Can a real-valued function on the sphere have exactly 2 critical points which are not antipodal?

Question

I came across this question while working on something quite different:

Can we build a $C^2$ function on the sphere of radius one, i.e. $f : S^{d-1} \mapsto \mathbb{R}, d \geq 3 $ such that it has exactly 2 critical points, $ x_{min} $ and $ x_{max} $ and such that $ x_{min} $ and $ x_{max} $ are not antipodal?

My intuition (weakly supported by numerically experimenting with a few functions) would be that it is not, and that as soon as we force $ \text{dist}(x_{min}, x_{max}) < \pi $, then at least one new critical point (a saddle?) will appear somewhere (then, if we further assume that we only have a finite number of critical points, I know that the numbers of minima, maxima and saddles needs to verify a certain relationship depending on the Euler characteristic of the sphere).

I suspect that there might be useful results regarding this in Morse theory; however, I am not familiar with it, so any references, examples or counter-examples are appreciated!

Answer 1

Take the function $f(x_1, \dots, x_{d}) = x_{d}$ defined on $\mathbb S^{d-1}$. Let $\phi : \mathbb S^{d-1 }\to \mathbb S^{d-1}$ be a diffeomorphism such that $\phi (0,\cdots, 0,1) = (0,\cdots, 0,1)$ and $y:=\phi^{-1}(0,\cdots, 0,-1) \neq (0,\cdots, 0,-1)$. Then the new function $g = f \circ \phi $ has two critical points $(0,\cdots, 0, 1)$ and $y$, and these two critical points are not antipodal.

For a slightly more concrete example: let $d=2$ and let $p = (0,\sqrt 2)$. Let $L^\pm$ be two lines in $\mathbb R^2$ touching $\mathbb S^1$ at $(\pm \sqrt 2^{-1}, \sqrt 2^{-1})$ respectively. rotating the line from $L^-$ to $L^+$ gives you a foliation of $\mathbb S^1$. One can find a function $f: \mathbb S^1 \to \mathbb R$ so that the level set corresponds to this foliation. $f$ would have exactly two critical points $(\pm \sqrt 2^{-1}, \sqrt 2^{-1})$, which is not antipodal.

Answer 2

I think that $f_{A,B}(C) = \frac{d(A,C)^2}{d(A,C)^2+d(B,C)^2}$ should do it.

This may help with the intuition:

Consider the Earth as $S^{d-1}$ with $d=3$. Let $f$ be a function that, given a point on Earth, gives the latitude. This fulfills the "exactly two critical points" requirement. The lines of longitudes are paths between the critical points, and they are parallel to the gradient. The lines of latitude are level curves of the function, and they are diffeomorphic to $S^{d-2}$ (they are not only diffeomorphic, but equal, but I'm phrasing it this way to make it more general). Imagine starting with a very small circle around the South Pole, and it expands out, then contracts back to the North Pole. At every point in time, there's a circle. I.e., there's some function $g$ of time that gives a circle as output. We can track a point on the expanding circle and it will follow a line of longitude, i.e. for each line of longitude, there is a function $h_l$ from time to the sphere.

We can derive $f$ from either $g$ or the collection $h_l$: given a point, we could find what latitude it's on, then find out when the circle expands to that latitude. Or we could find what longitude it's on, then look at the corresponding $h_l$ and find when it reaches that point. In other words, if we have a set of paths between A and B such that every point (other than A or B) is on exactly one path, and the paths are parameterized, then that defines an $f$. And if we have a set of disjoint, comprehensive subsets of $S^{d-1}$ that are each diffeomorphic to $S^{d-2}$ continuously indexed by real numbers, that also defines an $f$.

The latitude isn't going to be differentiable at the poles, but that can be addressed with a suitable transformation. For the requirement that the two critical points not be antipodal, just imagine sliding one of the poles around, dragging the lines of longitude along with it, and distorting the lines of latitude to accommodate. Or you can take any two points, and imagine drawing paths from one to the other, or expanding and contracting a manifold from one to the other. This can be generalized to any $d$.