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I'm reading Gilmore's Lie Group, Physics and Geometry and I'm trying to solve the exercise 1 at page 8, in which you have to find the full parametrization of a 2X2 $SL(2,R)$ matrix as a product of a symmetric matrix and a rotation one. While retrieving the rotation matrix is really simple, I'm struggling to find a suitable parametrization for the symmetric part. The book is claiming this is a 2-Manifold which can be parametrized by the two-sheeted hyperboloid $$z^2 - x^2 -y^2 = 1$$

Initially I thought to simply write the symmetric matrix as: $$\begin{bmatrix}z+x & y\\y & z-x\end{bmatrix}$$

which determinant is 1, but this is not a 2-Manifold, since showing 3 degree of freedom.

I've also tried the hyperbolic parametrization, i.e.: $$\begin{align} x=sinh(u)cos(v) \\ y=sinh(u)sin(v) \\ z=\pm cosh(u)\end{align}$$

, which is better but I cannot see how to put this explicitly in matrix form, to get the determinant = 1.

I'm sure it's related to the KAN decomposition of a $SL(2,R)$ matrix, somehow.

Any help is appreciated. Thanks in advance.

EDIT: I just realised that combining the 2 claims above we could write keep my symmetric 3-Manifold symmetric matrix replacing the hyperbolic relations and getting a symmetric matrix un u and v, but which solution to keep for the z? Choosing the + may work:

$$\begin{bmatrix}cosh(u)+sinh(u)cos(v) & sinh(u)sin(v)\\sinh(u)sin(v) & cosh(u)-sinh(u)cos(v)\end{bmatrix}$$

What do you think about this, please?

riccardoventrella
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1 Answers1

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The matrix you added in your edit is correct. The choice of $z=+cosh(u)$ and $z=-cosh(u)$ is exactly what makes the hyperboloid two-sheeted, since $cosh(u)$ is strictly positive.

Jonathan Trousdale
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  • Thank @ChainedSymmetry for your confirmation. However, my worry about the $\pm$ sign is still there. I mean, my final matrix is made by choosing $z=+cosh(u)$ in both z entries. But there is a twin matrix admitting $z=-cosh(u)$ as well. So I was puzzling which one is better or if they are equivalent. EDIT: probably what you meant is my matrix choice represents the upper fold, while the other one would make the bottom one. So probably there's not a unique matrix representing both? – riccardoventrella Jul 21 '19 at 17:41
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    That's right. You have to pick a sheet to do anything computationally, though I suppose you could represent both sheets in one matrix by writing $\pm cosh(u)$. – Jonathan Trousdale Jul 21 '19 at 18:31
  • It makes sense, thanks for your support – riccardoventrella Jul 21 '19 at 21:04
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    Why the first matrix try he did is not a 2-manifold? – Samuel Díaz Feb 28 '21 at 00:10