1
$\begingroup$

Let $\gamma$ be a closed path in the plane $\mathbb{C}$ and let $a\in \mathbb{C}$ which does not belong to the image of $\gamma$. The winding number (or index) is defined as $$I(\gamma, a)=\int_{\gamma}\frac{dz}{z-a}.$$ Cartan's book says that, if $\gamma$ is fixed, the map $a\mapsto I(\gamma, a)$ is locally constant because $dz/(z-a)$ is a closed form and for any closed form $\omega$,$$\int_{\gamma_1}\omega=\int_{\gamma_2}\omega$$ where $\gamma_1, \gamma_2$ are homotopic closed paths.

I know how to show the map $a\mapsto I(\gamma, a)$ is locally constant (for example differentiating under the integral sign), but I can't see how Cartan uses the fact above to show it.

$\endgroup$

1 Answer 1

0
$\begingroup$

I think what he is saying is that in $ \mathbb{C} - a $, the form $ \omega = \frac{dz}{(z- a)} $ is defined and closed. So then the integral $ \int_\gamma \omega $ only on the homotopy class of $ \gamma $ in $ \mathbb{C} - a $.

Then, moving $ a $ a little bit is the same as translating $ \gamma $ in $ \mathbb{C} - a $, which is homotopic. So it follows $ a \rightarrow I(\gamma, a) $ is locally constant.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .