Double centralizer theorem for ($\ell$-adic) Lie algebras $\DeclareMathOperator\GL{GL}$I'm reading the proof of Serre's open image theorem from his book "Abelian $\ell$-adic representations and elliptic curves". This is chapter IV Section 2.2 of the book. Let's assume that $E$ is an elliptic curve without CM over a number field $K$ and $\rho:G_K \rightarrow \GL(V_\ell)$ the associated Galois representation on $ V_\ell = T_\ell \otimes \mathbb{Q} $ where $ T_\ell $ is the Tate module. Then he proves that the image of the Galois representation is open in $\GL(V_\ell)$ w.r.t the $\ell$-adic topology. The only part that I don't understand from the proof is in the very beginning when he deduces that the ($\ell$-adic) Lie algebra of the image contains $\mathfrak{sl}_2$ from the fact that it's centralizer is $\mathbb{Q}_\ell$. Is he using some sort of classification for the Lie subalgebras here?
My main question is this: To me this seems like a Lie algebra analogue of the double centralizer theorem for simple subalgebras of central simple algebras. I was wondering if such a result exists, namely if from the centralizer of the Lie subalgebra being small we can deduce that the subalgebra itself is large in some sense. I'm not sure what should be the conditions and what's the right formulation but I feel like $\GL_2$ being reductive and $\operatorname{SL}_2$ being semi-simple might play a role here.
 A: $\newcommand{\g}{\mathfrak{g}}\newcommand{\sl}{\mathfrak{sl}}$The crucial point in the proof is that the absolute Galois group  $G_K$ acts irreducibly on $V_{\ell}$, which is based on a nontrivial Shafarevich finiteness theorem (Sect. 1.4). Applying this result to all finite algebraic extensions of $K$, one gets that the $\ell$-adic Lie algebra $\g_{\ell}$ of the image acts irreducibly (and faithfully) on $V_{\ell}$.
So,
$\g_{\ell}$ is an irreducible Lie subalgebra of $\mathrm{End}_{\mathbb{Q}_{\ell}}(V_{\ell})$, whose centralizer consists of scalars $\mathbb{Q}_{\ell}\mathrm{Id}$, i.e., the natural faithful  2-dimensional representation of $\g_{\ell}$ in $V_{\ell}$ is absolutely irreducible. Hence, $\g_{\ell}$ is reductive, i.e., splits into a direct sum
$$\g_{\ell}=\g_{\ell}^{0}\oplus c_{\ell}$$
of a semisimple Lie algebra $\g_{\ell}^{0}$ and the center $c_{\ell}$. The absolute irreducibility implies that $c_{\ell}$ is either $0$ or $\mathbb{Q}_{\ell}\mathrm{Id}$. In both cases $V_{\ell}$ is an absolutely irreducible representation of $\g_{\ell}^{0}$; in particular, $\g_{\ell}^{0}\ne \{0\}$. The semisimplicity of $\g_{\ell}^{0}$ implies that
$$\g_{\ell}^{0}\subset \sl(V_{\ell})\cong \sl_2(\mathbb{Q}_{\ell}).$$
Now it follows easily that
$\g_{\ell}^{0}= \sl(V_{\ell})$, because no proper Lie subalgebras of $\sl_2$ are semisimple.
