The degree condition in Llarull's theorem on scalar curvature rigidity
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arXiv
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| Auteurs principaux: | , |
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| Format: | Preprint |
| Publié: |
2025
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| _version_ | 1866910133624242176 |
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| author | Baer, Christian Zeidler, Rudolf |
| author_facet | Baer, Christian Zeidler, Rudolf |
| contents | 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. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2507_05459 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | The degree condition in Llarull's theorem on scalar curvature rigidity Baer, Christian Zeidler, Rudolf Differential Geometry 53C20, 53C24 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. |
| title | The degree condition in Llarull's theorem on scalar curvature rigidity |
| topic | Differential Geometry 53C20, 53C24 |
| url | https://arxiv.org/abs/2507.05459 |