4
$\begingroup$

Let $M$ be a (not necessarily compact)) smooth manifold.

1.Is there a smooth map $f:M\to \mathbb{R}$ and an open covering $\mathbb{R}=\cup U_{\alpha}$ such that each $f^{-1}(U_{\alpha})$ is homeomorphic to $\mathbb{R}^{n}$?

2.Is there a smooth map $f:M \to \mathbb{R}^{k}$, for some $k \in \mathbb{N}$ and an open covering $\mathbb{R}^{k}=\cup U_{\alpha}$ such that $f^{-1}(U_{\alpha})$ is a good cover for $M$?

$\endgroup$
3
  • $\begingroup$ Does your definition of "manifold" include "second countable"? $\endgroup$ Nov 2, 2014 at 23:47
  • $\begingroup$ @NateEldredge Yes I mean the standard definition of manifold.May be you are considering the long line? $\endgroup$ Nov 3, 2014 at 21:06
  • $\begingroup$ Right, some people consider the long line to be a manifold and others do not. $\endgroup$ Nov 4, 2014 at 2:08

2 Answers 2

3
$\begingroup$

For (1), the answer is almost always no; the combinatorial properties of open covers of $\mathbb{R}$ are far too restricted. Suppose that $M$ is compact and connected and such an $f$ and $\{U_\alpha\}$ exist. Let the image of $f$ be $[a,b]$. Each $U_\alpha$ is a union of disjoint intervals; by connectedness, only one of those disjoint intervals can intersect $[a,b]$. Restricting the codomain of $f$ to $[a,b]$, we may thus assume each $U_\alpha$ is a single interval. Now take a minimal subcollection of $\{U_\alpha\}$ that still covers $[a,b]$: this will be a finite sequence of intervals $I_k=(x_k,y_k)$ going from left to right such that only consecutive intervals intersect. That is, we have $$x_1\leq x_2\leq y_1\leq x_3\leq y_2\leq x_4\leq y_3 \leq\dots$$

Let $U$ be the union of the $f^{-1}(I_k)$ for $k$ odd and $V$ be the union of the $f^{-1}(I_k)$ for $k$ even. Then both $U$ and $V$ are disjoint unions of copies of $\mathbb{R}^n$, and they cover $M$. This implies that the LS-category of $M$ must be at most 1. In particular, for instance, this implies that the cohomology of $M$ can have no nontrivial cup products. If $M$ is not compact, a similar (but a little bit more difficult) argument can be made to reach the same conclusion.

$\endgroup$
1
  • $\begingroup$ what about the following weaker version :In first question we consider the weaker statment:' $f^{-1}(U_{\alpha})$ homeomorphic to an open subset of $\mathbb{R}^{n}$"? $\endgroup$ Nov 4, 2014 at 15:58
3
$\begingroup$

For 2., this might be nuking a mosquito, but here is an argument for existence of such $f$ assuming that $M$ is paracompact. WLOG, assume $M$ is connected (and hence second-countable). Use the Whitney embedding theorem to find an embedding $f: M \to \mathbb{R}^k$ as a closed submanifold of a Euclidean space; use this embedding to think of $M$ as a subset of $\mathbb{R}^k$. The Euclidean metric on $\mathbb{R}^k$ restricts to a Riemannian metric on $M$. For each point $p \in M$, find a ball $B_{r(p)}(p)$ in $\mathbb{R}^k$ such that $f^{-1}(B_{r(p)}(p)) = B_{r(p)}(p) \cap M$ is geodesically convex in $M$ (see here for example). The $B_{r(p)}(p)$ plus the complement of $M$ then form an open cover $U_\alpha$ of $\mathbb{R}^k$ for which the inverse images $f^{-1}(U_\alpha)$ form a good open cover.

$\endgroup$
4
  • 1
    $\begingroup$ I could be wrong, but I suspect this particular mosquito is genetically engineered to resist non-nuclear weaponry. In other words, I think the result is no easier than the Whitney embedding theorem. $\endgroup$ Nov 3, 2014 at 4:28
  • 1
    $\begingroup$ @PaulSiegel I suspect you're right. I confess that when I wrote those words, I was in the midst of deploying greater nuclear power by first putting a Riemannian structure on $M$, and then using the Nash embedding theorem to isometrically embed in Euclidean space. Then it occurred to me that the above would involve much less weaponry. :-) $\endgroup$
    – Todd Trimble
    Nov 3, 2014 at 5:58
  • $\begingroup$ @ToddTrimble Thanks for your very interesting answer. What about the following possible definition: $\endgroup$ Nov 3, 2014 at 21:36
  • $\begingroup$ For a manifold $M$, we define the minimum $k$ such that a good covering of $M$ can be obtained via pull back of an open covering of $\mathbb{R}^{k}$ for some smooth function $f:M\to \mathbb{R}^{k}$. Is this invariant some how related to some thing as LS category as in the answer of @Eric? As you said this minimum is less than $2n+1$. Thanks again for your interesting answer. $\endgroup$ Nov 3, 2014 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.