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In the notes of Vector Bundles and K-theory by Prof Allen Hatcher, on page 12 he proved a Proposition that for each vector bundle $E\to B$ with $B$ compact hausdorff there exists a vector bundle $E'\to B$ such that $E\oplus E'$ is the trivial bundle. Later in example 3.6 he showed that the compactness of $B$ is an important condition, otherwise canonical line bundle over $\mathbb{RP^\infty}$ would be a counter example.

My doubt is... Is this Proposition still valid for finite dimensional manifold??? Or can someone provide me a counter example of a finite manifold where it is not true.

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  • $\begingroup$ I think the title of this question should be changed. In addition to not being particularly informative regarding the topic of the question, it implies that Allen might have gotten something wrong in his K-theory book. As far as I can tell from the question and its answers, this isn't the case, so why not have the question reflect what's actually being asked? $\endgroup$ – Greg Friedman Oct 31 '15 at 7:42
  • $\begingroup$ @GregFriedman I am absolutely agree with you, The thing is that while posting the problem I didnt get any good title for this qes, and that is why I left it like that, If you have any suitable title for this doubt, then you are welcome to edit my post $\endgroup$ – Anubhav Mukherjee Oct 31 '15 at 11:44
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    $\begingroup$ How's that?.... $\endgroup$ – Greg Friedman Nov 2 '15 at 0:56
  • $\begingroup$ @GregFriedman thanks for this edit :) $\endgroup$ – Anubhav Mukherjee Nov 2 '15 at 2:44
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The proposition you refer to holds for any space homotopy equivalent to a finite dimensional CW complex. Here are the main points.

  1. The property that any vector bundle $E$ over a space has a complementary bundle $E^\prime$ is preserved by homotopy equivalences.

  2. Any finite dimensional CW complex is homotopy equivalent to a smooth manifold. (A classical result of Whitehead shows that any finite dimensional CW complex is homotopy equivalent to a locally finite finite dimensional CW complex. Then simplicial approximation gives a homotopy equivalent finite dimensional simplicial complex. Embed it into a Euclidean space, take a regular neighborhood, and smooth it.)

  3. The total space of any vector bundle $E$ over a smooth manifold $M$ smoothly embeds into a Euclidean space. The normal bundle bundle of $E$ restricted to $M$ is the desired complementary bunlde $E^\prime$.

  4. Note that any topological manifold is homotopy equivalent to a finite dimensional CW complex.

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    $\begingroup$ This can also be established without reducing to the manifold case. If $B$ is a $d$-dimensional CW complex then any rank $n$ vector bundle on $B$ can be embedded in a trivial bundle of rank $n+d$. The embedding can be constructed cell by cell in $B$. If $m\le d$ then any map from $S^{m-1}$ to the Stiefel manifold of linear embeddings of $\mathbb R^n$ in $\mathbb R^{n+d}$ can be extended to $D^m$. $\endgroup$ – Tom Goodwillie Oct 29 '15 at 16:40
  • $\begingroup$ @TomGoodwillie: this is nice. I did not think of Stiefel manifolds, and in particular did not know that $V_{n+d. n}$ is $(d-1)$-connected. $\endgroup$ – Igor Belegradek Oct 29 '15 at 16:57
  • $\begingroup$ More simply, the classifying map to the (approximate) universal bundle O(n+k)/e x O(k) ---> O(n+k)/(O(n) x O(k) can also be used to pull back the complementary bundle O(n+k)/O(n) x e ---> O(n+k)/O(n) x O(k), giving an inverse. $\endgroup$ – Robert Bruner Oct 29 '15 at 19:14
  • $\begingroup$ 4. is false. Take an infinite direct sum of tori. $\endgroup$ – Fernando Muro Nov 2 '15 at 5:58
  • $\begingroup$ @FernandoMuro: by a topological manifold in 4 I of course mean the finite dimensional one. $\endgroup$ – Igor Belegradek Nov 2 '15 at 11:51
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I think that the proposition still holds for smooth finite-dimensional manifolds. Here is a sketch of an argument.

First of all, for a bundle of rank $k$ the following statements are equivalent.

  • there exists a complementary bundle of rank $\ell$,
  • the classifying map $X\to BGL_k(\Bbbk)=G_k(\Bbbk^\infty)$ factors through $G_k(\Bbbk^{k+\ell})$.

Next, consider an $n$-dimensional smooth manifold, then $M$ admits a triangulation. The stars of its barycentric subdivision provide an open cover $\{U_0,\dots,U_n\}$ of $M$. Here each $U_i$ is the disjoint union of the stars of those vertices that come from the $i$-simplices of the original triangulation. In particular, each $U_i$ is a disjoint union of contractible open subsets of $M$. Fix trivialisations of $E|_{U_i}$ for each $i$. Then the transition maps $g_{ij}\colon U_i\cap U_j\to GL_k(\Bbbk)$ can be used to construct an explicit classifying map from $M$ to the $n$-fold join $(GL_k(\Bbbk)*\cdots*GL_k(\Bbbk))/GL_k(\Bbbk)$ in Milnor's construction of a classifying space.

Now, there is a homotopy equivalence $$\operatorname{colim}_n(\underbrace{GL_k(\Bbbk)*\cdots*GL_k(\Bbbk)}_{n+1\text {factors}})/GL_k(\Bbbk) \to\operatorname{colim}_\ell G_k(\Bbbk^{k+\ell})$$ between these two models for $BGL_k(\Bbbk)$. Because each of the joins on the left is a finite CW complex, the restriction ends up in a finite-dimensional Grassmannian. In particular, the classifying map $f$ above factors through one of them. Together with the observation above, that proves the claim.

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    $\begingroup$ Can one assume to have a finite triangulation? $\endgroup$ – მამუკა ჯიბლაძე Oct 29 '15 at 14:56
  • $\begingroup$ The triangulation will be finite if and only if $M$ is compact. But the argument works for infinite triangulations as well. I added a hint for the construction of the $U_i$. $\endgroup$ – Sebastian Goette Oct 29 '15 at 15:51
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    $\begingroup$ $\{U_0,...,U_n\}$ looks like finite to me, that's why I asked. I vaguely remember one uses locally finite covers in such cases... $\endgroup$ – მამუკა ჯიბლაძე Oct 29 '15 at 16:20
  • $\begingroup$ @მამუკაჯიბლაძე: Locally finite covers are good enough to produce classifying maps to the full Milnor model. But for this question, we want to end up in a finite subcomplex. $\endgroup$ – Sebastian Goette Oct 30 '15 at 15:53
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    $\begingroup$ Well, after the map lands in the finite subgrassmanian you may use cellular approximation to modify the cover and/or triangulation/cell structure of the source, but before that you have to deal with possibly infinite covers, no? $\endgroup$ – მამუკა ჯიბლაძე Oct 30 '15 at 15:59

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