...Prove Their Worth...

"Problems worthy of attack
prove their worth
by hitting back." - Piet Hein

A kind of running diary and rambling pieces on my struggles with assorted books, classes, and other things, as they happen. You must be pretty bored to be reading this...

Saturday, September 28, 2002

Heh. a*b does not generally equal b*a. By 'generally', I refer to cases where a and b are members of a general group - that is, they ain't necessarily the sensible numbers everyone meets in school, but more perverse things. So, ab != ba. Having drilled into myself with some exercises, I read this: [Uj(g)f](v) = f(g-1v). This somewhat intimidating (to me) equation comes up in the definition of 'representations of SU(2)'. SU(2) is the name of a group. It happens to just be a set of matrices satisfying some equations, but the point is that it's a group, and as such is a very ethereal beast. A lot of the time, what you really want to know is what a group does, in the 'real' world. Well, to do that, you need to figure out what the representation of a group is, for whatever 'world' you're living in. Hrm. This attempt at a metaphor is going nowhere fast.



Anyway. The interesting bit, as far as I'm concerned here, is the g-1 part. See, in proving that the equation given above does indeed define a representation, something rather strange is done (implicitly, which is worse): (a*b)-1 = b-1*a-1 is assumed to be a true statement. And remember, the beasties we're dealing with here are NOT commutative, a*b is NOT b*a. So what gives?



Well, it took me a while, but I've figured it out. See the a-1 part? Well, normally, the superscript means exponentiation. And for a normal, real number, which I'll call c, c-1 = 1/c. Simple enough. c-1 is also used to denote the inverse of c. This just means that c*c-1 = 1 -- the inverse of a number just means 'that by which you need to multiply to get 1'. And it works out: c*c-1 = c*1/c = 1. Everything's peachy. But that's with normal numbers.



We're not dealing with normal numbers. The -1 exponent here ONLY denotes 'inverse'. That is, a*a-1 = I, where I is basically a general version of the concept behind the number 1 (Meaning, a*I = a. Just like, say, 4*1 = 4.)



So. Let's take the inverse of a*b, and play with it. (ab)*(ab)-1 = I. That's by definition of 'inverse'. Now let's expand everything. a*b*a-1*b-1 is still supposed to equal I. But we've got a big problem. See, because ab != ba, we can't rearrange the terms in the above. We're stuck doing things in the order they appear. So, we've got a*(b*a-1)*b-1. The quantity in parentheses won't be equal to I, obviously, because the inverse of a ain't the inverse of b. So the stuff in parentheses will equal something we'll call c, and we'll have something like a*c*b-1, which obviously won't equal I! So everything blew up. The reason it blew up is that we expanded things wrong, thinking naively that (a*b)-1 = a-1*b-1. But look! If it was equal to b-1*a-1, everything would work out. The expansion would look like a*b*b-1*a-1=a*I*a-1 = a*a-1 = I, and as you can see everything lines up, and comes out to equal I, as it needs to.



If you stare at it long enough, it becomes clear that it has to be like this. Took me a while.



Anyway, I'm having a hell of a time understanding this business about representations of SU(2). I think I've got a handle on the spin-0 representation, but the spin-1/2 representation is giving me trouble (because of that damn inverse that shows up where I don't want it), and spin-1 and dual representations are another story all together...



My problem with spin-1/2, in brief. The book is asking me to show that for spin-1/2, elements of SU(2) act on C^2 by matrix multiplication - that is, spin-1/2 is the 'fundamental' representation. My trouble comes from having trouble seeing how the fuck else they could be acting on C^2. I mean, representations of SU(2) are defined by [Uj(g)f](v) = f(g-1v), where the f's are functions that are degree-2j homogenous polynomials, v's are elements of some vector space, in our case C^2, and U_1/2 being the name we're giving the representation. Sorta. My terminology is probably all fucked up. Anyway, the reps look like f(g-1v)! g^-1 is an element of SU(2), since g is. Elements of SU(2) are 2x2 matrices. v is just a vector, a 2x1 matrix. Well, you write two matrices down like that, and you're stating you're doing matrix multiplication. What the hell is there to show?



What's bothers me even more is that say we take j = 234. Well, the reps still look like f(g-1v). You're still doing matrix multiplication. But you can't have more than one fundamental representation, can you? That would rather contradict the notion of it being 'fundamental', wouldn't it? Or would it? I dunno. Of course, the functions f do change as j changes - they get hairier and hairier, rather rapidly. Hmm. As you can probably tell, I'm confused.


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