# 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

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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.

## References

- 1 Steven G. Krantz. , AMS Chelsea Publishing, Providence, Rhode Island, 1992.

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 |