15
$\begingroup$

Usually one proves the existence of good covers in compact manifolds by Riemannian methods: we pick an arbitrary Riemannian metric, prove that geodesically convex neighborhoods exist, that they are closed under finite intersections, and diffeomorphic to balls; this is, for example, the argument that Bott and Tu sketch in their book.

Is there a non-Riemannian approach to this?

While this is not necessary for most things, it is a nice fact that good covers can be found which realize the covering dimension bound.

Is there a differential-topological way to find them?

$\endgroup$
2
  • $\begingroup$ (I am writing notes on de Rham cohomology, would love to have good covers available, but would much prefer to avoid to go through the Riemannian detour...) $\endgroup$ Jul 13, 2012 at 19:00
  • $\begingroup$ Interesting question. Small comment: the Riemannian argument does not use compactness, but of course it guarantees existence of a finite good cover (in Bott&Tu's terminolgy). $\endgroup$ Jul 13, 2012 at 19:56

3 Answers 3

22
$\begingroup$

you don't really need a whole lot of Riemannian geometry to prove this. Embed the manifold into $\mathbb R^n$ by Whitney and look at very small charts around points given by orthogonal projections onto the tangent spaces. the transition maps will be arbitrary close to identity in $C^2$. that means that a small round disk in one chart will remain strictly convex in nearby charts (because if $f(x)=|x|^2$ and $\phi$ is a transition map such that $\phi-Id$ has small first and second derivatives then $f\circ \phi$ is still strictly convex and hence has convex sublevel sets). This is is all you need to conclude that all intersections are contractible. I guess since the above argument doesn't use any Riemannian geometry notions it should qualify as an answer to the second question?

Incidentally, does a good open cover always exist if a manifold is only topological?

$\endgroup$
2
  • $\begingroup$ Can I ask you to give an independent question consists your last statement? $\endgroup$ Nov 1, 2014 at 18:50
  • $\begingroup$ This proof can also be found in Andre Weil's paper pages 120-122 Sur les Theoremes de de Rham Comm.Math.Helvetici vol 26 pages 119-145 $\endgroup$ Jan 4, 2016 at 16:10
7
$\begingroup$

The answer to both questions is yes. Fix a triangulation of the manifold. For any vertex $v$ denote by $U_v$ the union of the relative interiors of all the faces of all dimensions that contain the vertex $v$. (Note: the vertex $v$ itself is a face containing $v$ and it coincides with its relative interior.) The set $U_v$ is open and contractible and the resulting open cover is good. Its nerve is is the simplicial set underlying the chosen triangulation. This cover answers both your questions.

$\endgroup$
11
  • 3
    $\begingroup$ Heh. COnstructing a triangulation from knowledge of a smooth structure alone is a longer detour, no? $\endgroup$ Jul 13, 2012 at 19:08
  • 2
    $\begingroup$ H. Whitney describes a simple triangulation procedure in his book Geometric integration theory. In the book by Singer & Thorpe Lecture Notes on Elementary Topology and Geometry they prove that DeRham cohomology is isomorphic to singular homology using triangulations and the trick I mentioned. There they work on triagulable manifolds to avoid the theorem about existence of traingulations. On the subject of DeRham theorem see also the undergraduate thesis below. nd.edu/~lnicolae/Fanoethesis.pdf $\endgroup$ Jul 13, 2012 at 20:02
  • 2
    $\begingroup$ A caveat: the procedure is simple, but the proof that it produces a triangulation is quite involved. Here is the procedure: embed the manifold in some vector space, choose a generic basis $(\xi_k)$ of the dual space and then look at the intersection of the submanifold with the hyperplanes $\xi_k\in \varepsilon\mathbb{Z}$. For $\varepsilon$ small these hyperplanes trace along the submanifold a polyhedral decomposition. This can be easily transformed into a triangulation. The hard part is to show that this procedure yields the promised polyhedral decomposition. $\endgroup$ Jul 13, 2012 at 21:09
  • 2
    $\begingroup$ Nothing in the original question says that the manifold is smooth or triangulable. $\endgroup$ Jul 28, 2012 at 6:35
  • 3
    $\begingroup$ @Misha The two page triangulation proof by Cairns (1961) is flawed: see more discussion here: mathoverflow.net/questions/139339/… $\endgroup$
    – Ramsay
    Jan 13, 2015 at 15:38
3
$\begingroup$

I think you can obtain a good cover of $C^2$ manifold (compact or not) from the charts/atlas definition and a little bit of topology (locally finite atlas and a relatively compact "shrinking" of it).

The very simple idea (akin to that in Vitali's answer) is that under a $C^2$ diffeomorphism between open subsets of euclidean $n$-space, the (pre-)image of a sufficiently small ball centered at a point will be convex, as soon as the curvature of its boundary "dominates" the second derivative of the diffeomorphism (or its inverse).

In formulas, if $\phi$ is the diffeomorphism, this boils down to the fact that the $C^2$ function $x\mapsto |\phi(x)-\phi(x_0)|^2$ has a positive definite hessian at $x_0$, hence is convex near $x_0$.

With a little more care, I think you can still conclude if $\phi$ is only $C^{1+Lip}$.

$\endgroup$
4
  • $\begingroup$ I think Mariano might object to using curvature explanations as being too Riemannian but the argument using convexity of the composite function (which was what I had in mind all along) doesn't formally use it although it of course amounts the same thing. $\endgroup$ Jul 14, 2012 at 14:23
  • 1
    $\begingroup$ Yes, this is essentially the same idea as yours, only replacing projections on tangent spaces by zooms of local charts, observing that zooming a $C^2$ diffeomorphism makes it look more and more linear. I think the main point is this avoids Whitney embedding (admittedly not so hard), and can be proved from the very definition of manifold. $\endgroup$
    – BS.
    Jul 14, 2012 at 15:07
  • $\begingroup$ you are quite right and even Whitney is not needed for this. $\endgroup$ Jul 14, 2012 at 15:36
  • $\begingroup$ This proof is in Demailly's book Complex Algebraic and Differential Geometry. $\endgroup$
    – Ben McKay
    Feb 19, 2019 at 12:49

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.