than mine) may have useful references? $\endgroup$ Commented Aug 15, 2015 at 20:41

1 Answer 1


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 $a_2\cdot a_6=a_8$. 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 $a'_4\cdot a_4=0$ consider two Schubert $4$-cells of $G(4,2)$: for basis $v_1,v_2,v_3,v_4\in\mathbb C^4$ they defined as sets of hyperplanes $\langle sv_1+v_2,\,tv_1+v_3\rangle$ and $\langle sv_2+tv_3+v_4,\,v_1\rangle$ for complex parameters $s,t$. Therefore cocycles $a_4$ and $a'_4$ are Poincare-dual for submanifolds $M':=\{W\in G(4,2):W\subset\langle v_2,v_3,v_4\rangle\}$ and $M:=\{W\in G(4,2):W\ni v_1\}$ (because intersection cell with corresponding submanifold is transversal). $M\cap M'=\emptyset$, thus $a'_4\cdot a_4=0$, $a_2\cdot a_4=0$. Also observe that self-intersection indices of $M$ and $M'$ are $1$, therefore $a_4^2=a_4'^2=a_8$.

So, $H^*(G(4,2))=\mathbb Z[a_2,a_4]/(a_2^5,\,\,a_2\cdot a_4,\,\,a_4^2-a_2^4)$.

  • $\begingroup$ Very nice explanation. $\endgroup$
    – King Khan
    Commented Feb 26, 2017 at 11:09
  • $\begingroup$ The first relation $a_2^5$ is generated by the other two, thus you do not need it. $\endgroup$ Commented Feb 25, 2021 at 15:27

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