...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...

Sunday, October 06, 2002

Referrer logs are wonderful entertainment*, though best enjoyed in moderation. See, every once in a while, there's Comedy Gold in them. For instance: in the last few days, according to said referrer logs, people have visited this page in the search for "hemmorhoid pictures" and "techinques [sic] to measure the penis". Now, thanks to this, I discovered that Google thinks that I am the #2 authority on the whole Interweb on "techinques to measure the penis". (If, for some reason, you want to verify that, note that bad spelling is important.)

I'm sorry to dissapoint, but I won't be posting any hemmorhoid pictures today, nor will I be writing about phallic metrics, or hairy balls (I've done that last in the past, and you can find it in my archives). Instead, I'll talk a bit about my current struggles with my reading.

I finally solved that problem with spin-1/2 representations of SU(2), though that's only true for a fairly liberal interpretation of the word 'solved'. Here's what I decided. Fact number one: If g is an element of a group, so is the inverse of g, g^-1. This is one of the basic properties of groups. Fact number two: for spin-1/2, j=1/2, 2j=1, and the space of polynomial functions on C^2 that are homogeneous of degree 2j then look like f(x,y) = x^1*y^0 + x^0*y^1 = x + y, optionally with constant coefficients thrown in just to be naughty. Now, the assertion is that for spin-1/2, SU(2) acts on C^2 by matrix multiplication - that is, take a vector in C^2, and feed it to an element of SU(2), and out comes another vector in C^2. Well, since the spin-j representations look like f(g^-1*v), where the f looks like the f(x,y) given above, and v is a vector in C^2, and g is in SU(2) (and therefore, g^-1 is too), everything comes out nicely. See, what I'd like is for g to just act directly on v, but that just ain't in the cards. But what is in the cards is that the inverse of g acts on v, and that's still 'SU(2) acting on C^2', and it's also very important to note that for spin-1/2, f(x,y) is of such a form that it makes sense to say that an element of SU(2) just acted on C^2 by matrix multiplication. But already for spin-1, f(x,y) looks like x^2 + xy + y^2, and you can't say that anymore.

Now, once I decided that, I could move to wrestle with spin-1 and all that. And it all actually kind of made sense. It made so much sense, I blitzed right through to the end of the material on Lie groups, and stopped when I saw Baez and Munian start talking about Lie algebras. However, I didn't stop to do any problems along the way, so I've still got things to work out. I'll probably talk about that later today. But for now, I'll just mention that one of the problem sets looks to be very interesting, but also hard: to work out some things about how SL(2,C) is the double cover of SO(3,1). Baez and Munian talk about how SU(2) is the double cover of SO(3), but leave the SO(3,1) stuff to the reader in exercises. Which, on the one hand, is frustrating and a bit scary, but is also very exciting, because this stuff is so very fascinating. (Bosons and fermions and cocycles and projective representations, oh my!)

* - but only if you don't have a life. Losers of the world, unite! Throw off the chains of capitalist oppression, and put down the running pig-dogs of your exploiters.


Post a Comment

<< Home