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

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): • "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) – ACuriousMind Jun 10 '20 at 22:21
• @ACuriousMind Thanks for the comment. It was an error. I edited the question now. – TheQuantumMan Jun 10 '20 at 22:53
• 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? – ACuriousMind Jun 10 '20 at 22:56
• @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 – TheQuantumMan Jun 10 '20 at 23:01
• If you look at e.g. Wikipedia's definition of a group representation there is no requirement that the vector space be finite-dimensional. – ACuriousMind Jun 10 '20 at 23:14

## 1 Answer

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.