sniffnoy | Yay disconnectedness

Without any good way to put a line over something in HTML, I'm just going to notate complex conjugate of z by z*.

Complex analysis final today. It included the following problem:

Ω⊆C open, connected, f holomorphic on Ω, f(Ω∩R)⊆R. Then f(z*)=f(z)* whenever z,z*∈Ω.

Now this was actually part d of a four-part problem, the parts that led up to it making it pretty clear that the way you were supposed to do this was note that f(z)=f(z*)* on Ω∩R, which is nondiscrete, and thus by analytic continuation f(z)=f(z*)* on Ω∩Ω*.

Problem: Ω∩Ω* is not necessarily connected. Consider a horseshoe-shaped region (--- is the real line):

XXX
X X         X X
--- becomes ---
X X         X X

So I figured, well, Ω∩Ω* is not necessarily connected, but I figured, it still has to work because each connected component must intersect the real axis. Now, why is that? I gave a false proof, crossed that out and gave another false proof, and finally gave a third proof which, I realized after handing my test in, was really just my original proof rewritten, and still just as false.

So I get back to Pierce and there's Lucas and Sam of Wallace and Dan of Shorey talking about the test, and they ask me how I did. And I say I did fine except for a technical detail on 2d, though it could be really said to be the entire problem. And they ask me what it was.

So it turns out none of them even realized that Ω∩Ω* doesn't even have to be connected. So the question becomes, why does each connected component have to intersect the real axis? They all start thinking about it, Dan reoffers my false proof, and then finally Lucas realizes that the connected components don't have to intersect the real axis at all, as follows:

X           X
XXX           X
--- becomes ---
  X           X
XXX         X

Well, this causes a bit of a stir, and gets us questioning if it's even true, and how in space we'd go about constructing a counterexample if it's not. We don't really know too many holomorphic functions on that set... or we do, but it can't be extendable to much of a larger set or the key property that Ω∩Ω* has components not intersecting the real line would be lost. Well. "If we could provide an isomorphism with the unit disk..." :P We do, after all, know a function which cannot be extended past the unit disk, Narasimhain gave us one (Σz^n!), as well as some more general theorems for finding such functions (though he didn't prove them), so that would give one not extendable past that region (hm, Wikipedia tells me that for *any* open set in C, you can get an analytic function not extendable past that set... interesting...), but 1. it probably wouldn't be messed up as we want it and 2. don't forget, we not only it to be very messed up, we need that it takes reals to reals, i.e. we need the isomorphism to take Ω∩R to D∩R (or equivalently just to some hyperbolic line in the disk). Not having an explicit isomorphism, that's something of a problem. And I don't think our proof of the Riemann mapping theorem gave us any extra results that would help us. Yay.

Lucas thought maybe taking branches of some multivalued function, but that has the problem that it's too extendable. (He also wanted to cut out a ray passing through Ω, forgetting f must be defined on all of Ω and not just Ω∩Ω*. :P )

So I'm going to guess that this statement is actually true, because he put it on the test, he just forgot about some "details" necessary to prove it. Also because the coming up with a counterexample is kind of ridiculous. (Though maybe we could get an explicit isomorphism to a similar region, say maybe based on quarter-circles... because as we saw you can get an explicit isomorphism to some regions that you wouldn't expect... actually right now I'm thinking I'd have an easier time trying to come up with a counterexample than trying to prove it. :P )

-Harry