# How do pullback and isothermal transformations of a Riemannian metric change curvature?

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.

• 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. – User X Aug 22 at 9:25
• So, is this true for closed surfaces? – User X Aug 22 at 9:30
• 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. – Travis Willse Aug 22 at 15:11
• 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. – Travis Willse Aug 22 at 15:14
• Yeah I have edited the text. Thank you again! – User X Aug 23 at 9:46