Suppose there is a vector bundle (smooth, with constant rank finite-dimensional fibres) over a (smooth, second-countable, Hausdorff, not necessarily connected) manifold $B$ of dimension $n$.

(a) Is it true, that the manifold B can be covered by a finite number of sets $U_1,\dots,U_N$ s.t. the vector bundle, restricted to $U_i$, is isomorphic to a trivial one for every $i=1,\dots,N$?

(b) If yes, can $N$ be taken to be $n+1$?

P.S. Some observations:

  1. It's proven in the book by Milnor and Stasheff, that every bundle allows a countable locally finite trivialization cover.
  2. Part (a) is obviously trivial for compact manifolds.
  3. It seems, that (b) is true if $B$ is an $n$-dimensional CW-complex. Proof: Denote with $B_k$ union of cells of dimension $0,\dots,k$. Prove by induction in $k$, that there are subsets $U_0,\dots,U_k$ of $B$, which cover $B_k$, s.t. the restriction of the bundle to each of them is trivializable. Start with case $k=0$: construct contractible neighbourhoods of each 0-cell, which do not intersect with each other. Take there union. Now to prove the claim for the next value of $k$ it is enough to construct a contractible non-intersecting neighbourhoods of each $X_\alpha=e_\alpha\setminus (U_0\cup \dots\cup U_{k-1})$. Call the desired neighbourhood with $V_\alpha$. First, note, that $X_\alpha$ is closed in $e_\alpha^k$ and doesn't intersect with its boundary $\partial e_\alpha^k$, so we can find its neighbourhood in $e_\alpha^k$, which doesn't intersect with $\partial e_\alpha^k$. This set is our candidate for $V_\alpha\cap B_k$. Extending it to an open set in $B_{k+1}$ can be done cell by cell: interpreting $e^{k+1}_\beta$ as a unit ball with the center in the origin, we can write every its point as $r\theta$, where $\theta\in S^k$ and $r\in [0,1]$. We include $r\theta$ in $V_\alpha\cap B_{k+1}$ iff $\theta$ is already there and $r>0.99$. Repeating this procedure we extend it to $B$.

Edit: Open sets $U_1,\dots, U_N$ are assumed to be open. I don't ask them to be connected.


6 Answers 6


The answer (to both questions (a) and (b)) is YES (assuming $B$ is a smooth manifold). A proof can be found on Walschap's book "Metric Structures in Differential geometry", p. 77, Lemma 7.1.

For the OP's convenience, here's a sketch of the proof. Choose an open cover of $B$ such that your vector bundle is trivial over each element. From general results in topology, this (and in fact any) cover of an $n$-dim manifold $B$ admits a refinement $\{ V_\alpha\}_{\alpha\in A}$ such that any point in $B$ belong to at most $n+1$ $V_\alpha$'s. Let $\{\phi_\alpha\}$ be a partition of unity subordinate to this cover and denote by $A_i$ the collection of subsets of $A$ with $i+1$ elements. Given $a=\{\alpha_0,\dots,\alpha_i\}\in A_i$, denote by $W_a$ the set consisting of those $b\in B$ such that $\phi_\alpha(b)\lt\phi_{\alpha_0}(b),\dots,\phi_{\alpha_i}(b)$ for all $\alpha\neq\alpha_0,\dots,\alpha_i$. Then the collection of $n+1$ open subsets $U_i:=\cup_{a\in A_i} W_a$ covers $B$ and is such that your bundle restricted to each $U_i$ is trivial.

  • $\begingroup$ I made purely formatting changes, to correct a conflict between MarkDown and MathJax. I hope I did not introduce any errors. $\endgroup$ Apr 19, 2012 at 3:42
  • $\begingroup$ I apologize for the weird formatting: my curly brackets seem not to work otherwise. I hope MathJax is rendering everything else without problems... $\endgroup$ Apr 19, 2012 at 3:44
  • $\begingroup$ @Theo: thanks! that's exactly what I needed... I'll look what you did so I finally learn to deal with this issue! Thanks again. $\endgroup$ Apr 19, 2012 at 3:45
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    $\begingroup$ Walschap's book is available in electronic form here: libgen.info/view.php?id=539769 $\endgroup$ Apr 19, 2012 at 8:51
  • $\begingroup$ @Renato: The problem, of course, is that Markdown and LaTeX both consider braces special, and so to escape them you need a backslash. Markdown goes first, sees \{ , and replaces it with { , which is all MathJax sees. The solution is to put the entire paragraph between <p> ... </p> tags, which Markdown interprets as "don't process anything in this section". The cost is that you don't get to use any MarkDown syntax within the paragraph, so to make something bold (for example) requires <b> or <strong> or other html. <p> also protects underscores from Markdown, of course. $\endgroup$ Apr 21, 2012 at 22:22

This should be a comment to the answer of Andreas Blass, but was too long.

The Lusternik-Schnirelmann category $\operatorname{cat}(X)$ of a space $X$ is the smallest number $k$ such that $X$ has a cover by open sets $U_1,\ldots , U_k$ which are contractible in $X$. This means the inclusions $U_i\hookrightarrow X$ are null-homotopic. Note that the $U_i$ need not be contractible themselves, or even connected (although each $U_i$ should be contained within a connected component $X_i$). Note also that $\operatorname{cat}(X)$ is greater than or equal to the minimum number of open sets needed to trivialize any vector bundle on $X$, by the bundle creeping lemma and the fact that any bundle over a point is trivial.

One of the first theorems about LS-category is that if $X$ is paracompact, then $\operatorname{cat}(X)\le \operatorname{dim}(X)+1$, where $\operatorname{dim}$ denotes the Lebesgue covering dimension (which for manifolds agrees with the usual dimension). So Andreas Blass is correct.

If you want the minimum number of sets in a trivializing cover for your bundle $E$, you can do better, using the notion of sectional category of a fibre bundle (also known as the Schwarz genus). The sectional category $\operatorname{secat}(p)$ of a fibre bundle $p\colon\thinspace E\to B$ is the smallest number $k$ such that $B$ has a cover by open sets $U_1,\ldots , U_k$ on each of which $p$ admits a continuous local section (that is, a continuous map $s\colon\thinspace U_i\to E$ such that $p\circ s =\operatorname{incl}\colon\thinspace U_i\hookrightarrow B$).

Then the minimum number of sets in a trivializing cover for the vector bundle $E\to B$ equals the sectional category of the frame bundle $F(E)\to B$.

Addendum: One can show using obstruction theory that for a $r$-connected manifold $B$, $$\operatorname{cat}(B)< \frac{\dim(B)+1}{r+1}+1.$$

So if, say, your manifold is simply-connected, you can find a trivializing cover with roughly half as many sets.

  • $\begingroup$ Very nice answer. I was looking for something like this. Thank you. $\endgroup$ May 27, 2020 at 12:00

From the reference above (Walschap's "Metric Structures in Differential geometry") it seems that, in order to construct the classifying map to a suitable Grassmannian $G(\mathbb{R}^N,k)$, where $k$ is the rank of your bundle $E$, one needs the existence of a finite trivializing cover.

Personally, I believe that such a perspective can be reversed, i.e., first construct a classifying map and then use it to prove the existence of a finite trivializing cover. Indeed, the classifying map can be constructed just by using the Gauss map associated with a Whitney embedding $E\subseteq\mathbb{R}^N$. Namely, the base manifold $M$ is embedded into $E$ via the zero section, and attached to any point $x$ of $M$ there is the vertical tangent space (i.e., the subspace of $T_xE$ tangent to the fiber through $x$). Apply then the Gauss map to this vertical tangent space, and you'll get an element of $G(\mathbb{R}^N,k)$. Since any bundle over $G(\mathbb{R}^N,k)$ admits a finite trivializing cover, so will any bundle pulled-back from it.

  • $\begingroup$ I think this is really the most conceptual answer. $\endgroup$
    – Ryan Reich
    Aug 7, 2013 at 6:16

I think what you're looking for (or rediscovering) is the concept of the Lusternik-Schnirelman category of a space - the minimum number of contractible open sets needed to cover the space. More precisely, you need the maximum LS-category of the components of your space.

  • 2
    $\begingroup$ @Andreas Blass : It's not necessary that the open sets be contractible, merely that each of their components are contractible. Once you allow disconnected components, then the OP has a correct proof that $(n+1)$ open sets are necessary for an $n$-dimensional manifold. $\endgroup$ Apr 19, 2012 at 2:23
  • $\begingroup$ @Andry Putman : I do allow a cover by disconnected open sets (I've just added this clarification to the question), but I didn't noticed my proof, you are referencing to. I have some proof for the case of CW-complexes, but Wikipedia says only that manifolds are homotopy equivalent to CW-complexes. $\endgroup$
    – Fiktor
    Apr 19, 2012 at 3:14
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    $\begingroup$ ALso, since the OP is asking for a local trivialisation of a given vector bundle (as opposed to all vector bundles), the connected components of the open sets don't need to be contractible, merely that the vector bundle restricts to be trivial over them. If he has proved that the manifold has a cover by (disjoint unions of) contractible opens, then this is even stronger than what the question requires. $\endgroup$
    – David Roberts
    Apr 19, 2012 at 3:16
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    $\begingroup$ @Fiktor : I guess that's true, but Kirby-Siebenmann proved that all manifolds have topological handle decompositions, and thus are homeomorphic to CW complexes. $\endgroup$ Apr 19, 2012 at 3:26
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    $\begingroup$ Actually, you specified that your manifolds are smooth, so there is no need to appeal to fancy things like Kirby-Siebenmann. Smooth manifolds can be triangulated, and thus are homeomorphic to simplicial complexes (better than just CW complexes!). $\endgroup$ Apr 19, 2012 at 3:28

(a) and (b) are also true for topological manifolds of dimension $n$, see pages 17-21 of

  • MR0336650 Greub, Werner; Halperin, Stephen; Vanstone, Ray: Connections, curvature, and cohomology. Vol. I: De Rham cohomology of manifolds and vector bundles. Pure and Applied Mathematics, Vol. 47. Academic Press, New York-London, 1972.

I wonder that this was not said before.

Take a triangulation of your manifold. Choose disjoint open balls around each 0-cell in the triangulation and set the union to be $U_0$ (a ball means here something diffeomorphic to an open ball).

For each 1-cell, choose an open ball such that together with the balls chosen before around the corners, the 1-cell is completely covered, and choose those balls to be disjoint. Take the union of those to be $U_1$.

Keep on going in the same way, until you reach the $n$-cells.

Now you have $n+1$ sets $U_0, \dots, U_n$, each consisting of disjoint unions of open balls and such that $U_0 \cup \dots \cup U_j$ covers all cells of dimension less or equal to $j$. In particular, the union of all these sets covers your manifold.

Your vector bundle must be trivial on each of the sets $U_j$, as each are disjoint unions of open balls.


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