Is there a non-constant function on the sphere that diagonalizes all rotations simultaneously?

Question

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 $SO(3)$.

Answer 1

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.

Answer 2

For a non-abelian group $G$ acting on a space $X$, such a function would have to be invariant under the commutator subgroup $[G,G]$ of $G$, and thus constant on the orbits of the commutator subgroup.

So the size of the orbits of the commutator subgroup gives a measure of how boring such a function must be. (One can probably prove that this is sharp - that for a compact group $G$ and a reasonable space $X$ a smooth function constant on $[G,G]$-orbits can be approximated by a sum of eigenfunctions - by reducing to Fourier analysis on $G^{ab}$ and various quotients of it.)

For $G= SO(3)$ and $X=S^2$ the whole space is a single orbit of the commutator subgroup so such orbits must be constant.

But for $SO(3)$ acting on other spaces, like $\mathbb R^3$, there are many more such functions.

Answer 3

As requested by the OP, I am making an answer from my comments to the question.

This is all standard material but I agree with him that it might be useful for somebody interested in this material.

Still, it might be better to migrate the whole stuff to math.SE. Please just vote for it if you think so.

Given any function $f$ on $\mathbb S^2$ with $\tau_A(f)=\lambda_Af$ for all $A\in\operatorname{SO}(3)$ as in the question, we will have$$f(Av)=\tau_{A^{-1}}(f)(v)=\lambda_{A^{-1}}f(v)$$for all $A$ and $v$, where we identify $\mathbb S^2$ with the set of all unit length vectors.

It follows that $f(ABv)=f(BAv)$ for all $A,B\in\operatorname{SO}(3)$ and all $v\in\mathbb S^2$.

This implies that $f$ is a constant function. Indeed given any unit vectors $v_1$, $v_2$, there is a $B\in\operatorname{SO}(3)$ with $v_2=Bv_1$. Let $v$ be a unit vector along the rotation axis of $B$, i. e. such that $Bv=v$; then there is an $A\in\operatorname{SO}(3)$ with $v_1=Av$. Then, $$ f(v_1)=f(Av)=f(ABv)=f(BAv)=f(Bv_1)=f(v_2). $$

This argument clearly depends on noncommutativity of $\operatorname{SO}(3)$, but not only on noncommutativity. Rather, declaring $AB=BA$ for all $A$, $B$ collapses $\operatorname{SO}(3)$ to the trivial group, i. e. $\operatorname{SO}(3)$ is a perfect group, which is stronger than just noncommutativity.

For example, the subgroup of $\operatorname{SO}(3)$ generated by all rotations around the $z$ axis together with the transformation sending $(x,y,z)$ to $(y,x,-z)$ (i. e. the $180^\circ$ rotation around the axis $(t,t,0)$) is noncommutative — it is the (continually) infinite dihedral group — but it admits a nonconstant eigenfunction $f(x,y,z)=z$.

For another example, the group of rotations of the unit cube is a noncommutative subgroup of order $24$ in $\operatorname{SO}(3)$ which has common eigenfunctions $f(x,y,z)=xyz$, $(x^2-y^2)(x^2-z^2)(y^2-z^2)$, $x^4+y^4+z^4$.

Just for fun, here are the level lines of these functions on $\mathbb S^2$, clearly exhibiting cubical symmetry:

$xyz$

$(x^2-y^2)(x^2-z^2)(y^2-z^2)$

$x^4+y^4+z^4$