Background and definitions.

Let $k$ denote a field complete with respect to a non-trivial non-archimedean norm. Let $R$ be the integers in $k$, and say $\pi\in R$ with $0<|\pi|<1$ ($\pi$ doesn't have to be a uniformiser and the maximal ideal of $R$ doesn't even have to be principal, but I don't know the answer to my question even if $k={\mathbb{Q}_p}$ and $\pi=p$). Let $V$ be a $k$-vector space, which will be infinite-dimensional in practice. Let me say that a sub-$R$-module $L$ of $V$ is a lattice if it has the following two properties:

(i) $kL=V$ (i.e. $L$ contains a basis for $V$), and

(ii) $L$ contains no line, i.e. if $v\in V$ and $kv\subseteq L$ then $v=0$.

[I'm aware that other people might use the world "lattice" to mean something else.]

Given $V$ and a lattice $L$ in $V$, one can put a topology on $V$; a basis of open sets is $v+\pi^nL$ for $v\in V$ and $n\in\mathbf{Z}$. We can even complete $V$ with respect to $L$; the completion is the projective limit of $V/\pi^n L$. The completion $\widehat{V}$ is also a topological vector space and there's a natural continuous and injective map $V\to\widehat{V}$ with dense image (injectivity follows from (ii) ).

The question.

Notation as above. Say $V$ is a $k$-vector space and we choose two lattices $L_1$ and $L_2$ in $V$, with $L_1\subseteq L_2$. Let $V_1$ denote $V$ with the topology induced from $L_1$ and let $\widehat{V}_1$ denote its completion. Similarly we define $V_2$ and $\widehat{V}_2$.

Because $L_1\subseteq L_2$, $L_2$ is open with respect to the $L_1$ topology, and the map $V_1\to V_2$ (identity on the underlying sets) is bijective and continuous. On the other hand there's certainly no reason for the induced map on completions $\widehat{V}_1\to\widehat{V}_2$ to be bijective.

Is it always injective though?


Here's an example. If $V=k[T]$ the polynomial ring, and $L_1=R[T]$, $L_2=R[T/\pi]$ then the completion corresponds to restriction of analytic functions from a disc to a smaller disc, and this map is far from bijective. It is injective though -- an analytic function on a big disc is determined by its values on a small disc. More generally if the $L_i$ are free $R$-modules one can get quite a concrete handle on the completions. But I would imagine that in general a lattice can be quite pathological as an $R$-module and I don't have some good examples to hand in order to test out my question in these cases.


Let us take $V=Q_p[T]$ and $L_2 = \oplus_{i \geq 0} Z_p \cdot T^i$ and $L_1 = Z_p \cdot p \oplus (\oplus_{i \geq 0} Z_p (T^i + p T^{i+1}))$. The element $x = 1 \cdot p - p \cdot (1+pT) + p^2 \cdot (T+pT^2) - \cdots $ belongs to the $p$-adic completion of $L_1$ and is nonzero in it, but its image in the $p$-adic completion of $L_2$ is zero. This should give you an example where your map $\hat{V}_1 \to \hat{V}_2$ is not injective.

  • $\begingroup$ Aah I was too quick to reject the possibility that the L_i could be free! Thanks a lot Laurent. $\endgroup$ – eric Feb 27 '14 at 23:57

This is not a complete answer, but a conditional one. If you assume that $ \bigcap_{N\in\mathbb N} \pi^N\left(L_2/\pi^n L_1\right)=0, $ or, what amounts to the same: $\bigcap_N(\pi^NL_2+ L_1)=L_1$, then the map is injective. It suffices to show that the map $\hat L_1\to \hat L_2$ is injective, where $L_j=\lim_{\leftarrow}L_j/\pi^n$. So let $l=(l_n)$ be in the kernel of that map, i.e., $l_n\in\pi^nL_2$, where $l_n\in L_1$ is only determined up to $\pi^nL_1$. Further $l_{n}\equiv l_{n+1}\mod \pi^{n}L_1$, so that $$ l_{n}=l_{n+1}+\pi^{n}\tilde l_1=\pi^{n+1}\tilde l_2+\pi^{n}\tilde l_1 $$ for some $\tilde l_1\in L_1$ and $\tilde l_2\in L_2$. Now we can modify $l_{n}$ so that $\tilde l_1$ is zero and thus get $l_{n}\in\pi^{n+1} L_2$, i.e., we have increased the power of $\pi$ by one. We can iterate this step to see that $l_n$ lies in the image of $L_1\cap\pi^NL_2$ in $L_1/\pi^n L_1$ for every $N\in\mathbb N$. As an element of $L_2/\pi^nL_1$, the element $l_n$ lies in $\pi^N(L_2/\pi^nL_1)$ for every $N$, so by our assumption, it must be zero.

  • $\begingroup$ Indeed, in Berger's example we have $p^NL_2+L_1$ contains $p^NT^N$ and hence $p^{N-1}T^{N-1}$... and hence 1, which is not in $L_1$. $\endgroup$ – eric Feb 28 '14 at 0:01

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