If two disks of different radii are homothetic, is the metric flat?

Question

$$\newcommand{\Vol}{\text{Vol}}$$ Let $(M,g)$ be a surface, and let $p \in M$.

Suppose that there exist $0<r_1<r_2<\text{inj}_p$ such that the geodesic balls $B_{r_1}(p)$ and $B_{r_2}(p)$ are homothetic. Is $g$ is flat?

Here $\text{inj}_p$ is the injectivity radius of $(M,g)$ at $p$, and I say that two Riemannian manifolds $(N,h), (\tilde N,\tilde h)$ are homothetic, if there exist a diffeomorphism $\phi:N \to \tilde N$ such that $\phi^* \tilde h =\lambda^2 h$, for some constant scalar $\lambda>0$.

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Edit:

I prove that a necessary condition is $$ \int_{B_{r_2}\setminus B_{r_1}} K_{ g} d\Vol_{ g}=0, $$ where $K_g$ is the Gaussian curvature of $g$.

Proof:

Suppose that $B_{r_1}(p)$ and $B_{r_2}(p)$ are homothetic, where $0<r_1<r_2<\text{inj}_p$. This is equivalent to the statement that $(B_{r_2}, \lambda^2 g) $ is isometric to $(B_{r_1},g) $.

Now, since $(B_{r_2}, \lambda^2 g) $ is isometric to $(B_{r_1},g) $, we have $$ \int_{B_{r_2}} K_{\lambda^2 g} d\Vol_{\lambda^2 g}=\int_{B_{r_1}} K_{ g} d\Vol_{g}. \tag{1} $$ Since $$ K_{\lambda^2 g}=\frac{1}{\lambda^2}K_{g}, \,\,\,\,d\Vol_{\lambda^2 g}=\lambda^2 d\Vol_{g} $$ Equality $(1)$ becomes $$ \int_{B_{r_2}} K_{ g} d\Vol_{ g}=\int_{B_{r_1}} K_{ g} d\Vol_{g},\tag{2} $$ so $$ \int_{B_{r_2}\setminus B_{r_1}} K_{ g} d\Vol_{ g}=0. $$

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Here is a proof $g$ is flat when assuming the stronger assumption that all the $B_r(p)$ are homothetic, for sufficiently small $r$.

Key point: Diameter and boundary length of sets scale in the same way under homothety.

Thus, if $B_r(p),B_r'(p)$ are homothetic, then the ratios between their diameter and boundary length are the same, i.e. $$ \frac{L(\partial B_r(p))}{\text{diam}(B_r(p))}=\frac{L(\partial B_r(p))}{2r} $$ is independent of $r$.

However, $$ \frac{L(\partial B_r(p))}{2\pi r}=1-\frac{K(p)}{12}r^2+o(r^4), $$ where $K$ is the Gaussian (scalar) curvature of $g$.

Thus $$ \frac{K(p)}{12}r^2+o(r^4) $$ is independent of $r$, which implies $K=0$.

Answer 1

I think that $g$ must be flat.

Let $a<1$, and suppose that $\phi:(B_r(p), g) \to (B_{ar}(p),g)$ is a homothety, i.e. $$ \phi:(B_r(p),\lambda^2 g) \to (B_{ar}(p),g) $$ is an isometry. Then $$ K_g(\phi(x))=K_{\lambda^2 g}(x)=\frac{1}{\lambda^2}K_{g}(x), $$ so $$ K_g(\phi^n(x))=\big(\frac{1}{\lambda^2}\big)^nK_{g}(x), $$ for any $x \in B_r(p)$.

Assume by contradiction that $K_{g}(x) \neq 0$ for some $x \in B_r(p)$. Since $\phi^n(x) $ lie in the compact set $B_{ar}(p)$ it has a (non-relabeled) subsequence $\phi^n(x) \to q$.

Since $\lambda<1$, $$ K_g(q)=\lim_{n \to \infty}K_g(\phi^n(x))=\lim_{n \to \infty}\big(\frac{1}{\lambda^2}\big)^nK_{g}(x)=\pm \infty, $$ which is a contradiction.