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Let $G$ be a Lie (or more general) group. Consider its continuous representation in a Banach space by isometries, i.e. preserving the Banach norm. Under what conditions this representation is equivalent to a unitary representation of a Hilbert space?

The notion of equivalence may depend on the context, e.g. an important example is when $G$ is a real reductive group, and the representation is admissible of finite length. In that case it is enough to understand the equivalence as isomorphism of Harish-Chandra modules.

The question has positive answer for finite dimensional representations. Indeed let $B$ be the unit ball of the (finite dimensional) Banach space. Let $E\subset B$ be the ellipsoid of maximal volume; such an ellipsoid is known to be unique and is called the John ellipsoid. By uniqueness, $E$ is invariant under $G$. Consider the Hilbert norm such that $E$ is its unit ball. Clearly it will be preserved by $G$, and hence our representation is unitary.

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  • $\begingroup$ Could you please be a bit more precise about what notion of equiavalence you have in mind? In particular, are you demanding that your Banach space is isomorphic (in the TVS category) to a Hilbert space? $\endgroup$
    – Yemon Choi
    Commented Jul 8, 2015 at 14:53
  • $\begingroup$ @YemonChoi: as I mentioned in the post, I do not have in mind any particular kind of equivalence. Also a priori (say in the case of reductive groups) I would not like to assume that the Banach space is isomorphic to Hilbert. $\endgroup$
    – asv
    Commented Jul 8, 2015 at 16:03
  • $\begingroup$ Probably I am confused because I think about Banach-space representations of arbitrary discrete groups more often than I do about representations of Lie algebras. Could you give me an example of an isometric representation of a group on some not-TVS-isomorphic-to-Hilbert Banach space, which is equivalent in your sense to a unitary representation of this group? $\endgroup$
    – Yemon Choi
    Commented Jul 8, 2015 at 16:44
  • $\begingroup$ @YemonChoi: As a first step, you may think about the usual isomorphism of representations you mentioned. However I do not insist to restrict to that notion. For example for representations of real reductive groups there is a notion of Harish-Chandra module which has no topology. Thus two Banach representations are called equivalent if they have isomorphic (in the usual sense) Harish-Chandra modules. $\endgroup$
    – asv
    Commented Jul 8, 2015 at 16:57
  • $\begingroup$ OK, thanks, I see from Wikipedia en.wikipedia.org/wiki/Harish-Chandra_module what you mean (but unfortunately can't help with the question) $\endgroup$
    – Yemon Choi
    Commented Jul 8, 2015 at 18:48

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