Timeline for Question on whether, "An entire function, nowhere zero, has an entire logarithm," holds for matrix-valued entire functions as well

14 events

when toggle format	what		by	license	comment
S Dec 27, 2023 at 5:35	history	suggested	Buzz	CC BY-SA 4.0	fixed LaTeX and typos
Dec 27, 2023 at 1:00	answer	added	Christian Remling		timeline score: 4
Dec 26, 2023 at 21:32	review	Suggested edits
S Dec 27, 2023 at 5:35
Dec 25, 2023 at 20:48	history	became hot network question
Dec 25, 2023 at 18:21	answer	added	Robert Israel		timeline score: 14
Dec 25, 2023 at 16:24	comment	added	Vik78		@ChristianRemling right, it was a typo on my part.
Dec 25, 2023 at 15:54	comment	added	Christian Remling		@Vik78: $A^{-1}$ is not the ($z$-)derivative of $\log A(z)$, and when trying to fix that, one runs into issues similar to these: en.wikipedia.org/wiki/Derivative_of_the_exponential_map
Dec 25, 2023 at 15:24	comment	added	Kanghun Kim		Aha. Now I get the idea.
Dec 25, 2023 at 15:18	comment	added	Vik78		if the determinant of the matrix is nowhere zero, then for fixed $a$ and varying $b \in \mathbb{C}$ the integral $F(b) = \int\limits_a^b A^{-1}(z)dz$ along some fixed choice of path from $a$ to $b$ is well-defined, and should be path-independent. Can we show this is a logarithm?
Dec 25, 2023 at 14:58	comment	added	Kanghun Kim		Both are intended.
Dec 25, 2023 at 14:34	comment	added	David E Speyer		Do you mean an entire function of many variables or of one variable?
Dec 25, 2023 at 13:16	history	edited	Kanghun Kim	CC BY-SA 4.0	added 8 characters in body
Dec 25, 2023 at 12:56	history	edited	Kanghun Kim	CC BY-SA 4.0	added 361 characters in body
Dec 25, 2023 at 12:47	history	asked	Kanghun Kim	CC BY-SA 4.0