Ciro Santilli @cirosantilli 34

Solution $Z$ for given $X$ and $Y$ of:where $e$ is the exponential map.

If we consider just real number, $Z=X+Y$, but when X and Y are non-commutative, things are not so simple.

Furthermore, TODO confirm it is possible that a solution does not exist at all if $X$ and $Y$ aren't sufficiently small.

This formula is likely the basis for the Lie group-Lie algebra correspondence. With it, we express the actual group operation in terms of the Lie algebra operations.

Notably, remember that a algebra over a field is just a vector space with one extra product operation defined.

Vector spaces are simple because all vector spaces of the same dimension on a given field are isomorphic, so besides the dimension, once we define a Lie bracket, we also define the corresponding Lie group.

Since a group is basically defined by what the group operation does to two arbitrary elements, once we have that defined via the Baker-Campbell-Hausdorff formula, we are basically done defining the group in terms of the algebra.

Elements of a Lie algebra can (should!) be seen a continuous analogue to the generating set of a group in finite groups.

For continuous groups however, we can't have a finite generating set in the strict sense, as a finite set won't ever cover every possible point.

But the generator of a Lie algebra can be finite.

And just like in finite groups, where you can specify the full group by specifying only the relationships between generating elements, in the Lie algebra you can almost specify the full group by specifying the relationships between the elements of a generator of the Lie algebra.

This "specification of a relation" is done by defining the Lie bracket.

The reason why the algebra works out well for continuous stuff is that by definition an algebra over a field is a vector space with some extra structure, and we know very well how to make infinitesimal elements in a vector space: just multiply its vectors by a constant $c$ that cana be arbitrarily small.

Like everything else in Lie groups, first start with the matrix as discussed at Section "Lie algebra of a matrix Lie group".

Intuitively, a Lie algebra is a simpler object than a Lie group. Without any extra structure, groups can be very complicated non-linear objects. But a Lie algebra is just an algebra over a field, and one with a restricted bilinear map called the Lie bracket, that has to also be alternating and satisfy the Jacobi identity.

Another important way to think about Lie algebras, is as infinitesimal generators.

Because of the Lie group-Lie algebra correspondence, we know that there is almost a bijection between each Lie group and the corresponding Lie algebra. So it makes sense to try and study the algebra instead of the group itself whenever possible, to try and get insight and proofs in that simpler framework. This is the key reason why people study Lie algebras. One is philosophically reminded of how normal subgroups are a simpler representation of group homomorphisms.

To make things even simpler, because all vector spaces of the same dimension on a given field are isomorphic, the only things we need to specify a Lie group through a Lie algebra are:Note that the Lie bracket can look different under different basis of the Lie algebra however. This is shown for example at Physics from Symmetry by Jakob Schwichtenberg (2015) page 71 for the Lorentz group.

As mentioned at Lie Groups, Physics, and Geometry by Robert Gilmore (2008) Chapter 4 "Lie Algebras", taking the Lie algebra around the identity is mostly a convention, we could treat any other point, and things are more or less equivalent.

Bibliography:

For this sub-case, we can define the Lie algebra of a Lie group $G$ as the set of all matrices $M\in G$ such that for all $t\in \mathbb{R}$:If we fix a given $M$ and vary $t$, we obtain a subgroup of $G$. This type of subgroup is known as a one parameter subgroup.

The immediate question is then if every element of $G$ can be reached in a unique way (i.e. is the exponential map a bijection). By looking at the matrix logarithm however we conclude that this is not the case for real matrices, but it is for complex matrices.

Examples:

TODO example it can be seen that the Lie algebra is not closed matrix multiplication, even though the corresponding group is by definition. But it is closed under the Lie bracket operation.

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.

This makes it clear how the Lie bracket can be seen as a "measure of non-commutativity"

Because the Lie bracket has to be a bilinear map, all we need to do to specify it uniquely is to specify how it acts on every pair of some basis of the Lie algebra.

Then, together with the Baker-Campbell-Hausdorff formula and the Lie group-Lie algebra correspondence, this forms an exceptionally compact description of a Lie group.

Based on the $L_x$,$L_y$ and $L_z$ derived at Lie algebra of $SO(3)$ we can calculate the Lie bracket as: