Anyone here familiar with deRham cohomology? I have a simple question: How do I show the second cohomology group of the plane is trivial. All the 2-forms are closed, so how do I show they are all exact?
Thought about using Green's, but not sure how to make it work.
So let a general 2-form be h(x,y) dx dy and a general 1-form be f dx + g dy. We see the 1-form becomes dg/dx-df/dy dx dy, so we need to solve dg/dx-df/dy = h(x,y).
It looks like Green's theorem to me. The analogous thing in R^1 is, if we want to get the form g(x) dx, to write f = integral from 0 to x g(u) du, but I don't see how to use Green's to do that here.
Can someone check my answer to my own question here?(math.stackexchange.com/questions/97128/…) I will not be able to sleep until I figure out this stupid technical issue.
@QED The idea is that things concerning close votes, reopening etc. will be easier to find, if they are in a single place? (As opposed to being in this chatroom hidden between many other messages.)
I talked with Uri yesterday, he thinks that I have enough for a thesis, even if there is not enough original research for a paper afterwards. I'm starting to write, and in the meantime I hope to finish at least one nontrivial theorem.
Just keep it going. But remember what counts is the contents, not the looks -- it's too easy to waste the time with layout tweaking. You'll need to keep that for the times when you don't make progress, so that you can cheat yourself into thinking you did something. Oh, well, but at least it looks a bit better now :)
First, Lauchli wrote in German. Second, his terminology was awkward in a nowadays standard. Third, he used Quine's atoms as atoms which make the construction harder.
In my result there is a slight improvement too, since I show consistency with an $\mathbf{DC}_\kappa$ for unbounded $\kappa$ (though bounded in every model, of course).
Better be safe than sorry... At least mention it somewhere :) Remember those German ministers whose name was like the inventor of the printing press...
@Potato Maybe you should say a few more words on "Note that for the points where they are defined, $dg_0=dg_1$ implying they differ by a closed 1-form"
(I don't think that what you say is what you intend)
So the domain of $g_0$ is the plane minus the half-line from (-1,0) on the real axis down to negative infinity, and the domain of $g_1$ is the plane minus the half line from (-1,0) to infinity.
So they are both defined on a domain consisting of two disjoint half planes.
I'm working on an exercise proving $f$ is continuous, where if $f$ is a real function with domain $\mathbb{R}$, with the property such that $f(a)<c<f(b)$, then $f(x)=c$ for some $x$ between $a$ and $b$. Also, for every rational $r$, the set of $x$ such that $f(x)=r$ is closed. There's a hint saying if $x_n\to x_0$ but $f(x_n)>r>f(x_0)$, then use this to find a contradiction (con't)
I do this by assuming $f$ is not continuous at some point $x_0$, so there is a sequence $x_n\to x_0$ such that $f(x_n)\not\to f(x_0)$. From this I could extract a subsequence such that $f(x_n)>f(x_0)$ for all $x_n$ in the subsequence.
The one thing I'm wary about is why one such rational $r$ should exist. What if the $f(x_n)$ of the subsequence get arbitrarily close to $f(x_0)$ from above? Would this contradict $f(x_n)\not\to f(x_0)$ for the $x_n$ in the original sequence somehow?
I am using the extra condition that the $x$ such that $f(x)=r$ is closed. I use the intermediate value condition to find a sequence of preimages of $r$ that approach $x_0$.
So $x_0$ is a limit point, and thus in that set, so I get $f(x_0)=r>f(x_0)$, a contradiction.
I was just curious why I could easily assume such $r$ existed.
@ZhenLin: come to think of it... I wanted to ask you for a long time already: Is your first name Zhen or Lin? I guess it's the latter but I'm not sure.
@ZhenLin I took an arbitrary sequence $x_n\to x_0$ but $f(x_n)\not\to f(x_0)$, but then took a subsequence $x_m$ such that $f(x_m)>f(x_0)$. Is it true that $\lim(f_m)\neq f(x_0)$ still? Because then I could just take $r$ between the limit and $f(x_0)$ like you suggested.
@t.b. Pretty well I think. I wrote it down at the end of my Rudin document, and I think it makes sense now. It's in my profile if you ever are bored and care to see.
@AsafKaragila I already have an acknowledgment overflow from you... :) By the way: if you include the socks thingie in your thesis. Be careful not to call the elements of your two points sets $a_n$ and $b_n$...
@tb Yeah, I wouldn't want a few people confusing internal and external elements again.
@tb Also, there is a major chapter about socks. I follow Goldstern's construction and the development both of us did over an email correspondence to the vector space with a basis that has a subspace without one.
@tb I have a lot written in various places (including my own tex files). After this semester ends, I don't have to take any more courses so I can sit through the nights and finish the document.
I'm a non-mathematican and non native speaker. I'm building kind of a wooden puzzle and got stuck. My problem is: I have squares whose 4 edges have x different key-and-slot-patterns. Each square looks the same. Now I can join different squares to each other as long as they don't share the same key-and-slot-pattern.
I prefer to see the keys and slots as a "color". That way each square has x colors whose edges can be joined to each other as long as their color differentiates. Joining may happen planar or perpendicular. In a first step I want to build a cube whose 6 faces consist out of 6 squares. I want to know how many different edge colors I need when building a) an ordinary cube b) a cube in cube system like the rubic's cube (3x3x3). Can anybody give me a tipp where to start?
@yunone It looks good. The only thing I think you could improve is to say explicitly where your points $x,y,r,s$ are supposed to live. Especially in the last paragraph
(and no problem, of course)
@prinzdezibel This looks like a good question to ask for the main site. Unfortunately our resident combinatorialist just headed out
@yunone "The alternative proof in Exercise 11 only requires that the range space be complete, and so the range space can be replaced by any complete metric space or Rk . The proof oļ¬ered here works if the metric space is compact, so the range space can be replaced by any compact metric space." -- Note that compact implies complete, so the second statement isn't quite necessary.
@tb Mike Spivey is strong, I am ok. =) There are others too (I'm not ranking them here).
@Srivatsan Thanks! I remember that from reading a bit about general topology, but it's never mentioned in Rudin. I was just trying to keep it in line with the book. I guess the proof is not hard using sequential compactness though, so maybe I should fix that.
Yeah, but you can do it for any complete metric space as range. The nested intervals (nested balls?) theorem only requires that. Completeness is of course necessary: try to extend $\mathbb{Q} \to \mathbb{Q}, x \mapsto x$ to a function $\mathbb{R} \to \mathbb{Q}$.
@Srivatsan but yunone mentioned sequential compactness?
Well, I never even held "principles" in my hands once, so I don't know what Rudin's doing there. But from his other books I'd be rather surprised if he didn't mention compactness.
@Srivatsan Ok, I've rewrote it to shift things down $1$. Isn't $[x]$ just the floor function, stated strangely? I don't think there should be much of a problem at the integers. $[4]$ is the integer such that $3<[4]\leq 4$. So $[4]=4$.
@yunone I see. I was very confused, my bad. So his $[x]$ is just the usual one. What about your statement $[n+] = n+1$ and $[n-]=n$? That is certainly off by $1$, right?
@Srivatsan Yeah, I wasn't thinking when I wrote that. I mentioned above that I've since fixed it to say $[n+]=n$ and $[n-]=n-1$. Don't know what I was thinking.
