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It is well known $\newcommand{\Diff}{\operatorname{Diff}}$ that the group $\Diff(S^1)$ of smooth diffeomorphisms of the circle behaves in many ways like $SL(2,\mathbb R)$. For example, if $S^1\subseteq \Diff(S^1)$ is the group of rotations then $\Diff(S^1)/S^1$ is an infinite dimensional complex domain (apparently this is due to Kirillov and Yuriev). This is analogous to the symmetric space $SL(2,\mathbb R)/S^1$ which is the complex upper half plane.

Now $SL(2,\mathbb R)$ has some very useful decompositions: the Iwasawa decomposition (or the $KAN$ decomposition) and the $KAK$ decomposition.

Does $\Diff(S^1)$ also have $KAN$ and $KAK$ decompositions?

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  • $\begingroup$ The question is unclear. To "have a $KAK$ or $KAN$ decomposition" in an abstract topological group means nothing without further specification. For $KAK$ one option is that $K$ is required to be compact and $A$ abelian, but the usual $KAK$ decomposition refers to a particular decomposition and not any decomposition with $K$ compact and $A$ abelian. $\endgroup$
    – YCor
    Aug 12, 2013 at 17:11
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    $\begingroup$ The group $Diff(S^1)$ is a smooth Lie group locally diffeomorphic to a Frechet vector space. I am looking for "reasonable" decompositions analogous to $KAK$ and $KAN$, but the question is intentionally stated vaguely from this viewpoint. Ideally, I would like $K$ to be the group of rotations. Also note that for loop groups there is a $KAN$ decomposition. $\endgroup$
    – Valerie
    Aug 12, 2013 at 17:29
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    $\begingroup$ I took the liberty of tweaking some formatting - I hope this is OK $\endgroup$
    – Yemon Choi
    Aug 12, 2013 at 20:18

1 Answer 1

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At least for the $KAN$ decomposition, there is a candidate.

Namely take $K$ the rotation group, $A$ the group of hyperbolic elements of $PSL(2,\mathbb{R})$ fixing $0$ and $\infty$, and $N$ the group of diffeos fixing $0$ and tangent to the identity there.

Then $G=KAN$ is the group $Diff_+(\mathbb{S}^1)$, and the decomposition of elements is unique, as is easily checked.

No idea yet for the $KAK$ decomposition.

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