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comment Segments of Voronoi Diagrams on smooth manifolds. Are they geodesics?
Nevermind: on a small enough neighborhood around each point of the bisector, the geodesics from the point cover all possible tangent directions. In particular, if the bisector contains all geodesics between points, then it is totally geodesic.
Aug
1
comment Segments of Voronoi Diagrams on smooth manifolds. Are they geodesics?
This paper only says that bisectors in $M$ are totally geodesic if and only if $M$ has constant curvature. This makes no difference for the original poster's question which was one-dimensional, but isn't it true that a bisector $B$ can be geodesic (the geodesic between any two distinct points in $B$ is contained in $B$) without being totally geodesic? I'm thinking, for example, of products.
Jul
29
awarded  Revival
Jul
29
answered Does anyone have an electronic copy of Waldspurger's “Sur les coef´Čücients de Fourier des formes modulaires de poids demi-entier”?
Jul
12
comment Generators of the graded ring of modular forms
Well, the result over $\mathbb{Z}[1/6N]$ implies the result over $\mathbb{Q}$ and therefore any field of characteristic $0$, so in particular it answers the original question (for the subring in even weight). More generally, one can apply flat base change; but I don't know how far we want to get into this in these comments.
Jul
12
comment Generators of the graded ring of modular forms
You may also want to refer to DZB's Proposition 11.3.1, since extends Theorem 9.3.1 to more general base rings. JV
Jul
12
comment Generators of the graded ring of modular forms
What are "minimal relations"? I think you just mean "relations", since you say "at most 12" anyway. JV
Apr
28
awarded  Nice Question
Apr
9
comment Noncommutative group of invertible ideals of a ring
@FernandoMuro The base ring $R$ is assumed to be commutative: it is a domain. So the field of fractions means what it usually does. (There are notions of rings of quotients for noncommutative rings, but they don't play a role here.)
Apr
9
comment Noncommutative group of invertible ideals of a ring
@Aurel: conventions differ, but you might ask only that $\mathcal{O}$ is contained in the left and right orders; if $I$ is invertible, then equality holds, so this doesn't matter for the question. (@FernandoMuro: It is also equivalent to take a fractional ideal $I$ to be a $\mathcal{O}$-sub-bimodule of $K \otimes_R \mathcal{O}$ of the form $I=cJ$ where $c \in K^\times$ and $J \subseteq \mathcal{O}$ is a two-sided ideal. Hence the ``fractional''.)
Apr
8
revised Noncommutative group of invertible ideals of a ring
deleted 90 characters in body
Apr
8
comment Noncommutative group of invertible ideals of a ring
Thanks Pace! (Something happened in the copy and paste, ugh.) The definition of invertible is what it must be to make this into a group: a fractional ideal $I$ is invertible if there exists a fractional ideal $J$ such that $JI=IJ=\mathcal{O}$. JV
Apr
8
revised Noncommutative group of invertible ideals of a ring
deleted 48 characters in body
Apr
8
asked Noncommutative group of invertible ideals of a ring
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