In an earlier post (Use Lie Sub-Groups of GL(3, R) for elastic deformation ? here), I mentioned polar decompositions as in F = RU where R in SO(3) & 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."
I have been trying to understand how a polar decomposition & 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 & 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.
Given all of this, I'm still unsure how these two decompositions are related.
Best Answer
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 $K = \{ g \in G \colon \theta(g)=g \}$ of $\theta$ is a maximal compact subgroup of $G$. For example, if $G=\operatorname{GL}(n,\mathbb{R})$, then $K=\operatorname{O}(n)$.
Next we observe that $\theta$ induces an involution (also denoted by $\theta$) at the Lie algebra level: explicitly, this is the map $X \mapsto -X^t$. If $\mathfrak{p}$ denotes the $-1$-eigenspace of this latter involution, then one has the following result.
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.
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 $g \in KAK$.
This hopefully alleviates your 3-terms-vs-2-terms issue.
I'm not aware of any relationship between the Iwasawa decomposition and the $KAK$ (polar) decomposition.