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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?

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  • $\begingroup$ (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...) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented 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). $\endgroup$
    – wildildildlife
    Commented Jul 13, 2012 at 19:56

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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?

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  • $\begingroup$ Can I ask you to give an independent question consists your last statement? $\endgroup$
    – Ali Taghavi
    Commented 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 $\endgroup$
    – Mohan Ramachandran
    Commented Jan 4, 2016 at 16:10
  • $\begingroup$ You don't need Whitney - any embedding in Euclidean space will work. You just need paracompactness and the existence of a proper smooth map to the line. If the manifold is compact, it's almost obvious. $\endgroup$
    – Ezra Getzler
    Commented Oct 21 at 10:27
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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.

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    $\begingroup$ Heh. COnstructing a triangulation from knowledge of a smooth structure alone is a longer detour, no? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jul 13, 2012 at 19:08
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    $\begingroup$ 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 $\endgroup$
    – Liviu Nicolaescu
    Commented Jul 13, 2012 at 20:02
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    $\begingroup$ 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. $\endgroup$
    – Liviu Nicolaescu
    Commented Jul 13, 2012 at 21:09
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    $\begingroup$ Nothing in the original question says that the manifold is smooth or triangulable. $\endgroup$
    – Greg Friedman
    Commented Jul 28, 2012 at 6:35
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    $\begingroup$ @Misha The two page triangulation proof by Cairns (1961) is flawed: see more discussion here: mathoverflow.net/questions/139339/… $\endgroup$
    – Ramsay
    Commented Jan 13, 2015 at 15:38
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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 $C^{1+Lip}$.

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  • $\begingroup$ 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. $\endgroup$
    – Vitali Kapovitch
    Commented Jul 14, 2012 at 14:23
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    $\begingroup$ 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.
    Commented Jul 14, 2012 at 15:07
  • $\begingroup$ you are quite right and even Whitney is not needed for this. $\endgroup$
    – Vitali Kapovitch
    Commented Jul 14, 2012 at 15:36
  • $\begingroup$ This proof is in Demailly's book Complex Algebraic and Differential Geometry. $\endgroup$
    – Ben McKay
    Commented Feb 19, 2019 at 12:49

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