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 $\mathbb Z_2$.

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    $\begingroup$ The only way to do this is to describe explicitly the attaching maps. K owing the number of cells is not enough. $\endgroup$ – Mariano Suárez-Álvarez Aug 14 '15 at 2:52
  • $\begingroup$ @MarianoSuárez-Alvarez How can one see what the attaching maps are? It seems hard to do explicitly, but perhaps I am just not thinking hard enough. $\endgroup$ – Potato Aug 14 '15 at 2:54
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    $\begingroup$ Well, try to figure it out from the description of the cells. I'm pretty sure both of your sources describe the cells, not only their number! $\endgroup$ – Mariano Suárez-Álvarez Aug 14 '15 at 2:56
  • $\begingroup$ @MarianoSuárez-Alvarez Do you know if the answer in the link at the end of my question is correct? I get that $H_2$ is nonzero, while it disagrees. I fear the error is on my end. $\endgroup$ – Potato Aug 14 '15 at 6:02
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    $\begingroup$ Super way late, but to address " (One can obtain the fundamental group of all Grassmannians in a different way, but that way doesn't help find the homology groups.)" - there are methods for computing the homology of homogeneous space just using the following information: $H$, $G$, and the inclusion of maximal tori $T_H\rightarrow T_G$. It boils down to a spectral sequence calculation, so there are sometimes extension problems, and it's hard to use practically if $G$ and or $H$ have torsion in their homology groups,but, for example, with coefficients in a field things can always be computed $\endgroup$ – Jason DeVito Feb 23 '17 at 19:22

Homology of the real grassmannian $G(2,4)$

The lovely accepted answer is an instructive lesson in the power of fiber bundle methods, and inspires me to learn more about spectral sequences. Thank you! (I admit I still need to ask why $G(2,4)$ is obviously orientable and why $H_3$ is not merely torsion but zero, and why $\tilde{G}$ is simply connected.) As a novice I also enjoyed thinking about this in the elementary context of the OP’s approach, by attaching cells. Here is the result, if it has interest for someone.

So consider $G(2,4)$ to be all projective lines in projective $3$ space and choose an auxiliary nested family consisting of a point $P$ lying on a line $L$ lying in a plane $\Pi$. Then $G(2,4)$ is the disjoint union of the following $6$ affine cells:

0) a $0$-cell consisting of the one line $L$. i) a $1$-cell consisting of the lines meeting $P$ and lying in $\Pi$, except for the line $L$. ii) a $2$-cell consisting of the lines in $\Pi$ but not meeting $P$. ii)’ another $2$-cell consisting of the lines meeting $P$ but not lying in $\Pi$. iii) a $3$-cell consisting of the lines meeting $L$ away from $P$ and not lying in $\Pi$. iv) a $4$-cell consisting of all lines not meeting $L$.

It already follows that each homology group has at most as many generators as the number of cells of that dimension, and that the Euler characteristic is $2$ = alternating sum of number of $k$-cells.

These $k$-cells are all isomorphic to copies of affine $k$-space. The corresponding closed cells are: the one line $L$ (a point), the lines in $\Pi$ meeting $P$ (a copy of $\mathbb{P}^1$), all the lines in $\Pi$ (a copy of $\mathbb{P}^2$), all the lines meeting $P$ (another copy of $\mathbb{P}^2$), all the lines meeting $L$ (a copy of $\mathbb{P}^1\times\mathbb{P}^2$ with one copy of $\mathbb{P}^1\times\{P\}$ collapsed to a point), and all lines in projective $3$-space, i.e. $G$ itself.

These closed $k$-cells are images of closed $k$-dimensional balls, by a map which is injective on the interior of the ball but not usually injective on the boundary sphere. The behavior on the boundary spheres determines the homology as follows.

We will construct $G$ topologically in stages, by sequentially mapping closed balls onto the closed cells in $G$, by attaching maps along their spherical boundaries, and compute how the homology changes at each step. We begin with the point represented by $L$, to which we attach a closed $1$-cell, by mapping the unit interval onto the $1$-cell of lines lying in $\Pi$ and meeting $P$, which is a copy of projective $1$-space, hence a circle, thus the two endpoints of the interval go to the same point, namely $L$. So now we have a copy of $\mathbb{P}^1$, namely a circle, so we know the homology is $\mathbb{Z}$ in dimensions $0$ and $1$, and zero elsewhere.

The closed $2$-cell in $G$ consisting of all lines in $\Pi$ is a copy of $\mathbb{P}^2$, so to parametrize it we attach a copy of a $2$-disc to the copy of $\mathbb{P}^1$ we already have constructed in $G$, by a map that wraps the boundary circle of the disc twice around the circle representing $\mathbb{P}^1$. When this has been done, we have attached a $2$-cell with boundary now equal to twice the existing copy of $\mathbb{P}^1$, so that the element of $H_1$ represented by that circle now becomes $2$-torsion, i.e. the copy of $\mathbb{P}^1$ that used to be a generator of $\mathbb{Z}$ in first homology, now generates only $\mathbb{Z}/2\mathbb{Z}$, the new first homology group. We have also attached a $2$-cell, so there is a chance we have introduced some two-dimensional homology, but in fact we have not, because the boundary of the $2$-cell is attached to a $1$-cycle that was not previously homologous to zero. So second homology is still zero.

Next we parametrize the second copy of $\mathbb{P}^2$, (the lines meeting $P$), by attaching another $2$-disc also with boundary equal to the same copy of $\mathbb{P}^1$. Now we have a union of two copies of $\mathbb{P}^2$ meeting along a common copy of $\mathbb{P}^1$. But now we do have some non trivial $2$- cycles, since now the boundary of this second $2$-disc has boundary which is already the boundary of the first attached $2$-disc. Thus we have just attached a $2$-cycle which generates now a copy of $\mathbb{Z}$ in second homology. But we could have attached the two copies of $\mathbb{P}^2$ in the other order, so now even though there is only a one-dimensional second homology group, either copy of $\mathbb{P}^2$ generates it, i.e. even though $\mathbb{P}^2$ has no second homology, the union of two copies of $\mathbb{P}^2$ joined along a common $\mathbb{P}^1$, has one-dimensional second homology, and either copy of $\mathbb{P}^2$ can be used to represent the generator.

Now we adjoin a $3$-ball to parametrize the set of lines meeting $L$, the disjoint union of the previous $2$-skeleton, and a copy of affine $3$-space, and we have to describe how the boundary $2$-sphere maps onto the existing $2$-skeleton. To picture this imagine a $3$-ball divided by its equator into two disjoint $2$-discs, and identify antipodal points of the equator only. This makes each hemisphere into a copy of $\mathbb{P}^2$, and the two $\mathbb{P}^2$'s are now joined along a common copy of $\mathbb{P}^1$, namely the equator modulo identifications. Since we have added a $3$-cell whose boundary is the sum of the two copies of $\mathbb{P}^2$, that sum has become zero in second homology. Since each of them represents the generator, we have killed twice the generator of second homology which has now become $\mathbb{Z}/2\mathbb{Z}$. We also adjoined a $3$-cell, so it is conceivable we have created some $3$-dimensional homology, but the boundary of that $3$-cell was twice a generator of the old $H_2 \cong \mathbb{Z}$, hence not zero, so we have not adjoined any $3$-cycles, so $H_3$ is still zero.

At last we adjoin a $4$-ball with boundary sphere mapping into the $3$-skeleton. This cannot increase the $H_3$ which is thus still zero. But the boundary $3$-sphere maps to zero in (the old $3$-dimensional) homology, so we have added a $4$-cycle, no multiple of which is a boundary, giving us $\mathbb{Z} \cong H_4$.

Thus $H_0 = \mathbb{Z}$, $H_1 = \mathbb{Z}/2\mathbb{Z}$, $H_2 = \mathbb{Z}/2\mathbb{Z}$, $H_3 = 0$, $H_4 = \mathbb{Z}$. This proves $G$ is orientable.

  • $\begingroup$ This is very nice! $\endgroup$ – Potato Feb 13 '17 at 21:59
  • $\begingroup$ thank you for the kind remark and the great question! It inspired me to learn a new beautiful computation. I have recently realized that the 6 kinds of reduced echelon forms of 2 by 4 matrices of rank 2 (corresponding to which columns are pivots) also give the disjoint stratification into 6 disjoint affine schubert cells, and I was looking to expand on this observation. (I.e. two matrices have the same reduced echelon form iff they have the same kernel, so the 2 by 4 ones of rank 2 classify 2 diml subspaces of R^4.) $\endgroup$ – roy smith Feb 14 '17 at 5:20

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 $H_2=\mathbb Z_2$.

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    $\begingroup$ It may be worth mentioning that $\tilde{G}(2,4)$ is actually diffeomorphic to $S^2\times S^2$. Then the covering map $\tilde{G}(2,4)\rightarrow G(2,4)$ is the map $S^2\times S^2\rightarrow S^2\times S^2/\sim$ where $(x,y)\sim \pm (x,y)$. $\endgroup$ – Jason DeVito Jun 21 '17 at 4:48

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