Index of the vector field $X(z) = p(z) \cdot e^{-\| z \|^2}$ projected on $\mathbb{S}^2$ Define the vector field $X:\mathbb{C} \to \mathbb{C}$ as $X(z) = p(z) \cdot e^{-\| z \| ^2}$ where $p$ is a polynomial. I'm trying to compute the index in $\infty$ of the vector field obtained projecting $X$ on $\mathbb{S}^2$.
This was my attemp: if I consider a sufficiently small circumference along a parallel, this is projected by stereographic projection on a circumference in $\mathbb{C}$ containing all the zeros of the vector field $X$ except $\infty$. Then is well known that the degree of the map $\frac{X(z)}{\lVert X(z) \rVert} = \frac{p(z)}{\lVert p(z) \rVert}$ is equal to the number of zeros (with
multiplicity) of the polynomial that are inside the circumference. Thus my claim is $ind_{X}(\infty) = deg(p)$.
Is my claim true? I'm not sure about the fact that the degree of the projection of $X$ normalized and restricted to a circumference is the same as the degree of $X$ normalized and restricted to the projection of the circumference.
 A: Let us define $\psi : \mathbb{C} \to \mathbb{S}^2$ the stereographic projection using the North Pole and $\varphi : \mathbb{C} \to \mathbb{S}^2$ the other one. Let $x_0$ be the North Pole and $x_1$ the South Pole.
The vector field induced in $\mathbb{S}^2$ and the one of which you want the index is the following:
$$ Z(y)=
\begin{cases}
0 \mbox{ if }z \mbox{ is the North Pole } \\
d \psi \circ X \circ \psi^{-1}(z) \mbox{ otherwise}
\end{cases}$$
Because you want to compute $ind_{Z}(x_0)$, let us find a parameterization of a neighborhood of $x_0$. We observe that we can take $\varphi$.
Thus by definition $ind_Z(x_0)=ind_{d \varphi ^{-1} \circ Z \circ \varphi}(0)$. The key point is that $\psi ^{-1} \circ \varphi = \frac{z}{\|z\|^2}=\frac{1}{\overline{z}}$. Thus we have reduced the problem to the following:
Consider the map $Y: \mathbb{C} \to \mathbb{C}$ defined in this way: $Y(z)=X(\frac {1}{\overline{z}})$ where $X(z) = p(z) \cdot e^{-\| z \| ^2}$ and $W:\mathbb{C} \to \mathbb{C}$ defined as $W(z)=\frac{1}{\overline{Y(z)}}$. Suppose $p(z)=z^n+a_{n-1}z^{n-1}+...+a_0$.
We want to compute $ind_W(0)=-ind_Y(0)$.
Now take $r >0$ sufficiently small such that $r^{-n}>|a_{n-1}|r^{-(n-1)}+...+|a_0|$.
Then define for $z \in \partial B(0,r)$.
$$H(t,z)=\frac{[t\overline{z}^{-n}+(1-t)p(\frac {1}{\overline{z}})]e^{-\frac{1}{r^2}}}{\|[t\overline{z}^{-n}+(1-t)p(\frac {1}{\overline{z}})]e^{-\frac{1}{r^2}}\|}=\frac{t\overline{z}^{-n}+(1-t)p(\frac {1}{\overline{z}})}{\|t\overline{z}^{-n}+(1-t)p(\frac {1}{\overline{z}})\|}$$
which is an homotopy between $\frac{p(\frac {1}{\overline{z}})}{\|p(\frac {1}{\overline{z}})\|} : \partial B(0,r) \to \mathbb{S}^1$ and $r^{-n}\overline{z}^{-n} : \partial B(0,r) \to \mathbb{S}^1$ which has degree $n$.
We conclude that $ind_X(\infty)=-n$.
