TITLE: Do vanishing characteristic classes of the tangent bundle imply a manifold is stably frameable?
QUESTION [16 upvotes]: 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?

REPLY [9 votes]: 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 
Non-stably trivial bundle with trivial characteristic classes
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.