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The question in a nutshell: Are the "best" spherical maps locally expanding?

Let $U \subseteq \mathbb{S}^2$ be an open subset of the round sphere.

Does there exist an $\epsilon >0$ such that for every smooth map $f:U \to \mathbb{R}^2$ satisfying $(1-\epsilon) |v| \le |df(v)| \le (1+\epsilon)|v|$ everywhere on $U$, $f$ is locally expanding in the sense that $|df_p(v)| \ge |v|$ for every $p \in U, v \in T_p\S^{2}$?

I want the statement to hold in a non-vacuous (non-trivial) way; if we take $\epsilon$ to be sufficiently small, then there are no maps satisfying the hypothesis at all. So, I would like $\epsilon$ to be in the admissible range.

I am fine with assuming $f$ is an immersion.

Some intuition and a failed attempt:

The intuition is that the sphere is "more cramped" than the Euclidean plane. Thus it seems reasonable(?) that "good embeddings" would expand.

I tried proving that a map $f$ which satisfies the hypothesis with very small $\epsilon$ should map geodesics to "almost geodesics", and then to use the fact that the rate of spread of geodesics in the sphere is smaller than in Euclidean space (this rate is governed by the curvature, via the Jacobi equation).

  • $\begingroup$ It seems to me that something is not quite correct with the question (or perhaps I misunderstand it). Indeed, suppose you've constructed a locally expanding map $f$. Suppose that $c$ is the minimal dilatation of tangent vectors by $df$, $c\ge 1$. Then the map $(1-\epsilon)c^{-1}f$ satisfies your conditions, but it is not locally expanding anymore... $\endgroup$ – Dmitri Panov Feb 19 at 14:31
  • $\begingroup$ Oh, it seems that you are right; If I understood you correctly, then what you say implies that if there exists a locally expanding map satisfying the dilation requirements, you can modify it to be non-expanding (while preserving the constraints)-thus such maps certainly do not have to be locally expanding. So, I agree now that this probably makes the question a bit ill-posed. I will think whether I can improve it-and if not will close. Thank you for this insight. $\endgroup$ – Asaf Shachar Feb 19 at 15:45

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