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For the $2 \times 2$ orthogonal group of matrices which for the $SO(2)$ group, there is only one free parameter in the group element and hence only one generator for the group. Which is,

$$ X_g = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right) $$

Now if this generator has to form Lie Algebra, it has to satisfy the Jacobi Identity and commutators. I don't understand how to do this with just one element.

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  • $\begingroup$ There is only one element(generator) in Lie-algebra, am not talking about the group. (No. of generators = No. of parameters in the group) $\endgroup$
    – user38249
    Feb 11, 2014 at 16:33

1 Answer 1

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Since the Lie group $SO(2)$ is abelian it has trivial Lie algebra, i.e., with zero Lie brackets. The "generator of rotations" is indeed $X_g$, which does not imply that the group has only one generator. Note that $$ X_g=\frac{d}{d\alpha}R(\alpha)\mid_{\alpha=0}, $$ where $R(\alpha)$ are the rotation matrices.

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  • $\begingroup$ Oh ok, thanks !! I think I have problem in the terminology "generator". Nevertheless, got the point $\endgroup$
    – user38249
    Feb 11, 2014 at 16:35

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