how to compute the cohomology ring of grassmannian G(4,2)

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