Hey my first question here, so I got an exercise but need some help. The questions is:

I have a domain $U\subset\mathbb{C}$, and curves $\gamma_j : [0, 1] \to U$, $j \in \{0, 1\}$ which are continuously differentiable.

They have the same start and end point, so $\gamma_0(0) = \gamma_1(0) = z_0$ and $\gamma_0(1) = \gamma_1(1) = z_1$. There is also a Homotopy between these points, $H : [0, 1] \times [0, 1] \to U$ with $H(0, \cdot) = \gamma_0$, $H(1, \cdot) = \gamma_1$, $H(\cdot, 0) = z_0$ and $H(\cdot, 1) = z_1$.

Show $\Gamma = \gamma_0 - \gamma_1$ is homologous to zero.

Here is the definition of homologous to zero:

If $\Gamma$ is in $U$ and if $Int(\Gamma) \subset U$, then $\Gamma$ is called homologous to zero. With $Int(\Gamma) = \{z\in\mathbb{C}\setminus[\Gamma]\mid ind_\Gamma(z) \neq 0\}$. $Int(\Gamma)$ is called Interior and $ind_\Gamma(z)$ is called winding number.

I probably have to show that $ind_{\gamma_0}(z) = ind_{\gamma_1}(z) \ \forall z \notin U$. Which probably implies $Int(\gamma_0 - \gamma_1) = Int(\Gamma) \subset U$ and therefore $\Gamma$ is homologous to zero.

Any ideas how to do this? Thanks in advance!


I think I got something but maybe there are some arguments missing:


Let $z\in\mathbb{C} \setminus U$. Since $H$ is continuously differentiable, the mapping $$j \mapsto ind_{\gamma_j}(z) = \frac{1}{2\pi i}\int_{\gamma_j} \frac{1}{\xi - z}d\xi = \frac{1}{2\pi i}\int_0^1 \frac{1}{H(j, t) - z}\frac{\partial^2 H(j,t)}{\partial j \partial t} dt$$ is continuously from $[0, 1]$ to $\mathbb{R}$. Also, the values from this map are in $\mathbb{Z}$. Therefore, the mapping is constant.

This implies $ind_{\gamma_0}(z) = ind_{\gamma_1}(z) \ \forall z \notin U$.

Which implies $Int(\gamma_0 - \gamma_1) = Int(\Gamma) \subset U$ and therefore $\Gamma$ is homologous to zero. $\Box$

  • $\begingroup$ I think the phrase you're looking for is "homologous to zero." $\endgroup$ – D. Brogan Jun 22 at 12:20
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    $\begingroup$ Your question does not make much sense. How is the subtraction $\gamma_0 - \gamma_1$ defined? How is $Int(\Gamma)$ defined? And "Is $\Gamma$ in $U$ and is $Int(\Gamma) \subset U$, so $\Gamma$ is called null-homolog" is not even a grammatical sentence, let alone a reasonable definition of what is called "homologous to zero" in English. $\endgroup$ – Lee Mosher Jun 22 at 12:48
  • $\begingroup$ @D.Brogan thank! I'll have a look maybe I'll find something regarding this :) $\endgroup$ – Xvid Jun 22 at 12:56
  • $\begingroup$ @LeeMosher sorry, english isn't my first language so I'm sorry for grammatical errors. Regarding your questions, we never explicitly defined $\gamma_0 - \gamma_1$. $Int(\Gamma)$ is an open subset, I have defined it in my post. Well, besides the grammatical error this is unfortunately the defintion my professor gave us. *I edited the post now, hopefully it's grammatically now a bit more correct :) $\endgroup$ – Xvid Jun 22 at 12:56
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    $\begingroup$ @Xvid Your proof doesn't work. See my above comment. I recommend to have a look at math.stackexchange.com/q/112679. $\endgroup$ – Paul Frost Jun 23 at 12:28

The discussion in the comments shows that some concepts need to be clarified. Here is a pragmatic approach.

You start with curves $\gamma_j : [0, 1] \to U$, $j \in \{0, 1\}$, which are continuously differentiable. Let us assume that the homotopy $H : I \times I \to U$ from $\gamma_0$ to $ \gamma_1$ is a continuous map which is partially differentiable with respect to the second coordinate (i.e. $\dfrac{\partial H}{\partial s}(t,s)$ exists for all $(t,s) \in I \times I$) and such that $\dfrac{\partial H}{\partial s} : I \times I \to \mathbb C$ is continuous. This implies that all intermediate curves $H_t = H(t,-)$, $t \in I$, are continuously differentiable. This requirement is a little bit weaker than assuming that $H$ is continuously differentiable (this concept would need further clarification).

Let us moreover agree that the winding number $ind_\gamma(z)$ is defined for closed piecewise continuously differentiable curves $\gamma$ and all $z \notin \gamma(I)$. The winding number is always an integer given by $$ind_\gamma(z) = \frac{1}{2\pi i} \int_\gamma \frac{1}{\zeta-z}d\zeta = \frac{1}{2\pi i} \int_0^1 \frac{\gamma'(s)}{\gamma(s)-z}ds .$$

Finally we regard $\Gamma = \gamma_0 - \gamma_1$ as a closed curve based at $z_0$. It is composed by $\gamma_0$ (path from $z_0$ to $z_1$) followed by the inverse of $\gamma_1$ (path from $z_1$ to $z_0$). Explicitly, $\Gamma(s) = \begin{cases} \gamma_0(2s) & s \le 1/2 \\ \gamma_1(2-2s) & s \ge 1/2 \end{cases} \quad$ .

Let us define a homotopy $G : I \times I \to U, G(t,s) = \begin{cases} H(t,s) & s \le 1/2 \\ \gamma_1(2-2s) & s \ge 1/2 \end{cases} \quad$ . This is a homotopy from $G_0 = \Gamma$ to $G_1 = \gamma_1 - \gamma_1$ such that all $G_t$ are piecewise continuously differentiable curves.

By definition $G(I\times I) \subset U$. It therefore suffices to show that for all $z \in \mathbb C \setminus G(I\times I)$ we have $ind_\Gamma(z) = 0$ because this implies that $Int(\Gamma) \subset G(I\times I) \subset U$.

For $z \in \mathbb C \setminus G(I\times I)$ we obviously have $ind_{G_1}(z) = \frac{1}{2\pi i} \int_{\gamma_1} \frac{1}{\zeta-z}d\zeta - \frac{1}{2\pi i} \int_{\gamma_1} \frac{1}{\zeta-z}d\zeta = 0$. We want to prove that $ind_\Gamma(z) = ind_{G_0}(z) = 0$. To do this, it has to be shown that the function $$\phi : I \to \mathbb Z, \phi(t) = \frac{1}{2\pi i} \int_0^1 \frac{G_t'(s)}{G_t(s)-z}ds$$ is continuous and therefore constant. I shall not give a proof but leave it as an excercise to you.


Thanks to everyone, I'll provide the solution from my teacher which seems, at least for me, a bit more easier to understand.

We have to show that $Int(\Gamma) = Int(\gamma_0 - \gamma_1) \subset U$, or in other words $ind_{\Gamma}(z) = ind_{\gamma_0 - \gamma_1}(z) = 0 \ \forall z \notin U$.

Let us define the mapping

$$\eta_s(z) : [0, 1] \to \mathbb{R}, s \mapsto ind_{\gamma_0 - \gamma_s}(z) \ \forall z \notin U$$

$$\rightarrow ind_{\gamma_0 - \gamma_s}(z) = \frac{1}{2 \pi i} \int_{\gamma_0 - \gamma_s} \frac{d\zeta}{\zeta - z}= \frac{1}{2 \pi i} \int_{\gamma_0} \frac{d\zeta}{\zeta - z} - \frac{1}{2 \pi i} \int_{\gamma_s} \frac{d\zeta}{\zeta - z}$$

The first term is obviously constant, as he doesn't vary. The second term can we rewrite as:

$$- \frac{1}{2 \pi i} \int_{\gamma_s} \frac{d\zeta}{\zeta - z} = - \frac{1}{2 \pi i} \int_0^1 \frac{\gamma_s^\prime(t)}{\gamma_s(t) - z} dt.$$

We know, from some previous Proposition from our lecture, that this is continuous and $\in \mathbb{Z}$, therefore this one is constant too.

$\Rightarrow \eta_s(z) = const.$

We only have to calculate now what the value of $\eta_s(z)$ for every $s\in [0,1]$ is (as it's constant).

Let $s = 0 \rightarrow \eta_0(z) = ind_{\gamma_0 - \gamma_0}(z) = 0$.

Therefore, $\eta_1(z) = ind_{\gamma_0 - \gamma_1}(z) = ind_{\Gamma}(z) = 0 \ \forall z \notin U$. $\Rightarrow Int(\Gamma) \subset U$ and therefore $\Gamma$ is homologous to zero.

With this, one can also easily come to the homotpy version of Cauchy's integral theorem.

It's probably not a perfect proof but this kinda works out for me, sorry for grammar errors

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    $\begingroup$ The proof is not correct. You consider the function $I(s) = -\frac{1}{2 \pi i} \int_{\gamma_s} \frac{d\zeta}{\zeta - z} = - \frac{1}{2 \pi i} \int_0^1 \frac{\gamma_s^\prime(t)}{\gamma_s(t) - z} dt$ and claim that it is continuous (which is certainly true) and an integer which is not true. An integral of the form $\frac{1}{2 \pi i} \int_{\gamma} \frac{d\zeta}{\zeta - z}$ only takes integer values for closed curves. $\endgroup$ – Paul Frost Jun 24 at 16:54
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    $\begingroup$ However, you can easily repair it. Although $I(s) = -\frac{1}{2 \pi i} \int_{\gamma_s} \frac{d\zeta}{\zeta - z}$ doesn't take integer values, it is continuous and therefore also $\eta_s$ is continuous. But $\eta_s$ takes integer values and thus is constant. $\endgroup$ – Paul Frost Jun 24 at 17:14
  • $\begingroup$ @PaulFrost thanks! Maybe I missed this argument but yeah make sense. $\endgroup$ – Xvid Jun 25 at 9:06

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