- #1
cianfa72
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- TL;DR Summary
- About the ##SL(2,\mathbb R)## Lie group parametrization as manifold
Hi,
consider the set of the following parametrized matrices
$$
\begin{bmatrix}
1+a & b \\
c & \frac {1 + bc} {1 + a} \\
\end{bmatrix}
$$
They are member of the group ##SL(2,\mathbb R)## (indeed their determinant is 1). The group itself is homemorphic to a quadric in ##\mathbb R^4##.
I believe the above parametrization is just a chart for the group as manifold. It makes sense only for ##a \neq -1## and this condition yields an open subset in ##\mathbb R^3##. On this open set the map is bijective.
That means the are matrices of ##SL(2,\mathbb R)## that cannot be parametrized by the above map (i.e. matrices with the element in the first row/column equal 0).
My question is: the fact that the above parametrization does not cover entirely the group manifold does not rule out in principle that ##SL(2,\mathbb R)## as manifold might be homeomorphic with ##\mathbb R^3##. Thanks.
consider the set of the following parametrized matrices
$$
\begin{bmatrix}
1+a & b \\
c & \frac {1 + bc} {1 + a} \\
\end{bmatrix}
$$
They are member of the group ##SL(2,\mathbb R)## (indeed their determinant is 1). The group itself is homemorphic to a quadric in ##\mathbb R^4##.
I believe the above parametrization is just a chart for the group as manifold. It makes sense only for ##a \neq -1## and this condition yields an open subset in ##\mathbb R^3##. On this open set the map is bijective.
That means the are matrices of ##SL(2,\mathbb R)## that cannot be parametrized by the above map (i.e. matrices with the element in the first row/column equal 0).
My question is: the fact that the above parametrization does not cover entirely the group manifold does not rule out in principle that ##SL(2,\mathbb R)## as manifold might be homeomorphic with ##\mathbb R^3##. Thanks.
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