# Do vanishing characteristic classes of the tangent bundle imply a manifold is stably frameable?

Suppose we know that all stable characteristic classes of the tangent bundle of a manifold $M$ vanish, i.e. the map $f:M\rightarrow BO$ stably classifying the tangent bundle is trivial on cohomology, i.e. $0=f^*: H^*(BO, A)\rightarrow H^*(M, A)$, for all coefficients $A$. Is it then possible to find a stable framing of $M$? In other words, can we lift $f$ to a map $\tilde{f}: M\rightarrow B\{1\}\simeq *$?

I know this is not the case for a general vector bundle (as discussed in Hatcher's "Vector bundles and K-theory", p.75-76), but I was wondering whether it might be true when restricting to the case where the bundle is the tangent bundle of a manifold. And how about if it is a normal bundle?

• A manifold of dimension $\le 7$ is stably frameable iff it has, in order, a stable orientation, a stable spin structure, and a stable string structure. The obstructions to these are $w_1, w_2$, and a characteristic class $\frac{p_1}{2}$ which is only defined given a stable spin structure. Your hypotheses imply that $w_1, w_2, p_1$ vanish, so it's possible that $\frac{p_1}{2} \in H^4(-, \mathbb{Z})$ is $2$-torsion. The smallest dimension this can happen in is $5$. But it might be easier to look for counterexamples in dimension $\ge 8$. – Qiaochu Yuan Feb 17 '16 at 16:09
• Indeed, thanks for the clarification. If someone has a list of characteristic classes of high-dimensional manifolds then that would be very helpful. – Renee Hoekzema Feb 17 '16 at 16:21
• Isn't every vector bundle a normal bundle? (And vice versa.) – Allen Knutson Feb 17 '16 at 18:03
• I don't think so. A normal bundle complements the tangent bundle of a manifold, so that the sum of the two is a trivial bundle. A normal bundle of dimension $k$ exists if and only if the manifold can be immersed in a Euclidean space with codimension $k$. See for example "Lectures on Immersion theory" by Ralph Cohen and Ulrike Tillmann. In particular, the characteristic classes of the normal bundle are dual to those of the tangent bundle. – Renee Hoekzema Feb 17 '16 at 18:28
• Stably, the questions for the normal and the tangent bundle are equivalent as the stable tangent bundle can be obtained from the stable normal bundle by composing with the canonical involution on $BO$ and vice versa. – archipelago Feb 17 '16 at 20:02

Suppose that $X^n$ is a finite complex and $V \to X$ a vector bundle which has trivial characteristic classes but is not stably trivial. Examples for such bundles were given in the answers to
The stable normal bundle of $S^{2n}$ admits a bundle map to $V$ (because $S^{2n}$ is stably parallelizable). This gives a map $f: S^{2n} \to X$, covered by a bundle map. By surgeries below the middle dimension, we can change $S^{2n}$ and $f$ to a map $f':M \to X$ which is covered by a bundle map $\nu_M \to V$ and such that $f$ is $n$-connected.
To see that $\nu_M$ is not stably trivial, note that since $X$ is $n$-dimensional and $f: M \to X$ $n$-connected, we can find a map $g: X \to M$ with $f \circ g \simeq id_X$, by basic obstruction theory.
But $g^* \nu_M \cong g^* f^* V \cong V$ is not stably trivial, and so $\nu_M$ isn't, either.