Let $\mathbb{V}$ be a representation of a Lie group $G$ and let $P \to M$ be a principal $G$-bundle with a principal connection. If $\mathbb{V}$ is finite-dimensional, then one can associate to this data an associated vector bundle $P\times_G \mathbb{V}$ with linear connection. I thought that basically the same construction should work also when $\mathbb{V}$ is an infinite-dimensional representation, but I haven't found any textbook that would not constrain itself to finite rank. All the textbooks concerning to infinite-dimensional differential geometry that I know of (Michor, Lang, Neeb) doesn't treat associated bundles and induced connections.


I now realize that it may not be as straightforward as it seems on a first glance. I want to, in fact, generalize a slightly more complicated construction -- the so called tractor connection induced by a Cartan connection.

Changing the notation a little bit, given a finite-dimensional Lie group $G$ with a closed subgroup $H$, I need to work with an infinite-dimensional vector space $\mathbb{V}$ which is a representation of $\mathfrak{g}$ and also a representation of $H$ (so I can form associated bundles to $H$-principal bundles) with these two representation being compatible. Practically, I am interested mainly in Harish-Chandra modules and their globalizations. I think I am also fine with just a "sort of connection" working on some dense subbundle of the associated bundle and so $L^2$-globalizations are also OK.

I can briefly describe the construction for $\mathbb{V}$ being finite-dimensional representation of $G$ if it is needed.

  • 1
    $\begingroup$ @Andrew: sure you know these things better than I do, but - as long as $M$ and $G$ are finite-dimensional - what is the problem with taking the same formula as in the finite-dim. case? $\endgroup$ Feb 13, 2012 at 10:17
  • 1
    $\begingroup$ Johannes: That case didn't occur to me! I assumed that $G$ was an infinite dimensional Lie group. Nonetheless, topology is still important and might behave a bit nastily. A standard situation is the space of $L^2$ sections of some fibre bundle. Then you run into the problem that, for example, $S^1$ doesn't act as nicely as it could on $L^2(S^1)$. Then I guess the explanation for the absence of this from the literature is that one can usually decompose the infinite representation into a sum of finite ones of different characters and study that collection instead of the single infinite one. $\endgroup$ Feb 13, 2012 at 10:47
  • 2
    $\begingroup$ r0b0t: Incidentally, what is the actual question here? $\endgroup$ Feb 13, 2012 at 14:22
  • 1
    $\begingroup$ rObOt: Most study of infinite dimensional representations of a semisimple (or reductive) Lie group involves additional hypotheses on the vector space and representation: Banach or Hilbert space, etc. Is it clear what effect this richer structure would have on your question? $\endgroup$ Feb 13, 2012 at 21:17
  • 1
    $\begingroup$ That's what I wanted to avoid as I hoped that these issues are already sorted out somewhere. Looks like I have to get my hands dirty. Thanks for comments. $\endgroup$ Feb 14, 2012 at 10:52

1 Answer 1


If you use the cocycle description of principal bundles (for finite dimensions) as in 18.7.4 of 1, you can describe the associated bundle using convenient calculus, since then you can flip coordinates freely. Inducing connections is then described in 19.8 (for associated fiber bundles) and in 19.10 (for associated vector bundles) of 1; use also 19.9 ("recognizing induced connections") where the cocycle description is spelled out.

Idea: You can carry over to infinite dimensions all constructions of finite dimensional differential geometry for which you have direct chart descriptions. But be very careful whenever you have to solve equations (ODE's, implicit functions, etc.)

1 Peter W. Michor: Topics in Differential Geometry. Graduate Studies in Mathematics, Vol. 93 American Mathematical Society, Providence, 2008. (pdf).


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.