A volume form on the sphere which gives equal areas to all hemispheres is the standard one?

Question

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 a scalar multiplication of the standard volume form induced by the round metric?

Answer 1

The answer is no. Let $f:\mathbb{S}^2 \to \mathbb{S}^2$ be the antipodal map $x \to -x$. Then we are looking at the space

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

This space is infinite dimensional. Here is a way to generate an arbitrary element in it: Take any form $\omega$ and define $\tilde \omega=\omega+f^*\omega$.

This is similar to the situation of inverse-invariant metrics on Lie groups.

(This happens whenever we want a symmetry of finite order, we can always average of pullbacked copies).