Ciro Santilli @cirosantilli 35

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.

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.

Every Lie algebra corresponds to a single simply connected Lie group.

The Baker-Campbell-Hausdorff formula basically defines how to map an algebra to the group.

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