Let $(X, g)$ be a (closed) Riemannian surface (or more generally a Riemannian manifold of any dimension). How do the following changes to $g$ change its curvature?

  1. Pullback by some diffeomorphism of $X$;
  2. Multiplication by a scalar function, i.e., replacing $g$ with $\lambda g$ for some $\mathbb R$-valued function $\lambda$.

More precisely, can we use these two operations to produce a metric with constant curvature?


Curvature is invariant under diffeomorphism in the sense that for a diffeomorphism $\phi$, the Riemann curvature tensor satisfies $$(\phi^* R)_p = R_{\phi(p)} :$$ tautologically, pulling back by a diffeomorphism cannot change the curvature up to diffeomorphism.

On the other hand, the behavior of curvature under a conformal rescaling, $$g(x) \mapsto \hat g(x) := \lambda(x) g(x) , \qquad \lambda > 0 ,$$ is a rich topic connected with several classical problems.

In dimension $2$, the Uniformization Theorem implies that every Riemannian surface admits a metric of constant scalar curvature, and in this dimension, sectional curvature and scalar curvature are essentially the same thing. For compact surfaces, the Gauss-Bonnet Theorem implies that the sign of the scalar curvature of such a metric is determined entirely by the genus.

In dimensions $\dim M \geq 3$ the notion of curvature is more complicated, and in particular constant scalar curvature does not imply constant sectional curvature. Now, curvature $R$ changes under conformal rescalings $g \mapsto \hat g$---this is no surprise, since curvature depends on derivatives of the metric---a straightforward calculation shows that the totally tracefree part $W$ of $R$, called the Weyl tensor (viewed as a $(1, 3)$-tensor) does not.

Any space with constant sectional curvature turns out to be (locally) conformally flat, that is, locally equivalent under a conformal transformation to the Euclidean metric, which has $R = 0$ and thus $W = 0$. So, if there is a conformal factor $\lambda$ such that $\hat g = \lambda g$ has constant sectional curvature, it must have been the case that the Weyl tensor $W$ of the original metric $g$ is zero. For $\dim M \geq 4$ this is a very restrictive condition, and general metrics do not satisfy it. (For $\dim M = 3$, the story is similar but we always have $W = 0$, even when the metric is not conformally flat; in this case, conformal flatness is instead characterized by the vanishing of another tensor, the Cotton(-York) tensor.)

Finally, we can ask whether for any metric $g$ there is always a $\lambda$ such that $\hat g = \lambda g$ has constant scalar curvature. We saw above that for $\dim M = 2$ the answer is yes. If we restrict attention to compact manifolds of $\dim M \geq 3$, we're posing the famous Yamabe Problem. The answer is yes, but the proof is nontrivial, and it was only achieved in 1984.

  • $\begingroup$ Thank you for your detailed answer! I did not make myself clear - actually by scalar I mean a scalar function, precisely as you explained in the remark - and this is exactly what I want. $\endgroup$ – User X Aug 22 at 9:25
  • $\begingroup$ So, is this true for closed surfaces? $\endgroup$ – User X Aug 22 at 9:30
  • $\begingroup$ You're welcome, I'm glad you found it useful. If you really mean that $\lambda$ is a scalar function, you should modify the text of the question. The notation $\lambda \in \Bbb R$ implies that $\lambda$ is a constant. $\endgroup$ – Travis Willse Aug 22 at 15:11
  • 1
    $\begingroup$ And yes, it is true for surfaces, both closed and otherwise. This is an immediate consequence of the Uniformization Theorem, which is a relatively old result by differential geometry standards (it dates to the first decade of the 20th C., IIRC). I've modified my answer to include mention of it. $\endgroup$ – Travis Willse Aug 22 at 15:14
  • $\begingroup$ Yeah I have edited the text. Thank you again! $\endgroup$ – User X Aug 23 at 9:46

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