# Why are infinitesimal rotations commutative, whereas finite rotations are not?

Infinitesimal rotations commute and every finite rotation is the composition of infinitesimal rotations which should logically mean they also commute; but they don't. Why?

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Infinitesimal rotations don't commute exactly if you're accurate enough. An infinitesimal rotation may be written as $$\exp( i a A )$$ where $a$ is an infinitesimal "angle" and $A$ is a combination of generators. Such an object doesn't commute with the analogous object $\exp(ibB)$ in general. Instead, $$\exp(iaA) \exp(ibB) = \exp(ibB) \exp(iaA) \exp(-ab [A,B] + O(a_i^3))$$ where $[A,B]=AB-BA$ is the ordinary "commutator" of operators i.e. the generators (of the bases "vectors" $A,B$ of the Lie algebra associated with the Lie group). The equation above may be verified by carefully expanding the exponentials on both sides to the second order in $a$ or $b$, ignoring cubic and higher-order terms, but being careful about the ordering of $A$ and $B$ etc.
The failure of the infinitesimal rotations to commute is only expressed by a smaller angle $ab$ which is second order but the accumulation of these $O(a_i^2)$ terms is what makes finite rotations "obviously noncommuting". Why? Because if you want to interchange $N$ copies of $\exp(iaA)$ in $\exp(iNaA)$ with $M$ copies of $\exp(ibB)$ in $\exp(iNbB)$, you need to make $MN$ similar permutations, so assuming that $Ma$ and $Nb$ are finite, the factors of $MN$ (large) and $ab$ (small) cancel and you get a finite difference between the products written in the opposite orders.
If you want to play around with this numerically, you can use Wolfram Alpha; here, you see that two rotations of an angle $a$ around the X and Z axis don't commute, but leave a remainder of order $a^2.$ – Gerben Oct 31 '11 at 0:11