I am trying to study the representation of the Weil group from the book "The Local Langlands Conjecture for $GL(2)$". I have some problem with the topology of this group.

Let $F$ be a non archimidian local field, so a finite extension of $\mathbb{Q}_p $ for a prime $p$. If $\overline{F}$ is an algebraic closure of $F$ and $F_{\infty}$ is the maximal unramified extension of $F$ in $\overline{F}$, we have an exact sequence $$1 \to \mathcal{Gal}(\overline{F}/F_{\infty})\to \mathcal{Gal}(\overline{F}/F)\to \mathcal{Gal}(F_{\infty}/F) \to 1.$$

Since ${\mathcal{Gal}(\overline{F}/F})\cong \lim \mathbb{Z}/m\mathbb{Z}$ and in $\mathbb{Z}/m\mathbb{Z}$ we have the Frobenius element, we have an element $\Phi_F=(\Phi_m)_m\in \lim \mathbb{Z}/m\mathbb{Z}$, called geometric profenious substitution where $\Phi_m$ is the inverse of the Frobenius element in $\mathbb{Z}/m\mathbb{Z}$. We define the Weil group as the topological group that satisfies the exact sequence $$1 \to I_F \to W_F \to <\Phi_F> \to 1$$ with $I_F=\mathcal{Gal}(\overline{F}/F_{\infty})$. The topology of $W_F$ is the topology in which $I_F$ is open, $I_F$ in $W_F$ has the same topology of $I_F$ in $\mathcal{Gal}(\overline{F}/F)$ and the map $\iota:W_F \to \mathcal{Gal}(\overline{F}/F)$ is continuous.

I am trying to understand who are the open subgroup of $W_F$. Of course, since $\iota:W_F \to \mathcal{Gal}(\overline{F}/F)$ is a continuous map, all the open subgroups of $\mathcal{Gal}(\overline{F}/F)$ are open subgroups of $W_F$. So all the groups of the form $W_F \cap \mathcal{Gal}(\overline{F}/E)$ with $E/F$ finite extension are open subgroups of $W_F$. In the book there is a (not proved) proposition that classify the open subgroup of finite index of $W_F$: they are of the form $W_F \cap \mathcal{Gal}(\overline{E}/E)$ for some finite extension $E$ of $F$, so the ones that I declared before.

Of course we have other open subgroup, infact $I_F$ has to be open (it is also closed) since $W_F/I_F$ is discrete, but in $\mathcal{Gal}(\overline{F}/F)$ it is closed but not open.

I would a better classification of the open subgroup of $W_F$ and their connection with $I_F$. For example, in the book that I mentioned in the first part of my question, when it takes a smooth representation $(V,\pi)$ of $W_F$, it assume that the open subgroup that fix an element $v \in V$ is of the form $I_F \cap \mathcal{Gal}(\overline{E}/E)$ for some finite extension $E/F$, and not only an open compact subgroup of $W_F$.

I am sure that there is a deep connection with the fact that $$W_F=<I_F,\sigma>$$ where $\sigma$ is a lift of $\Phi$. But it is not clear to me the connection between the open of $W_F$ and $I_F$.


1 Answer 1


Note that in a topological group $G$, any subgroup $H$ containing an open subgroup $U$ is itself open: we can write $H$ as a disjoint union of $hU$ for a set of coset representatives for $H/U$, and each $hU$ is open. Thus a subgroup $H \subseteq W_F$ is open if and only if $H \cap I_F$ is open in $I_F$.

If it helps, you can choose a lift $\sigma \in W_F$ of $\Phi$ to write $W_F$ as a semidirect product $I_F \rtimes \mathbf Z$. Under this identification, $W_F$ has the product topology on the underlying set $I_F \times \mathbf Z$ of $I_F \rtimes \mathbf Z$. Indeed, $W_F$ has a neighbourhood basis given by opens $Uw$ for $w \in W_F$. Writing $w$ as $(i,f)$ for $i \in I_F$ (inertia part) and $f \in \mathbf Z$ (power of Frobenius), we get $$Uw = (U,1)(i,f) = (Ui,f)$$ (even with the semidirect product $(n_1,h_1)(n_2,h_2) = (n_1\ {^{h_1}n_2},h_1h_2)$, which is why I'm using right cosets instead of left cosets). These form a basis for the product topology on $I_F \times \mathbf Z$ since the groups $Ui$ for $i \in I_F$ and $U \subseteq I_F$ an open subgroup form a basis for the topology on $I_F$.

A third way to see what open subgroups $U \subseteq W_F$ look like is by looking at their image in $\mathbf Z$. This will be a subgroup of $\mathbf Z$, hence either $0$, in which case $U \subseteq I_F$, or a finite index subgroup, in which case $U$ has finite index in $W_F$, hence comes from an open subgroup in $\operatorname{Gal}(\bar F/F)$.

Finally, note that every open subgroup $U$ of $I_F = \operatorname{Gal}(\bar F/F^\text{nr})$ comes from an open subgroup of $\operatorname{Gal}(\bar E/E)$ for some finite unramified extension $F \subseteq E \subseteq F^\text{nr}$. Indeed, $U$ corresponds to a finite field extension $F^{\text{nr}} \to L$, and a generator for $L$ over $F^{\text{nr}}$ can already be defined over a finite subextension $F \subseteq E \subseteq F^{\text{nr}}$.


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