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

Since it is a Reinhardt domain it can be represented by plotting it on the plane $|z|\times |w|$ 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 $|z|=|w|$, 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 $\overline{U}$ (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 $\overline{U}$ does contain zero while $U$ itself does not.

Title	Hartogs triangle
Canonical name	HartogsTriangle
Date of creation	2013-03-22 14:31:08
Last modified on	2013-03-22 14:31:08
Owner	jirka (4157)
Last modified by	jirka (4157)
Numerical id	5
Author	jirka (4157)
Entry type	Example
Classification	msc 32T05