Why is every l-adic Galois representation conjugate to one over the l-adic integers? [closed]

Why is every l-adic Galois representation $$G_{\mathbb{Q}_p}\rightarrow GL_n(\mathbb{Q}_{l})$$ conjugate to one over the l-adic integers? $$G_{\mathbb{Q}_p}\rightarrow GL_n(\mathbb{Z}_{l})$$

closed as off-topic by Venkataramana, Mikhail Bondarko, Olivier, Joël, Ryan BudneyOct 12 '15 at 19:13

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• because (1) $G(\mathbb Q _p)$ is compact, (2) $GL_n(\mathbb Z _l)$ is maximal compact, and every compact subgroup of $GL_n(\mathbb Q _l)$ can be conjugated into $GL_n(\mathbb Z _l)$ – Venkataramana Oct 11 '15 at 9:09
• See math.u-psud.fr/~fontaine/galoisrep.pdf (in particular the discussion following Definition 1.7). – David Loeffler Oct 11 '15 at 9:39
• @JakobD.Hüwer: Indeed, that's very funny that they summon a lattice by giving its definition. Ad the existence of a lattice: let $V=\mathbf Q_\ell^n$ endowed with its $\ell$-adic norm and, for $x\in V$, set $\nu(x) = \sup_{g\in G} \| g\cdot x\|$. Then $\nu$ is a G-invariant norm on $V$ with values in $\ell^{\mathbf Z}$. A classical lemma (see Weil, Basic Number Theory) then implies that $V$ has a basis $(e_i)$ consisting of vectors of $\nu$-norm equal to $1$. It generates the sought-for lattice. – ACL Oct 11 '15 at 11:03
• @JakobD.Hüwer: That discussion shows precisely that for any homomorphism $\rho: G \to GL_n(\mathbf{Q}_\ell)$, i.e. any $\mathbf{Q}_\ell$-representation of $G$ in Fontaine and Ouyang's language, we can find a $G$-invariant $\mathbf{Z}_p$-lattice; a basis of that lattice then gives a homomorphism $G \to GL_n(\mathbf{Z}_\ell$ conjugate to $\rho$. Isn't that precisely your question? – David Loeffler Oct 11 '15 at 12:52
• @DavidLoeffler: Now I understand it - except for one thing: How do we know that G/H is finite? OK, we know that H is a normal subgroup and G is profinite, but this does not mean that H need to be of finite index. It might be trivial. – jdh Oct 11 '15 at 13:33

Recall that all ${\mathbb Z}_l$-lattices in the ${\mathbb Q}_l$-vector space ${\mathbb Q}_l^n$ are conjugate in ${\rm GL}_n ({\mathbb Q}_l )$. Since the ${\rm GL}_n ({\mathbb Q}_l )$ stabilizer of the standard lattice ${\mathbb Z}_l^n$ is ${\rm GL}_n ({\mathbb Z}_l )$, all lattices in ${\mathbb Q}_l^n$ have open stabilizer in ${\rm GL}_n ({\mathbb Q}_l )$. Moreover a finite sum of ${\mathbb Z}_l$-lattices is again an ${\mathbb Z}_l$-lattice.
Now take $\rho$ : $G_{{\mathbb Q}_p}\longrightarrow {\rm GL}_n ({\mathbb Q}_l )$ a continuous Galois representation. Let $L$ be any ${\mathbb Z}_l$-lattice in ${\mathbb Q}_l^n$. Then its stabilizer $H$ in $G_{{\mathbb Q}_p}$ is open and the coset set $G_{{\mathbb Q}_p}/H$ is finite, since $G_{{\mathbb Q}_p}$ is profinite. Now
$$M := \sum_{g\in G_{{\mathbb Q}_p}/H} g.L$$
is a Galois stable lattice in ${\mathbb Q}_l^n$. We have proved that a conjugate of $\rho (G_{{\mathbb Q}_p})$ lies in ${\rm GL}_n ({\mathbb Z}_l )$.