14
$\begingroup$

Motivating examples:

  1. Let $V$ be a real vector space with Haar measure $dv$. The fourier transform induces the following topological isomorphism: $$H^s(V,dv) \cong L^2(V^*,(1+|v^*|^2)^sdv^*)$$ The LHS being the $W^{s,2}$-type Sobolev space and the RHS a weighted $L^2$ space on the dual.

  2. Consider $S^1$ a circle with haar measure $d\theta$. Fourier transform Induces a topological isomorphism: $$H^s(S^1,d\theta) \cong l^2(\mathbb{Z},(1+|n|^2)^s dn)$$ Here the LHS is again the sobolev space and on the RHS $dn$ is the counting measure on $\mathbb{Z}$ (This example can be generalized to any finite dimensional torus by suitably modifying both sides).

Notice we get the usual Fourier transform isomorphism by putting $s=0$.

Question: Is there a generalization of the above to non-abelian Fourier transforms on (locally?) compact lie groups? (assuming of course the existence of a Plancherel measure).

Elaboration:

Suppose $G$ is a compact (connected) lie group (for simplicity), let $dg$ denote the Haar measure and $d\mu$ the Plancherel measure on $\hat{G}$. Does there exist a function $q: \hat{G} \to \mathbb{R}$ (a weight, hopefully $q$ will be just the eigenvalue of the Caisimir) s.t. the non-abelian Fourier transform induces a topological isomorphism

$$H^s(G,dg) \cong \{ T \in\hat \oplus_{\pi \in \hat{G}} End(V_{\pi}): [ \pi \mapsto ||T_{\pi}||_{HS}] \in L^2(\hat{G},q^sd\mu) \}$$

$\endgroup$
4
  • 2
    $\begingroup$ How are you defining Sobolev space (on the line or the circle, let alone on the Lie group)? $\endgroup$
    – Yemon Choi
    Commented Mar 7, 2018 at 2:32
  • 3
    $\begingroup$ There are certainly results saying that sufficiently smooth functions on compact Lie groups have non-abelian Fourier transforms that decay at certain rates; I think there is an old paper of Sugiura on this theme $\endgroup$
    – Yemon Choi
    Commented Mar 7, 2018 at 2:34
  • 1
    $\begingroup$ Since $G$ is a compact manifold, you can define Sobolev spaces using charts and pseudodifferential operators. $\endgroup$
    – mcd
    Commented Mar 7, 2018 at 10:37
  • 1
    $\begingroup$ There are many Paley Wiener Theorems for various spaces obtained by quotienting out Lie groups by subgroups. I don't understand the details well enough (by far) to judge if any of those cover your case but if you somehow hadn't heard the term Paley-Wiener before it might be a good search engine query. $\endgroup$
    – Vincent
    Commented Mar 7, 2018 at 11:35

1 Answer 1

1
$\begingroup$

As a qualitative answer: although I do not know any cite-able sources, it is pretty clear that for compact (connected) real Lie groups a notion of Sobolev space parallel to that on products of circles and/or $\mathbb R^n$ can be formulated. One needs the decomposition of the biregular repn as $\bigoplus_\pi \pi\otimes \pi^\vee$, the parametrization of irreducibles by highest weights, the Weyl dimension formula for dimension of $\pi$ in terms of its highest weight, and the fact (e.g., in Stein-Weiss for orthogonal groups acting on spheres, but the same proof works generally) that the sup of sup-norm to $L^2$ norm on a $K$-stable space is proportionate (depending on normalization of measure) to the square root of the dimension of the space. And expression of eigenvalue of Casimir in terms of the highest weight.

Putting these together and renormalizing as desired with hindsight, it is merely a technical extension of the usual "global" Sobolev-inequality arguments to obtain analogues.

$\endgroup$
2
  • $\begingroup$ Eventually i'm really interested in the case where $G$ is non-compact real reductive. Do you know what happens in this case? At least whether one should expect the same kind of results? $\endgroup$ Commented Mar 10, 2018 at 19:25
  • $\begingroup$ The non-compact case is vastly more complicated, I think, for the reason(s) that essentially none of the "happy" control that we have in the compact case carries over to the non-compact case. $\endgroup$ Commented Mar 10, 2018 at 20:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .