# Morse Theory and Electrostatic Problem: Why a finite number of critical points?

So it states in Guillemin and Pollack:

Let $x_1, \ldots x_4$ be points in general position $R^3$ (that is they all don't lie in a plane.) Let $q_1, \ldots, q_4$ be electric charges placed at these points. The potential function of the resulting electric field is: $$V_q = \frac{q_1}{r_1} + \ldots + \frac{q_4}{r_4}$$ where $r_i = |x - x_i|$. The critical points of $V_q$ are called the equilibrium points of the electric field, and an equilibrium point is non-degenerate if the critical point is. Prove that for "almost every" $q$ the equilibrium points of $V_q$ are non-degenerate and finite in number.

They provide the hint to show that "the map: $R^3 - \{x_1, x_2, x_3, x_4\} \to R^4$ with coordinates $r_1$, $r_2$, $r_3$, r_4". I am somewhat sure that this means:

\begin{align} r_1 &= \frac{q_1}{|x - x_1|} \\ r_2 &= \frac{q_2}{|x - x_2|} \\ r_3 &= \frac{q_3}{|x - x_3|} \\ r_4 &= \frac{q_4}{|x - x_4|} \end{align}

Calculating $dV(q)_x$ gives (let $b(x,y,z) = (x-x_1)^2 + (y-x_2)^2 + (z-z_2)^2$):

$$\begin{bmatrix} q_1(x_1 - x) & q_1(x_1 - y) & q_1(x_1 - z) \\ q_2(x_2 - x) & q_2(x_2 - y) & q_2(x_3 - z) \\ q_3(x_3 - x) & q_3(x_3 - y) & q_3(x_3 - z) \\ q_4(x_4 - x) & q_4(x_4 - y) & q_4(x_4 - z) \end{bmatrix} b(x,y,z)^{-3/2}$$

The matrix is non-zero everywhere (well, on the weird domain) so $dV(q)_x$ must be injective and therefore the map is an immersion. They then say to use Exercise 21, the results of which is:

Let $\phi : X \to R^N$ be an immersion. Then for almost every $a_1, \ldots a_N$, $a_1\phi_1 + \ldots + a_N\phi_N$ is a Morse function.

Using this result we see that $V_q$ is an immersion so almost every point must be $q$ that is in the equilibrium must also be non-degenerate as a property of Morse functions. Any hints, or reasons why is it finite?