Suppose $G $ is a Lie group and $g $ its Lie algebra. Further let $H $ be a complex Hilbert space. I'm interested in the infinite dimensional case. The question concerns the relationship between unitary representations of $G $ on $H $ and representation of $g $ on $H $.

I would like sources defining what it means to have a representation of $g $ on $H $. In the finite dimensional case one gives a Lie algebra homomorphism $r: g\to \text {GL}(H) $. In the infinite dimensional case one wants to consider unbounded operators and it's not clear to me how to deal with commutation relations for those in a general setting.

If the Lie-exponential map is surjective, one may hope to construct a Lie group representation from an algebra representation. If the exponential map is not surjective, one can still hope to get representations by adding certain representatives in an ad-hoc manner. This is done in physics when considering representations of certain symmetry groups. What is the proper way to deal with infinite dimensional representations in this setting? In what cases can Lie group representations be constructed from Lie algebra representations by exponentiation (of unbounded operators in the case of interest) and what representations may be missed by this approach?


1 Answer 1


I can refer to two classical theorems by Garding and Nelson from the 1950s to answer your questions: A finite dimensional Lie-Algebra $\frak{l_{G}}$ with a generic generator A always admits a representation on a complex separable Hilbert space by (essentially) self-adjoint operators (generically $\mathcal{A}$) on a (common for all generators) dense everywhere invariant domain (this is called the Garding domain of the representation) whenever $\exp (i \mathcal{A} t)$ is a representation of the corresponding uniparametric subgroup of $\frak{G}$, $\exp (iAt)$.
Viceversa, given all generic $\mathcal{A}$ on the Garding domain of a Lie algebra $\frak{l_{G}}$, then one obtains by exponentiation a representation of the universal covering group of $\frak{G}$ by unitary operators acting on the whole Hilbert space containing the Garding domain.

The Garding domain is invariant, so all commutators of the $\mathcal{A}$'s are well defined.


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