# What is obstructing two stably-isomorphic vector bundles from being isomorphic?

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 $rank(\xi_i)=dim(X)$.)

• Don't forget that you are speaking about the first obstruction 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! – Alex Degtyarev Feb 14 '15 at 21:43
• Ah, you're right. I forgot that the Euler class in odd dimensions is not always 0 (I'll edit the question to fix that statement), but it is 2-torsion since odd-rank bundles admit orientation reversing isomorphisms. Thus if $H^n(X;\mathbb{Z})$ has no 2-torsion (like $X=S^n$) then the Euler class of any bundle will be 0, and so it still won't work as an obstruction in general. – William Feb 15 '15 at 12:13
• Side note: another way of seeing that the Euler class is 2-torsion for odd-rank bundles is by comparing the integral cohomology of $BSO(2n+1)$ with its cohomology with coefficients in an integral domain $\Lambda$ containing $\frac{1}{2}$. It's known that $H^*(BSO(2n+1);\Lambda)$ is a polynomial algebra on generators $p_1,\dots,p_n$ concentrated in degrees $4*$, so $H^{2n+1}=0$; integrally, you get that algebra plus the image of the Bockstein homomorphism, which only contains elements of order 2. So without knowing how to define the Euler class, one can still find that it must be 2-torsion. – William Feb 15 '15 at 12:23
• @w.gollinger Would you happen to know some (pedagogical) reference for the assertions in your last comment? I would like to learn a bit about this (specifically, I would like to understand if/why for an oriented rank 3 bundle, the Euler class is the Bockstein of $w_2$). – Danu Sep 4 '17 at 21:45
• I found a reference: The document titled "Characteristic classes" by Brunner, Catanzaro and May, available online, contains a discussion of these facts. – Danu Sep 6 '17 at 14:04

This is only a comment, but slightly too long. Let me discuss the case $n=5$, as an example of what the additional information beyond the Euler class could look like.

The classification of oriented real rank 5 bundles over CW-complexes of dimension $\leq 5$ was given in

• M. Cadek and J. Vanzura: On the classification of oriented vector bundles over $5$-complexes. Czech. Math. Journal 43 (1993), 753-764.

The main result is that the map $$[X,BSO(5)]\to H^2(X,\mathbb{Z}/2)\oplus H^4(X,\mathbb{Z}/2)\oplus H^4(X,\mathbb{Z}):$$ $$E\mapsto (w_2(E),w_4(E),p_1(E))$$ is injective if and only if both the following conditions are satified:

• Condition (A): $H^4(X,\mathbb{Z})$ does not have non-trivial elements of order $4$

• Condition (B): $Sq^2 H^3(X,\mathbb{Z}/2)=H^5(X,\mathbb{Z}/2)$.

In particular, this result explains how much isomorphism is determined by characteristic classes. In the case of the $5$-sphere, bundles are stably isomorphic if and only if they have the same characteristic classes. In the case of the $5$-sphere, problems come from Condition (B) - I think that there is a secondary characteristic class there which takes values in $H^5/Sq^2 H^3$ and which can detect the tangent bundle. Maybe this can be found in the paper of Peterson and Stein on secondary characteristic classes, but I do not have access to check right now.

Further information on comparison of isomorphism vs. stable isomorphism in dimensions up to $8$ should be in

• L.M. Woodward. The classification of orientable vector bundles over CW-complexes of small dimension. Proc. Roy. Soc. Edinburgh 92A (1982), 175-179.
• Thank you for the response Matthias, I've been looking at the references you suggested. They are providing some insight into what kinds of ideas and arguments have been considered for the problem of vector bundle classification, namely this "Woodward method." Of particular interest is how much of a role cohomology operations play here in identifying relations between characteristic classes, and maybe in order to use these ideas in higher dimensions it's necessary to have a general picture of how the Steenrod algebra and the Pontryagin squares act on the cohomology of BSO(n). – William Feb 18 '15 at 10:33

I know, I am late to the party (as always) but I could provide some more information. In this article I show that two $$5$$-dimensional vector bundles with $$w_2=w_4=0$$ over a spin $$5$$-manifold which are stably isomorphic, are isomorphic if and only if their generalized Kervaire semi-charactersitics agree. Moreover it is possible to show that if $$w_4\neq 0$$ then the vector bundle is uniquely determined by its stable class.

I think, it should be possible to generalize this to higher dimensions.