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.
The claim as stated is false. What is true is that if $U \subset \mathbb{R}^3$ is open, then $H_1(U;\mathbb{Z})=0$ if and only if $H^1_{dR}(U)=0$. Basically, the error lies in statement $2.$
So let's first talk about why the claim is false. It is known that there exists an embedding $K$ of the disk $D^2$ in $S^3$ such that $S^3-K$ is not simply connected. However, every embedding of any disk of any dimension in any sphere of any dimension is such that the complement is acyclic, i.e. has vanishing (reduced) singular homology. (You can see this in the chapter about Jordan's curve theorem in Bredon's book Topology and Geometry.) We can embed $S^3-K$ in $\mathbb{R}^3$ by picking a stereographic projection based on an element of $K$, and this gives us our open set that serves as a counterexample.
However, the claim that $H_1(U;\mathbb{Z})$ is torsion-free is true, and with that the result I mentioned follows, as you seem to observe in the question. If I recall correctly, a proof goes as follows: embed $U$ again in the sphere and let $C$ be the complement. By Alexander duality (also in Bredon's book),
$$H_1(U;\mathbb{Z})\simeq \check{H}^1(C;\mathbb{Z}). $$
Now let $X$ be any space. By the universal coefficients theorem, $H^1(X;\mathbb{Z}) \simeq \mathrm{Hom}(H_1(X);\mathbb{Z})$, since the $\mathrm{Ext}$ part vanishes as $H_0$ is free abelian. On the other hand, $\mathrm{Hom}(H_1(X);\mathbb{Z})$ is torsion-free. Thus, $H^1(X;\mathbb{Z})$ is torsion-free. It follows that $\check{H}^1(C;\mathbb{Z})$ is a direct limit of torsion-free abelian groups, and therefore is also torsion-free. (If I recall correctly, one way to see this is by noting that over $\mathbb{Z}$ this is equivalent to being flat, and flatness is preserved by direct limits.) So $H_1(U;\mathbb{Z})$ is torsion-free.
Some further points:
- In the comments, Moishe Kohan mentions that the wikipedia article first assumes that $U$ is an open subset of $\mathbb{R}^3$ and then switches mid-proof to being an open subset of $\mathbb{R}^2$. In $\mathbb{R}^2$, it is true that $H_1=0 \iff \pi_1=0$, and thus by making minor adaptations it is true that $\pi_1=0 \iff H^1_{dR}=0$. This is due to the fact that an open subset of $\mathbb{R}^2$ must have a free fundamental group, and thus $H_1$ is the free abelian group with the same number of generators of the $\pi_1$ by Hurewicz's theorem. This, of course, does not exempt the sloppiness on the wikipedia article.
- The result that $H_1=0 \iff H^1_{dR} =0$ does not hold for general open subsets of Euclidean spaces. The simplest example is obtained by embedding $\mathbb{R}P^2$ in some $\mathbb{R}^n$ and taking a tubular neighbourhood of it. This will have trivial $H^1_{dR}$, but $H_1=\mathbb{Z}_2$. This also shows that the result is not true as soon as $n=4$, since $\mathbb{R}P^2$ embeds in $\mathbb{R}^4$.
Best Answer
Well, if you allow the fact that $H_1$ is the abelianization of $\pi_1$ (Hatcher 2A.1), then
$$\mathrm{Hom}(\pi_1(X),\mathbb Z) \cong \mathrm{Hom}(H_1(X), \mathbb Z).$$
If $H_1(X)$ is finitely generated and if $\mathrm{Hom}(H_1(X), \mathbb Z)$ is zero, then $H_1(X)$ is finite. Since $X$ is a finite $CW$ complex, $H_1$ is definitely finitely generated.
We then want to show that $\mathrm{Hom}(\pi_1(X), \pi_1(S^1))$ is trivial.
Proposition 1B.9 in hatcher's text proves that since $S^1$ has a contractible universal cover,$$\mathrm{Hom}_{\mathrm{Grp}}(\pi_1(X),\pi_1(S^1)) \cong [X,S^1] =0$$
where the $[X,S^1]$ are based maps.