Can 3d rotations of functions with parity symmetry be mapped to a plane with real projective plane boundary conditions

Question

If I have function $f(\theta,\phi)$ such that $f(\theta,\phi)=f(\pi-\theta,\phi+\pi)$, then can this function be mapped to $\mathbb{R}P^2$, say on a square with appropriate boundary identifications such that rotations of the function on the sphere are mapped to translations on the plane?

Answer 1

When can define $\Bbb RP^2$ as $[0,\pi]^2 /\sim$, where $(0, \phi) \sim (\pi, \pi - \phi)$ and $(\theta, 0) \sim (\pi - \theta, \pi)$ (and everything else is equivalent only to itself). So we must have $f(0,\phi) = f(\pi, \pi - \phi)$ and $f(\theta, 0) = f(\pi - \theta, \pi)$ for $f$ to push forward to $\Bbb RP^2$. This is the only requirement. Any $f$ defined on $[0,\pi]^2$ that satisfies those two equations pushes forward to a map $\hat f$ on $\Bbb RP^2$. And if $f$ is continuous, then so is $\hat f$.

If $$f(\theta,\phi)=f(\pi-\theta,\phi+\pi)$$ then setting $\phi = 0$ gives exactly $f(\theta, 0) = (\pi - \theta, \pi)$. Setting $\theta = 0$ gives $$f(0,\phi) = f(\pi, \phi + \pi)$$which requires that $$f(\pi,\pi-\phi) = f(\pi, \phi + \pi)$$Unlike the other equation, this doesn't correspond to the requirement for $f$ to push forward. But note that it doesn't contradict it either. If $\phi \in [0,\pi]$, then $\phi + \pi \notin [0,\pi]$, so it doesn't matter what the value of $f(\pi, \phi + \pi)$ is.

So as long as your function $f$ also satisfies $f(0,\phi) = f(\pi, \pi - \phi)$, it will push forward. If $f$ does not also satisfy this condition, it doesn't push forward.