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I want to prove that $\xi$ is a rank $k$ vector bundle over an $n$ dimensional CW complex $X$ such that $k>n$. Then $\xi$ is trivial iff $\xi \oplus \varepsilon^1$ is trivial. Here $\varepsilon^1$ is the trivial bundle of rank $1$.

Note that one direction is trivial. The non trivial direction is if $\xi \oplus \varepsilon^1$ is trivial implies $\xi$ is trivial.

To prove this I want to use classifying spaces. Let $BO(k)$ be the classifying space of all rank $k$ vector bundles. Now let the bundle $\xi$ be given by the map $$f:X \to BO(k)$$ But we are given that $\xi \oplus \varepsilon^1$ is trivial. This is just pullback of the map $i \circ f$ where $i$ is the map $$i: BO(k) \to BO(k+1)$$ So now since $\xi \oplus \varepsilon^1$ is trivial $i \circ f$ is null homotopic. Now we want to show that $f$ is null homotopic. I am struck here. Any help will be appreciated. Thank you.

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  • $\begingroup$ In the CW structure of $BO(k)$, what are the dimensions of the cells that are added to get $BO(k+1)$? Can you use CW approximation to conclude that the map $f$ is homotopic to a map factors through $BO(k)$? $\endgroup$
    – Thomas Rot
    Feb 27, 2017 at 12:25
  • $\begingroup$ @ThomasRot I was trying to use the CW structure description as given in Milnor's Characteristic Classes but I was unable to figure out how to do this. Could you please give some hints? $\endgroup$
    – happymath
    Feb 27, 2017 at 20:13

1 Answer 1

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You can prove this using the long exact sequence associated to the pointed fiber sequence

$$ S^k = O(k+1)/O(k) \stackrel{j}{\to} BO(k) \stackrel{i}{\to} BO(k+1) $$

Namely, fix a basepoint $x_0$ in $X$ which is a $0$-cell. Then we have an exact sequence of pointed sets

$$ [X,S^k]_* \stackrel{j_*}{\to} [X,BO(k)]_* \stackrel{i_*}{\to} [X,BO(k+1)]_* $$

where $[Z,W]_*$ denotes the set of pointed homotopy classes of pointed maps from $Z$ to $W$, and the basepoint of $[Z,W]_*$ is the class of the constant map (the constant is of course the basepoint of $W$).

Using your notation, you have that $i \circ f$ is nullhomotopic. Since $BO(k+1)$ is path connected, we can assume that it is homotopic to the constant map $c$, where the constant is the basepoint of $BO(k+1)$. Moreover, since we chose $x_0$ to be a $0$-cell of $X$, we can assume that the homotopy preserves the basepoints (using the homotopy extension property).

This shows that $i_*[f] = [c]$ and $[c]$ is the basepoint of $[X,BO(k+1)]_*$, so there exists $g \colon X \to S^k$ pointed such that $[f]=j_*[g]=[jg]$. Since the dimension of $X$ is smaller than $k$, the map $g$ is nullhomotopic by cellular approximation, and by the same argument as before, it is pointed nullhomotopic.

Hence $f$ is pointed nullhomotopic, in particular nullhomotopic, and so $\xi$ is trivial.

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  • $\begingroup$ thank you for the answer. I could follow the whole answer except for the fact as to why $S^k$ is the fiber. My understanding of fiber sequence is that the fiber of $i:BO(k) \to BO(k+1)$ is the pull back of the path fibration on $BO(k+1)$. So I am confused as to how you concluded that $S^k$ is of the same homotopy type. It would be really great if you could clarify this. Thank you $\endgroup$
    – happymath
    May 4, 2017 at 19:58
  • $\begingroup$ The map $BO(k) \to BO(k+1)$ is induced by the group monomorphism $O(k) \to O(k+1)$ that sends a size $k$ square matrix $A$ to the size $k+1$ square matrix with two diagonal blocks, $A$ of size $k$ and the matrix $(1)$ of size $1$. That is, the first row and column is the vector $(1,0,\ldots,0)$ and the rest is $A$. In general a group monomorphism $H \to G$ induces a fibration $BH \to BG$ and the fiber is $G/H$, the space of left cosets. So in this case, we have $O(k+1)/O(k)$. $\endgroup$
    – Goa'uld
    May 4, 2017 at 20:59
  • $\begingroup$ Now $O(k+1)$ acts on $S^k$ by matrix multiplication, if you multiply an orthogonal matrix by a vector in $\mathbb{R}^{k+1}$ of norm one, the result again has norm one. This action is transitive, given any two unitary vectors, there is an orthogonal matrix that carries one to the other, this is easy to see with linear algebra. And the stabilizer of the element $(1,0,\ldots,0)$, that is, the subgroup of orthogonal matrices that fix this element coincides with those block matrices I mentioned above, that copy of $O(k)$ inside $O(k+1)$. Therefore $S^k$ is homeomorphic to $O(k+1)/O(k)$. $\endgroup$
    – Goa'uld
    May 4, 2017 at 21:05
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    $\begingroup$ You can find it in the book "Cohomology of finite groups" by Adem and Milgram. It is Lemma 1.11 in Chapter II. $\endgroup$
    – Goa'uld
    May 5, 2017 at 15:39
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    $\begingroup$ An online reference is this page on Omar Antolín Camarena's website. $\endgroup$
    – Michael Albanese
    May 19, 2017 at 16:54

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