# Why do infinitesimal rotations commute but finite rotations do not?

by MuIotaTau
Tags: commute, finite, infinitesimal, rotations
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P: 15,204
 Quote by voko I am not sure what you mean by that. Is that that adding ##2 \pi## to any rotation angle is essentially a no-operation?
Basically, yes. You said that "3D rotations and 3D vectors can be mapped 1:1." (Post #15). That's simply not true. The mapping from 3D rotations to vectors is one to many, not one to one.

 You are correct in saying that by "represented as/by a vector" I meant "represented by three numbers" in general and a single axis rotation in particular, where the vector-ish representation has a geometrical meaning as well.
There's a difference between "as a vector" and "by a vector", at least to me. The former implies all the baggage that goes along with vectors. In particular, commutativity of vector addition and distributivity of scalar multiplication don't hold for rotations in RN (rotations in 2D space excepted). The latter merely means I can take advantage of Euler's rotation theorem to describe a rotation in three space. (Note well: That N parameters are needed to describe a rotation in N dimensional space only applies to rotation in three dimensional space. Only one parameter is needed to describe a rotation in two dimensional space, six are needed to describe a rotation in four dimensional space, and in general N*(N-1)/2 are needed to describe a rotation in N-dimensional space.)
Thanks
P: 5,869
 Quote by D H Basically, yes. You said that "3D rotations and 3D vectors can be mapped 1:1." (Post #15). That's simply not true. The mapping from 3D rotations to vectors is one to many, not one to one.
I do not see why this needs to be the case. Rotations about a fixed axis, or 2D rotations, can be mapped to real numbers 1:1, at least formally. You admit that as well. Hence rotations in 3D can be mapped to 3D vectors 1:1. I do not think we are required to map rotations to vectors modulo ##2 \pi ##. To avoid confusion, I again mean the mapping with the rotation group structure induced by that of the 3D vector space, not the SO(3) structure.

 There's a difference between "as a vector" and "by a vector", at least to me. The former implies all the baggage that goes along with vectors. In particular, commutativity of vector addition and distributivity of scalar multiplication don't hold for rotations in RN (rotations in 2D space excepted). The latter merely means I can take advantage of Euler's rotation theorem to describe a rotation in three space.
I think I have explained by now that what I mean involves mapping between the two sets that does not preserve the SO(3) structure and I do not think we disagree in any significant way on any detail of that. But I find your distinction between "by" and "as", as well as the reference to the "vectorial baggage" puzzling. Vectors do not have any intrinsic SO(3) baggage, hence I do not see why a mapping that does not preserve that is somehow less vectorial than one that would.

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