Differential Forms and Applications by do Carmo - Chapter $6$ - Lemma $1$.

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?

$$\textbf{EDIT 1:}$$

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

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

• 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. – Ted Shifrin Nov 29 '18 at 17:27
• @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 $2$. – George Dec 1 '18 at 12:05
• 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 $\partial\triangle$. – Ted Shifrin Dec 1 '18 at 18:11
• @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. – George Dec 1 '18 at 19:41
• (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 $\triangle\subset M$). – Ted Shifrin Dec 1 '18 at 20:04