bio | website | math.dartmouth.edu/~jvoight |
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location | Dartmouth College | |
age | ||
visits | member for | 4 years, 11 months |
seen | Jan 9 at 1:44 | |
stats | profile views | 534 |
Oct 10 |
comment |
Factorisation of local quaternionic zeta functions
What is $q$? It better be $q^n=\mathrm{nrd}(I)$: the reduced norm is not always a square. |
Oct 9 |
awarded | Yearling |
Oct 9 |
answered | Factorisation of local quaternionic zeta functions |
Sep 10 |
awarded | Popular Question |
Jul 2 |
awarded | Curious |
Jun 4 |
awarded | Nice Question |
Apr 9 |
accepted | Infinite dimensional simple algebras of finite degree |
Apr 7 |
asked | Infinite dimensional simple algebras of finite degree |
Mar 12 |
comment |
Totally real points on curves
Cool, thank you very much! |
Mar 11 |
asked | Totally real points on curves |
Mar 11 |
answered | Explicit isomorphism for quaternion algebras over $\mathbb{Q}$? |
Mar 10 |
awarded | Yearling |
Mar 9 |
awarded | Revival |
Mar 9 |
answered | An application of Strong Approximation |
Mar 7 |
answered | Integral elements of quaternion algebras with predescribed properties |
Dec 10 |
comment |
Finiteness conditions and Veronese subrings
The hypothesis that $R$ is a domain is crucial. Take $R=k[x_2,x_3,x_5,x_7,\dots]/I$ where $I$ contains every binomial $x_ix_j-x_2^{(i+j)/2}$ for $i,j \geq 3$ odd. Then $G=\mathbb{Z}/2\mathbb{Z}$ acts $R$ by $x_j \mapsto (-1)^j x_j$; the invariants are $R^G \cong k[x_2]$, but $R$ is not finitely generated (has generators in every odd degree $\geq 3$). $R$ has plenty of zerodivisors, e.g., $x_5(x_5-x_2x_3)=0$. |
Nov 28 |
awarded | Commentator |
Nov 28 |
comment |
Is there an algorithm to compute efficiently the dessin d'enfant from a Belyi pair?
If instead you are given the finite index subgroup (or equivalently, the permutation triple), then it is much easier to write down the combinatorial-topological data given by the dessin as a graph: it is explained in Chapter 4 of Girondo, Gonzalez-Diez "Introduction to Compact Riemann Surfaces and Dessins d'Enfants", and presumably elsewhere: you just read off the dessin from the monodromy. In fact, you can do this in a conformally correct way on the desired surface: see another one of my preprints (arxiv.org/abs/1311.2081), where there are lots of pictures. |
Nov 28 |
comment |
Is there an algorithm to compute efficiently the dessin d'enfant from a Belyi pair?
Hopefully our preprint (arxiv.org/abs/1311.2529) addresses your question. If you are given equations, you "just" need to do some numerical homotopy: choose a base point and trace the preimages of loops on the curve around 0, 1, oo. This has been implemented by Kroeker (github.com/jakobkroeker/HMAC) and Bartholdi (github.com/laurentbartholdi/img). I don't think Bertini or PHCPack will give this to you directly, though they are similar in spirit. |
Nov 27 |
awarded | Nice Answer |