In Bushnell and Henniart, The Local Langlands conjecture for GL(2), there is a proposition on p. 184 in which they prove the following:

Let $F$ be a non-archimedean local field, $\mathcal W_F$ its Weil group, and $\tau$ an irreducible smooth representation of $\mathcal W_F$. Then, if $\tau(\mathcal W_F)$ is finite, it can be extended to an irreducible smooth representation of $\Omega_F = Gal(\overline F/F)$.

In the proof they fix a Frobenius element $\Phi\in\mathcal W_F$ and write any element $\omega\in \Omega_F$ as $\omega = \Phi^a\sigma$ (for some $a\in \hat{\mathbb Z}$ and $\sigma\in I_F$). My questions is: which element of $\Omega_F$ is meant by $\Phi^a$? Of course, there is an exact sequence $1\to \mathcal I_F\to \Omega_F\to Gal(F^{\text{unr}}/F) = \hat{\mathbb Z}\to 1$ but how is $\Phi$ used to cook up an element of $\Omega_F$ in the fiber of an element $a\in \hat{\mathbb Z}$?

Thanks in advance!

  • $\begingroup$ I think they secretly take a splitting $\hat{\mathbb{Z}}\to\Omega_F$, and $\Phi_a$ is the image of $a\in\hat{\mathbb{Z}}$ under that splitting. I also think this is what Keenan Kidwell explained in more detail. $\endgroup$
    – GH from MO
    May 11, 2015 at 19:23
  • 2
    $\begingroup$ The choice of $\Phi$ is equivalent to the choice of a splitting, since $\Omega_F$ is profinite and $\widehat{\mathbf{Z}}$ is the ``free profinite group on one element." In my answer I'm making the resulting splitting homomorphism (semi-)explicit. $\endgroup$ May 11, 2015 at 19:43

1 Answer 1


If $G$ is any profinite group, and $a\in\widehat{\mathbf{Z}}$, then for any sequence $(a_n)$ of integers converging in $\widehat{\mathbf{Z}}$ to $a$, the sequence $(g^{a_n})$ converges in $g$ to an element $g^a$ which is independent of the chosen sequence. Thus one can take $\widehat{\mathbf{Z}}$-exponents in a profinite group, and the usual rules apply, e.g., $g^{a+b}=g^ag^b$, $(g^a)^b=g^{ab}$. This is what is meant by the notation $\Phi^a$. It is defined via a limit in the absolute Galois group (of course if $a\in\mathbf{Z}$ it's the usual $a$-th power of $\Phi$).


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