Consider vector bundles on connected paracompact topological spaces. Such a vector bundle $E$ on $X$ is said to be invertible if there exists some other bundle $F$ whose sum with $E$ is trivial: $E\oplus F \simeq \epsilon ^N $. The terminology "invertible" (used by Tammo tom Dieck for example) comes from K-theory and is not so weird as it looks:in $\tilde K(X)$ the class of $F$ is indeed the additive inverse of that of $E$. If all vector bundles on $X$ are invertible, then every class (=virtual bundle) in $\tilde K(X)$ is represented by an actual bundle, which is rather nice.

Now, every vector bundle is invertible if $X$ is compact or is a differentiable manifold or even a topological manifold or even a subspace of some $\mathbb R^n$ or even a space of finite combinatorial Lebesgue dimension or even ... [Please correct me if I'm wrong: this is an interpretation/synthesis of what I read, sometimes between the lines, in several places.]

So one might optimistically hope that every vector bundle on a paracompact connected space is invertible: after all, what could go wrong? Here is what.

Consider $X=\mathbb {RP}^{\infty}$ (infinite dimensional real projective space) and the tautological line bundle $\gamma$ on $X$. Its total Stiefel-Whitney class is $w(\gamma)=1+x \in H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$, where $x$ is the first Stiefel-Whitney class of $\gamma$ . If $\gamma$ had an inverse vector bundle $F$ we would have $w(\gamma) w(F)=1$ and this is impossibl since $w(\gamma)=1+x$ is not invertible in the cohomology ring $H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$ ( a polynomial ring in one indeterminate over $\mathbb Z /2)$.

This leads me to ask the question:

**If a vector bundle on a connected paracompact space has a total Stiefel-Whitney class invertible in its cohomology ring, does it follow that the bundle itself is invertible?**

isinvertible. The problem is that its inverse has infinite degree so you can't get afinitedimensional bundle as the inverse. It does, however, have an infinite dimensional inverse: L^\perp. However, once you allow infinite dimensional inverses,everythinghas an inverse as E + H = H where H is a Hilbert space. (mumble, mumble, numerable cover, mumble, mumble, contractibility of unitary group, mumble, mumble) – Andrew Stacey Oct 18 '10 at 12:59thestandard textbook in the field. A variant terminology I have seen in Osborn is "stably invertible". As an algebraic geometer I am quite aware of the terminology "invertible sheaf","invertible module",... used in a mutiplicative sense: that's why I spelled out the definition I had in mind. And, last but not least, I like your terminology "complementable" and wish it good luck! – Georges Elencwajg Oct 18 '10 at 16:08