# Which Riemannian Manifolds are scale-free?

Let us call a Riemannian manifold $(M,g)$ scale-free if for any real positive scalar $\lambda$, $(M,g),(M,\lambda g)$ are isometric.

$\mathbb{R}^n$ with the standard metric $g$ is scale-free. (Via the map $\phi(x)=\sqrt \lambda x$).

Which other Riemannian manifolds are scale-free? In particular, is it true that every flat non-compact manifold is scale-free?

Are there any non-flat scale-free manifolds?

Remarks:

(1) As pointed out by Phillip Andreae, a compact manifold can never be scale-free since scaling the metric changes the volume.

(2) Since the scalar curvature scales by a reverse proportion ($\tilde{K} = \lambda^{-1} K$ if $\tilde{g} = \lambda g$) and it's an isometric invariant this imposes a limitation on the class of scale- free manifolds. (For example no manifold of constant nonzero sectional curvature can belong to it).

It sounds to me that the values of the sectional curvature are supposed to "spread uniformly" in some sense. (If the manifold is compact the volume of the points with specific value $k$ of the scalar curvature depends only on $\text{sign}(k)$?)

Also, what changes if we require only existence of 2 different scales to be isometric?

• Note that a compact manifold (e.g., a torus) can never be scale-free since scaling the metric changes the volume. – Phillip Andreae Jul 20 '15 at 14:56
• Not every flat non-compact manifold is scale free. The flat cylinder $S^1\times \mathbb{R}$ isn't scale free because the length of the shortest closed geodesic changes when you scale. Is there any scale free Riemannian manifold other than flat $\mathbb{R}^n$? – Jason DeVito Jul 20 '15 at 16:47

Here's a way to construct a lot of non-flat examples. Let $(\widehat M,\widehat g)$ be any Riemannian manifold whatsoever, and define a metric $g$ on $M = \mathbb R^+\times \widehat M$ by $$g = dt^2 + t^2 \hat g,$$ where $t$ is the standard coordinate on $\mathbb R^+$. Then for each $\lambda>0$, the map $\phi_\lambda\colon M\to M$ given by $\phi_\lambda (t,x) = (t\sqrt{\lambda},x)$ is an isometry from $(M,\lambda g)$ to $(M,g)$.
Note that these metrics can be flat: For example, if $(\widehat M,\widehat g)$ is the standard round $n$-sphere, then $(M,g)$ is isometric to $\mathbb R^{n+1}\smallsetminus\{0\}$ with its Euclidean metric. If $(\widehat M,\widehat g)$ is a circle of radius less than $2\pi$, then $(M,g)$ is a flat cone. But typically, $(M,g)$ will not be flat.
I think there's a reasonable chance that every example is either flat $\mathbb R^n$ or of this type, but I don't see a way to prove it.