5
$\begingroup$

I am trying to understand the non-commutative analysis for nilpotent Lie groups, so I've been reading Corwin's and Greenleaf's book on the representation theory of nilpotent groups and going through examples of the orbit method.

My problem is that I would like to apply the Fourier transform to the Lie algebra generators, and so basically I only need the unitary representations of the Lie algebra. We can simply differentiate the Lie group representations of one parametric subgroups, but as I was trying to go into more difficult examples, the calculations became quiet tedious very fast, because they rely (at least the way it's done in the book) on the Campbell-Hausdorff formula to construct the induced representations.

So my question is: are there any constructions of irreducible unitary representations of nilpotent Lie groups/algebras, that do not use the Campbell-Hausdorff formula or that are more friendly for high-dimensional cases?

Thanks.

$\endgroup$
2
$\begingroup$

One friendly class comes from parabolic induction. So in case your nilpotent algebra is nilradical of a parabolic subgroup, there is a general formula that is straightforward to apply. See Theorem 1.3 of Algebraic analysis on scalar generalized Verma modules of Heisenberg parabolic type I.: $A_n$-series. This case deals with Verma modules which are unitarizable only in very special case (where the nilradical is actually Abelian), but more or less the same combinatorics should work even in the real case. Indeed, it seems that this approach can be generalised considerably, see later works of the first author.

$\endgroup$
  • $\begingroup$ Sorry I didn't respond immediately. I've looked back then at what kind of geometries one can get and I wasn't completely satisfied because there are only few rank two distributions associated to such geometries (contact and Cartan if I remember correctly). I was hoping maybe for some not necessarily free rank 2 things. But nevertheless this is the best answer I've got. Thank you =) $\endgroup$ – Ivan Nov 28 '17 at 16:33

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.