# Why are smooth irreducible representations of the Weil group finite dimensional?

I am trying to understand the proof in Bushnell and Henniart - The local Langlands conjecture for $$\operatorname{GL}_2$$ of the fact (§28.6, Lemma 1) that smooth irreducible representations of the Weil group are finite dimensional. I have several doubts about the proof. The proof is:

Let $$(\rho,V)$$ be a smooth irreducible representation of the Weil group $$W_F$$ of a non Archimedean local field $$F$$. Let $$v\in V$$, $$v \neq 0$$. $$V$$ is irreducible, so the elements $$\rho(x)v$$ with $$x \in W_F$$ generate all $$V$$. Since $$V$$ is smooth, $$v$$ is fixed by an open subgroup of $$W_F$$ and so by an open subgroup of the inertia group $$I_F$$. These groups are of the form $$J=I_F \cap \Omega_E$$ where $$\Omega_E$$ is an algebraic closure of a finite extension $$E$$ of $$F$$. WLOG we can suppose $$E/F$$ Galois and so $$J$$ normal in $$I_F$$ and so in $$W_F$$. Since it is normal in $$W_F$$, then it is in the kernel of the representation. Now, let $$\phi$$ be a Frobenius element in $$W_F$$. It acts by conjugation on the finite group $$I_F/J$$, so some power of $$\phi^d$$ acts trivially. Thus $$\rho(\phi^d)$$ commutes with $$\rho(W_F)$$ and so, by Schur, $$\rho(\phi^d)$$ must be scalar. Hence, the vectorial space generated by $$\rho(x)$$ is finite dimensional and so is $$V$$.

My doubts are

1. Why being normal in $$W_F$$ implies being in the kernel?
2. I am not totally sure about what we are doing with the Frobenius element to say that $$\rho(\phi^d)$$ acts trivially and hence we can use Schur.
3. The conclusion: how is it enough?

1. The statement we are using is: Let $$H$$ be a subgroup of a group $$G$$ acting on a representation $$V$$. Assume $$V$$ is irreducible, $$H$$ fixes a nonzero vector $$v\in V$$, and $$H$$ is normal. Then $$H$$ lies in the kernel of $$V$$.
Proof: Since $$H$$ is normal, the set of vectors fixed by $$H$$ is a subrepresentation. By assumption this is nonzero and hence since $$V$$ is irreducible it is all of $$V$$.
1. Since the action of $$\phi^d$$ by conjugation on $$I_F/J$$ is zero and $$\rho: I_F \to \operatorname{Aut}(V)$$ factors through $$I_F/J$$, the action of $$\rho(\phi^d)$$ by conjugation on $$\rho(I_F)$$ is zero, i.e $$\rho(\phi^d)$$ commutes with $$\rho(I_F)$$. Since $$\rho(\phi^d)$$ commutes with $$\rho(\phi)$$, and $$\phi$$ and $$I_F$$ generate $$W_F$$ so that $$\rho(\phi)$$ and $$\rho(I_F)$$ generate $$\rho(W_F)$$, $$\rho(\phi^d)$$ commutes with $$\rho(W_F)$$.
2. The statement we are using is: Let $$H$$ be a finite-index subgroup of a group $$G$$ acting on a representation $$V$$. Assume $$V$$ is irreducible, $$H$$ has finite index, and every element of $$H$$ acts as scalars. Then $$V$$ is finite-dimensional.
Proof: Take any nonzero $$v\in V$$. Fix coset representatives $$r_i$$ for the cosets of $$G/H$$. Then for $$g\in G$$, $$gv$$ is a scalar multiple of $$r_i v$$ for some $$i$$. Hence the $$r_i v$$ span $$G v$$, but $$Gv$$ generates a nontrivial $$G$$-invariant subspace of $$V$$ and thus $$G v$$ generates $$V$$, so the $$r_i v$$ span $$G v$$.
• Thank you very much, I have only another one doubt. I understood the proposition that you are using in order to conclude, but it isn't clear to me who is the subgroup of $W_F$ of finite index you are referring to. We know that $\phi^d$ acts as a scalar, so all its power act as a scalar, but what after that? Commented Jul 17 at 15:43
• @Mario $\langle \phi^d, J \rangle$. Commented Jul 17 at 15:47