TITLE: Smooth functions on sphere
QUESTION [21 upvotes]: Let $u$ be a smooth function defined on the unit sphere $S^2$. Assume $u$ has two local maxima, two local minima, and two saddle points (a total of 6 critical points). Does there exist a plane $P$ passing through the origin such that $P\cap S^2$ contains at least three points $x_1,x_2,x_3$ with $\nabla u(x_i) \cdot n =0$, $i=1,2,3$, where $n$ is a vector normal to the plane $P$?

REPLY [3 votes]: EDIT As DimitriPanov points out Claim 1' is false and this leads to an unfixable gap.  I'll leave the answer in case someone finds it useful. 
Ironically, it looks like I can salvage the proof only for any Morse functions $u$ with at most 4 critical points. 
First some notation.
For any $\mathbf{v}\in \mathbb{S}^2$ let $D(\mathbf{v})=\{\mathbf{y}\in \mathbb{S}^2: \mathbf{y}\cdot \mathbf{v}\leq 0\}$ be the hemisphere on the side of the plane normal to $\mathbf{v}$ that does not contain $\mathbf{v}$.  Let $C(\mathbf{v})=\partial D(\mathbf{v})$ be the great circle at the boundary of $D(\mathbf{v})$ whose (outward) normal is $\mathbf{v}$.  Let $P=\{p: \nabla u (p)=0\}$ be the set of critical points of $u$.  As $u$ is Morse, this is a finite set with an even number of points.
Assumption: Let us now assume that $\nabla u \cdot \mathbf{v}$ has at most two zeros on $C(\mathbf{v})$ for all $\mathbf{v}$. 
I claim this assumption will lead to a contradiction and hence prove what you wanted, but for strictly fewer critical points than you requested.
Let $\Omega\subset \mathbb{S}^2$ be the set of $\mathbf{v}$ so that $\nabla u \cdot \mathbf{v}$ is never zero on $C(\mathbf{v})$ and let $\Omega'$ be the set of $\mathbf{v}$ so that $\nabla u \cdot \mathbf{v}$ is both positive and negative on $C(\mathbf{v})$.  Notice both $\Omega$ and $\Omega'$ are open sets. 
The assumption implies that, for $\mathbf{v}\in \Omega'$, $\nabla u\cdot \mathbf{v}$ vanishes at exactly two points on $C(\mathbf{v})$.  It also ensures that if $\mathbf{v}\not\in \Omega\cup \Omega'$ then $\nabla u\cdot \mathbf{v}$ has a sign away from either one or two points.
Claim 1: (This has been corrected thanks to DmitriPanov's observation). Suppose $\mathbf{v}$ is such that $C(\mathbf{v})\cap P=\emptyset$. If $D(\mathbf{v})\cap P$ contains an even number of points, then $\mathbf{v}\in \Omega'$. 
Proof: This follows from the Poincare-Hopf index theorem applied to the vector field $\nabla u$. Basically, the standing assumption ensures the boundary contributes  $1$ (when $\mathbf{v}\not\in\Omega'$) and critical points contribute either $1$ (for maximum or minima) or $-1$ (for saddle points).  The claim follows by (for instance) working mod 2.
Claim 1': Suppose $\mathbf{v}$ is such that $C(\mathbf{v})\cap P=\emptyset$. If $\mathbf{v}\in \Omega'$, then either $D(\mathbf{v})\cap P$ contains an even number of points or both $D(\mathbf{v})\cap P$ and $D(-\mathbf{v})\cap P$ contains at least three points of $P$.
Proof: Poincare-Hopf tells us that if $\nabla u$ contributes $0$ mod 2 along  $C(\mathbf{v})$, then $D(\mathbf{v})\cap P$ contains an even number of points.  Now suppose, $\nabla u$ contributes $1$ mod $2$ along the boundary.   Let $\mathbf{T}:C(\mathbf{v})\to \mathbb{R}^3$ be the unit tangent to $C(\mathbf{v})$ compatible with the orientation.  Write along $C(\mathbf{v})$
$$
\nabla u(p)=r(p) \cos \theta(p) \mathbf{T}(p)+r(p)\sin \theta(p) \mathbf{v}
$$
where here $\theta:C(\mathbf{v})\to \mathbb{S}^1$ and $r>0$.
The standing assumption ensures that $\sin \theta(p)=0$ only at two points $p_-$ and $p_+$ and $\mathbf{v}\in \Omega'$ means that, up to relabelling $\sin(\theta(p))<0$ on $C_-(\mathbf{v})$, the region between $p_-$ to $p_+$, (relative to the orientation) and $\sin(\theta(p))>0$ on $C_+(\mathbf{v})$, the region between  $p_+$ and $p_-$.  One verifies (e.g., by drawing a graph) that $\nabla u$ contributes $0$ to Poincare-Hopf if and only if $\cos(\theta(p_+))=-\cos(\theta(p_-))$ (e.g. $\theta(p_+)=0, \theta(p_-)=\pi$ and) while $\nabla u$ contributes $1$ when $\cos(\theta(p_+))=\cos(\theta(p_-))$ (e.g. $\theta(p_+)=\theta(p_-)=0$). 
In this latter case (which is the one we are interested in), the fact that $\nabla u$ is a gradient vector field, implies there are at least two points in both $C_-(\mathbf{v})$ and $C_+(\mathbf{v})$ where $\sin \theta$ vanishes.  In fact, one the two points will be a local maximum of $u|_{C_\pm (\mathbf{v})}$ and the other will be a local minimum of $u|_{C_\pm (\mathbf{v})}$.  Let $q_+$ be the local maximum in $C_-(\mathbf{v})$ and $q_-$ be the local minima in $C_+(\mathbf{v})$.  By analayzing level sets one obtains corresponding local maxima and minima of $u$, $q_-'$ and $q_+'$ in $D(\mathbf{v})$ ** Edit: This reasoning only works if $q_-$ and $q_+$ are actually absolute maximam on $C(\mathbf{v})$ -- i.e. if $q_-$ only a local minima can't guarantee existence of $q_-'$. **. That is, $D(\mathbf{v})$ contains at least two critical points of $u$ and so, by Poincare-Hopf, at least three points.  This argument is symmetric in $\mathbf{v}$ and $-\mathbf{v}$ so $D(-\mathbf{v})$ also contains at least three critical points of $u$.
Returning to the main argument:
Let $\Gamma=\bigcup_{p\in P} C(\mathbf{v}(p))$ where $\mathbf{v}(p)$ is chosen so $\mathbf{w}\in C(\mathbf{v}(p))$ if and only if $p\in C(\mathbf{w})$. Our standing assumption and the lower bound on the number of cirtical points ensures $C(\mathbf{v}(p))\neq C(\mathbf{v}(p'))$ for $p \neq p'$ as if $C(\mathbf{v}(p))=C(\mathbf{v}(p'))$ then $p$ and $p'$ are equal or antipodal.  In the later case, one can find a $C(\mathbf{w})$ containing $p,p'$ and a third element $p''$ so $\nabla u(p'')\cdot \mathbf{w}=0$ -- contradicting the standing assumption (see User4966's argument below).  Moreover, the standing assumption implies that $C(\mathbf{v}(p_1))\cap C(\mathbf{v}(p_2))\cap C(\mathbf{v}(p_3))=\emptyset$ for $p_1, p_2, p_3$ distinct.  Indeed, if $\mathbf{w}\in C(\mathbf{v}(p_1))\cap C(\mathbf{v}(p_2))\cap C(\mathbf{v}(p_3))$ then $p_1,p_2, p_3\in C(\mathbf{w})$ so $\mathbf{w}$ violates the standing assumption.
Let $\Omega''=\mathbb{S}^2 \backslash \overline{\Omega'}$ so $\Omega''$ is open.  Clearly, $\Omega\subset \Omega''$ (though they might not be equal).
Claim 2: If $P$ has at most $4$ points then, both $\Omega'$ and $\Omega''$ are non-empty and $\mathbb{S}^2\backslash \Gamma=\Omega'\cup \Omega''$ and each component of $\Omega'$ is contained in an open hemi-sphere.
Proof: 
Pick a $C(\mathbf{w})$ so $C(\mathbf{w})\cap P=\{p\}$. That is,  so $\mathbf{w}\in C(\mathbf{v}(p))\subset \Gamma$ but $\mathbf{w}\not\in C(\mathbf{v}(p'))$ for any other $p'\in P$.    By perturbing $\mathbf{w}$ slightly to $\mathbf{w}'$ one can ensure $C(\mathbf{w}')\cap P=\emptyset$ and $D(\mathbf{w}')\cap P=D(\mathbf{w})\cap P$ or $D(\mathbf{w}')\cap P=(D(\mathbf{w})\cap P)\backslash \{p\}$.  Depending on which of the two cases occur $D(\mathbf{w}')\cap P$ contians either an even or odd number of points.  Furthermore, as $P$ has at most $4$ elements $D( \mathbf{w}')$ and $D(-\mathbf{w}')$ can't both contain at least three points of $P$.  Hence, by Claim 1 and Claim 1',  we can make $\mathbf{w}'$ lie in either $\Omega'$ or in $\Omega''$. That is $\mathbf{w}\in \partial \Omega'$ and also $\mathbf{w}\in \partial \Omega''$.  It follows that $\Omega'$ and $\Omega''$ are non-empty and $\mathbf{w}\not\in \Omega'\cup \Omega''$. The second part of the claim follows as each component of $\Omega'$ lies in one of the components of $\mathbb{S}^2 \backslash C(\mathbf{v}(p))$ for some $p\in P$.
However we also have
Claim 3: There exists a connected component of $\Omega'$ that contains two antipodal points.
Proof:
Let $F:\mathbb{S}^2 \to \mathbb{R}$ be defined by
$$
F(\mathbf{v})=\int_{C(\mathbf{v})} \nabla u \cdot \mathbf{v}.
$$
Clearly, $F$ is a continuous odd function.  Our standing assumption ensures that $F^{-1}(0)\subset \Omega'$.  I claim that there is a connected component $Z$ of $F^{-1}(0)$ that contains two antipodal points -- see the second answer to this question for a proof (I didn't check this in detail but it seems intuitively clear).  This implies that there is such a component for $\Omega'$ and proves the claim.
Since Claim 2 and Claim 3 are in contradiction the standing assumption is false, which gives what you want when $u$ has at most 4 critical points.