Why do actions become commutative as we reduce the 'size' of it?

Question

If we have a physical object, then infinitesimal rotations have commutative behaviour while finite rotations have anti-commutative behaviour. I find this very intriguing as to why as we take smaller and smaller actions the actions start becomes commutative whilst for large actions it's not. So, due to this phenomena rotations become vectors as when we take differential rotations.

Reference from a physics book 'Fundamentals of physics' by Resnick, Halliday, Walker:

Answer 1

They don't.

(I will use essentially the same expression as Chrystomath's answer, but express the conclusions differently.)

Call the generators of rotations around $x$ and $y$ axes $X$ and $Y$, respectively, and consider rotations by angles $\alpha$ and $\beta$ The noncommutativity of two rotations $e^{\alpha X}$ and $e^{\beta Y}$ is expressed by the commutator $[\alpha X,\beta Y]$ (and higher commutators) in the Baker-Campbelll-Hausdorff formula, and for noncommuting elements, this is never zero.

What is true is that in the limit of rotation angles going to zero, the commutator goes to zero faster than the individual rotations. However, this is analogous to a first-order expansion of a function, $f(x+\epsilon)=f(x)+\epsilon f'(x)+\frac12\epsilon^2 f''(x)+\dotsm$. In the limt $\epsilon\to0$, the second-order term vanishes faster than the first-order one. In that sense, you could say that "locally, all functions are linear", but that is a misleading way to express the first-order approximation.

Answer 2

This is because small rotations are relatively close to being the identity function which commutes with everything. Lets look at two dimensional rotations to get a feel for why this is.

A rotation matrix with respect to the canonical basis is of the form $\pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta} = \cos \theta I + \sin \theta J$ with $J=\pmatrix{0 & -1 \\ 1 & 0}$. In this case, because $\theta$ is small, $\sin \theta$ is approximately $x$ near zero, the contribution of $J$ to the rotation is also small. We can also see that $\cos \theta$ should be close to $1$ when $\theta$ is near zero so the rotation as a whole is approximately $I$.

Answer 3

This is true not just of rotations but any Lie group actions. 'Small' actions, meaning those that are near the identity, can be written as exponentials, $R=e^{X}$. For example, for rotations, $X$ is a skew-symmetric matrix.

The Baker-Campbell-Hausdorff formula states $$e^{X}e^{Y}=e^{X+Y+[X,Y]/2+\cdots}=e^{Y}e^{X}E $$ If $X$, $Y$ are small then $E=e^{[X,Y]+\cdots}$ is almost the identity, because it consists of second order terms, so the actions related to $X$, $Y$, almost commute.