Timeline for Curves on K3 and modular forms
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 13, 2014 at 14:59 | history | edited | Jim Bryan | CC BY-SA 3.0 |
added 46 characters in body
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| Jun 13, 2014 at 14:56 | comment | added | Jim Bryan | @Qiaochu Yuan: the only other case I've thought about along these lines is covers of $\mathbb{RP}^2$ which is amusing: since the only connected covers are the degree 1 cover and the universal cover in degree two, $F=q+\frac{q^2}{2}$ and so we learn that $\exp (q+\frac{q^2}{2})$ is the generating series which counts order 2 elements in the symmetric group. (It is a fun exercise to derive this combinatorially). | |
| Jun 11, 2014 at 7:54 | vote | accept | IBazhov | ||
| Jun 11, 2014 at 5:02 | comment | added | Qiaochu Yuan | Interesting! I think it's worth mentioning how this argument changes if $\mathbb{Z} \oplus \mathbb{Z}$ is replaced by an arbitrary finitely generated group $G$. In this case the subgroup counting gets more complicated: one must take a sum over index-$d$ subgroups $H$ of the coefficients $\frac{1}{|\text{Aut}(G/H)|}$. If $H$ is normal, and in particular if $G$ is abelian, then this automorphism group is isomorphic to $G/H$ and hence the coefficient is $\frac{1}{d}$, but in general this automorphism group is isomorphic to $N_G(H)/H$ which is smaller. | |
| Jun 10, 2014 at 23:01 | history | edited | Jim Bryan | CC BY-SA 3.0 |
edited body
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| Jun 10, 2014 at 18:38 | comment | added | Jim Bryan | This answer was pretty long, so I put this as a comment. The above has nothing to do with the K3. The appearance of modular forms in the generating function for the number of curves on K3 is still somewhat mysterious, although there reasons from physics to expect that the generating function for the number of BPS states has modular properties. | |
| Jun 10, 2014 at 18:37 | history | answered | Jim Bryan | CC BY-SA 3.0 |