Back to Business

So, one holiday down, two more to go, I guess.  New Year's ironically straddles two years, so maybe we should say 3/2 for people who don't like to reduce (like me).  A big thanks to G & S for letting me stay for the weekend, etc.  I won't bother itemizing.  Now it's back to that weird stretch of the year that falls between major holidays, where you try to cram more learning into two weeks after just having been off for so long.  Good luck, professors! I'm pretty sure most students are just coasting at this point and won't fully regain consciousness until some time around January 4th.

I won't bore you with the details of my own holiday travels, since if you are reading this blog there's a significant chance you were there, and if you weren't, suffice it to say that it was fairly uneventful.  As for the shopping season which is upon us, I will add my usual request: don't buy me anything, and if you absolutely must throw your money at something, there are plenty of good causes out there, such as this guy, who probably has a donation link somewhere, but I couldn't find it in the first couple minutes of searching, so I gave up.

As for myself, I have research to do, although I just recently finished up what is basically my main problem, so now I have some other unrelated problem to work on.  It's kind of nice because it is so different and that gives me a break.  It also lets me think about functors like Cº: Top -> Rng and Spec: Rng -> Top.  Specifically, it is well-known that if you start with a topological space X, and look at the set Cº(X), the real-valued (although complex is fine) continuous functions on it, you get a ring, with multiplication and addition defined pointwise, i.e., (f + g)(x) = f(x) + g(x) and (fg)(x) = f(x)g(x).  In fact, it's a commutative ring with unity since R is.  It's also well-known, though by fewer people, that you can construct Spec of this ring to get a topological space, and that this space contains a "copy" of the original space.  Specifically, MaxSpec Cº(X) = X.

This isn't really crazy once you know the definitions.  Spec S for some ring S is the set of prime ideals of S, and it's easy to verify that px = { ƒ € Cº | ƒ(x) = 0 } is a prime ideal, and is in fact maximal. Because of the compact Hausdorff nature of X, it turns out that these are the only maximal ideals, and pushing through definitions the map that sends x to px is continuous and open, and so is a homeomorphism onto its image (there are other prime ideals, so Spec Cº(X) is actually more stuff than just X).  If you take away these nice properties of the underlying space, it may cause a failure of that map to be a homeomorphism.

As an example, if you take X = {1, 2} with the indiscrete topology, i.e., the only open sets are X and ø , then the only continuous functions will be constant functions, meaning that p1 = p2 = {0}, and you can see that this functor X -> MaxSpec Cº(X) sort of "forgets" that there were two points because from one point of view, there really weren't two points to begin with, since they couldn't even be separated by open sets.

That's just a dumb example I thought up on a train or something, but there are all sorts of still useful spaces that fail to be compact or even Hausdorff.  The usual topology on the real numbers, for example, gives a non-compact space. They are, however, locally compact, and even Lindelof, so maybe there's a hint.  The result I'm looking for is not actually homeomorphism, but that's a post for another day.  Ok, congrats if you got this far!