# What are the $p$-adic representations of $\hat{\mathbb{Z}}$ ?

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.

• I'm not entirely right either: instead of "stabilises a lattice" I should say it induces an isomorphism from a lattice to itself. Commented Nov 4, 2010 at 18:40
• If the image of 1 lands in some conjugate of GL(n,Z_p) then some finite power will land in the kernel of the map from this group to GL(n,Z/pZ) and then you're fine because this is pro-p and so there's a map from Z_p in. So the condition I suggest is sufficient. Commented Nov 4, 2010 at 18:47
• The condition I suggest is also necessary. Choose a random lattice in Q_p^n. Choose a continuous representation of Z-hat. Continuous image of compact is compact so the subgroup generated by the images of this lattice is a lattice and that's the one which is preserved by 1 and -1. Ok so I think we're done. Commented Nov 4, 2010 at 18:48
• Sorry, forgot to add the condition that the constant term must be invertible. Commented Nov 4, 2010 at 19:04
• @Kevin: yes I think so! Just to recap: the continuous image of a compact set is compact so the image has to land in a compact subgroup of $GL_n(Q_p)$ which we know is conjugate to a subgroup of $GL_n(Z_p)$. Conversely if $M = card(GL_n(F_p))$ and $f(1)$ is (say) in $GL_n(Z_p)$ then for every $k$, the image of $p^{k-1}M Z$ is in $1+p^k M_n(Z_p)$ so the map from $Z$ extends by uniform continuity. To relate this to Torsten's answer, it remains to check that a matrix is conjugate to an element of $GL_n(Z_p)$ iff its char poly has integral coeffts. Commented Nov 4, 2010 at 19:24

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.