# If any two triangles of equal area can be mapped via affine maps, what can we say about the geometry?

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 $$T^*M \otimes f^*TM$$.

• 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. May 24, 2021 at 16:20
• 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. May 24, 2021 at 17:12