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Are there Riemannian metrics other than the standard metric induced from the euclidean space on $S^2$ such that the sectional curvature is equal to 1 everywhere? Or is this the unique Riemannian metric of constant curvature (up to scaling)?

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  • $\begingroup$ If $\phi:S^{2} \to S^{2}$ is a diffeomorphism other than a rotation and $g$ is the round metric, then $\phi^{*}g$ is isometric to $g$ (via $\phi$) and therefore has constant curvature $1$, but is distinct from $g$ as a Riemannian metric. However, @mollyerin's answer is "morally" correct, in that if $\lambda$ is real, then up to isometry there is a unique complete, simply-connected Riemannian $n$-manifold of constant sectional curvature $\lambda$. $\endgroup$ – Andrew D. Hwang Jan 14 '15 at 2:56
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The standard metric is the only metric of constant sectional curvature $1$ on $S^2$.

More generally, for any positive integer $n$ and any $\lambda \in \mathbb{R}$, there is a unique simply connected Riemannian manifold of dimension $n$ and constant scalar curvature $\lambda$ (which therefore looks like a scaling of either $S^n, \mathbb{R}^n$, or hyperbolic space $H^n$ depending on the sign of $\lambda$).

This is a standard fact in Riemannian geometry, not entirely trivial to prove but not terribly complicated either. Any introductory Riemannian geometry text should have a proof; for instance, it's given as Theorem 4.1 in Chapter 8 of do Carmo and Corollary 10 in section 10.2 of Petersen.

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  • $\begingroup$ Thank you very much, i will take look at these! $\endgroup$ – Pete Jan 14 '15 at 11:50

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