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Suppose we set $\Omega = \mathbb R^3 \setminus \{ (x,y,0)\ :\ x^2+y^2 = 1 \}$: then $\Omega$ is not simply connected, for there exists a class of loops that are not contractible to a point, that is the equivalence class (w.r.t. $\Omega$-homotopy) with representative $\gamma_\varepsilon(t) = (1+\varepsilon\cos t, 0, \varepsilon \sin t) $, for some $0<\varepsilon<2$.

Ideally, it is then possible to find a closed form $\omega : \Omega \to (\mathbb R^3)^*$ such that the integral on any representative of this class of curves does not vanish, and even find one such form $\omega_\alpha$ for all $\alpha \in \mathbb R$ so that said integral ends up being equal to $\alpha$. One example would be the form corresponding to the magnetic vector field generated by a loop of wire $\Omega^\mathtt c$ in which a stationary current flows, where it is the current intensity that determines the value of the integral by Ampère's law. However, it is very hard to find the analytic expression of this form at every point in space.

Can you provide an explicit example of a closed form $\omega_\alpha : \Omega \to (\mathbb R^3)^*$ such that $$\int_{\gamma_\varepsilon} \omega_\alpha = \alpha?$$

giobrach
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    HINT: Try to write down the analog of $d\theta$ in the Euclidean plane where now $\theta$ is the angle around a circle normal to the $S^1$ (measured, say, from the horizontal axis). – Ted Shifrin Jun 30 '19 at 01:48
  • @Ted Shifrin If I understand your hint correctly, for any $\alpha$, I am asked to consider the $1$-form $\omega_\alpha=\frac{\alpha}{2\pi}\frac{1}{(\sqrt{x^2+y^2}-1)^2+z^2}\Big(-\frac{zx}{\sqrt{x^2+y^2}}dx-\frac{zy}{\sqrt{x^2+y^2}}dy+(\sqrt{x^2+y^2}-1)dz\Big)$. The integral along $\gamma_1$ should be $\alpha$, and, if I didn't mess up my calculation, the form is closed. However, for any $\alpha$ the form is only defined on $\Omega$ minus the $z$-axis. Could you please elaborate your hint a bit? That would be very much appreciated. – Max Demirdilek Feb 26 '23 at 07:57
  • @M.C. This form is actually not well-defined on the $z$-axis, but we can fix that very locally and then your approach is on the money. See my answer. – Ted Shifrin Feb 27 '23 at 02:40

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Topologically, what's going on here is that $\Bbb R^3$ retracts not to a torus with core $x^2+y^2=1$, $z=0$, but to the "torus with no hole" $S=\left(\sqrt{x^2+y^2}-1\right)^2+z^2=1$ (with the inner hole pinched to the origin). This surface has one-dimensional first homology, generated by a circle $x^2+z^2=1$, $y=0$. It is clear that there is a continuous retraction, but I haven't seen how to write down a $C^1$ retraction without a good deal of work. So let's just use this as motivation.

Because of symmetry about the $z$-axis, I'm going to work in cylindrical coordinates. We all know the usual closed-but-not-exact $1$-form on $\Bbb R^2- \{0\}$, i.e., $\dfrac{-y\,dx+x\,dy}{x^2+y^2}$. Pulling this back in the obvious way to $S$ by the map $(r-1,z)\rightsquigarrow (x,y)$, we get the form $$\omega = \frac{-z\,dr + (r-1)dz}{(r-1)^2+z^2}.$$ This form is not defined on the given circle, but has a problem when $r=0$, as $dr$ is not well-defined on the $z$-axis. On the surface $S$ (which is, admittedly, not a manifold at the origin), there is no problem, as $(r-1)dr + z\,dz = 0$ implies $dr = 0$ when $r=z=0$.

This suggests that we should replace $r$ with $\bar r$, with $d\bar r = 0$ when $r=0$, but with $r\rightsquigarrow \bar r$ smooth and a diffeomorphism away from $r=0$. In fact, with a standard bump function construction, you can arrange that $\bar r = r$ for $r\ge\delta>0$. Now take $$\bar\omega = \frac{-z\,d\bar r + (\bar r-1)dz}{(\bar r-1)^2+z^2}.$$ It follows that $\bar\omega$ is smooth, well-defined off our circle, and still closed.

Now, provided $\epsilon<1-\delta$, we will have $\bar\omega = \omega$ on the circle $\gamma_\epsilon$. (By the way, I think the OP should have had $0<\epsilon\le 1$. I don't know where the $2$ came from.) Then it is standard that $\displaystyle\int_{\gamma_\epsilon}\omega = 2\pi$.

Ted Shifrin
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