# Null-homologous cycles in an open set

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

• Do you know what the winding number is? – rawbacon Nov 12 '19 at 18:41
• no @Levi, i just need a concrete definition of a null homologous cycle and I should be fine. – KarinaMath Nov 12 '19 at 19:03
• Thank you so much. A definition like this is exactly what I needed. @Levi – KarinaMath Nov 12 '19 at 19:09
• @Levi Do you also have a definition of null homotopic curves in the same set? – KarinaMath Nov 12 '19 at 19:12
• $\gamma$ 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. – rawbacon Nov 12 '19 at 19:16

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.
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.