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Let $(M,g)$ be a two-dimensional compact surface, endowed with a Riemannian metric.

Fix $s>0$, and suppose that for any two geodesic triangles $A,B$ having area $s$, there exists an affine onto map $f:A \to B$, where I say $f$ is affine if $\nabla df=0$. (equivalently, $f$ maps parametrized geodesics to parametrized geodesics. Here $\nabla=\nabla^{T^*M} \otimes \nabla^{f^*TM}$).

I assume $s<<\text{Area}(M)$ is very small, so there are a lot of triangles of area $s$.

What can we say about the metric $g$? Does it have to be flat? Are there any restrictions on its curvature?

I do not require $f$ to be the restriction of an affine map $M \to M$; (I think this is a stronger requirement than the existence of "local" or piece-wise affine maps. e.g. for the flat torus, globally we only have $SL_2(\mathbb{Z})$.)


Edit:

I believe that the assumption means that that there a lot of affine maps locally $M \to M$; perhaps we can translate this into showing $M$ is flat.

In fact, if $\nabla^{T^*M} \otimes \nabla^{f^*TM}$ has zero curvature, then $M$ is flat. And 'many affine maps' means roughly 'many parallel sections of $T^*M \otimes TM$ '-- although not exactly, since for every $f$, $df$ is a section of different vector bundle, which is $T^*M \otimes f^*TM$.

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  • $\begingroup$ First, note that if the surfaces are both $\mathbb{R}^2$ with the flat metric, then any area-preserving affine map satisfies your property. I'm pretty sure that you can show that the converse holds locally. This, I believe, would show that the existence of a global map implies that both domain and range are affine flat and the map is locally affine. $\endgroup$
    – Deane
    May 24, 2021 at 16:20
  • $\begingroup$ Indeed, that is what I had in mind. The only question is how to prove that...I guess I might ask on MO in a few days, if there won't be any answer here. $\endgroup$ May 24, 2021 at 17:12

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