Confusion about the index and degree of vector fields

Question

These are the definitions I am using:

1) Let $v: \mathbb{R}^n \to \mathbb{R}^n$ be a vector field. For a regular value $y$ of $v$, we define the degree $$ \deg(v) = \sum_{x \in v^{-1}(y)} \operatorname{sign}(\det J(x)) $$ where $J$ is the Jacobian of $v$.

2) Suppose $v$ has an isolated zero at $x$. Choose a ball of radius $r>0$ about the isolated zero such that no other zero is contained in the ball. The index of the zero is defined to be the degree of the map $v(x+rw)/\|v(x+rw)\| : S^{n-1} \to S^{n-1}$.

Now, my questions. The vector field $$ v = (x^2 - y^2, -2xy) $$ has an isolated singularity at the origin. It has degree $-2$, easily seen by computing the Jacobian and looking at the pre-image of $(1,0)$.

From looking at plots of the field and also the answer to this question, I believe that in this case the index of the point 0 should also be $-2$. The answer to the linked question explicitly claims that this is the case. However I have some confusion about this.

If we compute the Jacobian of $v/\|v\|$, we find the determinant is everywhere $0$. This tells me that either $v/\|v\|$ has no regular values (which is impossible, because it contradicts Sard's theorem?) or else is not surjective, so there is some point with empty pre-image which is trivially regular, and hence the index of $v$ is $0$.

Question 1: What is the index of the zero at $0$ for this field $v$?

Consider the map $$ w = (x^2 - y^2 +x, -2xy+y) $$ Then the Jacobian of $w$ at the origin has determinant $1$. There is a theorem on p37 of Milnor that says the index must then be 1, however when I compute the Jacobian of $w/\|w\|$ the determinant is again everywhere $0$.

Question 2: What is the index of the zero at $0$ for the field $w$? It should be $1$, but the Jacobian of the map $w/\|w\|$ has determinant 0.

Is using the Jacobian when computing the degree of $v/\|v\|$ the wrong thing to do? I'd appreciate if someone could clear this up for me, and especially if they could work through the computation of the index of $v$ or $w$.

Edit: My other concern is that the sum of the indices of all isolated zeros of a vector field $v:\mathbb{R}^n \to \mathbb{R}^n$ must be equal to the Euler characteristic of $\mathbb{R}^n$, which is 1. This doesn't seem to be the case with what I'm doing. Help?

Answer 1

Your question really doesn't make very much sense, because it seems to be based on multiple confusions. But perhaps I can try to clear up some of these confusions.

First, your definition of the index at an isolated zero $x$ is incorrect, in that you ignored the appropriate domain of the map. Choose a small radius $r>0$ such that $x$ is the only zero in the closed ball around $x$ of radius $r$. The index of $v$ at $x$ is defined to be the degree of the map $f : S^{n-1} \to S^{n-1}$ defined by the formula $$f(w) = \frac{v(x+rw)}{||v(x+rw)||} $$

Now that this is cleared up, you write: "I am under the impression that the index of this vector field at $0$ should also be $-2$...". But this statement indicates a domain/range confusion:

1. The degree of $\nu$ is defined for a regular value $y$, and being a "value" of $\nu$ you should think of $y$ as a point in the range of the function $\nu$.
2. On the other hand, the index of $\nu$ is defined for an isolated zero $x$, and being a "zero" of $\nu$ you should think of $x$ as a point in the domain of the function $\nu$.

It makes no sense to ask whether an invariant of a point in the domain and an invariant of a point in the range are equal. For example, $x$ could be just one point out of a multiplicity of points forming the set $f^{-1}(y)$: What if $x$ is not an isolated zero of the vector field, so the index is not even defined? What if $x$ is one of several isolated zeroes of the vector field, all of which could have different indices?