Ciro Santilli @cirosantilli 34

Like the special orthogonal group is to the orthogonal group, $SO(m,n)$ is the subset of $O(m,n)$ with determinant equal to exactly 1.

This is a good first concrete example of a Lie algebra. Shown at Lie Groups, Physics, and Geometry by Robert Gilmore (2008) Chapter 4.2 "How to linearize a Lie Group" has an example.

We can use use the following parametrization of the special linear group on variables $x$, $y$ and $z$:

Every element with this parametrization has determinant 1:Furthermore, any element can be reached, because by independently settting $x$, $y$ and $z$, $M_{00}$, $M_{01}$ and $M_{10}$ can have any value, and once those three are set, $M_{11}$ is fixed by the determinant.

To find the elements of the Lie algebra, we evaluate the derivative on each parameter at 0:

Remembering that the Lie bracket of a matrix Lie group is really simple, we can then observe the following Lie bracket relations between them:

One key thing to note is that the specific matrices $M_x$, $M_y$ and $M_z$ are not really fundamental: we could easily have had different matrices if we had chosen any other parametrization of the group.

TODO confirm: however, no matter which parametrization we choose, the Lie bracket relations between the three elements would always be the same, since it is the number of elements, and the definition of the Lie bracket, that is truly fundamental.

Lie Groups, Physics, and Geometry by Robert Gilmore (2008) Chapter 4.2 "How to linearize a Lie Group" then calculates the exponential map of the vector $x{M}_x+y{M}_y+z{M}_z$ as:with:

TODO now the natural question is: can we cover the entire Lie group with this exponential? Lie Groups, Physics, and Geometry by Robert Gilmore (2008) Chapter 7 "EXPonentiation" explains why not.