Timeline for Calculation of Cartier-Manin matrix
Current License: CC BY-SA 3.0
14 events
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| Sep 16, 2016 at 21:18 | comment | added | Alexey Milovanov | @FelipeVoloch so, for such class of plane curves (where we understand that singularities are nice) we can find in poly$(g, \log q, p)$ time Cartier-Manin matrix and hence we can find the number of rational points on its modulo $p$! This is interesting. | |
| Sep 16, 2016 at 20:43 | comment | added | Felipe Voloch | @Alexey Yes, I meant the singularities of $f=0$. If the singularities are nice (e.g. double points) then it's an easy linear algebra problem. If they are complicated, I don't know. | |
| Sep 16, 2016 at 20:08 | comment | added | Alexey Milovanov | @FelipeVoloch Do you mean the singularities of $f(x, y) =0$? Let us assume that we know. | |
| Sep 16, 2016 at 19:56 | comment | added | Felipe Voloch | @Alexey Do you know the singularities? | |
| Sep 16, 2016 at 16:40 | comment | added | Alexey Milovanov | @FelipeVoloch if our curve is plane (it is given as $f(x,y)=0$) can we chose a suitable basis for the regular differentials? | |
| Sep 16, 2016 at 16:38 | history | edited | Alexey Milovanov | CC BY-SA 3.0 |
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| Sep 16, 2016 at 9:25 | comment | added | Felipe Voloch | @Alexey I think the main problem is how is the curve given to you. If I have a plane model and a basis for the regular differentials in terms of this plane model ("adjoint curves") then the formula in my paper with Stöhr might have the running time you want (I haven't checked details, but it shouldn't be hard). But the purely geometric question of putting the curve in the right form and computing the adjoints is something I don't know much about, so I can't really say how easy or hard it is. | |
| Sep 16, 2016 at 8:18 | comment | added | Alexey Milovanov | @FelipeVoloch Could you answer a simpler question: if somebody will give us this matrix ( and a basis), could we verify that it is indeed the Cartier-Manin for this basis in time poly($p, \log q, g)$? | |
| Aug 23, 2016 at 11:29 | comment | added | Axel Stäbler | This is also partially addressed in J. Gonzalez, "Hasse-Witt Matrices For The Fermat Curves Of Prime Degree" Tohoku Math J, 1997 | |
| Aug 21, 2016 at 3:59 | comment | added | Felipe Voloch | @JoeSilverman Thanks! The paper in my comment gives a formula and a way to calculate the Cartier-Manin matrix (aka Hasse-Witt matrix) for any curve, given a (possibly singular) plane model. But I don't know if it can be turned into an algorithm with the same running time as the one cited in the question. | |
| Aug 21, 2016 at 0:08 | comment | added | Joe Silverman | @FelipeVoloch That looks more like an answer than a comment! | |
| Aug 20, 2016 at 23:16 | comment | added | Felipe Voloch | K.O. Stöhr, J. F. Voloch, A formula for the Cartier operator on plane algebraic curves. J. Reine Angew. Math. 377(1987), 49-64. | |
| Aug 20, 2016 at 21:57 | history | edited | Alexey Milovanov | CC BY-SA 3.0 |
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| Aug 20, 2016 at 21:47 | history | asked | Alexey Milovanov | CC BY-SA 3.0 |