Timeline for How to compute the Betti numbers of S-D for a surface S and a divisor D?
Current License: CC BY-SA 3.0
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| Feb 17, 2014 at 13:06 | history | edited | user9072 |
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| Mar 19, 2012 at 10:07 | comment | added | Francesco Polizzi | What it is true is that $\sum (-1)^i b_i(S-D)=\sum (-1)^i b_i(S) - \sum (-1)^i b_i(D)$. In fact, the topological Euler-Poincaré characteristic is additive, as you can see by taking a triangulation of $S$ that contains a triangulation of $D$ as a sub-triangulation (racall that any smooth, complex projective variety can be triangulated). | |
| Mar 19, 2012 at 9:24 | comment | added | rose | Using the long exact sequence it is easy to represent the Euler number of S-D by the Euler numbers of S and D, but we still cannot represent the Betti numbers of S-D by the Betti numbers of S and D, right? At least we cannot find a universal formula... | |
| Mar 19, 2012 at 9:07 | comment | added | Dan Petersen | This is clearly false, think about $i=0$... In general you would study this by means of the long exact sequence for relative cohomology and the isomorphism $H^i(S;D) = H^{4-i}(S-D)^\vee$. | |
| Mar 19, 2012 at 8:58 | history | asked | rose | CC BY-SA 3.0 |