# monodromy theorem

Let $C(t)$ be a one-parameter family of smooth paths in the complex plane with common endpoints^{} ${z}_{0}$ and ${z}_{1}$. (For definiteness, one may suppose that the parameter $t$ takes values in the interval^{} $[0,1]$.) Suppose that an analytic function^{} $f$ is defined in a neighborhood^{} of ${z}_{0}$ and that it is possible to analytically continue $f$ along every path in the family. Then the result of analytic continuation does not depend on the choice of path.

Note that it is *crucial* that it be possible to continue $f$ along all paths of the family. As the following example shows, the result will no longer hold if it is impossible to analytically continue $f$ along even a single path. Let the family of paths be the set of circular arcs (for the present purpose, the straight line is to be considered as a degenerate case of a circular arc) with endpoints $+1$ and $-1$ and let $f(z)=\sqrt{z}$. It is possible to analytically continue $f$ along every arc in the family except the line segment^{} passing through $0$. The conclusion of the theorem does not hold in this case because continuing along arcs which lie above $0$ leads to $f({z}_{1})=+i$ whilst continuing along arcs which lie below $0$ leads to $f({z}_{1})=-i$.

Title | monodromy theorem^{} |
---|---|

Canonical name | MonodromyTheorem |

Date of creation | 2013-03-22 14:44:35 |

Last modified on | 2013-03-22 14:44:35 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 8 |

Author | rspuzio (6075) |

Entry type | Theorem |

Classification | msc 30F99 |