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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):
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    $\begingroup$ "Transformations of scalar fields under a general coordinate transformations are generated by differential operators $L_{\mu\nu}=x_\mu\partial_\nu-x_\nu\partial_\mu$." - what do you mean by this? Where have you heard it? (The claim is wrong: The group of diffeomorphisms ($\cong$ "general coordinate transformation") is in general infinite-dimensional, but the algebra of $L_{\mu\nu}$ is clearly finite-dimensional) $\endgroup$ – ACuriousMind Jun 10 '20 at 22:21
  • $\begingroup$ @ACuriousMind Thanks for the comment. It was an error. I edited the question now. $\endgroup$ – TheQuantumMan Jun 10 '20 at 22:53
  • $\begingroup$ Alright. So, what exactly is your question? The differential operators $L_{\mu\nu}$ are linear operators upon the vector space of scalar fields. So they constitute an (infinite-dimensional) representation of the Lorentz group. Is your problem that $n$ is not finite here? $\endgroup$ – ACuriousMind Jun 10 '20 at 22:56
  • $\begingroup$ @ACuriousMind Yes, I think so. A representation should be an element of the general linear group, so I can't see exactly how a differential operator fits this description, although now I suspect that the answer might be trivial $\endgroup$ – TheQuantumMan Jun 10 '20 at 23:01
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    $\begingroup$ If you look at e.g. Wikipedia's definition of a group representation there is no requirement that the vector space be finite-dimensional. $\endgroup$ – ACuriousMind Jun 10 '20 at 23:14
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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.

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