The length of shortest curve dividing topological sphere into two parts of same area

Question

$(S^2,g)$ is a Riemannian manifold, where $S^2$ is homeomorphic to the 2-dimensional sphere. And the area of $(S^2,g)$ is $4\pi$. $\gamma$ is the shortest curve dividing $S^2$ into two parts of equal area. Denote the length of $\gamma$ as $L(\gamma)$. Then, how to show $$ L(\gamma)\le 2\pi $$ I just feel it is right. But I don't know how to prove. If the proof is complex, tell me the reference is enough, thanks.

PS(2023-7-24): Please don't click the close. The example of psl2Z is not a counter-example, mollyerin have explain it.

Answer 1

If the metric admits a fixed-point-free involution, your estimate is equivalent to Pu's inequality. Otherwise this seems to be open, unless I am overlooking some recent literature. An estimate can be found in

Papasoglu, Panos. Cheeger constants of surfaces and isoperimetric inequalities. Trans. Amer. Math. Soc. 361 (2009), no. 10, 5139–5162.