I would like to compute $\pi_1$ and the integral homology groups of the real Grassmannian $G(2,4)$. (This is a question on an old qualifying exam.) The hint for the computation of $\pi_1$ is to put a CW structure on the space. Once we have this and have determined the attaching maps, we can also get the integral homology, so that is how I like would to proceed (if possible). (One can obtain the fundamental group of all Grassmannians in a different way, but that way doesn't help find the homology groups.)
Some discussion of the cell structure of Grassmannians appears in Milnor and Stasheff's Characteristic Classes and in Schwartz's Differential Topology and Geometry. From these sources, I have learned that we have a cell structure with
- 1 $0$-cell
- 1 $1$-cell
- 2 $2$-cells
- 1 $3$-cell
- 1 $4$-cell.
Further, we know that all attaching maps are 2 to 1, so using cellular homology, the coefficients in images of the cells under the boundary maps must be $0$ or $2$. But I don't see how to determine what those coefficients are. How can we proceed?
I note that a purported answer is given (without justification) here.
Writing down the matrices corresponding to the Schubert cell decomposition gave me the following boundary maps for cellular homology. I use $A_i$ to denote the cells, with a prime to distinguish the cells of dimension 2. $$d A_1 = 0$$ $$d A_2 = 0$$ $$ d A_2'=2A_1$$ $$ d A_3 = 2 A_2$$ $$ d A_4=0.$$ This gives an answer that agrees with the link, except at $H_2$. I get $\mathbb Z/2\mathbb Z$ here, while the link claims $H_2$ is zero. Who is correct? It seems like I am. If we use the universal coefficients theorem on the homology groups in the link, we get cohomology groups that contradict Poincaré duality. We agree that $H_1=\mathbb Z_2$, but if $H_2 = 0$, then $H^3= 0$, when duality says it should be $\mathbb Z_2$.
It's important to keep in mind that there exist oriented and non-oriented grassmannians (depending on you have fixed orientation of subspace or not).
For oriented grassmannian $\widetilde G(2,4)$ we can consider $S^1$-fibration $V(2,4)\to\widetilde G(2,4)$, where $V(2,4)$ is a Stiefel manifold. As it easy to see, $V(2,4)\simeq T_0S^3$ (unit tangent vectors of $3$-sphere), and since $TS^3$ is trivial, $V(2,4)\simeq S^3\times S^2$.
So, we have a fibration $S^3\times S^2\to\widetilde G(2,4)$ with a fiber $S^1$. Spectral sequence for this fibration shows for $\widetilde G(2,4)$ that $H_0=H_4=\mathbb Z$, $H_2=\mathbb Z^2$ and $H_1=H_3=0$.
And for non-oriented grassmannian $G(2,4)$ we have a $2$-covering $\widetilde G(2,4)\to G(2,4)$, therefore $\pi_1(G(2,4))=H_1(G(2,4))=\mathbb Z_2$. Note, $G(2,4)$ is orientable, so $H_4=\mathbb Z$. By Poincare duality we have $H^3=\mathbb Z_2$, and universal coefficient theorem gives $\mathbb Z_2=H^3=\mathrm{Hom}(H_3,\mathbb Z)\oplus \mathrm{Ext}(H_2,\mathbb Z)$; therefore $H_3=0$ and $H_2=\mathbb Z_2$.