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How is having commutativity or not having it useful for in practice? (whether it is for linear algebra, rings, basic airthmetic etc)

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If you have $aba^{-1}$, you know it's just b. Commutativity makes things much simpler. Or another example. Say you have abcdabcabcacacababa. If you don't have commutativity, it's hard to know what's going on. If you do, just count the number of each and you know

$abcdabcabcac = a^4 b^3 c^4 d$

That's much simpler. Just to be clear on that, if these all commute, then $ababab$, $a^3 b^3$, $b^3 a^3$, $bababa$, $bbaaab$, and many more are all the same thing. So, if you get any of them, you just simplify them down to the easiest one to work with.

Or, for a very simple example, if you're trying to multiply numbers and you have $9 \times 4 \times 1/3$. If I do that in my head, I first flip the order so that $9 \times 1/3 = 3$ and then multiply that by 4 to get 12. I could also do $9 \times 4 = 36$, so the answer is $36 \times 1/3 = 12$, but that's a bit harder. And, that generalizes to problems where "the other way" actually is much harder.

By the way, this is why commutativity is useful. Now, because everything is simpler, you get lots of extra things. For example, the fundamental theorem of finitely generated abelian groups gives us a very nice structure for any finitely generated abelian group.

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In geometry, non-commutativity is what allows us to have curvature.

Think about the following operations:

  • Walking north 1 mile. Call this operation $n$.

  • Walking west 1 mile. Call this $w$.

To describe a sequence of such operations, write them in order from right to left (like function composition). So $wn$ is the operation "walk north 1 mile, then walk west 1 mile".

If you live in Flatland, $wn = nw$; they commute. If you live on the earth (a sphere, approximately), they do not! (It may help to imagine trying it near the poles, where the effect is most extreme.) It's precisely the curvature of the earth that causes this effect. On a cylindrical planet, they'd commute again (a cylinder is effectively "flat"); on a hyperbolic planet, they wouldn't.

Indeed, in Riemannian geometry, the very definition of curvature is in terms of Lie brackets, which are nothing but glorified commutators; they measure failure (of vector fields) to commute with one another.

(Geometers will complain that my $w$ doesn't move along a geodesic in the spherical case, but I think it still gets the point across. If you can see an improved example, I'd be interested.)

A further exercise: Think about driving a car, where $d$ is the operation "drive the car forward", and $r$ is "turn the steering wheel right". Convince yourself that $d$ and $r$ do not commute.
Since your profile says you live in the UK, consider the operation $$d^{-1}r^{-1}d^{-1}rdrdr^{-1}.$$ What are the practical applications of this operation? How would that change if $d$ and $r$ did commute? (Examples like this are what make sub-Riemannian geometry and control theory interesting subjects.)

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This is an amazing answer, inspriational! Is commutativity in geometry with two dimension movements then because there is symmetry in the triangles of the vectors they sum too? Is a curved surface with 2 dimensions you work on actually implying the function of a 3rd dimension that is not written explicitly? A cylinder is flat cause the 3rd dimension does not come in at all times? (with the driving) So commutativity is also really tied to the problem considered..? (sheepishly) I what is Riemannian geometry, i have never understood it, should I as it as a question? Thanks! –  Vass Apr 14 '11 at 14:47

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