# Do all manifolds have a densely defined chart?

Let $M$ be a smooth connected manifold. Is it always possible to find a connected dense open subset $U$ of $M$ which is diffeomorphic to an open subset of R$^n$?

If we don't require $U$ to be connected, the answer is yes: it is enough to construct a countable collection of disjoint open "affines" whose union is dense, and this is not terribly difficult.

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I assume you also want M to be connected? –  Qiaochu Yuan Nov 25 '10 at 1:08
Yes, connected. –  Kevin Nov 25 '10 at 1:17
At least for compact manifolds, Morse theory gives you a CW structure (I don't know what happens for noncompact manifolds. I think they also have a CW structure but I'm not sure). Is there some reason you can assume you have one top dimensional cell? If so, that should give you your dense chart. –  Jason DeVito Nov 25 '10 at 1:36

If the manifold is compact, the answer is yes. The geometric idea is to put a small balloon in the manifold, and to inflate it. Since the manifold is compact, you eventually fill the manifold, so you're left with the interior of the balloon (an open ball) with identifications on the boundary, so it's dense.

You can make this precise by putting a Riemann metric on the manifold and using the Hopf-Rinow Theorem.

The relation between my answer and Jason's comment would be to take a handle decomposition of the manifold and to consider a maximal tree in the dual 1-skeleton as giving a "tree-like ball".

edit: responding to Qiaochu's comment, I believe the answer is affirmative for connected, non-compact manifolds as well. The technique of the proof has to be adapted some. Step 1: construct a proper Morse function on $M$, meaning a Morse function $f : M \to [0,\infty)$ such that it has only one local minimum -- which is the global minimum, $0$, and demand that $f^{-1}[0,c]$ is connected and compact for all $c\geq 0$. Step 2: you construct the charts on $f^{-1}[c_i,c_{i+1}]$ where $c_1<c_2<\cdots$ are an increasing sequence $\lim_{n\to\infty} c_n =\infty$ which are regular values of $f$. Step 3: paste the charts together along little arcs connecting them, pairwise.

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Do you know what the answer is for, say, noncompact surfaces? –  Qiaochu Yuan Nov 25 '10 at 2:01

Depending on what you consider a manifold, the long line may be a counterexample.

And for a non-connected manifold, surely the answer is no? Take your favorite smooth manifold, and take a disjoint union of more than $\mathfrak{c}$ copies of it. Again, unless your definition of "manifold" rules this out (by assuming separability, etc).

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The question is about connected manifolds, and I think when most people say "smooth manifold" they mean second-countable, etc. –  Qiaochu Yuan Nov 25 '10 at 2:15
I think he is considering only second countable Hausdorff manifolds. Also he edited the question adding that M must be connected (As Qiaochu pointed while I was writing). –  Nuno Nov 25 '10 at 2:16