**INTRODUCTION**. I am teaching a course in Harmonic Analysis. In class, very often I find myself stressing out the fundamental property that the functions  

$$
e\_n(x)=\exp(2\pi i n x), \quad \text{where }\quad x\in \mathbb R / \mathbb Z$$
are simultaneous eigenfunctions of all translations. By this I mean that, defining
$$\label{1}\tag{1}
\tau\_y f(x):=f(x-y), \quad \text{for }x, y\in\mathbb R/ \mathbb Z$$
it holds that
$$\label{2}\tag{2}
\tau\_y e\_n=\lambda\_{y,n}e\_n, \quad \text{for all }y\in\mathbb R/\mathbb Z.$$




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**QUESTION**. Let us now replace $\mathbb R/\mathbb Z$ with the sphere $\mathbb S^2$. By analogy with (1), define
$$
\tau\_A f(\omega):=f(A^{-1}\omega), \quad \text{for }A\in \mathrm{SO}(3),\ \omega\in\mathbb S^2.$$
Is there a non-constant $f\in L^2(\mathbb S^2)$ such that
$$\label{3}\tag{3}
\tau\_Af=\lambda\_A f, \quad \text{for all }A\in \mathrm{SO}(3)?$$




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**REMARKS**. From a more general standpoint, the simultaneous diagonalization property \eqref{2} of $e\_n$ is a common feature of all locally compact abelian groups. Indeed, if $G$ denotes such a group, then we can define
$$
\tau\_g f(h):=f(g^{-1}h), \quad g,h\in G,\ f\in L^2(G), $$
yielding a unitary representation of $G$. And since $G$ is abelian, all irreducible subrepresentations must be one-dimensional, or in other words, they must be eigenspaces of all $\tau\_g$ simultaneously. This is precisely the diagonalization phenomenon observed above in the case $G=\mathbb R / \mathbb Z$.


The present question occured to me while I was trying to find a concrete, pedagogical example of the failure of this simultaneous diagonalization in the non-abelian case.




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**ANSWER**. I can actually provide an answer to the question, which unfortunately I find unsatisfactory for reasons that I will explain below. The answer is the following.


We claim that no non-constant function $f$ satisfying \eqref{3} exists. To prove this, introduce the projectors
$$ P\_{n} f(\omega):=\int\_{\mathbb S^2} f(\nu)L\_n(\omega\cdot \nu)\ d\sigma(\nu),$$
where $L\_n$ is the Legendre polynomial of degree $n$, with an appropriate normalization (the exact value of which is irrelevant here). For each $n\in \mathbb N\_{\ge 0}$, $P\_n f$ is a spherical harmonic of degree $n$. It is clear that $P\_n$ commutes with $\tau\_A$ for all $A\in\mathrm{SO}(3)$, so by \eqref{3}
$$
\tau\_A\left( P\_n f\right) = P\_n \tau\_Af= \lambda\_AP\_n f. $$
So, if a non-constant $f$ satisfying the property \eqref{3} existed, we could find a spherical harmonic $Y\_n=P\_n f$ of degree $n>0$ that satisfies the same property. But this cannot be, for in that case $Y\_n$ would span an irreducible subrepresentation of the space $\mathbb{Y}\_n$ of spherical harmonics of degree $n$ and we know that $\mathbb{Y}\_n$ is irreducible.


The only possibility is that $f$ is a constant, for in that case $P\_n f=0$ except for $n=0$. $\Box$




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**CONCLUDING REMARK**. The answer provided above is unsatisfactory for my teaching purposes, since it is too technical and requires the introduction of several concepts, such as spherical harmonics and irreducible subrepresentations. Ideally, I would like an answer that is geometrical in nature and which directly exposes the role of the non-Abelianity of $SO(3)$.



Is there a non-constant function on the sphere that diagonalizes all rotations simultaneously?
For $\mathrm{C}^1$-functions, the argument can be reduced to the circle case: Write the sphere as a union of circles meeting at the poles. On each circle the considered function is equivariant under all rotations, therefore its restriction to the circle must be of the form $t\mapsto C\exp(i\lambda t)$ for some real numbers $C,\lambda$. Since the circles meet at the poles and the function is continuously differentiable, the $C$'s and $\lambda$'s agree for all the circles. To see that $\lambda$ must be zero, apply the argument again with a different choice of poles, for example a pair of antipodal points in the original equator.


