What does it mean to be a real Lie group

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What does it mean to be a real Lie-group ?

For example it is said that $SU(N)$ is a real Lie-group. While for example for $SU(2)$ the 2 dimensional matrix-representation consists of the Pauli matrices (which are complex). What does that mean for this group ? Does it imply some extra structures ?

addendum: for a Lie-group we write the group-elements as exponents:$$g=\exp\left(\sum_a\alpha_a(x)t_a\right),$$with $t_a$ the generators of the group (for the $SU(N)$-example this would be traceless anti-hermitian matrices) and $\alpha_a(x)$ the different parameters.

Does being a real Lie-group imply that the $\alpha_a(x)$ are real ?

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A "real Lie group" means, simply, a Lie group, which by definition is a differentiable manifold equipped with a group operation that is differentiable and has differentiable inversion map.

A "complex Lie group" means a complex manifold equipped with a group operation that is complex differentiable and has complex differentiable inversion map.

If one were to take a group like $GL(n,\mathbb{C})$, which is a complex Lie group, and then to write a sentence like "$GL(n,\mathbb{C})$ is a real Lie group", this is an example of a forgetful functor. In it's most natural definition, $GL(n,\mathbb{C})$ is a complex Lie group. But, every complex Lie group is a real Lie group. You simply forget the imaginary unit "i" and treat a complex number $x+iy$ as a pair of real numbers $(x,y)$. Forgetting "i" lets you treat $\mathbb{C}^n$ as $\mathbb{R}^{2n}$, and it lets you treat complex differentiable maps between open subsets of $\mathbb{C}^n$ as differentiable maps, in the ordinary sense, between open subsets of $\mathbb{R}^{2n}$.

So to answer your question "Does it imply some extra structure?" --- no indeed, it implies less structure.

Where things get fun is where you ask questions like this:

  • Do there exist two complex Lie groups which, after applying the forgetful functor, become isomorphic real Lie groups?
  • Does there exist a real Lie group which is not isomorphic, as a real Lie group, to some complex Lie group with the forgetful functor applied?

The answers are "yes" and "yes", and the requisite examples can be found in textbooks on the subject.