Hartogs triangle

A non-trivial example of domain of holomorphy that has some interesting non-obvious properties is the Hartogs triangle which is the set

 $\{(z,w)\in{\mathbb{C}}^{2}\mid\lvert z\rvert<\lvert w\rvert<1\}.$

Since it is a Reinhardt domain it can be represented by plotting it on the plane $\lvert z\rvert\times\lvert w\rvert$ as follows.

Figure 1: Hartogs triangle

It is obvious then where the name comes from. To see that this is a domain of holomorphy, then given a boundary point we wish to exhibit a holomorphic function on the whole Hartogs triangle which does not extend beyond that point. First note that on the top boundary $z$ is anything and $w=e^{i\theta}$ for some $\theta$, so $f(z,w)=\frac{1}{w-e^{i\theta}}$ will not extend beyond $(z,e^{i\theta})$. Now for the diagonal boundary this is where $\lvert z\rvert=\lvert w\rvert$, that is $z=e^{i\theta}w$ for some $\theta$, so $f(z,w)=\frac{1}{z-e^{i\theta}w}$ will do not extend beyond $(e^{i\theta}w,w)$.

One of the many properties of this domain is that if $U$ is the Hartogs triangle, then it is a domain of holomorphy, but if we take a sufficently small neighbourhood $V$ of $\bar{U}$ (the closure of $U$), then any function holomorphic on $V$ is holomorphic on the polydisc $D^{2}(0,1)$ (just fill in everything below the triangle in Figure 1). So if $V$ does not include all of $D^{2}(0,1)$ then it is not a domain of holomorphy. This is because a Reinhardt domain that contains zero (the point $(0,0)$) is a domain of holomorphy if and only if it is a logarithmically convex set and any neighbourhood of $\bar{U}$ does contain zero while $U$ itself does not.

References

• 1 Steven G. Krantz. , AMS Chelsea Publishing, Providence, Rhode Island, 1992.
Title Hartogs triangle HartogsTriangle 2013-03-22 14:31:08 2013-03-22 14:31:08 jirka (4157) jirka (4157) 5 jirka (4157) Example msc 32T05