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In this theorem why if $\pi'$ is an unbranched covering, then its restriction is unbranched? And what is the meaning of ideal points?enter image description here

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  • $\begingroup$ What is your definition of unbranched? If $\pi'$ is unramified above every point of $X'$ and $U \subseteq X'$, then $\pi'$ is unramified above every point of $U$. $\endgroup$ – Viktor Vaughn Oct 20 '20 at 18:42
  • $\begingroup$ @Richard D.James Oh,that's right, actually I confused by the proof.could you please help me with the part that I marked? Thanks a lot. $\endgroup$ – Mathgreek Oct 20 '20 at 19:29
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I hope this would help you.

Generally, Let $\pi: E \to X$ be a covering of a topological space $X$, which is locally path-connected. For an open set $U$ consider $\pi^{-1}(U)$; then the restriction of $\pi$ to each of the connected components of $\pi^{-1}(U)$ is a covering map for $U$.

To prove this, you should notice that $E$ also becomes a locally path-connected space, and due to this, the connected components of $\pi^{-1}(U)$ are open sets in $E$.

Considering the above lemma and the fact that $\pi^{\prime}$ is an unbranched covering would give that the restriction of $\pi^{\prime}$ is an unbranched covering.


Now for your second question, $Y^{\prime}$ does not contain the points needed to make the branched covering and $Y$, so we have to add them. That's why it is called ideal points, I think.


I think when we add the ideal points to our surface and construct the new covering map, generally speaking, we have a branched covering, meaning that we don't know whether it is branched or unbranched. An example of this is that you consider an unbranched covering, which is also proper, as the identity map from a Riemann surface to itself and remove some points from it and then do the following instructions to again get the identity map, which is unbranched.

The compatibility of the charts comes from the fact that you can analytically extend the maps $\zeta_{av}$ to the points $p_{av}$.

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  • $\begingroup$ Thank you very much,but I'm not very sure why when we add these points ,we have a branch covering? $\endgroup$ – Mathgreek Oct 21 '20 at 19:36
  • $\begingroup$ and for checking compatibility of charts for y, is it true to say because on V* we have the first chart is biholomorphic to the complex structure of y',and for p(av) the first chart is not defined we conclude compatibility of charts for y? $\endgroup$ – Mathgreek Oct 21 '20 at 20:13
  • $\begingroup$ @Hanieh, I have edited my answer. I hope it will solve your ambiguities. $\endgroup$ – Amirhossein Oct 23 '20 at 5:38

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