# What does it mean to be a real Lie group

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 ?

• It means that multiplication and inversion are smooth maps: en.wikipedia.org/wiki/Lie_group – Ayman Hourieh Jun 4 '14 at 20:25
• @AymanHourieh, does it also imply conditions on the complex conjugates of the elements ? – Nick Jun 4 '14 at 20:26
• @Nick: complex conjugation is something specific to elements of $\text{SU}(n)$ thought about as a matrix Lie group in a particular way. It's not a feature of general Lie groups. – Qiaochu Yuan Jun 4 '14 at 20:32
• If you have a complex Lie group, then you can also make it real. – Christopher A. Wong Jun 4 '14 at 20:32
• @ChristopherA.Wong, that's by searching for an equivalence with other real Lie-groups then ? I was wondering of the reality of the Lie-group also imposes conditions on the $\alpha_a(x)$ (see updated question). – Nick Jun 4 '14 at 20:34

## 1 Answer

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.

• $SU(N)$ is not complex Lie group, so there is no forgetful functor in sight here, really :-) For one thing, it has odd real-dimension half of the time. – Mariano Suárez-Álvarez Jun 5 '14 at 17:29
• Just replace $SU(n)$ with $SO(3,1)_o$ for instance. (It is a complex Lie group but not in a completely obvious way.) – Moishe Kohan Jun 6 '14 at 20:53
• @MarianoSuárez-Alvarez: Oops. Blah. I picked a different and hopefully correct example. – Lee Mosher Jun 7 '14 at 14:41
• @LeeMosher, thanks for the complete answer! In some course of mine it was claimed that because of the fact that $SU(2)$ is a Lie-group we could define a complex conjugation as: $i\sigma_2$, which yields a complex conjugate as operator. But I don't see how this might be related ? – Nick Jun 7 '14 at 15:53
• @Nick: If you can formulate a solid question along those lines, it might make a good new question. – Lee Mosher Jun 7 '14 at 18:32