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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?

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    $\begingroup$ 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. $\endgroup$ Feb 23, 2020 at 12:54
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    $\begingroup$ 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 $\endgroup$
    – Conrad
    Feb 23, 2020 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).

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    $\begingroup$ I can't answer that question, since it is a complex connected Lie group. $\endgroup$ Feb 23, 2020 at 11:00
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    $\begingroup$ and it is not abelian. $\endgroup$ Feb 23, 2020 at 12:31
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    $\begingroup$ you forgot compact $\endgroup$
    – Conrad
    Feb 23, 2020 at 13:10
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    $\begingroup$ $SU(2)$ 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. $\endgroup$ Feb 23, 2020 at 17:07
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    $\begingroup$ @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. $\endgroup$
    – YCor
    Feb 23, 2020 at 22:54

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