Ciro Santilli @cirosantilli 35

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.

Group of rotations of a rigid body.

Like orthogonal group but without reflections. So it is a "special case" of the orthogonal group.

This is a subgroup of both the orthogonal group and the special linear group.

Intuition, please? Example? mathoverflow.net/questions/278641/intuition-for-symplectic-groups The key motivation seems to be related to Hamiltonian mechanics. The two arguments of the bilinear form correspond to each set of variables in Hamiltonian mechanics: the generalized positions and generalized momentums, which appear in the same number each.

Seems to be set of matrices that preserve a skew-symmetric bilinear form, which is comparable to the orthogonal group, which preserves a symmetric bilinear form. More precisely, the orthogonal group has:and its generalization the indefinite orthogonal group has:where S is symmetric. So for the symplectic group we have matrices Y such as:where A is antisymmetric. This is explained at: www.ucl.ac.uk/~ucahad0/7302_handout_13.pdf They also explain there that unlike as in the analogous orthogonal group, that definition ends up excluding determinant -1 automatically.

Therefore, just like the special orthogonal group, the symplectic group is also a subgroup of the special linear group.