Read e-book online First Steps in Several Complex Variables: Reinhardt Domains PDF

By Marek Jarnicki and Peter Pflug

ISBN-10: 3037190493

ISBN-13: 9783037190494

This booklet offers a accomplished advent to the sphere of numerous advanced variables within the surroundings of a really distinctive yet uncomplicated type of domain names, the so-called Reinhardt domain names. during this approach the reader could examine a lot approximately this zone with out encountering too many technical problems. bankruptcy 1 describes the basic notions and the phenomenon of simultaneous holomorphic extension. bankruptcy 2 provides a reasonably entire dialogue of biholomorphisms of bounded (complete) Reinhardt domain names within the dimensional case. The 3rd bankruptcy offers a category of Reinhardt domain names of life for an important sessions of holomorphic features. The final bankruptcy bargains with invariant capabilities and provides particular calculations of a lot of them on sure Reinhardt domain names. various routines are integrated to aid the readers with their figuring out of the fabric. additional effects and open difficulties are extra that could be invaluable as seminar themes. the first objective of this ebook is to introduce scholars or non-experts to a couple of the most learn components in numerous complicated variables. The e-book presents a pleasant invitation to this box because the merely prerequisite is a simple wisdom of research.

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Extra resources for First Steps in Several Complex Variables: Reinhardt Domains (EMS Textbooks in Mathematics)

Sample text

If E(S) = 0, then we put S =e:=C". 5 (d) will show that Ds is connected and, therefore. DS is really a domain. 42 Chapter 1. 2. Let K C C" be a Reinhardt compact set and let r > 0. Put K(r) := U F(a. r) aEK and observe that K(r) is also a Reinhardt compact (EXERCISE). Then there exists a 9 E (0, 1) such that for every a E IR" with Kt"l C C"(a) we have max lzal < 91a1 ZEK max lzal, zEK(r) where at := latI +... + lan1. Proof. Observe that if z = (z i , ... , z") E K and a j0 < 0, then 1z10 l > r.

0)=Cj-1 . X(0)xC"-i n = 0). Z n) E C n : Z, ... Z Chapter 1. 6. ) (a) Let A be a Reinhardt set. Then int(log A) = log(int A). log A = log A. Consequently, for any set B C IR", we have log(int exp B) = int(log exp B) = int log(exp B) = int B. g. if X is a convex domain), then D := int exp X is a fat Reinhardt domain with log D = X. Conversely, if G is a Reinhardt domain with log G = X, then int G = D. In fact, if log G = X, then G \ Yo = exp X. Hence, d = exp X, and finally, int G = D. ,yn) E A n IR;o, I E [0,11.

Consider any two-dimensional real space P C R" with L C P and X' := X n P 9& 0. We have Ep (X') = {0}. Let C' be the open cone in P generated by X'. Obviously, C' = C n P. Hence C' + L C C' and the proof is reduced to the case n = 2. In the case n = 2 we only need to observe that if E (X) = (0), then there exist two different half-planes I41,0, 142,0 with X C I ',0 n k2,0; a contradiction. Consequently, there exists a fl° E (R"), such that C n (X E IR" : (x, f°) = 0) = (0). Indeed, we use induction on n.

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First Steps in Several Complex Variables: Reinhardt Domains (EMS Textbooks in Mathematics) by Marek Jarnicki and Peter Pflug

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