Poincare dual of unit circle

Question

I'm trying to self-study Differential Forms in Algebraic Topology by Bott and Tu. I've come across this exercise:

Show that the closed Poincare dual of the unit circle in $ R^2-\{0 \} $ is zero, but the compact Poincare dual is the nontrivial generator $ \rho(r)dr$ of $ H_c^1(R^2- \{0 \}) $ where $ \rho(r) $ is a bump function with integral $ 1 $.

I tried to show the first part by letting $\omega=f(r,\theta)dr+g(r,\theta)d\theta$ and using the inclusion of the unit circle into $R^2-\{0\}$: $ i(\theta)=(1,\theta)$. I got $$\int_{S}i^{*}\omega=\int_0^{2\pi}(f(1,\theta)d(1)+g(1,\theta)d\theta)=\int_{0}^{2\pi} g(1,\theta)d\theta$$

If $g$ is a bump function with value $1$ on the unit circle and compact support, the integral is nonzero. I must be making a mistake. What is it? How can I solve the exercise then?

Thank you

Answer 1

EDIT 2: Oh, I think I finally understood your reasoning. You argued as follows: assume $\eta_S$ vanishes. Then the integral also has to always vanish. But for your $\omega$ it does not. The problem with this reasoning is that your $\omega$ doesn't come from any class in $H^1_c({\bf R^2 - \{0\}})$. That is, it is impossible to extend $\omega$ to the whole plane such that it both stays compactly supported and will be closed. I hope this clears up the confusion.

To actually solve the exercise you then have to argue as follows: take any closed 1-form $\omega$ with compact support and show that the corresponding integral vanishes for it. In fact, since we know how the cohomology group looks like (it's generated by the form mentioned in the second part of the exercise), it's enough to prove it for that form. And there it is obvious, since that form doesn't have any $d \theta$ part (this is basically the original explanation I gave before the edits).

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EDIT: A compact Poincare dual of a $k$-dim submanifold $S$ is a $(n-k)$-form with compact support $\eta_S$ such that $$\int_S i^* \omega = \int_M \omega \wedge \eta_S$$ holds for every $\omega \in H^k(M)$. In particular, if we plug a 1-form $\omega = \eta_S$ into the equation we'll get $0$. I think this is what you are trying to do and the reason you are not getting zero is because you didn't plug $\eta_S$ in there but some random form $\omega$. The fact that it contains a bump form is irrelevant. Not every bump form is a Poincare dual!

The same discussion applies for the closed Poincare dual, only it will not be always possible to plug $\eta_S$ into the equation because it is only required to hold for $\omega \in H^{k}_c(M)$ and $\eta_S$ might be non-compact.

I hope it makes some sense. In case I'm still not quite getting what you were trying to do, post a comment again.

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I think you mixed the two parts of the exercise. In the first part, you are to consider generic 1-form. In the second, the bump 1-form. But in your derivation you have plugged the bump 1-form in place of the generic one.

Let me try to provide some explanation. The bump 1-form is of course closed since $$d \left( \rho(r) dr \right) = \partial_r \rho(r) dr \wedge dr = 0.$$ In the standard cohomology, it is also exact because $$ \rho(r) dr = d\sigma(r)$$ where $\sigma(r) = \int_0^r \rho(s) ds$ is a smooth step function that vanishes near the origin and equals $1$ outside some compact set. Therefore, the non-compact Poincare dual is zero. But $\sigma(r)$ doesn't have compact support and thus $\rho(r) dr$ is not exact and so represents a non-trivial element of the cohomology with compact supports.