Second cohomology group of twice-punctured plane

Question

I am looking for pure calculus solution for computing the second cohomology class $H^2(\mathbb{R}^2-\{p , q\}, \mathbb{R})$ i.e. plane with two points removed.

I know that every closed $2$-form is exact for $\mathbb{R}^2-\{p , q\}$, but I cannot find the explicit $1$-form for every $2$-form that under the exterior derivative produces the $2$-form. I want to "integrate" the $2$-form and get the $1$-form by "d" operator. Note you have to somehow integrate "around" the points.

This is a problem from page 19 of Bott and Tu. I am trying to read the book on my own which would not be brought up in May if this was a Spring course.

Answer 1

The plane with two points removed retracts to a wedge of circles. Start at $p$ and gradually unfold , i.e., make the hole about $p$ larger. Do the same with $q$ simultaneously. Maybe doing this in a square first will help. Then, by homotopy invariance, the homology of $\mathbb R^2 -$ {$p,q$} is that of a wedge of two circles. For the top homology alone, you just need to decide if $\mathbb R^2 -$ $p,q$ is orientable or not.

If you prefer something more computational, you may want to use the Mayer-Vietoris sequence, and it would help if you accept that the plane minus a point, equivalently, the plane minus a disk retracts to $S^1$, and you know the (co)homology of $\mathbb R^2$. I can only think of one decomposition for using Mayer-Vietoris.

Answer 2

$H$ below denotes de Rham cohomology, so this is a pure calculus proof. I assume only that you know that $H^2(\mathbf R^2)=0$, which itself follows from $H^1(\mathbf R)=0$ by the Künneth decomposition.

First, it suffices to calculate $H^2(\mathbf R^2-{p})$, because $$(\mathbf R^2-{p}) \cap (\mathbf R^2-{q}) = \mathbf R^2-\{p,q\}$$ and $$(\mathbf R^2-{p}) \cup (\mathbf R^2-{q}) = \mathbf R^2;$$ since $H^2(\mathbf R^2) = 0,$ the Mayer-Vietoris sequence gives an isomorphism $$H^2(\mathbf R^2-\{p,q\}) = H^2(\mathbf R^2-{p}) \oplus H^2(\mathbf R^2-{q}).$$

Now, remark that $$\mathbf R^2-{0} = (\mathbf R^2-(X)) \cup (\mathbf R^2-(Y))$$ where $(X),(Y)$ are the axes. Remark also that

$$(\mathbf R^2-(X)) \cap (\mathbf R^2-(Y)) = \mathbf R^2-(X)\cup (Y)$$

and this last open set is diffeomorphic to the disjoint union of $4$ copies of $\mathbf R^2$, hence its $H^2$ vanishes. Thus, Mayer-Vietoris again identifies $H^2(\mathbf R^2-0)$ with $$H^2(\mathbf R^2-(X))\oplus H^2(\mathbf R^2-(Y))$$ But again, each of $\mathbf R^2-(X)$ and $\mathbf R^2-(Y)$ is diffeomorphic to 2 copies of $\mathbf R^2$, so the $H^2$ vanishes. Therefore $H^2(\mathbf R^2-p)$ vanishes, and so does $H^2(\mathbf R^2-\{p,q\})$.

Answer 3

I don't think that constructions from the Mayer-Vietoris sequence or Kuenneth formulas or other devices would be in the spirit of the first portion of Bott and Tu. If you, like me, worry about these things, then remember that if we're going to do freshman calculus---and we need to---it is perfectly legitimate to choose our coordinates. We can always translate, rotate, and scale coordinates to place our points where we want them. It is very nice, then to have $P:(x=-1, y=1)$ and $Q:(x=1,y=-1)$, and, as I said, this can be done passively just by choosing our coordinates to our own liking. We know that all the 2-forms are closed because we are at the highest grade for this space. An ordinary 2-form might be $g(x,y) dx \wedge dy$, perhaps. So, why can't we just do freshman calculus and consider for our own pleasure $\omega= -(\frac{1}{2})(\int_0^y g(x, \eta) d\eta) dx + (\frac{1}{2})(\int_0^x g(\xi, y)d\xi)dy$? It is a very nice 1-form, I think, and it is one of very many that will have $d$ leading us where we want to go. We really only need a sort of slightly disguised version of the fundamental theorem for a single variable, and all is well. So, all the 2-forms are exact, and the cohomology vanishes. Only the words are more confusing than freshman year, nothing more. I think this is sort of the spirit of Bott for the 2-grade.

Answer 4

I just wanted to write down an answer that didn't use the Mayer-Vietoris sequence explicitly, although it is certainly just this underneath. I however found it useful to try to get some deeper insight on the inner workings of the theorem. The general lesson that learnt from this, which is of course a simple corollary of the Mayer-Vietoris sequence, is that if a $k$-form $\omega$ can be written as an exact form $\text{d}\alpha$ on $U$ and $\text{d}\beta$ on $V$, and $U\cap V$ is such that every closed $(k-1)$-form is exact, then $\omega$ can in fact be written as an exact form on $U\cup V$.

The procedure in order to do this is the following. First, note that the difference between the two expression of $\omega$ on $U\cap V$ is closed, since $\text{d}(\alpha-\beta)=\omega-\omega=0$. The assumption on $U\cap V$ implies then that this difference is exact. In other words, there is a $(k-2)$-form $\gamma$ such that $\alpha=\beta+\text{d}\gamma$ on the intersection. Of course, we would like this to extend to all of $U\cup V$. In order to do so, we choose a partition of unity $\{\rho_U,\rho_V\}$ subordinate to $\{U,V\}$. We can then define the $(k-2)$-forms $\rho_U\gamma$ on $V$ and $\rho_V\gamma$ on $U$. We can then slightly modify our $(k-1)$-forms to $\alpha'=\alpha-\text{d}(\rho_V\gamma)$ and $\beta'=\beta+\text{d}(\rho_U\gamma)$. They are still representations of $\omega$ on their respective domains since they differ from the original ones by closed forms but now have a very nice property: they agree on the intersection $$\alpha'-\beta'=\alpha-\beta-\text{d}((\rho_V+\rho_U)\gamma)=\alpha-\beta-\text{d}\gamma=\alpha-\alpha=0.$$ They can thus be extended to a single $(k-1)$-form $\mu$. Moreover $\omega=\text{d}\mu$.

The intuition here is clear. The function $\rho_V$ vanishes far away from $V$, such that $\alpha'$ coincides with $\alpha$ there. However, as we are about to leave $U$ and enter $V$ it becomes $1$, such that $\alpha'=\beta$ there. In other words, $\alpha'$ smoothly interpolates between $\alpha$ and $\beta$. Moreover, it does so in a very special way which is where the hypothesis enters. Note that we could've still done this smoothing directly on the difference of our two forms $\rho_V\text{d}\gamma$. However, unlike $\text{d}\gamma$, this is not closed.

One thus only need to consider a cover of $\mathbb{R}^2\setminus\{p,q\}$ by two half-spaces, each surrounding only one of the removed points and intersecting on a strip, which would satisfy the conditions above $H^1(U\cap V)=0$. The proof that any 2-form is exact on each of these opens can be found in Ted Shiffrin's answer in how to compute the de Rham cohomology of the punctured plane just by Calculus?. Alternatively, one can further decompose each of these opens following Bruno Joyal's excellent answer.