The specific situation is the following:

Let $n>0$ be a natural number, let $X$ be a finite CW complex of dimension $n$, and let $\xi_0,\xi_1$ be oriented real vector bundles of rank $n$ over $X$ such that $\epsilon\oplus\xi_0 \cong \epsilon\oplus \xi_1$ (where $\epsilon$ is the trivial rank 1 bundle). Here the vector bundles $\xi_i$ are considered as maps $X\to BSO(n)$ and the operation "$\epsilon\oplus\cdot$" is interpreted as compostion with the natural map $BSO(n)\to BSO(n+1)$.

The stable isomorphism asserts that there is a homotopy $H:X\times I\to BSO(n+1)$ between the stabilized bundles, and in oder to determine if $\xi_0\cong \xi_1$ one is lead to the commutative diagram

\begin{array}{ccc} X\times\{0,1\}&\to &BSO(n)\\ \downarrow & & \downarrow \\ X\times [0,1]&\to &BSO(n+1) \end{array}

where the upper map restricted to $X\times\{i\}$ is $\xi_i$. The obstructions to lifting $H$ in this diagram find themselves in the groups $H^k(X\times[0,1],X\times\{0,1\};\pi_{k-1}(S^n))$, and so (using the suspension iso) the only potentially non-zero obstruction, say $o(\xi_0,\xi_1)$, lies in $H^n(X;\mathbb{Z})$.

So the question itself is "Can we identify this obstruction with something familiar?" (Maybe restricting to the case of $X$ a manifold or Poincare complex is easier)

In the case where $n$ is even, a guess would be $o(\xi_0,\xi_1)=e(\xi_1)-e(\xi_0)$, the difference of the Euler classes. This is certainly a necessary condition, and at least in the case of $X=S^2$ it is also sufficient.

For $n$ odd this guess is definitely not correct: the euler class of any oriented odd-rank real vector bundle is 2-torsion, so must vanish for any rank $n$ bundle over $S^n$; but the tangent bundle of $S^n$ is stably-trivial and not actually trivial for $n\neq 1,3,7$. In this case it's not as clear what a description of the obstruction should be.

(Remark: if $rank(\xi_i)>dim(X)$, then $\xi_i\cong\xi'_i\oplus\epsilon^k$ where $\xi'_i$ has rank equal to $dim(X)$ (this can be shown with an obstruction argument using the connectivity of Stiefel manifolds). Then by another obstruction argument one can show that in fact $\xi'_0\oplus\epsilon \cong \xi'_1\oplus\epsilon$, and so the case $rank(\xi_i)>dim(X)$ is handled by the case $rank(\xi_i)=dim(X)$.)

firstobstruction only! Euler class is defined (and is potentially non-zero) in odd dimensions as well, so you shouldn't rule it out just because it does not work for tangent bundles of closed manifolds! $\endgroup$ – Alex Degtyarev Feb 14 '15 at 21:431more comments