Directly calculating $H^1_{\mathrm{sing}}(S^1;\mathbb Z)$

Question

While playing around with DeRham and singular cohomology, I decided to try and find out what an element $[\omega]\in H^1_{\mathrm{sing}}(S^1;\mathbb Z)$ would actually look like. The closure condition on $\omega\in C_{\mathrm{sing}}^1(S^1;\mathbb Z)$ means that $\langle\omega,\gamma\rangle\in\mathbb Z$ is independent of the parametrization of the path $\gamma:[0,1]\to S^1$ (up to orientation) and that $\langle\omega,\gamma_1\cdot\gamma_2\rangle= \langle\omega,\gamma_1\rangle+\langle\omega,\gamma_2\rangle$ where $\gamma_1\cdot\gamma_2$ denotes concatenation of paths. I'd like to have $\omega$ behave like the winding number, and give $\langle\omega,\gamma_0\rangle=1$ for $\gamma_0(t)=e^{2\pi i t}$ the standard parametrization of $S^1$, but I don't see how this can be done without having $\langle\omega,\gamma\rangle\notin\mathbb Z$ for some $\gamma$. So if a generator $\omega$ doesn't behave like a winding number, what does it look like?

Answer 1

The idea of winding numbers is good, but the issue is that there is not a good notion of what a winding number is for something other than a cycle. How often does the arc from $1$ to $i$ wind around the circle?

Apologies for the long answer that follows; this is because it tries to do two things: tell you what a generator actually looks like, and tell you how to better capture your idea of winding numbers.

---

What one usually does is the following. This tells you how to write down generators of cohomology, but not how to write down geometric generators.

We have the map $d: C_1(S^1) \to C_0(S^1)$. Its image, $B_0(S^1)$, is free abelian (as any subgroup of a free abelian group is free abelian). Therefore, there is some isomorphism $$C_1(S^1) \cong Z_1(S^1) \oplus B_0(S^1)$$ so that the map $d: C_1(S^1) \to B_0(S^1)$ is simply projection to the second factor. This is the first painful step for those of us who like geometric representatives, because this isomorphism is highly non-explicit (you are choosing, for each generator of $B_0(S^1)$, a corresponding chain whose boundary is that generator).

Now we have a further surjective homomorphism $Z_1(S^1) \to H_1(S^1) \cong \Bbb Z$. Again, because the codomain is free abelian, we may write $Z_1(S^1) \cong H_1(S^1) \oplus B_2(S^1)$.

Thus we have an isomorphism $$\varphi: C_1(S^1) \cong B_2(S^1) \oplus H_1(S^1) \oplus B_0(S^1),$$ respecting all of the differentials we care about. This is horribly non-explicit, but given this, we may construct the desired cocycle $f_\varphi$ by $$f_\varphi(\sigma) = \pi_2 \varphi(\sigma),$$ where this notation means we follow the above isomorphism and then project to the $H_1(S^1)$ factor.

---

Here is my favorite notion of geometric representative of an integral cohomology class. A slightly more general version of this captures all cohomology classes which arise from the $\text{Hom}(H_n(M;\Bbb Z), \Bbb Z)$ factor in the universal coefficient theorem. The other factor would arise instead as a linking number construction and require some unpleasant choices.

As I mentioned in the comments, these should be defined on a subcomplex $C$ of geometric nature, so that the inclusion $C \to C_*(M;\Bbb Z)$ is an isomorphism on homology.

To capture the idea of winding number, we'll do the following: pick a point $x \in S^1$ arbitrarily. Let the subcomplex $C^x_* \subset C_*(M;\Bbb Z)$ consist of those chains $\sigma: \Delta^k \to M$ so that $\sigma$ is a smooth map, and so that both $\sigma$ and $\partial \sigma$ are transverse to $x$. This means that if $k < \dim M$, the map $\sigma: \Delta^k \to M$ should not have $x$ in its image (and similarly for $\partial \sigma$ when $k = \dim M$). If $k = \dim M$, the map $\sigma: \Delta^n \to M$ should have $\partial \sigma$ disjoint from $x$, and for every $p \in \Delta^n$ with $\sigma(p) = x$, we should have $d\sigma_p: \Bbb R^n \to T_x M$ surjective. Similarly for $k > \dim M$.

First, this complex $C^x_* \subset C_*(M;\Bbb Z)$ is quasi-isomorphic to the whole complex. This is essentially because of the transversality theorem: given any smooth manifold $N$ and map $f: N \to M$, we may modify $f$ by a homotopy to be transverse to $x$; and if $f$ is already transverse to $x$ along some closed subset of $N$, we may keep the homotopy fixed along that subset. This allows us to take cycles in $M$ and obtain homologous cycles lying in $C^x$; similarly if $v \in C^x_*$ is a cycle which is a boundary in $C_*(M;\Bbb Z)$, we may modify the bounding chain by a homotopy, fixed on the boundary, so that in fact $v$ is a boundary in $C^x_*$.

Now if $M$ is oriented, I can define a cocycle $\deg_x: C^x_n \to \Bbb Z$, given by $$\deg_x(\sigma) = \#\left(p \in \Delta^n \mid \sigma(p) = x\right),$$ where this sum is weighted by signs: if $d\sigma_p: \Bbb R^n \to T_x M$ is an oriented isomorphism, count $p$ positively, and if not, count it negatively.

I claim that $\deg_x$ defines a class in $H_n(\text{Hom}(C^x, \Bbb Z)) = H^n(M;\Bbb Z)$, and in fact, this is the usual "fundamental class" of $M$. The picture is that you restrict to those $n$-chains whose boundaries do not cross over $x$, and then $\deg_x(\sigma)$ just counts (with sign) the number of points in the cycle that actually do cross over $x$. This is how one makes sense of the winding number when there is boundary flying around: pick a basepoint to check how often we "wind around at", and make sure the boundary doesn't get in the way.

Note that no matter what $x$ we choose, for a simplex $\sigma: [0,1] \to S^1$ with $\sigma(0) = \sigma(1) \neq x$, we have $\deg_x(\sigma) = \text{Wind}(\sigma)$.