Let $u$ be a smooth function on the sphere, and for each $y \in \mathbb{S}^2$, let $R_y$ be the $180^\circ$ rotation about the vector $y$. For each pair $(x, y) \in \mathbb{S}^2 \times \mathbb{S}^2$, define the function $$ f(x, y) = \left(u(x) - u(R_y(x)), \nabla u(x) - R_y(\nabla u(R_y(x)))\right). $$ As noted by Pietro Majer in the comments, $R_y(x)=2(x.y)y-x$. I am interested in determining whether there exist points $x, y \in \mathbb{S}^2$ with $y\neq \pm x$ such that $f(x, y) = 0$. Essentially, this inquiry seeks pairs of distinct points $x_1 = x$ and $x_2 = R_y(x)$ where the function $u$ takes the same value and the gradients at these points are equal, modulo the rotation that maps $x_1$ to $x_2$.

Define $f: \mathbb{S}^2 \rightarrow \mathbb{R}^2$ by $$f(x)=\left(u(x)-u(-x), |\nabla u(x)|-|\nabla u(-x)|\right).$$ Then it follows from Borsuk–Ulam theorem that there exists $x_0$ where $f$ vanishes. So one can always find two distinct points where the norm of the gradient are equal. Noe also that at the antipodal points, the gradients are both also orthogonal to $x$. By relaxing the constraint in the above sense and not requiring the two points to be antipodal, I am looking for two points whose gradients as well as the values are equal in the above sense.

There is a generalization of Borsuk–Ulam theorem to product spaces, but it does not seem to apply: file:///Users/new-imachome/Downloads/1475179848%20(2).pdf

More generally, I am looking for a rotation such that the function $v(x)=u(x)-u(R(x))$ and $\nabla v(x)$ vanishes at two distinct points on the sphere. $R$ could be any rotation, and not necessarily a $180$ degree rotation about a vector. Note that when the rotation is a 180 degrees rotation, then if function $v(x)=u(x)-u(Rx)$ vanishes at $x_0$, then it also vanishes at $Rx_0\neq x_0$, and we automatically get two points where $v$ and it's gradient vanishes.

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    $\begingroup$ Isn’t $R_y$ involutory so that $R^{-1}_y=R_y$? I understand that $R_y(x)=2(x\cdot y)y-x$ for $x,y$ in $\mathbb S^2$, is this correct? $\endgroup$ Commented Apr 24 at 15:10
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    $\begingroup$ @PietroMajer Yes, $R_y$ is involutory. Your formula for $R_y(x)$ is correct. I will include it in the body of the question. $\endgroup$ Commented Apr 24 at 15:59
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    $\begingroup$ @KhashF You are right. I revised the question. $\endgroup$ Commented Apr 24 at 17:48
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    $\begingroup$ @KhashF Yes, indeed I am looking for a rotation such that the function $v(x)=u(x)-u(R(x))$ and $\nabla v$ vanishes at two distinct points on the sphere. $R$ could be any rotation, and not necessarily a $180$ degree rotation about a vector. $\endgroup$ Commented Apr 25 at 15:59
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    $\begingroup$ @PietroMajer Yes, I want $v$ to vanish at two distinct points. I have not received the answer I am looking for, provided solutions answers the problem under additional assumptions which makes the question a completely different question. $\endgroup$ Commented Apr 26 at 12:14

1 Answer 1


Now that I have a better understanding of what the question entails, I am going reorganize my answer. We are interested in points where a function of form $v(\mathbf{x}):=u(\mathbf{x})-u(R\mathbf{x})$--here $R\in{\rm{SO}}(3)$ denotes a rotation--and its gradient vanish simultaneously; i.e. points $\mathbf{x}_0\in\Bbb{S}^2$ with $u(\mathbf{x}_0)=u(R\mathbf{x}_0)$ and $\nabla u(\mathbf{x}_0)-R^{\rm{T}}\nabla u(R\mathbf{x}_0)=\mathbf{0}\Leftrightarrow R\nabla u(\mathbf{x}_0)=\nabla u(R\mathbf{x}_0)$. Of course, certain trivial cases should be excluded, like $R$ being the identity or more generally, cases where $R\mathbf{x}_0=\mathbf{x}_0$ is allowed: Otherwise, one can take $\mathbf{x}_0$ to be a critical point and any rotation fixing the unit vector $\mathbf{x}_0$ works. Hence I am going to focus on cases where $R\mathbf{x}_0\neq\mathbf{x}_0$. The question is if there exists $R\in{\rm{SO}}(3)$ and two distinct points on the sphere for which these equations hold.

Question) Are there two distinct points $\mathbf{x}_0,\mathbf{x}_1\in\Bbb{S}^2$ and a rotation $R\in{\rm{SO}(3)}$ such that $R\mathbf{x}_i\neq\mathbf{x}_i$, $u(\mathbf{x}_i)=u(R\mathbf{x}_i)$ and $R\nabla u(\mathbf{x}_i)=\nabla u(R\mathbf{x}_i)$ for $i\in\{0,1\}$?

I am going to prove the existence of one such point, and discuss how the same idea can potentially be applied to the general question. But before that, a remark regarding the special case of $180^\circ$-rotations $R=R_{\mathbf{y}}$ which are mentioned at the beginning of the question: If $R=R_{\mathbf{y}}$, since this is an involution, once we find one desired point $\mathbf{x}_0$, $R_{\mathbf{y}}\mathbf{x}_0$ would work as $\mathbf{x}_1$. Notice that $R_{\mathbf{y}}\mathbf{x}_0\neq\mathbf{x}_0$ amounts to $\mathbf{y}\neq\pm\mathbf{x}_0$ (such a condition also appears at the beginning of the question). An even more special case is when $R(\mathbf{x})=-\mathbf{x}$. I am going to say a few words about the antipodal case at the end.

Claim) There exist $R\in{\rm{SO}}(3)$ and $\mathbf{x}_0\in\Bbb{S}^2$ such that $R\mathbf{x}_0\neq\mathbf{x}_0$, $u(\mathbf{x}_0)=u(R\mathbf{x}_0)$ and $R\nabla u(\mathbf{x}_0)=\nabla u(R\mathbf{x}_0)$.

Proof of the Claim) Let $C$ be a connected component of a regular fiber of $u:\Bbb{S}^2\rightarrow\Bbb{R}$. Thus $C$ is homeomorphic to a circle. We deduce that the function $|\nabla u|:C\rightarrow (0,\infty)$ cannot be injective. So there exist points $\mathbf{x}_0\neq\mathbf{x}_1$ on $C$ with $|\nabla u(\mathbf{x}_0)|=|\nabla u(\mathbf{x}_1)|>0$. And of course $u(\mathbf{x}_0)=u(\mathbf{x}_1)$ since these points belong to the same level set. Now, given that we are dealing with the gradient of a function defined on the unit sphere, we have the following pairs of orthogonal unit vectors in $\Bbb{R}^3$: $$ \mathbf{x}_0\perp \frac{\nabla u(\mathbf{x}_0)}{|\nabla u(\mathbf{x}_0)|}, \quad \mathbf{x}_1\perp \frac{\nabla u(\mathbf{x}_1)}{|\nabla u(\mathbf{x}_1)|}. $$ Hence there exists a unique element $R$ of ${\rm{SO}}(3)$ with $R(\mathbf{x}_0)=\mathbf{x}_1$ and $R\left(\frac{\nabla u(\mathbf{x}_0)}{|\nabla u(\mathbf{x}_0)|}\right)=\frac{\nabla u(\mathbf{x}_1)}{|\nabla u(\mathbf{x}_1)|}$. The latter equality can be written as $R\nabla u(\mathbf{x}_0)=\nabla u(R\mathbf{x}_0)$ because $|\nabla u(\mathbf{x}_0)|=|\nabla u(\mathbf{x}_1)|$.

The partial solution above allows us to reformulate the question, at least for regular points of $u$, as:

Question$^\prime$) Let $$ \mathbf{M}:=\{(\mathbf{x},\mathbf{y})\in\Bbb{S}^2\times\Bbb{S}^2\mid\mathbf{x}\neq\mathbf{y},u(\mathbf{x})=u(\mathbf{y}),|\nabla u(\mathbf{x})|=|\nabla u(\mathbf{y})|>0\}; $$ and define $F:\mathbf{M}\rightarrow{\rm{SO}}(3)$ by setting $F(\mathbf{x},\mathbf{y})$ to be the unique element of ${\rm{SO}}(3)$ under which $\mathbf{x}\mapsto\mathbf{y}$, $\nabla u(\mathbf{x})\mapsto\nabla u(\mathbf{y})$ and $\mathbf{x}\times\nabla u(\mathbf{x})\mapsto\mathbf{y}\times\nabla u(\mathbf{y})$. Is $F$ non-injective?

(Keep in mind that if two distinct pairs $(\mathbf{x}_0,\mathbf{y}_0)$ and $(\mathbf{x}_1,\mathbf{y}_1)$ belong to the same fiber $F^{-1}(R)$, then, as desired, $R\mathbf{x}_i\neq\mathbf{x}_i$, $u(\mathbf{x}_i)=u(R\mathbf{x}_i)$ and $R\nabla u(\mathbf{x}_i)=\nabla u(R\mathbf{x}_i)$ for $i\in\{0,1\}$ where $\mathbf{x}_0\neq\mathbf{x}_1$. Also notice that $F(\mathbf{y},\mathbf{x})=F(\mathbf{x},\mathbf{y})^{-1}$, thus it is also enough to show that the range of $F$ contains an involution.)

Given that the subset $\mathbf{M}$ of $\Bbb{S}^2\times\Bbb{S}^2$ is defined by two constraints, it should generically be a 2D "object" (except in special cases like $u$ being constant). So I struggle to see any immediate argument showing that the map $F$ from $\mathbf{M}$ to the 3D group ${\rm{SO}}(3)$ cannot be injective.

The antipodal case) It is possible to find antipodal points belonging to same level set whose gradient vectors have the same projection along a given direction.

To see this, it suffices to apply the Borsuk–Ulam to $$ \mathbf{x}\in \Bbb{S}^2\mapsto \left(u(\mathbf{x}),\langle\nabla u(\mathbf{x}),\mathbf{p}\rangle\right)\in\Bbb{R}^2. $$ where $\mathbf{p}\in\Bbb{S}^2\subset\Bbb{R}^3$ is an arbitrary unit vector. The theorem implies the existence of $\mathbf{x}_0\in\Bbb{S}^2$ for which $u(\mathbf{x}_0)=u(-\mathbf{x}_0)$ and moreover, the projections of $\nabla u(\mathbf{x}_0),\nabla u(-\mathbf{x}_0)\in\Bbb{R}^3$ along the line parallel to $\mathbf{p}$ passing through the origin coincide.

A bit more can be said in view of the fact that $\nabla u(\mathbf{x}_0),\nabla u(-\mathbf{x}_0)$ both belong to the subspace ${\rm{T}}_{\mathbf{x}_0}\,\Bbb{S}^2={\rm{T}}_{-\mathbf{x}_0}\,\Bbb{S}^2$ which is the orthogonal complement of $\mathbf{x_0}$ in $\Bbb{R}^3$. For instance, if $\mathbf{p}$ is so that $u(\mathbf{p})\neq u(-\mathbf{p})$, then the unit vectors $\mathbf{x}_0,\mathbf{p},-\mathbf{p}$ are distinct. Hence vectors $\nabla u(\mathbf{x}_0),\nabla u(-\mathbf{x}_0)$ are subjected to two independent constraints: they are orthogonal to $\mathbf{x}_0$ and have the same projection along $\mathbf{p}$; thus they differ by a multiple of $\mathbf{p}\times\mathbf{x_0}$.

  • $\begingroup$ Thank you KhashF, but this is not what I want. $\endgroup$ Commented Apr 25 at 12:13
  • $\begingroup$ Thank you KashF, but the problem asks for a rotation $𝑅$ such that $𝑣=𝑢(𝑥)−u(𝑅𝑥)$ and its gradient vanishes at two distinct points. In your proof it is only guaranteed to vanish in one point. $\endgroup$ Commented Apr 26 at 4:18
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    $\begingroup$ @MathLearner Your question was changed multiple times so a bit hard to keep track. Initially, you asked for $u(\mathbf{x})=u(R_{\mathbf{y}}\mathbf{x})$ and $\nabla u(\mathbf{x})=R_{\mathbf{y}}\nabla u(R_{\mathbf{y}}\mathbf{x})$ where $\mathbf{y}\neq\pm\mathbf{x}$ to exclude the trivial case of $R_{\mathbf{y}}\mathbf{x}=\mathbf{x}$. Then you said $R_{\mathbf{y}}$ is allowed to be any rotation. Rewriting for a general rotation, the goal is: $u(\mathbf{x})=u(R\mathbf{x})$ and $\nabla u(\mathbf{x})=R^{\rm{T}}\nabla u(R\mathbf{x})$. Meaning a single common zero of $v,\nabla v$, not two. $\endgroup$
    – KhashF
    Commented Apr 26 at 4:35
  • $\begingroup$ @KashF, The question never changed. I just added extra details. The part that I asked for a function $v$ with two points has never changed, probably you did not pay attention. The point is that of the rotation is 180 degrees, then we will automatically get two points with such properties. Indeed $v=u(x)-u(Rx)$ would vanish both at $x_0$ and $R(x_0)$. The rotation does not need to have 180 degrees, but I still want two points. This has never changes in my writing. $\endgroup$ Commented Apr 26 at 12:08
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    $\begingroup$ Thank you anyways. If I don't get any other response, I will accept your answer. $\endgroup$ Commented Apr 26 at 14:22

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