Consider the vector field $X : \mathbb{C} \times \mathbb{R} \longmapsto \mathbb{C} \times \mathbb{R}$ such that $X(z,t) = (z^2,t)$.
I'd like to prove that $X$ has only one zero of index $2$. This seems reasonable since $\mathbb{S}^{1} \longmapsto \mathbb{S}^{1}$ the power to the square has degree $2$ and basically I'm doing nothing on the second component. But I'm having trouble dealing with computation.
It's clear that $(0,0)$ is the only zero here, hence isolated. But I have to compute $\text{ind}_{X}((0,0))$ this is by definition the degree of the map from a ball centered in the zero, with any radius in this case to $\mathbb{S}^{2}$ in this case since we are in $\mathbb{R}^{3}$ (But it's not clear to me how I'm seeing $\mathbb{S}^{2}$ here as subset of $\mathbb{C} \times \mathbb{R}$ ) which is $X(z,t) \longmapsto \frac{X(z,t)}{\lvert\lvert X(z,t)\rvert\rvert}$.
How to compute the degree of $X(z,t) \longmapsto \frac{X(z,t)}{\lvert\lvert X(z,t)\rvert\rvert}$ ? It's not even clear to me wether I should think this map as a composition or not, and try to guess who the regular values are.
The definition of degree I'm using is : Let $M$ be compact $N$ connected with dim$M=$dim$N$, both without boundary, orientable and oriented. Take $f : M \longmapsto N$ of class $C^{\infty}$ then deg($f$) := $\sum\limits_{x \in f^{-1}(y)} \text{sgn}df_x$ for $y \in \text{RegVal}(f)$.
The index of an isolated zero $p$ is defined as the pullback through a parametrization of the manifold of the index on an open set of $\mathbb{R}^{n}$ so I'm going to give this definition : $\text{ind}_X(p) := \text{deg}(f_{r})$, where $f_r : \partial B(p,r) \longmapsto \mathbb{S}^{n-1}$ with $f_r(q) := \frac{X(q)}{\lvert \lvert X(q) \rvert \rvert}$ for every $r \in (0,\epsilon)$ given $\epsilon$ the radius of the ball where there are no other zeros.
Any help will be appreciated, thanks in advance.
We can take $r=1$. Then your definition of the index is $$ind_X(0)=deg(f_1)$$ where $f_1: S^2\to S^2$ is given by $$f_1\left(se^{i\theta}, t\right) = \frac{\left(s^2e^{2i\theta}, t\right)}{\sqrt{s^4+t^2}},$$ where $s$ and $t$ are real positive numbers so that $s^2+t^2=1$. Now you must find a regular value to compute the degree of $f_1$. I suggest you use $(1,0)$. The inverse image are the two points $(-1,0)$ and $(1,0)$.
Is this clear so far? Can you take it from here?