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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?

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Minor comment: the cohomology ring of RP^\infty should really be thought of as a power series ring, not a polynomial ring, so 1 + x is invertible. The problem is that its inverse has infinite degree so you can't get a finite dimensional bundle as the inverse. It does, however, have an infinite dimensional inverse: L^\perp. However, once you allow infinite dimensional inverses, everything has an inverse as E + H = H where H is a Hilbert space. (mumble, mumble, numerable cover, mumble, mumble, contractibility of unitary group, mumble, mumble) –  Loop Space Oct 18 '10 at 12:59
    
Dear Andrew, you are right but I knew this ( although I am not quite sure about the fourth mumble...). I am intersted in finite rank bundles exclusively. –  Georges Elencwajg Oct 18 '10 at 13:06
    
I suspected that you might, which is why I left it as a comment. –  Loop Space Oct 18 '10 at 13:12
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I don't like the notation "invertible bundle", because it clashes with "invertible sheaf" in algebraic geometry (which is not a sheaf whose -reduced?- Grothendieck group additive inverse has a "genuine" representative). I would call such objects "complementable" bundles instead. –  Qfwfq Oct 18 '10 at 14:04
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I agree that "invertible" is not an ideal terminology, but again it is used in tom Dieck's new book Algebraic Topology, which may very well become,along with Hatcher's magnificent homonymous tome, the standard 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

1 Answer 1

up vote 19 down vote accepted

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.

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Thank you for this very interesting answer, Charles. –  Georges Elencwajg Oct 18 '10 at 16:12
    
I've added to my answer, giving a counterexample to the Pontryagin class version of the statement. –  Charles Rezk Oct 18 '10 at 16:23
    
Wow! Not only is your answer great but it improves from minute to minute! It will take me a little time to digest, since I didn't know this method of using representation theory, but I have a fantastic teacher now...Thanks again, Charles. –  Georges Elencwajg Oct 18 '10 at 16:57
    
Well, I keep removing errors each time too. The claim that "bundles over BG" are the same as G-representations is too strong; it's not true, as was already known to Atiyah in 1961. –  Charles Rezk Oct 18 '10 at 18:07
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Here's a more direct proof of what I need. Claim: If $V$ and $W$ are real vector bundles over $BG$ of dimensions $m$ and $n$ (with $G$ any finite group), then $V\oplus W$ trivial implies $V$ trivial. Proof: a trivialization of $V\oplus W$ gives a map $BG\to G(m+n,m)$, where $G(m+n,m)$ is the Grassmannian of $m$-planes in $R^{m+n}$. By Miller's theorem (aka the Sullivan conjecture), any map from $BG$ to a finite CW complex (such as the Grassmannian) is homotopic to a constant map. Thus $V$ and $W$ are equivalent to trivial bundles. –  Charles Rezk Oct 18 '10 at 18:32

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