# Index of zero of a vector field

From reading http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Popa.pdf, I understand that the index of any non-degenerate zero (defined using degree theory) of a vector field on a manifold must be $\pm 1$. I have two questions.

1. If a zero is degenerate, can one still use degree theory to define its index? For example, it seems that in the answer to this question: zero of vector field with index 0, the degree definition wouldn't make sense, because the map $$S^1\rightarrow S^1$$ would be undefined at the top and bottom of the circle. Here the degree is defined by counting the number of revolutions a vector would make in going around a small circle around the zero. Is there an analogue of this for higher dimensions?

2. What is an example of a vector field with a zero of index other than $\pm 1$, say 2? What definition of index is used to compute this?

Many thanks.

• The answer at math.stackexchange.com/questions/614255/… has a flaw. The index at a zero is defined only for isolated zeros. May 30, 2017 at 12:31
• The answer at math.stackexchange.com/questions/614255/… was amended. The first part of the "answer" is still not good but the second part is ok. May 30, 2017 at 14:58

The definition: Let $X: U \subset {\mathbb R}^2 \to {\mathbb R}^2$ be a vector field. To define the index we assume that the zero $z \in U$ is isolated, i.e. that there is an $r>0$ so that if $0<|w-z|<r$, then $X(w) \neq 0$. Therefore, the unit vector field $Y:=X/|X|$ is well-defined on the punctured disc $V:= \{w: 0<|z-w|<r\}$. A map from $S^1=\{w: |w|=1\}$ to itself is defined by $w \to Y((r/2)\cdot w)$. (Here I multiplied by $r/2$ to get $w$ into $V$, but it is a fact that multiplication by any number less than $r$ will give same degree.) The "index" is defined to be the (topological) degree of this map. (I strongly recommend Milnor's short book "Topology from the differentiable viewpoint" for a discussion of degree.)
Each self-map $f: S^1 \to S^1$ of the circle lifts to a map from $\tilde{f}:{\mathbb R} \to {\mathbb R}$. If $\tilde{f}$ is monotone increasing, then the cardinality of $f^{-1}(\theta)$ does not depend on $\theta$. This cardinality is the degree of $f$.
For example, the restriction of the vector field $(x,y) \mapsto (x^2-y^2,2xy)$ defines a self-map $f$ of the circle. By introducing the change of coordinates $x = \cos(\theta)$ and $y=\sin(\theta)$, we lift the map to a map $\tilde{f}:{\mathbb R} \to {\mathbb R}$. By the double angle formulas in trigonometry, we see that $\tilde{f}(\theta)= 2 \theta$. The map $\tilde{f}$ is monotone increasing, and the number of points in $f^{-1}(\theta)$ equals $2$ regardless of which $\theta$ we choose.
Thus, the vector field $(x^2-y^2, 2xy)$ has index 2.
If the lift $\tilde{f}$ of some self-map $f$ were monotone decreasing, then the degree of $f$ would equal the negation of the number of pre-images. For example, the vector field $(x^2-y^2, -2xy)$ has index -2.
The example of degree 2 can be understood geometrically by considering a vector field on the sphere described as follows: Choose a line that is tangent to the sphere at the north pole. For each plane containing this line, the intersection of each plane with the sphere is a circle. These circles `foliate' the sphere. By choosing vectors tangent to the circles, one forms a vector field on the sphere. If one looks down at the north pole, then this vector field looks like $(x^2-y^2, 2xy)$. (A picture of the vector field appears on page 33 of Milnor's book.) Since the Euler characteristic of the sphere equals 2, the