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Nov 4, 2011 at 12:39 vote accept One_math_boy
Nov 3, 2011 at 23:34 comment added Jack Huizenga You can replace "line bundle" with divisor everywhere, and change $L$'s to $D$'s. If you're not comfortable with this correspondence yet, now would be a great time to learn it. I'd suggest Rick Miranda's book on algebraic curves as a good place to get familiar with it, as it is made very explicit there. I'm also implicitly using several facts that are covered in the first chapter of Arborello, Cornalba, Griffiths, and Harris "Geometry of Algebraic Curves," particularly the fact that a general degree $d$ divisor $D$ has $\dim |D| = \max(0,d-g)$.
Nov 3, 2011 at 21:09 comment added One_math_boy Sorry, but if You have time, can You write the proof above without using the definition of (general?) linear bundle, but only the definition of divisors? I'm beginner in algebraic geometry, so I don't understand formally every step in the proof yet. My main book is Shafarevich's "Basic algebraic geometry", the book of Hartshorne is too comprehensive to me at this moment. Also all links to the used propositions, lemmas etc. will be very useful. Thanks.
Oct 31, 2011 at 15:05 history edited Jack Huizenga CC BY-SA 3.0
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Oct 31, 2011 at 13:24 comment added David Lehavi One can take L to be the sum of g+1 distinct Weierstrass points; in which case you have two nice generating sections: the g+1 chosen points, or their complimentary points.
Oct 31, 2011 at 11:21 comment added Jack Huizenga Just the dimension of $|L|$; I've removed this notation.
Oct 31, 2011 at 11:21 history edited Jack Huizenga CC BY-SA 3.0
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Oct 31, 2011 at 9:06 comment added rita What is $r(L)$?
Oct 31, 2011 at 0:38 history edited Jack Huizenga CC BY-SA 3.0
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Oct 31, 2011 at 0:26 history edited Jack Huizenga CC BY-SA 3.0
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Oct 30, 2011 at 23:26 history edited Jack Huizenga CC BY-SA 3.0
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Oct 30, 2011 at 23:08 history edited Jack Huizenga CC BY-SA 3.0
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Oct 30, 2011 at 22:56 history edited Jack Huizenga CC BY-SA 3.0
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Oct 30, 2011 at 22:38 history answered Jack Huizenga CC BY-SA 3.0