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I had been naively thinking that $GL_n(\mathbb C)$,$SL_n(\mathbb C)$,$U_n(\mathbb C)$ and $SU_n(\mathbb C)$ being the complex analogs of $GL_n(\mathbb R)$ etc should be complex Lie groups.

But wiki says that while $GL_n(\mathbb C)$ is indeed a complex Lie group, $SU_n(\mathbb C)$ is not.

  1. What makes the general linear group a complex Lie group but the special unitary group not one?
  1. What can be said about the special linear and unitary group over $\mathbb C$?
Angry_Math_Person
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  • The definition of a Lie group. Although $SU_n(\Bbb C)\subseteq GL_n(\Bbb C)$, it is not a complex Lie group, i.e., multiplication and inverse are not complex differentiable - see wikipedia.
    • – Dietrich Burde Feb 28 '22 at 09:38
  • I know the lie algebra is the trace zero skew hermitian matrices. So the dimension of that space will be $n^2 -1$? Okay that seems like a way to immediately say that it is not a complex lie group. But can one give a more topological answer? Or is it the case that any 2n dim matrix real Lie group is actually a complex lie group? and dimension argument is the only way? – Angry_Math_Person Feb 28 '22 at 09:39
  • @Angry_Math_Person you are correct that $SU_n(\mathbb{C})$ is of real dimension $n^2-1$. The necessary condition is indeed that the real dimension has to be even. But note that for odd $n$, the dimension $n^2-1$ is even. However $SU_n(\mathbb{C})$ is never a complex Lie group. General case requires a more sophisticated argument. – freakish Feb 28 '22 at 09:40
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    Wikipedia: A ''complex Lie group'' is a Lie group over the complex numbers; i.e., it is a complex manifold that is also a group in such a way that $G \times G \to G, (x, y) \mapsto x y^{-1}$ is holomorphic. And, is this true for $SU_n(\Bbb C)$? Which justification do you have for saying, "naively it should be"? I don't see this. – Dietrich Burde Feb 28 '22 at 09:41
  • What about $SL_n$? Here we do get even real dim. But how can we say/not say it is a complex lie group? – Angry_Math_Person Feb 28 '22 at 09:45
  • @DietrichBurde You are right. I was naively replacing $\mathbb R$ by $\mathbb C$ in my head and hoping it would work out. It seems to be much more complicated than that. – Angry_Math_Person Feb 28 '22 at 09:48
  • For $SL_n$ we also can consider the "complexification", see for example the answers to this post. – Dietrich Burde Feb 28 '22 at 09:57