# Image of continuous Galois representation constructed by Deligne corresponding to a eigenform.

It is a well known theorem that given a eigneform $$f$$ of weight $$k$$ and level $$N$$, Deligne, Serre, Shimura and others have constructed continuous $$\ell$$-adic Galois representation $$\rho_f : \mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\to \mathrm{GL}_2(E)$$ such that the trace and determinant of the image of Frobenius at prime $$p$$ different from $$\ell$$ and not dividing $$N$$ satisfies nice properties which connects it to the Fourier coefficients of $$f$$. But I have seen far too many times that an expert while giving a talk says that we can take the representation $$\rho_f : \mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q}) \to \mathrm{GL}_2(\mathcal{O}_E)$$. What is the reason of this switch from $$\mathrm{GL}_2(E)$$ to $$\mathrm{GL}_2(\mathcal{O}_E)$$ ? Does it have something to do with the compactness of $$\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$$ ?

Thanks for the help !!

• What is your $E$? A p-adic local field? It's a "fun fact" that for a continuous finite dimensional Galois representation $\rho : \operatorname{Gal}(\overline{\Bbb Q}/\Bbb Q) \to \operatorname{GL}_2(E)$ there is a $\operatorname{Gal}(\overline{\Bbb Q}/\Bbb Q)$-equivariant $\mathcal{O}_E$-lattice in your $E$-vector space (so you get another representation $\operatorname{Gal}(\overline{\Bbb Q}/\Bbb Q) \to \operatorname{GL}_2(\mathcal{O}_E)$). Edit: just read that your $E$ is an $\ell$-adic local field. Commented Feb 1, 2021 at 11:21

This follows from the compactness of $$\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$$. Let $$G$$ be the image of $$\rho_f$$. The point is that, since $$G$$ is compact, there is a $$G$$-stable lattice in $$\Lambda \subset E^2$$ and $$\mathrm{Aut}(\Lambda)\cong \mathrm{GL}_2(\mathcal O_E)$$
Indeed, $$G$$ is a compact subgroup of $$\mathrm{GL}_2(E)$$ so, in particular, its denominators are bounded.
Consider $$\Lambda = \sum_{g\in G}g\mathcal O_E^2$$, an $$\mathcal O_E$$-module preserved by $$G$$. Then $$\mathcal O_E^2\subseteq\Lambda\subseteq \pi^{-k}\mathcal O_E^2$$ for some $$k$$, where $$\pi$$ is a uniformiser of $$E$$. It follows that $$\Lambda$$ is a free $$\mathcal O_E$$-module and hence there exists $$h\in\mathrm{GL}_2(E)$$ such that $$h\mathcal O_E^2 = \Lambda$$. Then $$h^{-1} G h\subset \mathrm{GL}_2(\mathcal O_E)$$.