# Are manifolds with the same constant scalar curvature locally isometric?

I was reading through Wald's General Relativity book and he mentions in chapter 5 that if two manifolds have the same dimension, metric signature, and constant curvature, they are locally isometric. I believe I should be interpreting "curvature" here as the sectional curvature, for which I can a proof in Lee's Intro to Riemannian Manifolds in Corollary 10.15. If I assume constant scalar curvature I am not so sure. For instance, if the manifolds are 2 dimensional, this follows from Minding's Theorem. But if they are 3-dimensional then I doubt this claim is true. For instance, any solution to the Einstein vacuum equations $$R_{ab} = 0$$ will have constant $$0$$ scalar curvature, yet the Schwarzschild metric cannot be locally isometric to flat space because the norm of its Riemann tensor, the Kretschmann invariant, is nowhere vanishing.

If this is correct then I can accept Wald's claim since he is working with isotropic 3-dimensional foliations of spacetime, and an isotropic 3-dimensional manifold has constant sectional curvature. Otherwise, maybe all vacuum solutions are locally flat?

• I can write a full answer in the morning, but if you consider a Riemannian product $S^2\times S^3$ where each factor has a round metric, then you can easily create examples of non-locally isometric spaces with the same scalar curvature by varying the radii of the spheres. Aug 30, 2022 at 3:30

I will write $$S^n(r)$$ for the round $$n$$-sphere of radius $$r$$. It is well known that the scalar curvature is a constant $$\frac{n(n-1)}{r^2}$$.

Now, for two positive real numbers $$r,s$$, consider the Riemannian product $$S^2(r)\times S^2(s)$$. This has scalar curvature $$\frac{2}{r^2} + \frac{2}{s^2}$$.

A relatively uninspired calculation shows that the scalar curvature has the constant value $$2$$ if and only if $$r = \frac{s}{\sqrt{s^2-1}}$$. (The value of $$2$$ was picked just to make this formula nice - nothing that follows depends critically on this choice.) In particular, there are infinitely many distinct pairs $$(r,s)$$ which give the same scalar curvature.

However, I claim that "most" of these pairs $$(r,s)$$ give different local isometry types. Specifically, the unordered pair $$\{r,s\}$$ controls the local isometry type. More precisely, I claim:

Proposition If $$S^2(r)\times S^2(s)$$ and $$S^2(r')\times S^2(s')$$ are locally isometric, then $$(r,s)$$ is a permutation of $$(r',s')$$.

Proof: We will assume without loss of generality that $$r\geq s$$ and that $$r'\geq s'$$.

Now, at every point of $$S^2(r)\times S^2(s)$$, the maximum sectional curvature of a two-plane is obtained by a $$2$$-plane tangent to the $$S^2(s)$$ factor, and it has sectional curvature $$\frac{1}{s^2}$$. Since local isometries preserve maximum curvature at a point, we must have $$1/s^2 = 1/s'^2$$, which easily implies $$s = s'$$. Moreover, the existence of a local isometry implies the scalar curvatures match, so $$\frac{2}{r^2} + \frac{2}{s^2} = \frac{2}{r'^2}+\frac{2}{s'^2}$$. As we already know $$s = s'$$, this simplifies to $$\frac{2}{r^2} = \frac{2}{r'^2}$$, which then implies $$r= r'$$. $$\square$$.

One can play a similar game with $$S^m(r)\times S^n(s)$$ as long as both $$m,n\geq 2$$. In particular, there are infinitely many such examples in each dimension $$4$$ or larger. However, one can not play this game with, say, $$n=1$$, basically because $$S^1(s)$$ is locally isometric to $$S^1(s')$$ even if $$s\neq s'$$. So the above approach won't work in dimension $$3$$.

However, there are three dimensional examples: the Berger spheres. A Berger sphere is obtained as follows: start with a round $$3$$-sphere $$S^3(r)$$ and the shorten the metric in the directions tangent to the Hopf fibration. If these directions are shortened by a factor of $$t < 1$$, then one can compute the resulting metric is homogeneous, and that, in an appropriate orthonormal basis of $$T_p S^3$$, that $$K(X,Y) = K(X,Z) = r^2t^2$$ while $$K(Y,Z) = r^2(4-3t^2)$$.

It follows that the scalar curvature is constant with value $$r^2(8-2t^2)$$. In particular, one can find infinitely many pairs $$(r,t)$$ with the same scalar curvature. However, note that for different choices of $$t$$, the ratio $$\frac{\max{\text{sectional curvature}}}{\min{\text{sectional curvature}}} = \frac{4-3t^2}{t^2}$$ is a local isometry invariant. In particular, $$t$$ is determined by the local isometry tyrpe, so there are infinitely many examples of different local isometry types but the same constant scalar curvature.