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):
enter image description here

  • 1
    $\begingroup$ "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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.