In one of my particle physics textbooks, I came across the statement that U(1) and SO(2) are locally equivalent. I don’t really know what it means. I know a bit of group theory and that $U(1)$ is just the group of $1$-D unitary transformations and $SO(2)$ the $2$-D proper rotation group.

Could anyone explain the statement in both a formal and intuitive way? Thanks!

  • 4
    $\begingroup$ Are you sure it said something like that involving those two groups? Usually people talk about $SU(2)$ and $SO(3)$ being locally equivalent... $\endgroup$ Feb 19, 2013 at 20:41
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    $\begingroup$ The two groups are quite different: $U(1)$ is abelian and $1$-dimensional, while $SO(3)$ is non-abelian and of dimension $3$. There is no sensible sense in which the are locally equivalent. $\endgroup$ Feb 19, 2013 at 20:41
  • $\begingroup$ Sorry, it should have been SO(2) instead of SO(3). $\endgroup$
    – Funzies
    Feb 19, 2013 at 20:46
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    $\begingroup$ In that case, you can eaiy describe each of the group and see that they are not only «locally equivalent», whatever that may mean, but actually the same. $\endgroup$ Feb 19, 2013 at 20:48
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    $\begingroup$ What Mariano says. The isomorphism is $$e^{i\phi}\mapsto \pmatrix{\cos\phi&-\sin\phi\cr\sin\phi&\cos\phi}.$$ $\endgroup$ Feb 19, 2013 at 21:18

1 Answer 1


Both $U(1)$ and $SO(2)$ are geometrically the circle $S^1$.

$U(1)$ is the set of complex numbers (or $1\times 1$ complex matrices) such that $z\bar{z} = \bar{z}z = 1$. That is, the set $e^{i\theta}$ for real $\theta$.

$SO(2)$ is the set of rotations in $\mathbb R ^2$. The rotations $r_\theta$ form a group under composition.

These two groups are isomorphic, and the isomorphism is just the one described by Jyrki: for a complex number $z$ in $U(1)$ the corresponding element in $SO(2)$ is the rotation by the angle which is the argument of $z$.


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