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?
Best Answer
The answer is no. Let $G$ be a cyclic group of order $n$ not divisible by $2$, let $V$ be an irreducible $2$-dimensional representation of $G$, and consider the associated vector bundle $EG\times_G V\to BG$ (which I'll call $V$ again). Then $V$ has trivial Stiefel-Whitney classes since $H^q(BG;\mathbb{Z}/2)=0$ if $q>0$, but $V$ can have non-vanishing Pontryagin class (we have $H^*(BG;\mathbb{Z})=\mathbb{Z}[[x]]/(nx)$ where $x\in H^2$, and $p(V)=1-a^2x^2$ with $a\in \mathbb{Z}/n$ depending on the original representation.) Since the Pontryagin class satisfies Whitney sum up to $2$-torsion, this gives a counterexample: the virtual bundle $-V$ is does not come from a vector bundle, since $p(-V)=(1-a^2x^2)^{-1}$.
Now you ask, what if we also require that the Pontryagin classes are (finitely) invertible? There's probably a counterexample here too, though I don't have one at hand.
Added later. Here's one. (I hope: I keep needing to fix it.) If $G$ is a finite $p$-group, then the "Borel construction" defines a bijection between of: the set $\mathrm{Rep}_n(G)$ of isomorphism classes of real $n$-dimensional representations of $G$ into the set $\mathrm{Bun}_n(G)$ of isomorphism classes of real $n$-dimensional vector bundles over the classifying space $BG$. In representation theory, there are no additive inverses, so non-trivial bundles over $BG$ which come from representations cannot have inverses. So it's enough to find a non-trivial representation $V$ whose characteristic classes all vanish.
Let $G$ be a cyclic group of order $p^2$, where $p$ is an odd prime, generated by an element $\sigma$. Let $L$ be the $1$-dimensional complex representation given by $\sigma|L=e^{2\pi i/p}$, and write $V$ for the real $2$-dimensional vector bundle underlying the complex line bundle $EG\times_G L\to BG$. It appears that the Pontryagin classes vanish: up to signs, the total Pontryagin class is the total Chern class of $V\otimes \mathbb{C}\approx L\oplus \overline{L}$, and we compute $c(V)=c(L)c(\overline{L}) = (1+px)(1-px) = 1-p^2x^2 = 0$.
(The fact about the bijection $\mathrm{Rep}(G)$ into $\mathrm{Bun}(G)$ is non-trivial; it follows from a theorem of Dwyer and Zabrodsky ("Maps between classifying spaces", LNM 1298). I don't know a more elementary proof, but a condition such as "$G$ is a $p$-group" is probably necessary.)
End addition.
There are obstructions in $K$-theory to "inverting" bundles. There are exterior power operators $\lambda^k:KO(X)\to KO(X)$, such that if $[V]$ is the $K$-class of an actual bundle $V$, we have $\lambda^k([V])=[\Lambda^k V]$, the class of the exterior power bundle. The formal sum $\Lambda_t(x)\in KO(X)[[t]]$ given by $$ \Lambda_t(x)= 1+\lambda^1(x)t+\lambda^2(x)t^2+\dots $$ has a Whitney formula ($\Lambda_t(x+y)=\Lambda_t(x)\Lambda_t(y)$), coming from the usual decomposition $\Lambda^nV=\sum \Lambda^iV\otimes \Lambda^{n-i}V$. Furthermore, if $x=[V]$ is the class of an honest $n$-dimensional bundle, we must have $\lambda^i([V])=[\Lambda^iV]=0$ for $i>n$, so that $\Lambda_t(x)$ is polynomial.
In this case, trivial bundles don't have trivial $\Lambda$-class; instead, writing "$n$" for the trivial $n$-plane bundle, we have $\Lambda_t(n)=(1+t)^n$. Thus, an isomorphism $V\oplus W= n$ implies an identity $\Lambda_t([V])\Lambda_t([W])=(1+t)^n$.
So an even stronger form of your question is: if $V$ is a vector bundle (over a nice space) such that $\Lambda_t([V])/(1+t)^n \in KO(X)[[t]]$ is a polynomial, must it follow that there exists a bundle $W$ such that $V\oplus W\approx n$? Again, I don't have a counterexample here.