5
$\begingroup$

Let $h:\mathbb{S}^2 \to \mathbb{R} $ be a smooth function satisfying:

  1. $h(-x)=-h(x)$

  2. For every hemisphere $A \subseteq \mathbb{S}^2$, $\int_{A}h\text{Vol}_{\mathbb{S}^2}=0$, where $\text{Vol}_\mathbb{S}^2$ is the standard volume form on the sphere.

I am interested in finding an elementary proof that $h=0$. (Without relying on the invertibility of the Funk transform, see details below on the connection of this to the problem above).

Edit: As commented by Alex Degtyarev, if we assume $h$ is everywhere non zero, then it's trivial. (In that case $h$ has constant sign, so assumption $1$ alone immediately implies $h=0$).

Motivation:

Let $\omega$ be a 2-form on $\mathbb{S}^2$ with the property that the induced area of all the hemispheres is the same.

I want to find an elementary proof that $\omega$ is invariant under the antipodal map, i.e $f^*\omega=\omega$, where $f(x)=-x$.

Denote $$V=\{ \omega \in \Omega^2(\mathbb{S}^2) \, | \, \int_{A}\omega=\int_{A}f^*\omega \, \text{ for every hemisphere $A \subseteq \mathbb{S}^2$} \},$$

$$W=\{ \omega \in \Omega^2(\mathbb{S}^2) \, | \, \omega=f^*\omega \}.$$

We want to prove $V \subseteq W$. Let $\omega \in V$, and define $\tilde \omega:=\omega-f^*\omega$. Since $V$ is a vector space, closed under the operation $\omega \to f^*\omega$, we have $\tilde \omega \in V$. Note that $f^*\tilde \omega=-\tilde \omega$, and that we need to show $\tilde \omega=0$.

Thus, the problem is equivalent to the following:

Let $\omega \in V$, satisfy $f^*\omega=-\omega$. Then $\omega=0$.

The assumptions imply $\int_A \omega=0$ for every hemisphere $A$. Writing $\omega=h\text{Vol}_{\mathbb{S}^2}$, we obtain the formulation of the question as stated in the beginning.

Edit: If we assume $\omega$ is non-degenerate (i.e everywhere non-zero), then the question becomes trivial: In that case $h$ has a constant sign, hence must be zero due to the property $h(-x)=-h(x)$.

It turns out that using flows by Killing fields, one can reduce this problem to the invertibility of the Funk transform, but this is a non-elementary result which I prefer to avoid.

(Essentially the idea is that if $\int_A\omega=0$ on any hemisphere, then $\int_{A}L_X\omega=0$ for every Killing field $X$). For details see here and here.

$\endgroup$
6
  • 1
    $\begingroup$ This seems to follow from "positive" and $h(-x)=-h(x)$. $\endgroup$ Commented Nov 13, 2017 at 8:20
  • 1
    $\begingroup$ Thanks. You are right of course, in that case the question is trivial. However, I am also interested in the case where $h$ can change sign (the volume form can be zero at some points). I have edited the question to make this clear. Thanks again for your observation. $\endgroup$ Commented Nov 13, 2017 at 8:29
  • 1
    $\begingroup$ You also want to say "where $A$ is any hemisphere", I think. $\endgroup$ Commented Nov 13, 2017 at 8:40
  • 1
    $\begingroup$ @WillSawin Can you please elaborate? I am guessing you are thinking of presenting $\omega$ as a combination of weighted eigenfunctions (which form a basis in $L^2$ or something...), but how do you know that the eigenfunctions $\omega_i$ satisfy $f^*\omega_i=-\omega_i$. (Perhaps I misunderstood your comment). $\endgroup$ Commented Nov 13, 2017 at 8:50
  • 1
    $\begingroup$ Because $f$ commutes with the Laplacian, we can decompose $\omega$ into joint eigenfunctions of $f$ and the Laplacian. $\endgroup$
    – Will Sawin
    Commented Nov 13, 2017 at 10:16

1 Answer 1

4
$\begingroup$

This is Lemma 6.2 in

Gonzalez, Fulton B.; Kakehi, Tomoyuki, Dual Radon transforms on affine Grassmann manifolds, Trans. Am. Math. Soc. 356, No. 10, 4161-4180 (2004). ZBL1049.44001.

(actually, the lemma is for arbitrary dimension, so the special functionology might be simplified further for $\mathbb{S}^2$)

For convenience here is the Lemma (complete with proof):

enter image description here

$\endgroup$
2
  • $\begingroup$ "Without relying on the invertibility of the Funk transform" seems to be the crucial part of the question. Of course, your convolution is not exactly that, but I would say that the invertibility of the Funk transform is easier (just because the roots of orthogonal polynomials interlace and all odd polynomials vanish at $0$, so no even one does). $\endgroup$
    – fedja
    Commented Nov 13, 2017 at 18:09
  • $\begingroup$ @fedja My claim is that the argument given here is reasonably self-contained, and does not require any knowledge of Funk transform (or its existence). It seems unlikely that one can get the result without the representation theory (though I would be happy to be proven wrong...) $\endgroup$
    – Igor Rivin
    Commented Nov 13, 2017 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.