What is the index of this vector field? $$(x,y)\in \mathbb R^2 \mapsto (y^2 - x^2 , -2xy)$$
My Attempt
$V(0,1)=(1,0)$
$V(1,0)=(-1,0)$
$V(-1,0)=(-1,0)$
$V(0,-1)=(1,0)$
$V(1,1)=(0,-2)$
I plotted the vector field to see what is going on here. Not sure how to proceed. The definition we use in class is:
Definition. Let $A \subseteq \mathbb R ^m$ be an open and let $V:A \rightarrow \mathbb R ^m$ be a smooth vector field with an isolated critical point $c \in A$. The index $\iota(V;c)$ of $V$ at $c$ is the degree of the function
$$ \partial \bar{V_\iota (c)} \rightarrow^{\phi_\epsilon} S^{m-1}$$ $$x \mapsto \frac{V_x}{|V_x|}$$
Where $\epsilon >0$ is small enough so that the vector field $V$ vanishes nowhee on $V_\epsilon (c)$\ $\{c\}$.