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What does it mean to say that an algebra is complete with respect to a filtration? Thanks!

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2 Answers 2

A filtration induces a topology on the algebra, by providing a basis of open neighbourhoods of 0. The algebra is then complete if the associated topological space is complete. Concretely, if the filtration is called ${\mathfrak{F}}^i$, indexed by natural numbers, where the subalgebras are decreasing as the index increases, say, then you can define a sequence $a_i$ in the algebra to converge if, after possibly replacing it by some translate, it satisfies that for any $i$ there exists an $N\in\mathbb{N}$ such that $a_n\in {\mathfrak{F}}^i$ for all $n\geq N$. You can define Cauchy sequences analogously and the algebra is complete if every Cauchy sequence converges.

Local rings are a prominent example, where the filtration is given by powers of the maximal ideal.

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Let $G$ be a filtered commutative group and

$$ \dots \subset F^{s+1} \subset F^s \subset \dots $$

its decreasing filtration. Quoting Boardman "(...) we topologize $G$ by taking the cosets $x + F^s$ of $F^s$ for all $x \in G$ and all $s$ as basic open sets (...) A Cauchy sequence $n \mapsto x_n$ is one in which $x_m − x_n \mapsto 0$ as $m, n \mapsto \infty$." And $G$ is said to be complete if every Cauchy sequence has a limit.

This condition can be written in more algebraic terms as follows: for each $s$, consider the exact sequence

$$ 0 \longrightarrow F^s \longrightarrow G \longrightarrow G/F^s \longrightarrow 0 $$

and apply $\varprojlim_s$ to it to obtain the long exact sequence

$$ 0 \longrightarrow \varprojlim_s F^s \longrightarrow G \longrightarrow \varprojlim_s G/F^s \longrightarrow \varprojlim_s{}^1F^s \longrightarrow 0 \ . $$

Then the filtration is complete if and only if $\varprojlim_s^1F^s = 0$.

Moreover: a filtration is called Hausdorff if so is the topological space $G$. In Hausdorff spaces convergent sequences have unique limits. This is easily seen to be equivalent to the fact that $\varprojlim_s F^s = \bigcap_s F^s = 0$.

So in $G$, with the topology induced by the filtration, every Cauchy sequence has a unique limit if and only if $G \cong \varprojlim_s G/F^s$. This last group is called the completion of $G$. (Beware: Weibel in his introduction to Homological Algebra uses the term with a different meaning; for him, "complete" filtration means complete and Hausdorff.)

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I was just about to recommend Boardmans paper. It is such a great intro to the above. (but maybe not for SS's, a great second look though) –  Sean Tilson Dec 9 '10 at 2:44

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