0
$\begingroup$

I read from some physics books that the 2D rotation can be expressed in exponential form:

$$R(\theta) = e^{X(\theta)}, \theta \in R $$

In physics, X relates to the momentum of a particle, and is defined to be the generator of the rotation. Some authors also state that X is the generator of the rotation group .

Is that statement correct, and if it is, can we consider the rotation group ($N\ge2$) as a cyclic group. As if I understand correctly, only cyclic groups have a generator.

$\endgroup$
4
  • 1
    $\begingroup$ A group is called cyclic if all elements are powers of one single element, i.e., if it can be generated by a single element. This answers your question. $\endgroup$ Nov 3, 2020 at 12:36
  • 1
    $\begingroup$ As @DietrichBurde says, cyclic groups are groups generated by a single element. But be careful, in some languages, "cyclic" means "cyclic and finite". And there is another word for "cyclic infinite". (For example in French, "cyclique" means that the group is also finite, and we call "monogène" for an infinite cyclic group). So maybe the answer can depend on which language you consider :) $\endgroup$ Nov 3, 2020 at 12:43
  • $\begingroup$ @TheSilverDoe: Do you mean that there is more than 1 definition of cyclic group? $\endgroup$
    – Rekkhan
    Nov 3, 2020 at 13:41
  • 1
    $\begingroup$ @Rekkhan In English no. In English, "cyclic group" means that the group is generated by a single element. I just wanted to underline the fact that if you are not English, maybe "cyclic" in your language means another thing ! $\endgroup$ Nov 3, 2020 at 13:42

1 Answer 1

3
$\begingroup$

In case of a rotation group, this is called (in mathematics) an infinitesimal generator and is not a generator in the sense you mention.

For a cyclic group, there is a generator $G$ such that the elements of the group are $G^k$ with $k \in \mathbb Z$. For a rotation group, there is an "infinitesimal generator" $X$ such that the elements of the group are $e^{tX}$ with $t \in \mathbb R$.

====

If $R(t) = e^{tX}$ is the rotation group, then the derivative is $$ R'(t) = \lim_{t \to 0}\frac{R(t) - I}{t} = X $$ (the identity matrix $I = R(0)$) and that is the sense it is an "infinitesimal generator". We have $$ R(t) = I + tX + O(t^2) $$ so if $t$ is very small, then rotation is approximately $I+tX$. So $X$ is the "tangent direction" to the rotation.

$\endgroup$
3
  • $\begingroup$ Thank you, but another question is whether the infinitesimal generator X is a group element or not? And can I consider the SO(2) as an cyclic group, where the generator is an infinitesimal rotation? namely we can apply a "large number" of infinitesimal rotation to obtain a finite rotation. (I deleted my previous reply) $\endgroup$
    – Rekkhan
    Nov 3, 2020 at 13:53
  • 1
    $\begingroup$ No, in general $X$ does not belong to the Lie group, but to the tangent space -- the corresponding Lie algebra. $\endgroup$
    – GEdgar
    Nov 3, 2020 at 16:16
  • $\begingroup$ Thank you very much. $\endgroup$
    – Rekkhan
    Nov 4, 2020 at 7:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .