Topology on Completion

Question

If G is a topological group the completion is defined as the space of all cauchy sequences and defining an equivalence relation on the space as mentioned in the Atiyah-Macdonald Commutative Algebra book. The topology is not defined there . I mean is there any canonical way to topologize the completion ? P.s in case of rationals , the completion is reals . And the topology in rationals are given by ordered topology which canonically defines a topology on reals . What about the general case ?

Answer 1

I'm sorry, I thought you didn't know how to put a metric on the completion.

You take $U$ a subset of the Cauchy sequences and set $$U\text{ open}\iff \forall x\in U,\forall (x_n)_n\text{ Cauchy sequence which represents }x,\ \exists N\in \mathbb{N},\ \forall n\geq N, \phi(x_n)\in U$$ where $\phi(x_n)$ is the constant sequence of value $x_n$, like defined in Atiyah-MacDonald.

When your base space is metric, this topology coïncides with the metric topology of your completion as I described it in my first answer.

It's not a very useful definition, as he then goes to define particular topologies such as that defined by a sequence of subgroups: $$G=G_0\supseteq G_1\supseteq\ldots\supseteq G_n\supseteq\ldots$$ you can define a pseudo-metric $d$ as follows:

first set $v(\alpha)=\max \{n\in \mathbb{N}\ |\ \alpha\in G_n\}$, a pseudo-valuation and then $$d(x,y)=2^{-v(x-y)}$$

It defines the same topology as the one described in the book. It's a metric iff the topology is Hausdorff i.e.: $$\bigcap_{n\in \mathbb{N}} G_n=\{0\}$$

You can then prove the same way as I did in my answer above that you can construct a completion which is pseudo-metric. Of course you can't actually have an embedding, unless it is Hausdorff.

In classical examples you get $G_n=p^n\mathbb Z$, with $p$ prime, or $G_n=X^n k[X]$ and the valuations are $v_p(n)=\max \{ k\ |\ p^k|n\}$ and $v_0(P(X))=\max \{ k\ |\ X^k|P(X)\}$.

In a general way you define the $I$-adic topology on a $A$-module $M$ with $I$ n ideal of $A$, by setting $I^nM$ to be a fundemantal base of neighbourhoods of $0$. Then you gaet neighbourhoods for other points by translation.

When Atiyah-MacDonald do the Artin-Rees lemma this has an interpretation in terms of this topology:

You take $M'$ to be a submodule of $M$. Then you put the $I$-adic topology on them. However, since $M'\subset M$ you can put the induced topology on $M'$ aswell. The Artin-Rees lemma gives you a condition for these two topologies to be equivalent.

Answer 2

If you have a general metric space $(X,d)$ you can actually create a complete space $(E,\delta)$ such that $X$ can be embedded isometrically in it and such that its image is dense in $E$.

You construct $E$ in this way:

Step 1: Set $$\mathcal{E}=\{x=(x_n)_n\ |\ x\text{ is a Cauchy sequence}\}$$ and you quotient this by the following equivalence relation $\sim$: $$(x_n)\sim (y_n)\iff x_n-y_n \rightarrow 0$$

so you get $$E=\mathcal{E}/\sim$$

Step 2: put a metric on it. For two elemnts $x,y$ in $E$, with representatives, two sequences $(x_n),(y_n)$ in $\mathcal{E}$ you set $\delta_n=d(x_n,y_n)$, then you prove that $(d_n)$ is Cauchy in $\mathbb R$. You then set $\delta(x,y)=\lim \delta_n$. You need to prove that this limit is independent of the chosen representatives.

Step 3: Prove that $X$ can be embedded with the followng map $$\begin{aligned} i:\ X & \longrightarrow E\\ x & \longmapsto (x_n)_n mod \sim\text{ where }x_n=x,\ \forall n \end{aligned}$$ and that it is ismoetric: $\delta(i(x),i(y))=d(x,y),\ \forall x,y\in X$

Step 4: Prove that $X$ is dense in $E$. This is easy. Set $x\in E$ and a representative $(x_n)$ in $\mathcal{E}$ and then the sequence $(i(x_n))_n$ converges to $x$ in $(E,\delta)$.

Step 5: Prove that $E$ is complete. Set $(\alpha_n)_n$ a Cauchy sequence in $E$ and, by density, set $x_n\in E$ such that $\delta(\alpha_n,i(x_n))<\dfrac{1}{n}$. Then you prove that $(x_n)$ is Cauchy in $X$ and thus you have $\alpha$ it's class mod $\sim$ in $E$ which is the limit of $(\alpha_n)_n$.

Remark 1: You can prove up to isometric homeomorphism that this space is unique: it actually verifies a universal property (I henceforth suppose that $X$ is a subset of $E$, hence $i$ being the canonical inclusion):

For every complete space $(F,\delta')$ and every uniform continuous map $$f:\ X\longrightarrow F$$ ther exists a unique map $$g:\ E\longrightarrow F$$ which is uniformly continuous and $g\big|_X=f$. You can use this to prove that the completion space $E$ is unique up to isomorphism. In fact you can go further and prove it is unique up to isometric isomorphism. This is classic metric topology.

Remark 2: You needed to see that $\mathbb R$ was complete in order to define $\delta$, but $\mathbb R$ is constructed in a similar way from $\mathbb Q$ (with its absolute value, inherited by the natural order on it)which is a problem: how do you define $\mathbb R$? Convergence? Cauchy sequences?

One way yu can do is consider only rational $\varepsilon$-s in you definitions. In a similar way you get a complete space, and that inherits a ring structure from $\mathbb Q$: for example you take two reals $x,y$ with representatives $(x_n),(y_n)$ and set $x+y$ to be the class of $(x_n+y_n)_n$. You also get an order: set $x,x_n,y,_n$ as in the last sentence. Then you set $$\begin{aligned} (x_n)_n > (y_n)_n\iff \exists \varepsilon \in \mathbb{Q}_{>0},\ \exists N\in \mathbb{N}, \text{ such that}\\ \forall m,n\geq N,\ x_m\geq \varepsilon +y_n \end{aligned}$$ This order relation is consistent with $\sim$ and thus you have a total order in $\mathbb R$.

Remark 3: this proceedure can be extended to metric topological groups, and you still obtain a topological group.

Remark 4: lastly, there is actually a much faster way to do this. You set $\mathcal{B}(X,\mathbb{R})$ the set of bounded functions on $X$. You put the norm: $$\| f\|_{\infty}=\sup_{x\in X} |f(x)|$$ You prove it to be a Banach space, then you embed $X$ in the following way: you fix $a\in X$ arbitrarly and let $$\begin{aligned} \varphi:\ X & \rightarrow \mathcal{B}(X,\mathbb{R})\\ x & \longmapsto \{\varphi_x\ :\ y\mapsto d(y,a)-d(y,x) \end{aligned}$$ You prove that it is an isometric embedding and then set $E$ to be the closure of $\varphi(X)$ in $\mathcal{B}(X,\mathbb{R})$. It is complete since it's a closed subspace of a complete space, and you have $\varphi(X)$ dense in it by definition.