Though it's relatively clear that the characteristic classes do not characterise a vector bundle (and after looking through some books) I could not find an example of a vector bundle which is not stably trivial but whose characteristic classes (those which may be defined*) are all trivial. Could someone be so kind as to point out a reference for this?

*an example with trivial Stiefel-Whitney class would already be nice.


3 Answers 3


You will find an answer to your question in Hatcher's book project "Vector bundles an K-theory" (p. 75-76) (available on his homepage). Using the fact that $\pi_8(O(10))=\mathbb Z_2$, you can build a non- stably trivial vector bundle over the sphere $S^9$ (using the clutching function associated to the non-trivial homotopy class $S^8\rightarrow O(10)$ in $\pi_8(O(10))$). This vector bundle has all his Stiefel-Whitney and Pontryagin classes equal to zero.

The vanishing of $w_9$ follows from Wu's formula $w_9=w_1w_8+Sq^1(w_8)$.

  • $\begingroup$ Cool ! What means "representing" ? $\endgroup$ Nov 18, 2012 at 16:58
  • $\begingroup$ Could you include a proof that the Stiefel-Whitney classes vanish? $\endgroup$ Nov 18, 2012 at 17:02
  • $\begingroup$ Thanks! a shame it's not more clearly stated (I certainly read through those pages without seeing this implication)... $\endgroup$
    – ARG
    Nov 18, 2012 at 22:45
  • 3
    $\begingroup$ @IgorBelegradek The base space is $S^9$, so the only possible Stiefel-Whitney or Pontryagin class which could possibly be non-zero is $w_9$. This too vanishes, as the OP explained. $\endgroup$ Oct 7, 2015 at 20:49

Let $E=\gamma^1\otimes\mathbb{C}$ be the complexified tautological bundle over $X=\mathbb{P}^6(\mathbb{R})$, and set $F=4E=E\oplus E\oplus E\oplus E$. It is not hard to check that $c(E)=1+a$ and $w(E_\mathbb{R})=1+\bar a$ with $a\in H^2(X,\mathbb{Z})\cong\mathbb{Z}/2$ and $\bar a \in H^2(X,\mathbb{Z}/2)\cong\mathbb{Z}/2$ being the non-zero element. So all imaginable characteristic classes of $F$ and $F_\mathbb{R}$ vanish by the Whitney sum formula (note that $H^*(X,\mathbb{Z})=\mathbb{Z}[a]/2a=a^4=0$).

However, the group $\tilde K^0(X)$ of the classes of complex vector bundles on $X$ up to stable equivalence is cyclic of order 8 and is generated by $E-1$, see e.g. Karoubi, K-theory, corollary 6.47 from Chapter IV. So $4(E-1)=F-4\in \tilde K^0(X)$ is non-zero, and so $F$ is not stably trivial.


There is an old reference which gives low-dimensional examples. The paper

A. Dold and H. Whitney. Classification of oriented sphere bundles over a $4$-complex. Ann. Math.(2) 69 (1959), pp. 667--677

contains a classification of oriented vector bundles over CW-complexes of dimension $4$. Take a rank $n$ real vector bundle $\mathcal{E}$, $n\geq 5$, with trivial Stiefel-Whitney classes and trivial Pontryagin class over a $4$-dimensional CW-complex $X$. Then there is a cohomology class in $H^4(X,\mathbb{Z})$ which measures the difference between $\mathcal{E}$ and the trivial bundle, the class is $0\mod 2$ and annihilated by multiplication by $2$. Conversely, any such class can be realized by a bundle.

Therefore, a non-trivial bundle exists on a $4$-complex if there are classes which are $0\mod 2$ and annihilated by $2$. The obvious example of such a complex is the Moore space $M(\mathbb{Z}/4,4)$ obtained by glueing a $4$-cell to $S^3$ with a degree $4$ map. Note also that rank $n\geq 5$ on a $4$-complex is in the stable range, so the bundle obtained from the Dold-Whitney paper is stably non-trivial.

Edit: Since the question was tagged differential-geometry, I'll also give a manifold example. Take a free action of $\mathbb{Z}/4\mathbb{Z}$ on $S^5$ (which can be obtained from the embedding $S^5\subset \mathbb{C}^3$ and coordinatewise multiplication with suitable roots of unity). The quotient $X=S^5/(\mathbb{Z}/4\mathbb{Z})$, a lens space, is a $5$-dimensional smooth manifold with $H^4(X,\mathbb{Z})\cong\mathbb{Z}/4\mathbb{Z}$. This means that there is a non-trivial $SO(6)$-bundle with trivial characteristic classes on the $4$-skeleton, as follows from the Dold-Whitney paper. Since $\pi_5(BSO(6))=0$, obstruction theory implies that any $SO(6)$-bundle on the $4$-skeleton extends uniquely to a vector bundle on $X$. Again, rank 6 on dimension 5 is stable range, so the bundle constructed this way is stably non-trivial.


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