A wonky gravitational potential and its critical points

Take 4 points arranged in a square. The middle of the square is a critical point by symmetry, and the midpoint of the four sides of the square would be a critical point just for the two vertices it joins together, ignoring the other two. But the other two are little more than twice as far away, so eight times weaker force, so if you bring the point closer to the center of the square by an amount about 1/8 of the way to the other side, you will cancel out the force from the far pair from the force on the near pair. So there are 5 critical points for four points on a square, and this holds for all sufficiently fast-falling-off forces.

Using squares-of-squares, I believe it is easy to establish that the number of critical points is generically $n^2$. I believe it is a difficult and interesting mathematical problem to establish any sort of nontrivial bound on the number of critical points. The trivial bound is from the order of the polynomial equation you get, and it's absurdly large---- it grows like $2^n$.

EDIT: The correct growth rate

The answer for a general configuration is almost certainly N+C critical points (this should be an upper bound and a lower bound for two different C's --- I didn't prove it, but I have a, perhaps crappy, heuristic). For the case of polygons, there are N+1 critical points. For squares where the corners are expanded to squares, and so on fractally, the number of critical points is N+C where C is a small explicit constant. The same for poygons of polygons.

I found a nifty way of analyzing the problem, and getting some good estimates, but I want to know how good the mathematicians are at this before telling the answer. Perhaps you can ask this question on MathOverflow?

In this answer we analyze a generalization of OP's question. In the last section 6 we will argue heuristically that one should expect an upper bound of critical points of the form

$$c(M)~\leq~~5n-11 \qquad {\rm for} \qquad n\geq 3. $$

1) Let us identify the plane $\mathbb{R}^2\cong \mathbb{C}$ with the complex plane $z=x+\mathrm{i} y$. Let

$$Z:=\{z_1, \ldots, z_n\}\subseteq \mathbb{C}$$

be a set of $n$ different punctures in the complex plane, where $n\in\mathbb{N}$. Consider the punctured complex plane $M:=\mathbb{C} \backslash Z$ and the Riemann sphere $S^2:=\mathbb{C} \cup \{\infty\}$ with Betti numbers

$$b_0(M)~=~1, \qquad b_1(M)~=~n, \qquad b_2(M)~=~0,$$ $$b_0(S^2)~=~1, \qquad b_1(S^2)~=~0, \qquad b_2(S^2)~=~1,$$

and Euler characteristics

$$\chi(M)~=~b_0(M)-b_1(M)+b_2(M)~=~ 1-n, $$ $$\chi(S^2)~=~b_0(S^2)-b_1(S^2)+b_2(S^2)~=~ 2, $$

respectively.

2) Let $p>0$ and $k_1, \ldots, k_n >0$ be $n+1$ positive constants. Let the potential $V:M \to]0,\infty[$ and its extension $\tilde{V}: S^2 \to [0,\infty]$ be

$$V(z)~:=~\sum_{i=1}^n \frac{k_i}{p|z-z_i|^p},$$

$$ \tilde{V}(z)~:=~\left\{ \begin{array}{rcl} V(z) &{\rm for}& z \in M, \cr +\infty &{\rm for}& z \in Z, \cr 0 &{\rm for}& z \in\{\infty\}. \end{array} \right. $$

Let $$ c_0~:=~ \#{\rm minimum~pts}, \qquad c_1~:=~ \#{\rm saddle~pts}, \qquad c_2~:=~ \#{\rm maximum~pts}, $$

So

$$c_0(S^2)~=~c_0(M)+1, \qquad c_1(S^2)~=~c_1(M), \qquad c_2(S^2)~=~c_2(M)+n, $$

because $z=\infty$ is a minimum point and $Z$ are maximum points for $\tilde{V}$.

3) Define two positive functions $E, F:M \to\mathbb{R}_{+}$ as

$$E(z)~:=~\sum_{i=1}^n\frac{ k_i(x-x_i)^2}{|z-z_i|^{p+4}}~>~0,\qquad F(z)~:=~\sum_{i=1}^n\frac{ k_i(y-y_i)^2}{|z-z_i|^{p+4}}~>~0, \qquad z \in M. $$

The $2\times 2$ Hessian matrix $H$ for the potential $V$ is

$$H_{xx}~=~\sum_{i=1}^n k_i\frac{(p+1)(x-x_i)^2-(y-y_i)^2}{|z-z_i|^{p+4}} ~=~(p+1)E-F,\qquad z \in M, $$ $$H_{yy}~=~\sum_{i=1}^n k_i\frac{(p+1)(y-y_i)^2-(x-x_i)^2}{|z-z_i|^{p+4}} ~=~(p+1)F-E,\qquad z \in M, $$ $$H_{xy}~=~(p+2)\sum_{i=1}^n \frac{k_i(x-x_i)(y-y_i)}{|z-z_i|^{p+4}}, \qquad z \in M,$$

with positive trace

$$ {\rm tr}H ~=~H_{xx}+H_{yy}~=~p(E+F) ~=~p\sum_{i=1}^n \frac{k_i}{|z-z_i|^{p+2}}~>~0, \qquad z \in M.$$

and determinant

$$ {\rm det}H ~=~H_{xx}H_{yy}-H^2_{xy} ~=~\underbrace{(p^2+2p+2)EF}_{>0} - \underbrace{\left((p+1)(E^2+F^2)+H^2_{xy}\right)}_{>0}.$$

A sufficient condition for negative determinant is:

$$F> (p+1)E \qquad \vee \qquad E> (p+1)F\qquad\Rightarrow \qquad {\rm det}H<0. $$

The off-diagonal element $H_{xy}$ is bounded by

$$ \frac{|H_{xy}|}{p+2} ~\leq~\sum_{i=1}^n \frac{k_i|x-x_i||y-y_i|}{|z-z_i|^{p+4}} ~\leq~\frac{E+F}{2}. $$

4) The positive trace ${\rm tr}H>0$ implies that there cannot be any maximum points in $M$,

$$c_2(M)~=~0 \qquad \Leftrightarrow \qquad c_2(S^2)~=~n .$$

A minimum point $z$ must have positive determinant ${\rm det}H(z)>0$, while a saddle point $z$ must have non-positive determinant ${\rm det}H(z)\leq 0$. Generically, one may assume that all critical points $z$ in $M$ are non-degenerate ${\rm det}H(z)\neq 0$, so that $V$ is a Morse function. This implies that all critical points in $M$ are isolated points. Morse theory yields that$^{1}$

$$ c_2(M)-c_1(M)+c_0(M)~=~\chi(M)\qquad \Leftrightarrow \qquad c_2(S^2)-c_1(S^2)+c_0(S^2)~=~\chi(S^2),$$

or equivalently,

$$ c_1(M)-c_0(M)~=~n-1\qquad \Leftrightarrow \qquad c_1(S^2)-c_0(S^2)~=~n-2.$$

5) Lemma: The determinant ${\rm det}H(z)<0$ is negative sufficiently close to the set $Z$.

Sketched proof of Lemma: The open set

$$ U := \{z\in M \mid {\rm det}H(z)>0\} ~=~\cup_a U_a $$

consists of a number of connected components $U_a$. Since the restriction $V|_{U}$ is concave upwards, each connected component $U_a$ can contain at most one minimum point.

6) Up until now our analysis is rigorous. In the rest of the answer we speculate. Heuristically, there should exists a triangularization of the Riemann sphere $S^2$ with the set $Z$ as vertices, such that each face contains at most one minimum point. Assume that this is true. Let the number of faces, edges, and vertices be $f$, $e$, and $n$, respectively. Assume $n\geq 3$. Then

$$ f-e+n ~=~\chi(S^2)~=~ 2, \qquad 2e~=~\sum_{j\geq 3} j f_j~\geq~3f, \qquad f~=~\sum_{j\geq 3} f_j, $$

implies that $\frac{f}{2}~=~\frac{3f}{2}-f~ \leq~ e-f~=~n-2$, and hence,

$$c_0(M)+1~=~c_0(S^2)~\leq~f~\leq~ 2(n-2)\qquad \Rightarrow \qquad c_0(M)~\leq~ 2n-5.$$

Then an upper bound for the number of critical points in $M$ becomes

$$c(M)~=~ c_{0}(M) + c_{1}(M) + c_{2}(M) ~=~ 2c_{0}(M)+n-1~\leq~ 5n - 11. $$

--

$^{1}$ Naively, the Morse inequality implies that $c_{0}(M) \geq b_{0}(M)=1$, which is obviously wrong for $n=1$. This can be cured, if one takes boundary contributions in $M$ properly into account. The formulation in terms of the Riemann sphere $S^2$ doesn't have this problem, because it has no boundary.