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Let $\omega$ be a volume form on $\mathbb{S}^2$ with the property that the induced area (w.r.t $\omega$) of all the hemispheres is the same.

Is it true that $\omega$ is invariant under the antipodal map? i.e let $f(x)=-x$, does $f^*\omega=\omega$ hold?

The assumption implies that for any hemisphere $A \subseteq \mathbb{S}^2$, we have

$$ \int_{A}\omega=\int_{f(A)}\omega=\int_{A}f^*\omega=\frac{1}{2}\int_{\mathbb{S}^2} \omega. \tag{1}$$

Edit:

Lemma 1:

$\omega$ satisfies $(1)$ if and only if $\int_{A}L_X\omega=0$ for every Killing field $X$ and hemisphere $A$.

Proof:

$\Rightarrow$ Suppose $\omega$ satisfies $(1)$:

Let $\phi_t$ be the flow of a Killing field $X$ on $\mathbb{S}^2$, and let $A$ be a hemisphere. Since all the $\phi_t(A)$ are hemispheres we get

$$ \int_{\phi_t(A)}\omega=\int_{A}\phi_t^*\omega=\text{const},$$ so

$$ 0=\frac{d}{dt}|_{t=0} \int_{A}\phi_t^*\omega=\int_{A}\frac{d}{dt}|_{t=0}\phi_t^*\omega=\int_{A}L_X\omega.$$

$\Leftarrow$ Suppose $\omega$ satisfies $\int_{A}L_X\omega=0$ for every Killing field $X$ and hemisphere $A$:

Let $A$ be a hemisphere. We want to show $ \int_{A}\omega=\int_{-A}\omega$. There exist a Killing field $X$ s.t its flow takes $A$ to $-A$ at some time $t=t_0$. (i.e if $\phi_t$ is the flow, $\phi_{t_0}(A)=-A$). Now,

$$ \frac{d}{dt}|_{t=s} \int_{A}\phi_t^*\omega=\int_{A}\frac{d}{dt}|_{t=s}\phi_t^*\omega=\int_{A}\phi_s^*L_X\omega=\int_{\phi_s(A)}L_X\omega=0,$$ where the last equality is exactly the assumption.

Since $s$ was arbitrary, this implies $\int_{A}\phi_t^*\omega=\int_{\phi_t(A)}\omega$ is independent of $t$, so in particular $$\int_{A}\omega=\int_{\phi_0(A)}\omega=\int_{\phi_{t_0}(A)}\omega=\int_{-A}\omega,$$ as required.


Lemma 2:

The conditions in lemma 1 are equivalent to $\int_{C} i_X\omega=0$ for any great circle $C$ and Killing field $X$.

Proof:

Let $C$ be a great circle. $C$ bounds a hemisphere $A$. By Cartan's magic formula, $L_x\omega=d(i_X\omega)$, so

$$ \int_{A}L_X\omega=\int_{A} d(i_X\omega)=\int_{\partial A} i_X\omega=\int_{C} i_X\omega.$$

Since any hemisphere has a great circle for a boundary we are done.


Discussion:

Let $\tilde \omega$ be the standard round volume. Let $\omega=h\tilde \omega$ be an arbitrary form. $\omega$ is invariant iff $h(x)=h(-x)$.

By looking at a great circle $C$, we see that the condition in lemma 2 is equivalent to

$$ \int_C h|_{C}(\theta)\sin \theta=0, \int_C h|_{C}(\theta)\cos \theta=0.$$

This is because the space of Killing fields on $\mathbb{S}^2$ is $3$-dimensional, and one non-trivial Killing field always fixes the two hemispheres bounded by $C$, so we are effectively left with two equations.

Now, as mentioned by Anthony Carapetis, there is a solution which is not $\pi$-periodic: We can just take $h$ to be constant along latitude's, and for latitude $\theta$, set $h(\theta)=2+\sin(3\theta)$.

However, this solution cannot be lifted to a suitable candidate on the sphere.

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  • $\begingroup$ I wrote up a proof for this last night, but it was wrong. The differentiation idea tells you that $\int_C i_X \omega = 0$ for any great circle $C$ and Killing field $X$. Restricting attention to a single $C$, we can write $i_X \omega = h(s)ds$ for $s$ the usual coordinate on a circle, and this condition becomes $\int_0^{2\pi} h(s)\sin(s-t)ds = 0$ for all $t$. Sadly $h(s) = 2+\cos(3s)$ satisfies this but is not $\pi$-periodic, so this approach doesn't work (at least not by considering a single great circle.) Nice question. $\endgroup$ – Anthony Carapetis Oct 25 '17 at 7:10
  • $\begingroup$ Oops, that should be $\omega = g\,dA$ and $h = g|_C.$ $\endgroup$ – Anthony Carapetis Oct 30 '17 at 3:22
  • $\begingroup$ Thanks, I got essentially the same things. Here are some thoughts: (1) In fact the information from the differentiation is "all there is" in the sense that if the differentiation condition is satisfied, then the form indeed gives equal areas to all hemispheres (you just differentiate at a suitable time). (2) Considering point (1), your nice counter example for $h$ along a single great circle can be lifted to a counter example on the entire sphere, by taking it to be constant on latitudes, no? $\endgroup$ – Asaf Shachar Oct 30 '17 at 7:11
  • $\begingroup$ (3) This leaves two more natural questions: Is every factor $h$ constant on latitudes? and is the quotient space finite dimensional? You can see my edits in the question regarding all the points above. I would be happy to hear what you think. Thanks again. $\endgroup$ – Asaf Shachar Oct 30 '17 at 7:12
  • $\begingroup$ I don't follow (2) - if you really mean $\theta$ to be latitude/polar angle (which ranges from $0$ to $\pi$) then $\sin(3 \theta)$ isn't differentiable on the sphere, and is antipodeally invariant. If you mean longitude/azimuthal angle (so that the restriction to the equator is my example on the circle), then this gives the north and south hemispheres equal measure, but if you choose an "eastern" hemisphere containing 2 maxima of $h$ so that the corresponding "western" hemisphere contains only one, I think these hemispheres will have different measures. $\endgroup$ – Anthony Carapetis Oct 30 '17 at 7:24
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Any signed measure $\mu$ on the sphere which gives equal areas to all hemispheres is invariant under $x\mapsto -x.$

We can assume:

  • $\mu$ is odd, by taking the odd part $\mu'(B)=\frac{\mu(B)-\mu(-B)}2.$
  • (Optional, if $\mu$ is already assumed to be equal to a continuous function) $\mu$ is equal to a polynomial in $\mathbb R^3,$ i.e. a linear combination of spherical harmonics, by convolution by a polynomial on $SO(3)$ (using Riesz representation + Weierstrass approximation to ensure there is some polynomial that gives a non-zero convolution)
  • $\mu$ depends only on the latitude, by rotating to ensure the north pole is non-zero, and averaging over rotations fixing the north pole

The value on each $y$ co-ordinate give an odd continuous function $p:[-1,1]\to \mathbb R.$ We need to show that $p$ is identically zero.

By considering a great circle whose $y$-co-ordinate ranges from $-a$ to $a$, if I have calculated correctly, the Killing field argument implies the condition

$$\int_0^{2\pi} p(a\cos t) \cos t dt=0\qquad\text{ for all }0\leq a\leq 1.$$

The function on the sphere defined by $q(y)=p(y)y$ is even, continuous, and has identically zero Funk transform, so it is itself identically zero. This implies $\mu$ is zero. (In fact this argument doesn't even need any averaging. With the averaging, it is also possible to use the Abel transform - see the comments.)

Alternatively, since we can assume $p$ is analytic, suppose $p(x)=Cx^k+O(x^{k+2})$ for some odd $k.$ Then $\int_0^{2\pi}p(a\cos t)\cos t dt=Ca^{k}\int_0^{2\pi}\cos^{k+1}(t)dt + O(a^{k+2})$ as $a\to 0,$ so $C=0.$ This proves that $p$ is identically zero, hence so is the measure $\mu.$

And without the odd assumption, we get that $\mu$ must be even.

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  • $\begingroup$ Nice, I was trying to find some averaging trick but didn't spot that one! For the purposes of an easy proof in the smooth case, the need for the analytic approximation is bothering me - do you think there is a way to make this work for smooth $p$ by using the condition for all $a$ and not just near $a=0$, perhaps? $\endgroup$ – Anthony Carapetis Oct 30 '17 at 11:18
  • $\begingroup$ @AnthonyCarapetis: not sure, but I suspect it is possible to invert the transform $p\mapsto \hat p$ where $p$ is odd and $\hat p(a)=\int_0^{2\pi}p(a\cos t)\cos t dt$ - it might even be a change of variables from the Abel transform. In the other direction, the reduction could go further to the case that $\mu$ is a polynomial, i.e. a linear combination of spherical harmonics, not just analytic. $\endgroup$ – Dap Oct 30 '17 at 13:48
  • $\begingroup$ Interesting. Reducing the integral to $(0,\pi/2)$ using the symmetry and making the substitution $r = (a \cos t)^{-1}$ makes it look very much like the Abel transform of $p(r^{-1})$, though if my calculations are correct the power of $r$ on top of the fraction is different. $\endgroup$ – Anthony Carapetis Oct 30 '17 at 14:13
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    $\begingroup$ @AnthonyCarapetis: That substitution looks good to me! Since we're just testing for zero, we can absorb powers of $r$ into the function to be transformed. Powers of the transform variables ($a$ or $y$) can also be ignored. I think this argument just needs the property that if the Abel transform of $q(r)=p(1/r)/r$ is zero then $p(1/r)/r$ is zero. (Note $p(1/r)/r$ must be $O(r^{-2})$ since $p(0)=0.$) $\endgroup$ – Dap Oct 30 '17 at 14:59
  • $\begingroup$ @AnthonyCarapetis: I've edited my answer to mention this non-analytic argument, and that the Funk transform can also be used $\endgroup$ – Dap Oct 30 '17 at 17:35

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