Parametrization manifold of SL(2,R)

  • #1
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Main Question or Discussion Point

I'm reading a book on Lie groups and one of the first examples is on SL(2,R). It says that every element of it can be written as the product of a symmetric matrix and a rotation matrix, which I can see, but It also makes the assertion that the symmetric matrix can be parameterized by a hyperboloid with equation given by ## z^2 - x^2 - y^2 = 1##, while the rotation matrix is parameterized by a point on a circle. I guess the rotation matrix being parameterized by a point on a circle kind of makes sense intuitively, but I can't see how the hyperboloid can possibly parameterize the symmetric matrix. What am I missing here? If anyone can demonstrate the parametrization for the rotation matrix in a rigorous way aswell, I'd be glad.
 

Answers and Replies

  • #2
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I'm reading a book on Lie groups and one of the first examples is on SL(2,R). It says that every element of it can be written as the product of a symmetric matrix and a rotation matrix, which I can see, but It also makes the assertion that the symmetric matrix can be parameterized by a hyperboloid with equation given by ## z^2 - x^2 - y^2 = 1##, while the rotation matrix is parameterized by a point on a circle. I guess the rotation matrix being parameterized by a point on a circle kind of makes sense intuitively, but I can't see how the hyperboloid can possibly parameterize the symmetric matrix. What am I missing here? If anyone can demonstrate the parametrization for the rotation matrix in a rigorous way aswell, I'd be glad.
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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Thanks for your reply. I think I should've mentioned this but the book also states that both matrices are unimodular. So we could have
## 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.
 
  • #5
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I see, book would've done me a favor writing the equation in this second manner. Still, I guess I should've noticed that it was just a silly transformation. Thanks again for clarifying things.
 

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