# Trying to understand the topology of the Weil group for the local Langlands conjecture

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 $$W_F=$$ where $$\sigma$$ is a lift of $$\Phi$$. But it is not clear to me the connection between the open of $$W_F$$ and $$I_F$$.

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 $$F \subseteq E \subseteq F^{\text{nr}}$$.