The degree condition in Llarull's theorem on scalar curvature rigidity

TL;DR

This paper explores whether the degree condition in Llarull's scalar curvature rigidity theorem can be replaced by surjectivity, finding it only works in dimension two, but not for higher dimensions, with Ricci curvature replacing scalar curvature working universally.

Contribution

It demonstrates that the degree condition cannot generally be replaced by surjectivity in higher dimensions, but can in dimension two, and extends results to Ricci curvature.

Findings

Surjectivity replaces degree condition only in dimension two.

In dimensions three and higher, surjectivity does not imply the rigidity.

Replacing scalar curvature with Ricci curvature makes the condition sufficient in all dimensions.

Abstract

Llarull's scalar curvature rigidity theorem states that a 1-Lipschitz map $f: M\to S^n$ from a closed connected Riemannian spin manifold $M$ with scalar curvature $\mathrm{scal}\ge n(n-1)$ to the standard sphere $S^n$ is an isometry if the degree of $f$ is nonzero. We investigate if one can replace the condition $\mathrm{deg}(f)\neq0$ by the weaker condition that $f$ is surjective. The answer turns out to be "no" for $n\ge3$ but "yes" for $n=2$. If we replace the scalar curvature by Ricci curvature, the answer is "yes" in all dimensions.