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A text states every complex connected Lie group must be abelian. Now surely this must be an error and has to be wrong because the unitary group is certainly (has to be complex in general) complex except for the real subgroups such as real orthogonal group and it is NOT abelian. For an easy example $SU_2$ for which the generators are the 2 dim. Pauli spin matrices in which the usual representation has $\sigma_2=\sigma_y$ say as imaginary and then taking $\exp(\sqrt{-1}t_i\sigma_i)$ for arbitrary real scalars $t_i$,implicit repeated sum over $i$ $1$ to $3$ and all possible infinite products of such is certainly a complex Lie group and not abelian. Eg obviously the subgroups $\exp(\sqrt{-1}\sigma_{1\text{ or }3})$ has imaginary generators $\sqrt{-1}\sigma_{1\text{ or }3}$ in a familiar basis. So would we not all agree the statement is false?

José Carlos Santos
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user158293
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    My guess is it’s a gaff and they meant something else, or you didn’t read it correctly. For instance perhaps there should be a hypothesis about compactness. – Charlie Frohman Feb 23 '20 at 12:54
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    the result requires compact complex connected and the proof is one line - analytic functions on compact complex manifolds are constant by the maximum modulus theorem, but the adjoint representation is such (it is a matrix of analytic maps), hence it is constant, hence the group is abelian – Conrad Feb 23 '20 at 13:13

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It is not true that every complex connected Lie group must be Abelian; for instance, $SL(2,\Bbb C)$ is not Abelian. However, it is true that every complex, connected and compact Lie group must be Abelian (this is a consequence of the maximum modulus principle). And, yes, $SU(2)$ is not abelian. There is no contradiction here, since $SU(2)$ is a real Lie group but not a complex one (notice that its dimension as a real differentiable manifold is $3$; if it had the structure a complex differentiable manifold, its dimension as a real differentiable manifold should be even).

José Carlos Santos
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  • Why isn’t $SL_2\mathbb{C}$ a complex connected Lie group? – Charlie Frohman Feb 23 '20 at 10:56
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    I can't answer that question, since it is a complex connected Lie group. – José Carlos Santos Feb 23 '20 at 11:00
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    and it is not abelian. – Charlie Frohman Feb 23 '20 at 12:31
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    you forgot compact – Conrad Feb 23 '20 at 13:10
  • yea i forgot compact anyway my main point is that su2 is complex in every sense of the word and not abelian and is 2 dimensional - well it has representations in all even dimensions being j=1/2 odd spin – user158293 Feb 23 '20 at 13:35
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    $SU(2)$ is a real form of the complex group $SL_2\mathbb{C}$. Think of the real numbers inside the complex numbers, there is an antiholomorphic involution, complex conjugation, that has the real numbers as its fixed point set. Similarly, the involution that sends every element of $SL_2\mathbb{C}$ to the conjugate transpose of its inverse, has $SU(2)$ has its fixed point set. My comments are just a rigorous version of what you were trying to say. – Charlie Frohman Feb 23 '20 at 17:07
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    @user158293 $\mathfrak{su}(2)$ has real dimension $3$, so its complex dimension is $1.5$... Complex in every sense of the word, except the precise mathematical sense. – YCor Feb 23 '20 at 22:54
  • ok that was an overstatement . It's not complex in every sense if you are considering complex manifold which i don't really know exactly what it means but i guess a non real geometry? and can't be imbedded in a real rectangular or even a non-flat space with real coordinates of the same dimension? – user158293 Feb 24 '20 at 20:56
  • Actually, $SU(2)$ is isomorphic to the group of matrices of the form$$\begin{bmatrix}a & b & -c & d \ -b & a & -d & -c \ c & d & a & -b \ -d & c & b & a\end{bmatrix},$$with $a,b,c,d\in\mathbb R$ and $a^2+b^2+c^2+d^2=1$. – José Carlos Santos Feb 24 '20 at 21:14
  • is U(2) also considered a real lie algebra ? Since it is compact and connected and not abelian i guess according to the above it would have to be ? Is it correct that there is no such thing as U(2,c) since that would be the same as U(2) ? – user158293 Feb 24 '20 at 21:40
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    Actually, $U(2)$ is a Lie group, not a Lie algebra. And, yes, it is a real Lie group. – José Carlos Santos Feb 24 '20 at 21:46
  • Yea i knew that, my text error. So there is no such thing as U(2,c) lie group? since it is same as U(2)? What about the group of 2x2 matrices such that the transpose of complex conjugate times itself is any complex number of modulus 1? eg exp(sqrt(-1)*t) for any real t. Or would that just be direct product of U(2) and U(1)? and still a real lie group? – user158293 Feb 24 '20 at 22:47
  • I don't see how I would define $U(2,\mathbb C)$. – José Carlos Santos Feb 24 '20 at 22:48
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    yea i can't either other than same as U(2). I am trying to get a feel of what makes something a complex lie group because it isn't just because it's representation is complex. – user158293 Feb 24 '20 at 22:54
  • Yes, it is. ${}$ – José Carlos Santos Feb 24 '20 at 23:02
  • yea i figured that so(1,1) had to be a real lie group shortly after i asked along with all unitary groups though not real sure about all complex orthogonal groups but my guesstimate would be that they are real lie groups. What makes SL(2,C) a complex lie group and complex orthogonal group in 2 dimensions a real lie group? If by SL(2,C) one means only the determinant is equal 1? It seems to me there are an infinity of elements in SL(2,C) not in complex orthogonal group of 2 dimensional matrices but the complex orthogonal group whose det is 1,called SO(2,c)?, is a subgroup of SL(2,c). – user158293 Feb 25 '20 at 10:34
  • I have just read that O3(c) is a complex group but not sure about O2(c) ? nor SO3(c) for that matter? – user158293 Feb 25 '20 at 12:46
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    All of them are complex Lie groups. – José Carlos Santos Feb 25 '20 at 13:37