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: [U_{j}(g)f](v) = f(g^{-1}v). 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 [U_{j}(g)f](v) = f(g^{-1}v), 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^{-1}v)! 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^{-1}v). 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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