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- I
- Thread starter Wledig
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- #2

fresh_42

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Can you show the decomposition? I know of the Iwasawa decomposition into a product diagonal times upper triangle times rotation.

The rotation part is easy: A rotation matrix is of the form ##K = \begin{bmatrix}\cos \varphi & -\sin \varphi \\ \sin \varphi & \cos \varphi\end{bmatrix}## which is parameterized by ##\varphi \in [0,2\pi )##.

The Iwasawa decomposition is ##G=ANK## with ##A=\begin{bmatrix}e^t&0\\ 0 & e^{-t}\end{bmatrix}\; , \;N=\begin{bmatrix}1&x\\0&1\end{bmatrix}\,.## Of course I could consider ##AN## but this isn't symmetric.

- #3

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## A=\begin{bmatrix}a&c\\ c & b\end{bmatrix}\ ## and ## K = \begin{bmatrix}\cos \varphi & -\sin \varphi \\ \sin \varphi & \cos \varphi\end{bmatrix} ## so that the product AK yields a matrix with determinant equal to that of the symmetric matrix which is 1 by the unimodular condition. The big question is how to relate the symmetric matrix to the hyperboloid though.

- #4

fresh_42

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http://www.wolframalpha.com/input/?i=z^2-x^2-y^2=1

and

http://www.wolframalpha.com/input/?i=xy-z^2=1

is.

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