Holomorphic Maps and Functions (Pt. II)
Point of Post: This is a continuation of this post.
Let us now give some examples of holomorphic maps:
Let and be domains in . Then, a holomorphic map is just a holomorphic map using the standard definitions from complex analysis.
Consider the Riemann sphere and the map given by (where we understand and ) is an automorphism of . Indeed, to see that is actually holomorphic we consider the standard atlas for consisting of the charts and where . Now, it’s easy to see that and we quickly check that is just the identity map on , and thus evidently holomorphic. Now, it’s equally evident that and since is just the identity map on again, we see that is legitimately holomorphic. Since is bijective and we see that is evidently an automorphism.
More generally, we claim that if is a rational function (where we assume for convenience that and are coprime) then is a holomorphic map where we define and whenever . Indeed, let be the roots of (obviously meaning roots in and counted with multiplicity). Now, we can clearly find an open subset for which does not contain any or (if one wants to explicit take a closed disc around each mentioned points [not intersecting each other for visual simplicity] and then take the Riemann sphere minus these discs). So, it’s definitely clear that and that so that we need only check is holomorphic, but this is evident. Thus, it remains to prove is holomorphic at each , and at . Now, for each , choose a neighborhood which contains no root of , nor any other , nor containing . Clearly then we have that and so we can check holomorphicity at by checking that is holomorphic at . But, is given by
Now, evidently is holomorphic on and is holomorphic on because exists, and is equal to (this is Riemann’s Removable singularity theorem). Thus, it remains to show that is holomorphic at . This requires two cases. If , choose a neighborhood of for which does not contain any , and for which doesn’t (why can we do this?). Then, , and thus we need to check that is holomorphic. Now, is some neighborhood of , and so we see that on one has that
Now, since doesn’t contain any roots, we know that this is holomorphic on and since it’s continuous by construction, is holomorphic by Riemann’s Removable singularity theorem. If , then we choose a neighborhood of for which doesn’t contain any roots of , nor of , and for which doesn’t contain . Then, we have that we need only check the holomorphicity of which is just
By similar reasoning, this is holomorphic on and continuous (by construction) on , and thus holomorphic. Thus, we have shown that is holomorphic at every point of and thus is holomorphic.
It’s obvious that if is any Riemann surface and is a chart on , with connected. Then, is a biholomorphism .
New Concept, Old Theorems
We now prove some of the classic theorems of complex analysis still holds true for holomorphic maps between Riemann surfaces. The basic idea in most of the proofs is that local properties transfer from subsets of to Riemann surfaces, because they are locally the same. Global concepts then shall be pieced together via these local results, connectivity being an integral part of these theorems.
Probably one of the most shocking theorems concerning the mapping properties of analytic functions is the Identity Theorem, which says that if two holomorphic functions agree on a non-discrete subset of a domain, then they are necessarily equal! This tells us, for example, that any holomorphic function on the unit disc is determined by its values , or more generally that if is any convergent sequence of points, then we need only specify a function on . As one can probably guess, this is true for general Riemann surface.
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 Rudin, Walter. Real and Complex Analysis. New York: McGraw-Hill, 1966. Print.