Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In Chriss and Ginzburg's Complex Geometry and Representation Theory, they define a lattice in the following setting: let $A$ be a ring (I guess I only care about the case that $A$ is a commutative algebra over a field) and $t \in A$ a central non-zero divisor such that $\cap_{i \geq 0} t^iA = 0$. If $M$ is a module over the localization $A_t$, a lattice is a finitely generated $A$-submodule $L \subset M$ such that $A_t \cdot L = M$.

What is the geometric motivation behind this concept? In particular, why is it called a lattice? I would be happy with an explanation for e.g. $A = \mathbb{C}[t]$, but I think the authors introduce the notion in order to apply it to more interesting examples such as $A = \mathbb{C}[x,y,t]/(xy-t)$.

share|improve this question
    
I don't know of the origin of this use. But this use of the word 'lattice' (and its French equivalent 'reseau') can be found in Bourbaki's Commutative Algebra, Ch. VII, Section 4. Maybe the historical references could be useful. –  Keerthi Madapusi Pera Jun 15 '12 at 14:56

2 Answers 2

The normal definition of a lattice can be stated as: A lattice in a $\mathbb R$-module $M$ is a finitely generated $\mathbb Z$-submodule $L \subset M$ such that $\mathbb R \cdot L = M$,

Now, $\mathbb R$ is not a localization of $\mathbb Z$, but I believe this is where the name comes from.

If this is true, then the term "lattice" is not from the geometric meaning of this concept, but is just a convenient name for an algebraic structure that happens to be useful for unrelated reasons.

share|improve this answer
1  
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. –  Ben Wieland Jun 15 '12 at 1:44
    
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. –  Fred Rohrer Jun 15 '12 at 6:27
    
@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 $t$. –  Will Sawin Jun 15 '12 at 6:43
    
In my comment above, "$M=\mathbb{R}$" should of course be "$M$ is a $\mathbb{Q}$-vector space of finite dimension". –  Fred Rohrer Jun 15 '12 at 14:07

Wikipedia tells us that a lattice is a discrete subgroup of $\mathbb{R}^n$ that spans the whole vector space. This use of the word seems to have its origin in the study of constructions made of laths (OED: narrow strips of wood) that, ignoring boundary behavior, have two directions of discrete translational symmetry.

The original use of lattice has been generalized in several directions. One that is relevant to us is the following: Given a locally compact topological group $G$, a lattice is a discrete subgroup $\Gamma$ such that $G/\Gamma$ has finite volume. In the particular case when $G$ is a vector space over a characteristic $p$ local field like $\mathbb{F}_q((t))$, lattices in this sense are (up to some finite dimensional alteration) a sum of copies of $\mathbb{F}_q[t^{-1}]$.

In the setting of such local fields, or more generally, Tate vector spaces, the word lattice is also used for a sort of dual object, namely a compact subgroup. The two notions are sometimes distinguished with a prefix: $k[t^{-1}]$ is a $d$-lattice (with d for discrete) in $k((t))$, while $k[[t]]$ a $c$-lattice (with c for compact). $d$-lattices and $c$-lattices have finite dimensional intersection, and their sum is finite co-dimension in the total space. See e.g., BBE.

At any rate, if you set $A = k[[t]]$, so $A_t = k((t))$, and $M$ is a finite-dimensional $A_t$-vector space, then a lattice in your sense is the same as a $c$-lattice in the Tate vector space sense. I imagine they were generalizing from this idea.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.