Atlases Refining Covers, part 1

A bit late, but at least I got it up today!

Our first step in finding partitions of unity subordinate to a given cover $\{U_\alpha\}$ is actually to set up a nice atlas.

We want a countable differentiable atlas $(V_k,\phi_k)$ where the collection $\{V_k\}$ is a locally finite refinement of the cover $\{U_\alpha\}$ . Locally finite we just covered yesterday; recall that being a refinement means that each $V_k$ is contained in some $U_\alpha$ .

Getting more technical, we will also require that the image $\phi_k(V_k)\subseteq\mathbb{R}^n$ is $B_{3n}(0)$ — the open ball of radius $3n$ centered at the origin. That is, $\phi_k(V_k)=B_{3n}(0)$ consists of exactly those vectors with length strictly less than $3n$ . Further, we will define $W_k$ to be the inverse image $\phi_k^{-1}(B_1(0))$ of the ball of radius $1$ , and we will require that the collection $\{W_k\}$ is also a cover of the manifold $M$ .

Now, in order to pull this off, we actually need to add some technical restrictions to our definition of a manifold. Since $M$ is locally homeomorphic to $\mathbb{R}^n$ , and $\mathbb{R}^n$ is locally compact — each point has some open neighborhood with compact closure — the same is true of $M$ . We don’t however, know that $M$ is Hausdorff or second-countable, both of which we’ll need.

I want to give some counterexamples, showing how these conditions can fail, and why the pathologies they prevent don’t adhere to our intuitive notion of “manifold”. Luckily, both of them are one-dimensional, so they aren’t impossible to visualize.

First, we have the line with a doubled origin. Take two real lines and glue them together by identifying each nonzero number on each line with the corresponding nonzero number on the other line. But do not identify the two zeroes. What’s left is one line, but it has two zero points “on top of each other”. Every nonzero number clearly has a nice open neighborhood that looks just like a regular interval — just stay away from the zero points — and we can also set up patches that look like the interval $(-1,1)$ , one containing each of the two zeroes. These last patches are still open, and so every point has an open neighborhood homeomorphic to an interval.

But this space is not Hausdorff! Any two open sets, each containing one of the zeroes, must intersect, and yet these two points are not the same. If the doubled portion were an interval, we’d see the line fork in half on either side of the doubled section, and the forking point would clearly not have any neighborhood that looked like an interval, and so it would clearly not be a manifold. But when we shrink down and only double a point, there is no “forking point”, and we can’t use that to rule this case out. So instead we say that manifolds must be Hausdorff.

The other pathological example is the “closed long ray”. This is less obviously pathological, and second-countability is mainly around so that we can get countable sequences and series and such to make our lives easier down the road. Anyhow, to get our hands on it is sort of technical. We start with the half-open interval $[0,1)$ and the “first uncountable ordinal” $\omega_1$ . Actually, any uncountable well-ordered set will do [commenter Stevie Hair below isn’t so sure, and he has a good point] but the first one that arises is the most convenient.

Now, we take the product $\omega_1\times[0,1)$ and give it the “lexicographic order”. That is, we compare pairs $(o,x)$ and $(p,y)$ by first comparing $o$ and $p$ in the order from $\omega_1$ . If they’re different, we use that order. If they’re the same, though, we move on to compare $x$ and $y$ in the usual way. We then give it the “order topology”, similar to the way we constructed the topology on the rational numbers.

The upshot is that it’s like we’ve strung together an uncountable number of copies of the interval $[0,1)$ . Within each copy, obviously, it looks like $\mathbb{R}^1$ , and where we glue two copies together it does too, just like we have no trouble going from $[0,1)$ to $[1,2)$ . It’s even Hausdorff already, so the previous condition doesn’t rule this case out. The problem is that it’s not second-countable. There’s no countable collection of open sets that generate the whole topology. It’s simply too big to be described without bringing uncountable numbers of sets into the picture.

So, we add the Hausdorff and second-countable conditions to our definition of a manifold, and move forward.

March 10, 2011 - Posted by John Armstrong | Differential Topology, Topology

7 Comments »

Hi I’ve been following you for a little while now. Great blog!

Anyway. A couple of small mistakes. When talking about the line with doubled origin you start the second para “But this space is not homeomorphic!” which makes no sense. I think you meant Hausdorff rather than homeomorphic?

Also, when describing the closed long ray, you state that “any uncountable ordinal will do.” I’m not sure about that. If you pick a larger ordinal, then the point (\omega_1,0) doesn’t have a neighbourhood homeomorphic to the reals. Any neighbourhood has to “reach back” to a countable ordinal somewhere and will be “too big”. I think.

Cheers

Stevie

Comment by stevie hair | March 10, 2011 | Reply
Yes, of course I was partly asleep at the switch and got my h-words mixed up. As for the uncountable ordinals, you do have a point. The details of these pathological cases sometimes escape me — they’re unexpected, after all — but the overall construction still works.

Comment by John Armstrong | March 10, 2011 | Reply
[…] Now, armed with our two new technical assumptions, we can prove the existence of the refining covers we asserted yesterday. […]

Pingback by Atlases Refining Covers, part 2 « The Unapologetic Mathematician | March 11, 2011 | Reply
[…] as we’ve seen we can find a countable atlas ; we use for the coordinate maps since we’ll want the free. […]

Pingback by Partitions of Unity (proof) « The Unapologetic Mathematician | March 14, 2011 | Reply
Stevie’s right. Here’s why:

Suppose the construction works. Take a neighborhood of (\omega_1, 0). It will contain some point (\beta, x) with \beta countable, and therefore all the points (\beta + 1 + \alpha, 0), where \alpha ranges over all the countable ordinals. This is an uncountable well-ordered set, which transfers to such a set in R. But any well-ordered set in R is countable, because it’s possible to find an isomorphic set in Q (round the image of \alpha up to the “simplest” rational between \alpha and \alpha + 1). Contradiction: only \omega_1 works.

Comment by Chad | April 2, 2011 | Reply
- Sorry, obviously “neighborhood” above should be “open interval neighborhood”, otherwise the rest doesn’t follow.
  
  Comment by Chad | April 2, 2011 | Reply
[…] we know we can find some chart so that the slices are all integrable submanifolds of . By the assumption that is second-countable we can find a countable cover of consisting of these […]

Pingback by Foliations « The Unapologetic Mathematician | July 1, 2011 | Reply

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