Confused by introductory quantum group theory book...

[KipIngram]

Hi. I'm looking through this book:

 

https://www.math.columbia.edu/~woit/QM/qmbook.pdf

 

At the very top of page 8, the following is presented:

 

g1 . (g2 . f)(x) = g2 . f(g1^-1 . x)

= f(g2^-1 . (g1^-1 .x))

= ...

 

I understand what they did - they applied equation 1.3 from the previous page treating g1 as g in 1.3 and then applied 1.3 again treating g2 as g.

 

But it looks to me like it would also be valid to do this:

 

g1 . (g2 . f)(x) = g1 . [ (g2 . f)(x) ] = g1 . [ f(g2^-1 . x) ]

 

In other words, it looks like we can great f as a function acted on by group member g2, and move g2 inside first, OR treat g2 . f as a function acted on by group member g1 and move g1 inside first.

 

I suspect only one of these is valid (the second one, which is what's given in the book). But I'm not clearly seeing why.

 

Can anyone advise?

 

[Country Boy]

What you suggest would work if the operator group, G, were commutative. But G here is specifically said to be not comutative.

[KipIngram]

Well, I'm aware of that limitation. But it looks like they're treating g2.f as a function (call it u) and then applying a rule to g1.u. Why is not permissible to first apply that same rule to g2.f? They worked from the outside in - I'm wondering why you can't work from the inside out, being careful in both cases not to commute an operator applications.

 

I'm sure it does wind up breaking one of those simple rules, but it's just not clear to me how - it's not as obvious to me as ab != ba.

 

Thanks for the reply, btw - I'd nearly forgotten about this post.

[uncool]

Your reasoning is almost correct - but you miss that there is a switch there. Remember, the group acts on x, not on the argument of the function. In other words:

 

(g1 . (g2 . f)) (x) = (g1 . h)(x), where h(x) = f(g2^-1 x). Then by the definition, to apply the action of g1, we apply it to x, and get h(g1^-1 x) = f(g2^-1 g1^-1 x).

Edited June 8, 2017 by uncool

[KipIngram]

Thank you very much.