1
$\begingroup$

What are null homologous cycles in an open set $Ω$ in $\mathbb C$? What is also an example that is not null-homologous?

$\endgroup$
  • $\begingroup$ Do you know what the winding number is? $\endgroup$ – rawbacon Nov 12 '19 at 18:41
  • $\begingroup$ no @Levi, i just need a concrete definition of a null homologous cycle and I should be fine. $\endgroup$ – KarinaMath Nov 12 '19 at 19:03
  • $\begingroup$ Thank you so much. A definition like this is exactly what I needed. @Levi $\endgroup$ – KarinaMath Nov 12 '19 at 19:09
  • $\begingroup$ @Levi Do you also have a definition of null homotopic curves in the same set? $\endgroup$ – KarinaMath Nov 12 '19 at 19:12
  • $\begingroup$ $\gamma$ is null-homotopic if there exists a continuous map (a "homotopy") $H: [0,1] \times [0,1] \rightarrow \Omega$ such that (i) $H(s, 0) = \gamma(0)$, (ii) $H(s, 1) = \gamma(1)$, (iii) $H(0, t) = \gamma(t)$, (iv) $H(1, t)$ is a constant path. (Null-homotopic means homotopic to a constant path). Intuitively, it means that you can deform $\gamma$ inside $\Omega$ such that in the end you are left with just a point. $\endgroup$ – rawbacon Nov 12 '19 at 19:16
1
$\begingroup$

Let $\Omega \subset \mathbb C$. A closed curve $\gamma: [0,1] \rightarrow \Omega$ is null-homologous if the winding number of $\gamma$ around any point in $\mathbb C \setminus \Omega$ is zero. The winding number of $\gamma$ around $z_0$ is the number of times $\gamma$ wraps around zero. Using complex analysis, we can define the winding number $n(\gamma, z_0)$ of $\gamma$ around $z_0$ to be $$n(\gamma,z_0)=\frac{1}{2\pi i}\int_{\gamma}\frac{1}{z}\,dz.$$ It is not obvious that these two notions coincide (this has to do with homotopy-invariance of path integrals), but they sure do.

Here are some instructive examples.

  1. A constant path is null-homologous in all $\Omega$.
  2. The standard embedding of the circle $t \mapsto e^{2\pi i t}$ is null-homologous in $\mathbb C$ (trivially because there is no point in $\mathbb C \setminus \mathbb C$), but not null-homologous in $\mathbb C \setminus \{0\}$ (because it winds once around $0$).
  3. If you have seen path-homotopy then you might think that null-homologous and null-homotopic (being "continuously shrinkable to one point") are the same. It turns out that being null-homologous is actually weaker. Indeed, one can hang up a picture using two nails and a string such that removing any one of the nails causes the picture to fall down. Let $\Omega = \mathbb C \setminus \{-1, 1\}$. Let $\gamma$ be the following curve: trace an $\infty$-shape around these two points: once around $1$ counterclockwise, around $-1$ counterclockwise, then again around $1$ but clockwise, and then finally around $-1$ clockwise. Then $\gamma$ is null-homologous in $\Omega$, but not null-homotopic.
| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.