A continuous representation $\hat{\mathbb{Z}} \rightarrow GL\_n(\mathbb{Q}\_p)$ is determined by the image of $1$. But the image of $1$ does not always defines such a representation (consider for example the representation which sends $1$ on $p$ from $\mathbb{Z}$ to $GL\_1(\mathbb{Q}\_p)$). So my question is : what are the conditions on the image of $1$ ?


For example if $n=1$, then I know that $1$ must be sent on an element of $\mathbb{Z}\_p^\times$, but I don't know if the converse is true.


EDIT: Correction about the example.



What are the $p$-adic representations of $\hat{\mathbb{Z}}$ ?
I am going to write a community wiki answer here which people can vote up.
(See [this meta thread](http://mathoverflow.tqft.net/discussion/780) concerning the Mathoverflow user,
which bumps questions with no voted-up answer.)


Main result: A homomorphism $f: \mathbb Z \to GL\_n(\mathbb Q\_p)$ extends continuously to
$\hat{\mathbb Z}$ if and only if the image of $f$ can be conjugated into $GL\_n(\mathbb Z\_p)$.


Proof:
If $f:\hat{\mathbb Z} \to GL\_n(\mathbb Q\_p)$ is continuous, the image is compact, hence contained 
in a maximal compact subgroup, which can be conjugated into $GL\_n(\mathbb Z\_p)$.


Conversely, if $f:\mathbb Z \to GL\_n(\mathbb Q\_p)$ lands in a compact subgroup,
then the closure of the image is compact, hence profinite (any compact subgroup of $GL\_n(\mathbb Q\_p)$ is profinite), and hence $f$ extends to $\hat{\mathbb Z}$
(since $\hat{\mathbb Z}$ is precisely the profinite completion of $\mathbb Z$).
QED


As noted in the comments, to tell if a matrix (e.g. $f(1)$) can be conjugated into
$GL\_n(\mathbb Z\_p)$, one simply has to look at the characteristic polynomial,
and ask that all the coefficients lie in $\mathbb Z\_p$, with the constant term being
a unit. Thus to apply the theorem in practice, one simply computes the characteristic polynomial of $f(1)$ and see if its satisfies these conditions.


EDIT: Now actually made community wiki; sorry about that --- I thought I had already clicked the CW box,
but obviously not. (The point is that the above argument is just a rephrasing of what is in the comments.)


