Clarification about the index of singular points and curves on the sphere.

differential-topologysmooth-manifoldsVector Fields

I am having some problems understanding the concept of index of a singular point of a vector field on a Manifold (in particular on a 2-dimensional sphere) and some of its properties. And hope that you could help me clarify some aspects of it.

I think I understand intuitively the meaning of the index on the plane. You just have to calculate "How many times the field rotates" around a singular point. I also understand that the index of a curve does not change under continuous deformations of the curve, and you can assign an index to singular points by considering small loops around them. I also understand that the index of a curve will be equal to the sum of the indeces of the singular points that are within the curve. And that the index of a singular point will not change under diffeomorphisms. All of this is valid on the plane.

More formally given a vector field $\textbf{v}$ with coordinates $(v_1,v_2)$ you can define the form $\omega=\frac{v_2 dv_1-v_1 dv_2}{v_1^2+v_22}$ whenever the vector field is non-singular. Then the index of a curve $\gamma$ will be $ind(\gamma)= \frac{1}{2\pi} \oint_\gamma \omega$.

Then one can generalize the concept of index to a manifold since it is preserve under diffeomorphisms and is the integral of a differential form. But then I get confuse with some of its properties. To be more precise let me narrate and example.

Suppose you have a vector field on a sphere with some isolated singular points. And you find a point $P$ such that the field is non-singular on $P$. Like in the next figure (singular points- red):
Fig 1: A vector field on a sphere with many singular points (in red) and a non-singular point $P$

Then you can use stereo-graphical projection around $P$ and have every singular point on this map. Choosing as our curve $\gamma$ a circle with enough radius we can guaranty that every singular point is on the inside of the circle. And by the property that the index of a curve is the sum of the indeces of the singular points we know that the index of $\gamma$ will be the total sum of the indeces of the singular points. This is illustrated on the next figure.

Choosing appropriate coordinates we can have every singular point on a map. And with a circle of enough radius we can have every singular point within the circle.

Then we can pass the circle to the sphere and by continuous deformations transform it to a circle around the point $P$. But since the index of a curve is given by a differential form and will not change under continuous deformations we will preserve the index of the curve. This is illustrated on the next image.
The circle can be put on the sphere and deform to a circle around $P$

However if we now choose coordinates near the point $P$ we have a curve around a non-singular point. And so the index is zero. This is illustrated on the last image.

But in coordinates near $P$ the index of the curve is zero, because in the interior there are no singular points

In summary the index of the curve $\gamma$ must be zero, but it must also be the sum of the indeces of the singular points. But this result is false since it contradicts the Poincaré-Hopf theorem, which in the case of the sphere states that the sum of the indeces of the singular points must be equal to two.

So I have committed a mistake on my reasoning and I do not know where. Is the index of a curve on a manifold not preserve under continuous deformations? Only on the plane? Is the index of a curve dependent on coordinates chosen to calculate it? But isn't it the integral of a differential form and therefor independent of coordinates? If the index changes with coordinates, how can we interpret that the index of the same curve has different values depending on the chart? More general, what are the properties that are valid on the plane but not in a general manifold?

I am really clueless about this problem and hope that you could help me. Thank you!

Additional notes: I am reading the last pages of Vladimir Arnold's book on Ordinary Differential Equations, and he uses this example of the sphere to give a sketch of the proof of Poincare-Hopf theorem for the sphere, so that result is not known at that moment of the book. He also does not touch the subject to deeply so additional readings are welcome (:

He also writes "we can interpret the index of the circle on the first chart as 'the number of revolutions of the field $\textbf{v}$ with respect to the field $\textbf{e_1}$ ' during a traversal of the circle" But I have no idea what he is taking about.

Best Answer

The problem is that the $1$-form $\omega$ is not well defined globally on the set where $v\ne 0$: it depends on a particular choice of coordinates. Provided $U$ is a simply connected coordinate domain containing the image of $\gamma$, and $\omega$ is computed with respect to smooth coordinates on $U$, the expression $\frac{1}{2\pi} \oint_\gamma \omega$ will be equal to the sum of the indices of the singular points of $v$ inside $\gamma$. But if you do the computation with respect to a different coordinate domain, you'll get a different $\omega$, and different critical points might be inside $\gamma$, so the value of the integral expression will probably be different.

To address the question in your last paragraph: what's implicit in Arnold's remark is that "how many times a vector field rotates" can only be defined with respect to a particular coordinate chart. If you choose coordinates $(x^1,x^2)$ and let $\mathbf e_1$ denote the unit vector field in the direction of $\partial/\partial x^1$, then you can find a smooth angle function $\theta$ such that $$ \cos \theta(t) = \left\langle \frac{v(t)}{|v(t)|} , \mathbf e_1(\gamma(t))\right\rangle. $$ In other words, $\theta(t)$ represents a smooth choice of the angle between $v(t)$ and $\mathbf e_1(t)$. Then a computation shows that $\oint_\gamma \omega$ represents the total change in $\theta(t)$ as you traverse $\gamma$.

But of course this angle depends on $\mathbf e_1$, which in turn depends on your choice of coordinates.

Related Question