# Are spherical maps with low distortion locally expanding?

$$\newcommand{\SO}{\text{SO}(#1)}$$ $$\newcommand{\Hom}{\text{Hom}(#1)}$$ $$\newcommand{\R}{\mathbb{R}}$$ $$\newcommand{\S}{\mathbb{S}}$$

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).