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 $h=0$).


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.

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    $\begingroup$ This seems to follow from "positive" and $h(-x)=-h(x)$. $\endgroup$ Commented Nov 13, 2017 at 8:20
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    $\begingroup$ 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
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    $\begingroup$ You also want to say "where $A$ is any hemisphere", I think. $\endgroup$ Commented Nov 13, 2017 at 8:40
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    $\begingroup$ @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
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    $\begingroup$ 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


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

  • $\begingroup$ "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

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