Timeline for Monodromy group of 1-dimensional families of hyperelliptic curves
Current License: CC BY-SA 3.0
10 events
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| Dec 23, 2012 at 5:24 | answer | added | JSE | timeline score: 3 | |
| Nov 24, 2012 at 17:30 | history | edited | François G. Dorais |
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| Nov 19, 2012 at 8:42 | answer | added | Jun Lu | timeline score: 2 | |
| Nov 19, 2012 at 6:39 | answer | added | Venkataramana | timeline score: 3 | |
| Nov 10, 2012 at 17:54 | comment | added | Noam D. Elkies | When $h_m = c_m t^{2m}$, all nonzero choices of $t$ yield isomorphic curves. When each $h_{2m+1}(t) = -h_{2m+2}(t)$, the right-hand side of the defining equation $y^2 = \prod_i (x-h_i(t))$ is a polynomial in $x^2$ so the curve has an additional involution (the hyperelliptic involution $(x,y) \leftrightarrow (x,-y)$ must commute with the monodromy as well, but this doesn't help because that involution multiplies all the holomorphic differentials by $-1$). | |
| Nov 10, 2012 at 17:51 | comment | added | Jack | Thanks. I did not understand the reason why monodromy group is trivial if each $h_{m}$ is of the form $c_{m}t^{2m}$. Also by the involution you probably mean the hyperelliptic involution $(x,y)→(x,−y)$? | |
| Nov 10, 2012 at 17:36 | comment | added | Noam D. Elkies | Not in general. The monodromy group could even be trivial (e.g. if each $h_m(t)$ is a constant function, or of the form $c_m t^{2m}$ for some constants $c_m$). Or if each $h_{2m+1}(t) = -h_{2m+2}(t)$ the monodromy must commute with the involution $(x,y) \leftrightarrow (-x,y)$. I don't know if there's an algorithm that can be guaranteed to compute the monodromy in all cases. | |
| Nov 10, 2012 at 11:45 | history | edited | Jack |
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| Nov 10, 2012 at 11:36 | history | edited | Jack | CC BY-SA 3.0 |
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| Nov 10, 2012 at 11:08 | history | asked | Jack | CC BY-SA 3.0 |