Norm continuous infinite dimenisonal representation of a Lie group

Question

Given a Lie group G and an infinite dimensional Hilbert space $\mathcal{H}$. In the literature I have only encountered the two following notions of a representation $\pi$ of G on $\mathcal{H}$ :

1) $\pi$ : G $\to$ $B(\mathcal{H})$ is a group homomorphism (consider only the invertible bounded linear operators...) and $G \times \mathcal{H} \to \mathcal{H}$ is continuous.

2) $\pi$ : G $\to$ $B(\mathcal{H})$ is a continuous group homomorphism, where we consider the strong operator topology on $B(\mathcal{H})$.

I got the notion, that nobody seems interested in the case, when $\pi$ : G $\to$ $B(\mathcal{H})$ is a continuous group homomorphism with the usual operator topology on $B(\mathcal{H})$. If my impression is not correct, could you please give me some references that deal with this case or some examples? If my impression is correct, what are the reasons that one neglects this case?

Answer 1

Your question also makes sense for a Banach space and that is where the following results also apply. The Hilbert space case is then only a particular case where you can get some more information about the infinitesimal generators.

So let $G$ be a connected Lie group (the discrete pieces are sort of uninteresting...) For a Banach space $V$ a group morphism $U\colon G \longrightarrow B(V)$ into the (invertible) bounded operators on $V$ is norm continuous iff it is of the form \begin{equation} U_{\exp(t\xi)} = \sum_{r=0}^\infty \frac{t^n}{n!} A_\xi^n \end{equation} for all $\xi \in \mathfrak{g}$ (the Lie algebra of $G$) and with a continuous operator $A_\xi$, where the exponential series converges in the operator norm. The map $\xi \mapsto A_\xi$ is then a Lie algebra representation by continuous operators on $V$.

The proof of this fact is essentially as in finite dimensions so you can take you favorite differential geometry book/Lie group book and copy the proof. Note that such a representation automatically will also converge for complex $t$ and hence gives a representation of the complexified Lie group on the nose.

The reason why this is sort of representation is rather un-interesting is that in "real life" they simply don't show up. The simnplest example is perhaps the translation group acting on $L^2(\mathbb{R})$ for which the above statement is false. In "real life" the generators $A_\xi$ of the representation will typically be unbounded operators and the exponential series does not converge in a norm sense. Nevertheless, a big part of harmonic analysis deals with the question on which vectors of $V$ one can have a Lie algebra representation such that the exponential series will converge after applied to such a vector (then in the norm sense of $V$).

Answer 2

I agree with the sentiment of the other answer and the comments that restricting to norm-continuous representations cuts out a lot of interesting ones. So it's less that these aren't studied, but that any techniques that rely on being norm-continuous don't apply to many of the most interesting examples and so aren't used all that much. Of course, any norm-continuous representation defines a weaker one so using techniques in the other direction works fine.

The classic example that I know of (which is a generalisation of that in Stefan's answer) is of the action of a compact Lie group $G$ on the space of $L^2$-functions on $G$.

Here's a case where that comes up in differential topology.

1. For a smooth manifold, say $M$, the space of smooth loops $L M := C^\infty(S^1,M)$ is a very interesting space.
2. However, it's a bit complicated and many things can be simplified by restricting them to $M$ sitting inside $L M$ as the subspace of constant loops.
3. To make the above work, we have to consider not just constant loops but "infinitesimally small" loops. These have the form of a vector bundle over $M$ whose fibre is modelled on the (orthogonal) complement of the constant loops in $L \mathbb{R}^n$. This vector bundle provides the bridge between the "big" space $L M$ and the "little" space $M$.
4. Hilbert bundles are much simpler than arbitrary bundles, we can complete this to a Hilbert bundle modelled on (a finite codimension subspace of) $L^2(S^1,\mathbb{R}^n)$.
5. But Hilbert bundles are really too simple. A bare Hilbert bundle is trivial, so we need some extra structure involved to ensure that whilst simple, it isn't too simple.
6. That extra structure is the natural circle action which comes from rotating loops. As $M$ is "constant loops", rotating has no effect on $M$. But it does have an effect on "infinitesimal loops" and so acts on the fibres of the bundle. This action is precisely the circle action on $L^2(S^1,\mathbb{R}^n)$.
7. As we now have a fibrewise circle action on a Hilbert bundle, we can split that bundle up according to the representation of the circle. It turns out that this splits this bundle into finite dimensional pieces (that happen to be isomorphic to the complexification of the tangent bundle of $M$, but that's not important). Being finite dimensional, they are simple enough to be analysed but not so simple as to be trivial.
8. If just one of the factors was infinite dimensional, this whole story would collapse because then the whole bundle would carry no information. Thus being able to split into finite dimensional factors is crucial.
9. But as the bundle is itself infinite dimensional, to split in to finite dimensional pieces there must be an infinite number of such.
10. Such a splitting cannot happen if the representation is norm-continuous, for the very simple reason that if I can find orthonormal vectors $v_k$ such that $\lambda \in S^1$ acts on $v_k$ by multiplication by $\lambda^k$ then the map $S^1 \to U(H)$ must have $\|R_\lambda - R_\mu\| = \sup\lbrace|\lambda^k - \mu^k| : k \in \mathbb{Z}\rbrace$ which is $2$ for almost all $\lambda$ and $\mu$.

Answer 3

It is well-known that every locally compact group admits a continuous, faithful unitary representation on a Hilbert space $H$, where continuity is in the usual sense (strong topology on $U(H)$).

Now, let $G$ be a connected Lie group; $G$ admits a faithful {\it norm} continuous representation on some Banach space, if and only if $G$ is linear, i.e. admits a faithful representation on $\mathbb{C}^n$ for $n$ large enough. See: Luminet, Denis; Valette, Alain {\it Faithful uniformly continuous representations of Lie groups}. J. London Math. Soc. (2) 49 (1994), no. 1, 100–108.

So if you consider norm continuous representations you have not enough representations to separate points.

Answer 4

Just to make the point: natural repns of topological groups $G$ on $L^2(G)$ or on $L^2(H\backslash G)$ are basically never continuous in the norm topology. Only for $G$ discrete (or $H\backslash G$ discrete) is there continuity in the norm topology.

This is already illustrated completely in the case of the action of $G=\mathbb R$ on $L^2(\mathbb R)$. In effect, to prove continuity of this repn in the strong topology (weaker than the norm topology!), uniform continuity of compactly-supported functions is used, and a two-or-three epsilon argument.

Yes, this fact-of-nature might seem a little awkward, if one's repertoire of topological vector spaces weren't adequate to include the space of operators with strong topology which is not Frechet, but it is quasi-complete and locally convex, which suffices for integrations and such things.

In reality, all the arguments in repn theory do not need continuity in norm topology. The strong topology is not only natural, but works just fine.