A Question about covering map of punctured unit disk

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Let ##\alpha \in \mathbb{R}^*,\, p:\mathbb{H} \to \mathbb{D} \setminus \{0\}, \, p(z) = e^\frac{2 \pi i z}{|a|}##. I want to show that ##p## is a covering map but I dont't know how to make this. I think I need to start with an ##y \in \mathbb{D} \setminus \{0\}## and take an open disk ##D \subset \mathbb{D} \setminus \{0\}## to be a neighbourhood of ##y##. Now, because disk is open and connected, there exists a holomorphic branch of logarithm in ##D##. It's ok this start? How I can continue? Thanks!
 
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Sorry for my ignorance -- What is ##\mathbb{R}^*##? Is ##\mathbb{H}## a half-plane?
 
@FactChecker ##\mathbb{R}^* = \mathbb{R} \setminus \{0\}## and ##\mathbb{H}## is the upper half-plane
 
I think you may need to include more at the branch cut. I think that ##\mathbb{H}## does not include any of the real line. Don't you need to include the real line in the domain of ##p## in addition to ##\mathbb{H}## in order to cover ##\mathbb{D} \setminus \{0\}##? And doesn't the domain need to include an open set that includes the reals?

CORRECTION: My mistake. The positive reals in ##\mathbb{D} \setminus \{0\}## will be covered by the image of ##z=x+|\alpha|i##.
 
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