# Why U(1) and SO(2) are locally equivalent?

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!

• Are you sure it said something like that involving those two groups? Usually people talk about $SU(2)$ and $SO(3)$ being locally equivalent... Feb 19, 2013 at 20:41
• 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. Feb 19, 2013 at 20:41
• Sorry, it should have been SO(2) instead of SO(3). Feb 19, 2013 at 20:46
• 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. Feb 19, 2013 at 20:48
• What Mariano says. The isomorphism is $$e^{i\phi}\mapsto \pmatrix{\cos\phi&-\sin\phi\cr\sin\phi&\cos\phi}.$$ Feb 19, 2013 at 21:18

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 $z$.