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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.

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2 Answers 2

up vote 23 down vote accepted

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 $w_9=w_1w_8+Sq^1(w_8)$.

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Cool ! What means "representing" ? –  Alexander Chervov Nov 18 '12 at 16:58
Could you include a proof that the Stiefel-Whitney classes vanish? –  Igor Belegradek Nov 18 '12 at 17:02
Thanks! a shame it's not more clearly stated (I certainly read through those pages without seeing this implication)... –  Antoine Nov 18 '12 at 22:45

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.

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