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Suppose $m$ is a smooth Riemannian metric on $\mathbb{S}^2$, the uniformization theorem of surfaces tell us that $m$ is conformally equivalent to the standard round metric. Formally this says that there exists a diffeomorphism $f$ of $\mathbb{S}^2$ and a function $h: \mathbb{S}^2 \to \mathbb{R}$ such that $f^{*}(m) = e^{h}m_0$, where $m_0$ is the standard round metric and $f^{*}$ denotes pullback by $f$.

Observe that $f$ and $h$ are not unique. One can compose $f$ by Mobieus transformations and obtain different $f$ and $h$ that way. Nonetheless, one would expect that it is possible to find $f$ and $h$ that just depend on how my Riemannian metric $m$ is.

Formally, my question is: Given a smooth metric $m$ on $\mathbb{S}^2$. Can one obtain a pair $(f,h)$ uniformizing $m$ whose $C^r$ norm (measuring up to $r$ derivatives) depends (say polynomial, exponential, etc) on the $C^r$ norm of the metric $m$ (as a 2 tensor)?

Any result vaguely related to this question will be greatly appreciated.

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The $C^{k,\alpha}$ version of what you want is in theorem 8.4 of https://arxiv.org/abs/1510.07269, which is based on theorem 4.1 in https://arxiv.org/abs/1307.3167. The result must be well-known to experts but I do not know of an earlier source where this is done. The underlying issue is studying how the solution of the Beltrami equation depends on the dilatation. In Teichmüller theory either one works in a weak setting, or in the analytic one, and few people care about $C^k$ setting. I do not know how to keep the $C^k$ regularity, and I think this is impossible (due to the usual reason of why Hölder spaces are used for in elliptic regularity).

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You should have a look into the following paper (https://arxiv.org/pdf/1507.00798v1.pdf) by Hass and Koehl.

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