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?



Do vanishing characteristic classes of the tangent bundle imply a manifold is stably frameable?
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](https://mathoverflow.net/questions/112764/non-stably-trivial-bundle-with-trivial-characteristic-classes?rq=1)


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.


