Iwasawa Decomposition & Polar Decomposition related how ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T17:10:54Z http://mathoverflow.net/feeds/question/42652 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42652/iwasawa-decomposition-polar-decomposition-related-how Iwasawa Decomposition & Polar Decomposition related how ? John Craighead 2010-10-18T15:11:46Z 2010-10-19T12:36:40Z <p>In an earlier post (Use Lie Sub-Groups of GL(3, R) for elastic deformation ? <a href="http://mathoverflow.net/questions/40470/use-lie-sub-groups-of-gl3-r-for-elastic-deformation?" rel="nofollow">here</a>), I mentioned polar decompositions as in F = RU where R in SO(3) &amp; U in symmetric positive-semidefinite matrices. In response, I received the following comment: "The decompositions you mention are well-known in the theory of Lie groups, e.g., F=RU is called Iwasawa decomposition." </p> <p>I have been trying to understand how a polar decomposition &amp; Iwasawa decomposition are related. I have found the following definitions of both. Iwasawa: G=KAN for G in GL(n, R) where K = orthogonal matrices (rot'ns ?), A = positive diag. matrices &amp; N = upper triag. matrices with diag. entries = 1. Polar: G=KAK where G = semi-simple Lie group, K = maximal compact subgroup of G, A = abelian subgroup of G. For GL(n, R), I think K in SO(n) but unsure about A. The only abelian subgroup of GL(n, R) appears to be the set of all nonzero scalar matrices (i.e. scalar multiples of identity matrix). I'm also puzzled by G=KAK having 3 terms but F=RU only 2. </p> <p>Given all of this, I'm still unsure how these two decompositions are related. </p> http://mathoverflow.net/questions/42652/iwasawa-decomposition-polar-decomposition-related-how/42673#42673 Answer by Jim Humphreys for Iwasawa Decomposition & Polar Decomposition related how ? Jim Humphreys 2010-10-18T17:25:56Z 2010-10-19T12:36:40Z <p>Two mathematical sources worth consulting are Helgason <em>Differential Geometry, Lie Groups, and Symmetric Spaces</em> (Chapter IX.1) and a paper by Kostant available online from www.numdam.org <a href="http://www.numdam.org:80/numdam-bin/item?id=ASENS_1973_4_6_4_413_0" rel="nofollow">here</a>; reviewed in MR0364552 (51 #806) 22E45, Kostant, Bertram, On convexity, the Weyl group and the Iwasawa decomposition. Ann. Sci. Ecole Norm. Sup. (4) 6 (1973), 413–455 (1974). </p> <p>Both of these older sources put the various matrix decompositions into the framework of Lie groups and the complexified Lie algebras. There are certainly other discussions in the literature which might be helpful but don't come to mind immediately. Anyway, Helgason's viewpoint emphasizes the passage to the <em>symmetric space</em> <code>$G/K$</code> generalizing the space of symmetric matrices, while Kostant emphasizes decompositions within a Lie group.</p> <p>ADDED: In his answer Faisal points out more explicitly some of these connections with matrix decompositions. There is a lot of interplay among the Cartan, Iwasawa, and Bruhat decompositions, as shown in the above references. The polar decomposition is part of this story, though in a way it's less refined. All of the viewpoints here are important, but historically the Cartan decomposition has some primacy.</p> http://mathoverflow.net/questions/42652/iwasawa-decomposition-polar-decomposition-related-how/42724#42724 Answer by Faisal for Iwasawa Decomposition & Polar Decomposition related how ? Faisal 2010-10-19T00:29:41Z 2010-10-19T00:29:41Z <p>You can obtain the $G=KAK$ decomposition from a decomposition of the type $F=UR$. To avoid unnecessary complications, let's assume that our reductive group $G$ is a selfadjoint subgroup of $\operatorname{GL}(n,\mathbb{R})$. Then the map $g \mapsto g^{-t}$ is an involution of $G$, which is called the Cartan involution and is typically denoted by $\theta$. The first observation to make is that the fixed-point set <code>$K = \{ g \in G \colon \theta(g)=g \}$</code> of $\theta$ is a maximal compact subgroup of $G$. For example, if $G=\operatorname{GL}(n,\mathbb{R})$, then $K=\operatorname{O}(n)$.</p> <p>Next we observe that $\theta$ induces an involution (also denoted by $\theta$) at the Lie algebra level: explicitly, this is the map <code>$X \mapsto -X^t$</code>. If <code>$\mathfrak{p}$</code> denotes the $-1$-eigenspace of this latter involution, then one has the following result.</p> <blockquote> <p>The map $K \times \mathfrak{p} \to G$ given by $(k, X) \mapsto k e^X$ is a diffeomorphism.</p> </blockquote> <p>In particular, every $g \in G$ can be expressed as $k e^X$ for some $k \in K$ and $X \in \mathfrak{p}$. This decomposition is known as the Cartan decomposition; it is a generalization of the polar decomposition to $G$ (and is, I presume, the $F=UR$ decomposition stated in the OP). Indeed, if $G = \operatorname{GL}(n,\mathbb{R})$, then $\mathfrak{p}$ is just the set of symmetric matrices, and thus the set $\exp \mathfrak{p}$ consists of symmetric, positive semidefinite matrices.</p> <p>Now let $\mathfrak{a}$ denote a maximal abelian subspace of $\mathfrak{p}$. Then it can be shown that $A = \exp \mathfrak{a}$ is a closed abelian subgroup of $G$ with Lie algebra $\mathfrak{a}$. It can also be shown that $\mathfrak{a}$ is unique up to conjugacy via an element of $K$. That is to say, if $\mathfrak{a}'$ is another maximal abelian subspace of $\mathfrak{p}$, then there is a $k \in K$ such that $\text{Ad}(k) \mathfrak{a} = \mathfrak{a}'$. With this information we can obtain the decomposition $G=KAK$: given $g \in G$, one observes that $p=gg^t \in \exp \mathfrak{p}$, say $p=e^X$. Thus there is a $k \in K$ such that $\text{Ad}(k)X \in \mathfrak{a}$, and then $e^{-\text{Ad}(k)X/2}kg \in K$ (because it is fixed by $\theta$), whence $g \in KAK$.</p> <p>This hopefully alleviates your 3-terms-vs-2-terms issue.</p> <p>I'm not aware of any relationship between the Iwasawa decomposition and the $KAK$ (polar) decomposition.</p>