))
Degree of hyperplane divisor of map versus of image
Archived series ("Inactive feed" status)
When?
This feed was archived on December 09, 2017 16:01 (
Why? Inactive feed status. Our servers were unable to retrieve a valid podcast feed for a sustained period.
What now? You might be able to find a more up-to-date version using the search function. This series will no longer be checked for updates. If you believe this to be in error, please check if the publisher's feed link below is valid and contact support to request the feed be restored or if you have any other concerns about this.
Manage episode 185577011 series 1521141
Content provided by Random Stuffs. All podcast content including episodes, graphics, and podcast descriptions are uploaded and provided directly by Random Stuffs or their podcast platform partner. If you believe someone is using your copyrighted work without your permission, you can follow the process outlined here https://player.fm/legal.
Suppose phi:X to P^n is a holomorphic map with image Y being a smooth projective curve. Let H be a hyperplane in P^n. Then deg (phi*H) = (deg phi)(deg Y). (Recall deg Y is deg of intersection divisor of H on Y, denoted div(H)). This is because for each p, (phi* H)(p) is intersection multiplicity at phi(p) between H and im phi. For each phi(p), (div H)(phi(p)) is the same thing. Each term phi(p) in the second divisor correspond to deg phi terms in the first divisor (all points in the fiber). If D is very ample then phi is an isomorphism onto its image so have degree 1. As phi*H is linearly equivalent to X it's degree is deg X. So I'm that case Deg (D) = deg Y
…
continue reading
172 episodes
Share
