Complex polar decomposition and the complex orthogonal group

155 Views Asked by At

Consider $O_\mathbb{C}(n)=\{G\in Gl(n,\mathbb{C})|G^TG=I\}$. I want to show that each factor of the complex polar decomposition applied to some element in $O_\mathbb{C}(n)$ belongs to $O_\mathbb{C}(n)$.

As far as I understand, then the complex polar decomposition works by every $G\in Gl(n,\mathbb{C})$ can be decomposed as $G=UH$, where $U\in U(n)$ and $H\in Herm(n,\mathbb{C})$. So I guess I want to show that $XU$ and $XH$ belongs to $O_\mathbb{C}(n)$ for some $X\in O_\mathbb{C}(n)$. However I keep getting stuck.

For instance considering $XU$ we get

$$(XU)^TXU=U^TX^TXU=U^TU$$

but I don't see how even $U^TU=I$ as $U\in U(n)$

And likewise

$$(XH)^TXH=H^TX^TXH=H^TH=H^2$$

I don't see how this should equal $I$.

Am I misunderstanding something?

1

There are 1 best solutions below

0
On BEST ANSWER

It is important in the polar decomposition that $H$ is positive semidefinite. This is necessary for the uniqueness of polar decomposition.

I don't quite understand how you are using $X.$ I would argue as follows. Take $X\in O_{\mathbb C}(n)$ with polar decomposition $X=UH.$ Then

$$UH=X=X^{-T}=(UH)^{-T}=U^{-T}H^{-T}$$ where $M^{-T}$ means $(M^{-1})^T.$ The map $M\mapsto M^{-T}$ preserves unitariness and Hermitianness because it is a group homomorphism of $GL_{\mathbb C}(n)$ commuting with complex conjugation. Furthermore it preserves positive definiteness - one can show that the inverse of a positive definite matrix is positive definite. By the uniqueness of polar decomposition, $U=U^{-T}$ and $H=H^{-T},$ which means $U$ and $H$ are orthogonal.