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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:

$$I = \int_C \tau = \int_C d\varphi = 2\pi I,$$

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 $d\tau = 0$,

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

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.

enter image description here

$\textbf{EDIT 2:}$

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).

enter image description here

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

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

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  • $\begingroup$ The outward normals to $C_1$ and $C_2$ as the boundary of $\triangle$ point in opposite directions, and so the curves must be oppositely oriented to fit the definition of boundary orientation. $\endgroup$ – Ted Shifrin Nov 29 '18 at 17:27
  • $\begingroup$ @TedShifrin, what you mean by "outward normals to $C_1$ and $C_2$ as the boundary of $\triangle$"? I don't have that the codimension is equal to one, indeed, I don't know even if the manifold $M$ is embedded on an other manifold, I just know that the $M$ is a Riemannian manifold, oriented, compact of dimension $2$. $\endgroup$ – George Dec 1 '18 at 12:05
  • $\begingroup$ The annulus $\triangle$ inherits an orientation from the surface $M$. I'm talking about the boundary orientation of $C_1$ and $C_2$ as they comprise $\partial\triangle$. $\endgroup$ – Ted Shifrin Dec 1 '18 at 18:11
  • $\begingroup$ @TedShifrin, I understood what you said now, thanks, but I'm don't realize why the normals point in opposite directions. I edited my OP, tried explain better my doubt and I "ilustrated" why I think the normals have the same direction. $\endgroup$ – George Dec 1 '18 at 19:41
  • $\begingroup$ (a) The normal has to point out of $\triangle$, so it's the second picture. (b) But then, by definition of boundary orientation, the normal vector followed by the tangent vector of the curve has to be a "right-handed" basis (i.e., agreeing with the orientation on $\triangle\subset M$). $\endgroup$ – Ted Shifrin Dec 1 '18 at 20:04

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