Timeline for Problem in Rick Miranda: finding genus of a Projective curve
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Dec 19, 2016 at 2:04 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
edited body
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| Feb 28, 2011 at 16:52 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
deleted 6 characters in body
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| Feb 16, 2011 at 23:33 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
added 26 characters in body
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| Feb 14, 2011 at 17:52 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
added 2 characters in body
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| Feb 14, 2011 at 16:53 | comment | added | Sándor Kovács | Roy, you can certainly add a URL, or just tell people to go to your webpage. Cheers. :) | |
| Feb 14, 2011 at 16:27 | comment | added | roy smith | Jack, when I taught the course from this book, I introduced Milnor numbers of isolated singularities, without full proofs, to help us compute the change in the topological genus of curves under degeneration. I have written some of this stuff up in the course notes. Is there a way to attach a pdf file to comments or answers in this thread? | |
| Feb 14, 2011 at 14:49 | comment | added | roy smith | This nice answer illustrates also how one can visualize the genus (arithmetic if you like) of a curve of type (a,b) on a smooth quadric, namely as homologous to the union of a lines from one ruling and b from the other, hence a rectangle with g = (a-1).(b-1) holes. | |
| Feb 14, 2011 at 1:46 | comment | added | Sándor Kovács | @J.C.: Thanks for the compliment and the typo catching. I think I got all the renegades. Did I? | |
| Feb 14, 2011 at 1:44 | comment | added | roy smith | Jack, \ good question, and I had a similar one while teaching this, but maybe the relation between euler characteristic and genus should differ in the reducible case? maybe subtract one less than the number of components? i.e. 4-3 = 1? | |
| Feb 14, 2011 at 1:42 | comment | added | Sándor Kovács | @Jack: Your comment made me realize that there is a better way to present this solution. I edited my answer accordingly. Thanks for the contribution! | |
| Feb 14, 2011 at 1:41 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
Modified the presentation to avoid confusing argument.; added 2 characters in body; edited body; edited body
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| Feb 14, 2011 at 1:38 | comment | added | Jack Huizenga | Ah thank you, that picture makes more sense now. | |
| Feb 14, 2011 at 1:34 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
removed confusing statement about triangulations
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| Feb 14, 2011 at 1:31 | comment | added | Sándor Kovács | @Jack: I guess the comment "If in doubt, triangulate it" was a little too cavalier and perhaps more than a little confusing. I will try to write up an explanation of that later, but I have to run now. (But I will edit the answer before I do). | |
| Feb 14, 2011 at 1:26 | comment | added | Sándor Kovács | @J.C.: I suppose the reason Miranda chose $x_0x_3=2x_1x_2$ is that he wanted a smooth intersection. | |
| Feb 14, 2011 at 1:15 | comment | added | J.C. Ottem | Great answer! Makes me wonder why the question chose $x_0x_3=2x_1x_2$ instead of $x_0x_3=x_1x_2$. Anyways, there are some typos above (some $x_1$s should be $x_3$s). | |
| Feb 14, 2011 at 1:11 | comment | added | Jack Huizenga | I am sorry if I came off as overly critical. I was in particular remarking on his comment that "the topological genus is clearly 1. If in doubt, triangulate it." Triangulating the space, I get an Euler characteristic of 4, which does not suggest "genus 1" to me. What is the sense in which I should interpret this? Clearly the arithmetic genus solves all the problems, but then there's the work of relating arithmetic genus to topological genus. | |
| Feb 14, 2011 at 1:05 | comment | added | Emerton | Dear Jack, Sandor's answer is (of course) correct when understood in the appropriate sense. Rather than simply criticizing it, why don't you work on understanding what that sense is? Regards, Matthew | |
| Feb 14, 2011 at 0:47 | comment | added | Jack Huizenga | As a plane conic degenerates to a pair of intersecting lines, isn't the picture that a sphere degenerates to a wedge of two spheres? The former space has Euler characteristic 2, while the latter has Euler characteristic 3. Certainly the continuity argument works if fibers are smooth, but I don't buy that the topological genus plays nice with singular degenerations--only the arithmetic genus does. | |
| Feb 13, 2011 at 23:56 | history | answered | Sándor Kovács | CC BY-SA 2.5 |