generalisation of GL(3,R) polar decomposition

2011-07-13T14:32:20Z

Does polar decomposition work when the 'orthogonal' matrices are not orthogonal wrt to the identity ($O^TO=Id$), but wrt to some other symmetric matrix $K$ (i.e. $O^TKO=K$)?

Specifically, $GL(3,R)$ can be decomposed as $SO(3)\times Diag(3)\times SO(3)$, but can it also be decomposed as $SO(3)\times Diag(3)\times SO_K(3)$? (SO_K are the K-orthogonal matrices mentioned above)

Thanks for any help.

EDIT:

Sorry to resurrect this, but I've had a further thought. First of all thank you for you answers (over my head as they were, they're still appreciated). From what I gather the $SO_k \times Diag^+ \times SO_k$ decomposition works, but $SO \times Diag^+ \times SO_k$ does not?

What if we were to use the standard polar decomposition on $Q\in GL(3)$ so that $Q=RS$ for orthogonal $R$ and symmetric $S$, and then further decompose this symmetric matrix into a $SO_k \times Diag^+ \times SO_k$ product? This should be valid, since $S\in GL(3)$, right? So you'd end up with a matrix that looks like $SO\times [something]\times SO_k$. This is really what I was after when I first asked the question (I realise I could have been clearer about that). The something in the middle doesn't really matter to me.

Does that make any sense?

Thank you again.

Answer by Lior Silberman for generalisation of GL(3,R) polar decomposition

2011-07-13T22:37:16Z

Let $G = SL_3(\mathbb{R})$. Let $K = SO_3(\mathbb{R})$ and $H=SO_{2,1}(\mathbb{R})$ be the two orthogonal groups in three variables (one can think of $H$ as the image of $SL_2(\mathbb{R})$ under the symmetric square or adjoint representation). Let $A$ be the subgroup of diagonal matrices. Let $g\in G$ be arbitrary.

I think the problem is then whether either of $G = K^gAK$ and $G=H^gAK$ might hold.

The first assertion is false: I will show that $G = K^g AK$ iff $g\in AK$. The sufficiency is clear. For necessity let $S=G/K$ be the symmetric space with $x_0$ the point fixed by $K$. Then $F_0 = A\cdot x_0$ is the standard flat through $x_0$, and $K^g A K$ is the preimage in $G$ of the union of the set $K^g F_0$ of flats. Let $x_g = gx_0$ be the point fixed by $K^g = gKg^{-1}$. Then $g\notin AK$ is equivalent to $x_g \notin F_0$. Since $F_0$ is closed there is a ball $B_S(x_g,r)$ which is disjoint to $F_0$. This ball is $K^g$-invariant, so it is disjoint from the union of the $K^g$-translates of $F_0$ as well.

I'm not sure about the second assertion.

generalisation of GL(3,R) polar decomposition - MathOverflow

generalisation of GL(3,R) polar decomposition

Answer by Lior Silberman for generalisation of GL(3,R) polar decomposition