Htykuut

Let be $M$ a compact manifold $M$ which has a finite number of singular isotaled points and $I$ the index of a vector field around a singular isolated point.

The definition of the index $I$ of a vector field around a singular isolated point is the number of "turns" given by the vector field as we go along a simple closed curve around an isolated singularity. The $I$ is given by the following equation:

$$I = int_C tau = int_C dvarphi = 2pi I,$$

where $C$ is a simple closed curve which encloses an isolated singularity and $tau$ and $varphi$ are introduced on Lemmas $4$ and $5$ of the chapter $5$ of this book (if necessary, I can include the lemmas here).

I'm trying understand the following lemma in the do Carmo's book:

$textbf{Lemma 1.}$ The definition of $I$ does not depend on the curve $C$.

$textbf{Proof.}$ Let $C_1$ and $C_2$ be two simple closed curves around $p$ as in the definition of index. Assume first that $C_1$ and $C_2$ do not intersect adn consider the annular region $triangle$ bounded by $C_1$ and $C_2$. Let $I_1$ be the index computed with $C_1$ and $I_2$ be the index computed with $C_2$. By Stokes theorem and the fact that $dtau = 0$,

$$I_1 - I_2 = frac{1}{2pi} int_{C_1} tau - frac{1}{2pi} int_{C_2} tau = frac{1}{2pi} int_{triangle} dtau = 0$$

and this proves the Lemma in this case. If $C_1$ and $C_2$ intersect, we choose a curve $C_3$ that does not intersect both $C_1$ and $C_2$. By applying the above, we conclude that $I_1 = I_3 = I_2$. $square$

I didn't understood why the sign of the integral $int_{C_2} tau$ change for the curve $C_2$. I will tell what I thought about this. I know that the orientation of the boundary is induced by the manifold as do Carmo proved in his book (see here for a proof given by do Carmo), then the orientation of the boundary needs to be compatible with the orientation of the manifold $M$. I think the orientation of the annular region $triangle$ is the same of the $M$ since $M$ is oriented and the orientations of the curves $C_1$ and $C_2$ are induced by the orientation of the annular region $triangle$ since I want to use Stokes' theorem, then the orientation of the curves need to be compatible of the orientation of $M$. I suspect that's the reason why the sign of the integral around the curve $C_2$ changes, because I need the orientation of these curves to be compatible with the orientation of $M$ and if the curves has the same orientation, then at least one of these curves doesn't have an orientation compatible of $M$, but I can't see why this could be true. Am I correct in my thoughts? If I'm correct why at least one of these curves doesn't have an orientation compatible of $M$ if both has the same orientation?

Thanks in advance!

$textbf{EDIT 1:}$

Fix a "clockwise" orientation on a manifold $M$ of dimension $2$ and let be ${ T(C_1), N(C_1) }$ and ${ T(C_2), N(C_2) }$ orientations for the curves $C_1$ and $C_2$, respectively, then $N(C_1$ and $C_2$ has the same orientation or they have opposite orientations. I know that we would like these orientations be compatible with the orientation of $M$ since $M$ is oriented, but it seems to me that the $N(C_1)$ and $N(C_2)$ need on the same direction (left side of the image below) for the orientation of the curves be compatible with the orientation of $M$ (the "clockwise" orientation) while seems to me that the $N(C_1)$ and $N(C_2)$ can't have the same orientation (right side of the image below), because this would imply that $C_1$ or $C_2$ don't have the orientation of $M$. I know I'm wrong, but I can not figure out where I'm going wrong or what I'm forgetting.

$textbf{EDIT 2:}$

I think that I finally understood why the sign of $int_{C_2} tau$ changes in the proof of the Lemma $1$. Can someone confirm if my argument is correct?

Since $M$ is oriented, the annular region $triangle$ inherit the orientation of $M$.

On the one hand, observe that, seeing the curves $C_1$ and $C_2$ as $1$-dimensional submanifolds of the $2$-dimensional manifold $M$, we realize that the curves $C_1$ and $C_2$ has only two possibilites of orientation. Furthermore, observe that the orientations of $C_1$ and $C_2$ $textbf{inherited by the annular region $triangle$}$ are opposite. Indeed, if we see the annular region $triangle$ as a strip with only one couple of opposite edges identified, we realize that $C_1$ and $C_2$ can't have the same orientation when the orientations of them are $textbf{inherited by the annular region $triangle$}$, otherwise, $partial triangle$ wouldn't be orientable (see the picture below).

By the other hand, the union of the region enclosed by the curve $C_2$ and the annular region $triangle$ ($Q cup triangle$) inherits the orientation of $M$ and the region enclosed by the curve $C_2$ inherits the orientation of $M$, then the orientation of $C_1$ $textbf{inherited by $Q cup triangle$}$ and the orientation of $C_2$ $textbf{inherited by $Q$}$ are the same (observe that this is not contradict what we observe previously since the orientations of $C_1$ and $C_2$ obtained here were inherited by different regions of the annular region).

The point of the proof of the Lemma $1$ is that $C_2$ has the orientation $textbf{inherited by $Q$}$ by the definition of index of a vector field around an isolated singularity, while we need $C_2$ with the orientation inherited by $textbf{inherited by the annular region $triangle$}$, because these two orientations are opposite which was observed in the previous paragraphs.

Denote by $left( int_{C_1} tau right)^{Q cup triangle}$ the integral of $tau$ over $C_1$ with the orientation $textbf{inherited by $Q cup triangle$}$, $left( int_{C_2} tau right)^Q$ the integral of $tau$ over $C_2$ with the orientation $textbf{inherited by $Q$}$ and $left( int_{C_2} tau right)^{triangle}$ the integral of $tau$ over $C_2$ with the orientation $textbf{inherited by the annular region $triangle$}$,then the proof of the lemma can be rewritten as

$$I_1 - I_2 = frac{1}{2pi} left( int_{C_1} tau right)^{Q cup triangle} - frac{1}{2pi} left( int_{C_2} tau right)^Q = frac{1}{2pi} left[ left( int_{C_1} tau right)^{triangle} + left( int_{C_2} tau right)^{triangle} right] = frac{1}{2pi} left[ left( int_{C_1 cup C_2} tau right)^{triangle} right] = frac{1}{2pi} int_{partial triangle} tau = frac{1}{2pi} int_{triangle} dtau = 0$$