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.

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:

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.

The basic intuition for this is to start from the origin and make small changes to the function based on its known derivative at the origin.

More precisely, we know that for any base b, exponentiation satisfies:And we also know that for $b=e$ in particular that we satisfy the exponential function differential equation and so:One interesting fact is that the only thing we use from the exponential function differential equation is the value around $x=0$, which is quite little information! This idea is basically what is behind the importance of the ralationship between Lie group-Lie algebra correspondence via the exponential map. In the more general settings of groups and manifolds, restricting ourselves to be near the origin is a huge advantage.

Now suppose that we want to calculate $e^1$. The idea is to start from $e^0$ and then then to use the first order of the Taylor series to extend the known value of $e^0$ to $e^1$.

E.g., if we split into 2 parts, we know that:or in three parts:so we can just use arbitrarily many parts $e^{1/n}$ that are arbitrarily close to $x=0$:and more generally for any $x$ we have:

Let's see what happens with the Taylor series. We have near $y=0$ in little-o notation:Therefore, for $y=x/n$, which is near $y=0$ for any fixed $x$:and therefore:which is basically the formula tha we wanted. We just have to convince ourselves that at ${\lim }_{n\to \infty }$, the $o(1/n)$ disappears, i.e.:

To do that, let's multiply $e^y$ by itself once:and multiplying a third time:TODO conclude.