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Question : are the continuous characters of the form

  • $\eta : \mathbb{Z}_p^* \to \mathbb{Z}_p^*$, or
  • $\eta : (1+p\mathbb{Z}_p)^{\times} \to \mathbb{Z}_p^*$ (i.e., on the principal units in $\mathbb{Z}_p^*$)

well understood? Can such characters be classified in either case ?

I'm hoping to find an analytic classification ; i.e. to describe such characters as functions, or more precisely, how the functions $z\mapsto z^s$ for $s\in\mathcal{O}_{\mathbb{C}_p}$ 'sit' inside the set of characters $\eta : (1 + p\mathbb{Z}_p)^\times \to \mathbb{Z}_p^*$ (i.e., how 'far' is a generic character from some character of this type ?).

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Yes, these are very well-understood! Here's what they are. If $p$ is odd then $\mathbf{Z}_p^\times$ is a direct product of $\mu$, the subgroup of $p-1$th roots of unity, and $1+p\mathbf{Z}_p$, the principal units. A continuous character of the product is a product of continuous characters, so that reduces the first part to the second part. As for the second part, the principal units are topologically generated by $1+p$ so it suffices to say where $1+p$ should go. Note however that $1+p$ can't go to an arbitrary element of $\mathbf{Z}_p^\times$ because you need that if $(1+p)^{n_i}$ tends to 1 in $\mathbf{Z}_p$ then $s^{n_i}$ tends to 1 in $\mathbf{Z}_p$, where $s$ is the image of $1+p$. You can check that, for example, $s=-1$ does not have this property (because the $n_i$ can be even or odd and still tend to zero $p$-adically). But it's also not hard to check that $s$ has this property iff $s$ is a principal unit. I do this in Lemma 1 of my paper "On p-adic families of automorphic forms" here but this is most certainly standard and not due to me.

So in summary, for $p>2$, characters of the principal units biject with $1+p\mathbf{Z}_p$ non-canonically, the dictionary being "image of $1+p$", and characters of the full unit group biject with the product of this and the cyclic group of order $p-1$, that being the characters of $\mu_{p-1}$.

For $p=2$ the two questions are the same, and the same trick, appropriately modified, works. The group $1+4\mathbf{Z}_2$ is procyclic, generated by 5, and its characters biject with the principal units, the dictionary being "image of 5". For the full unit group the characters biject with the principal units product +-1, because $\mathbf{Z}_2^\times$ is just a product $\pm1\times (1+4\mathbf{Z}_2)$.

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  • $\begingroup$ Maybe it's a bit vague (and perhaps I should have said so to begin with -- I'll edit it in), but I'm hoping to be able to classify the characters analytically. It would be nice to really describe them as functions, or more precisely, how the functions $z\mapsto z^s$ for $s\in\mathcal{O}_{\mathbb{C}_p}$ 'sit' inside the set of characters $\eta : (1 + p\mathbb{Z}_p)^\times \to \mathbb{Z}_p^*$ (i.e., how 'far' is a generic character from some character of this type ?). $\endgroup$
    – xuros
    Commented Apr 27, 2010 at 11:37
  • $\begingroup$ Again this is easy. What did you try? Hint: if p>2 then log and exp give topological isomorphisms of topological groups 1+pZ_p = pZ_p. $\endgroup$
    – Kevin Buzzard
    Commented Apr 27, 2010 at 12:36
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    $\begingroup$ A warning is worth giving if you're going to use the ``$z^s$ for $s \in \mathcal{O}_{\mathbf{C}_p}$'' description. If $s \notin \mathbf{Z}_p$ then $\chi(z)=z^s$ takes values in $\mathcal{O}_{\mathbf{C}_p}^\times$, not necessarily in $\mathbf{Z}_p^\times$, so you might consider this larger class of characters. And for these more general characters $\chi$, then $\chi$ admits an "exponential" description ($\chi(z)=z^s$) if and only if $|\chi(1+p)-1|_p \leq p^{-1/(p-1)}$. This is related to the $p$-adic radius of convergence of the exponential power series $\exp(X)$ in the last comment. $\endgroup$
    – Jay
    Commented Apr 27, 2010 at 14:09

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