Timeline for Lattices in algebraic geometry

5 events

when toggle format	what		by	license	comment
Jun 15, 2012 at 14:07	comment	added	Fred Rohrer		In my comment above, "$M=\mathbb{R}$" should of course be "$M$ is a $\mathbb{Q}$-vector space of finite dimension".
Jun 15, 2012 at 6:43	comment	added	Will Sawin		@Fred: My normal definition was intended to, when in doubt, hew to the geometric picture of a lattice in space (space being usually taken to be $\mathbb R^n$, not $\mathbb Q^n$) rather than the algebraic definition, to make the intuition as clear as possible. $\mathbb R$ is indeed not quite $\mathbb Q$, but it is "like" $\mathbb Q$, being, for instance, its metric completion. Also note that, by the definition interpreted literally, $\mathbb Q$ doesn't work since only finitely many primes can divide $t$.
Jun 15, 2012 at 6:27	comment	added	Fred Rohrer		Dear @Will, if in your "normal definition" you replace the reals by the rationals (a ring of fractions!) then it has a closer relation with the usual definition (cf. Bourbaki AC.VII.4.1), i.e., in case $M=\mathbb{R}$ it is a special case thereof.
Jun 15, 2012 at 1:44	comment	added	Ben Wieland		A key example is $k((t))$, which is locally compact if $k$ is finite. This contains $k(t^{-1})$, which is discrete and cocompact, just like $\mathbb Z\subset \mathbb R$. But it also contains the dual object $k[[t]]$, which is compact and co-discrete. This second example fits the definition in the book.
Jun 14, 2012 at 23:57	history	answered	Will Sawin	CC BY-SA 3.0