Consistency of transformation of scalar fields with mathematical definition of a representation of Lie algebra and Lie group

Question

Transformations of scalar fields under a Lorentz group transformation are generated by differential operators $L_{\mu\nu}=x_\mu\partial_\nu-x_\nu\partial_\mu$.

On the other hand, a representation of a Lie group $G$ and algebra is defined as a homomorphism $\pi: g\in G \rightarrow \pi(g)\in GL(n,\mathbb{C})$ and $\psi: X\in \mathbb{g} \rightarrow \psi(X)\in \mathbb{gl}(n,\mathbb{C} )$, where $GL(n,\mathbb{C})$ denotes the General Linear Matrix group of $n\times n$ complex invertible matrices and $\mathbb{gl}(n,\mathbb{C})$ is its Lie algebra and $G$ is the group in question with $\mathbb{g}$ being its Lie algebra.

So, how can generators of a Lie algebra be represented by differential operators if representations of Lie algebras have the above definition (i.e. mappings to the Lie algebra corresponding to the group of $n\times n$ invertible matrices)?

EDIT: The first part was inspired by the following (from Freedman and Van Proyen's Supergravity, p.14):

Answer 1

That's not the definition of a representation of a Lie group - it's the definition of a finite dimensional representation of a Lie group.

More generally, a representation on a vector space $V$ is a group homomorphism $\pi:g\in G\mapsto \pi(g)\in\operatorname{Aut}(V)$, where $\operatorname{Aut}(V)$ is the set of automorphisms on $V$. If $V$ is finite dimensional, then $\operatorname{Aut}(V)\simeq GL(n,\mathbb R)$ or $GL(n,\mathbb C)$, but $V$ need not be finite dimensional, as is the case here.