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cv.complicated variables – The (measurable) Riemann mapping theorem Answer

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cv.complicated variables – The (measurable) Riemann mapping theorem

You misstated Riemann’s (authentic) theorem:
a definitive assumption is that your launch subset
is just related.

Both theorems can breathe thought-about as classification theorems
of Riemann surfaces. The Riemann authentic theorem says that each merely related province within the sphere, whose complement incorporates no less than 2 factors
is conformally equal to the
unit disk.

“Measurable Riemann theorem” says {that a} sphere outfitted with any Riemannian metric, matter to inescapable situation (that the Beltrami coefficient $mu$
has norm <1) is conformally equal to the Riemann sphere.

It has unostentatious corollaries {that a} aircraft or a disk outfitted with a Riemannian metric
satisfying the identical situation are conformally equal to the aircraft and disk respectively with the benchmark metric. (A disk with capricious Riemannian metric is a generalization of a merely related province in
the aircraft with the habitual metric).

The ancient, classical designation of the “Measurable Riemann theorem” was “Existence and uniqueness theorem for Beltrami equation”, or it was referred to as just by the designation of an
creator (Korn and Lichtenstein, or Morrey or Boyarski, relying on require situations, and the flavor of the one who refers). Boyarski’s contribution is the crucial proven fact that correctly normalized $f$ relies on $mu$ analytically.

The fashionable designation comes from the paper of Ahlfors and Bers, Riemann’s mapping theorem for variable metrics, Ann. Math., 72 2 (1960), 385-404, the place they restated the outcome of Boyarski within the spirit that I outlined above, and emphasised this analytic dependence on $mu$. Besides $|mu|_infty<1$,
no situation on $mu$ is imposed besides that it’s Lebesgue measurable, and the phrase “measurable” within the designation of the concept comes from this truth.

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