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Atiyah Patodi and Singer [Spectral asymmetry and Riemannian geometry III] write that if $E$ is a complex flat bundle (non holomorphic, just smooth and complex) on a compact manifold $X$ (more generally a CW-complex) there must be some positive integer $m$ such that the direct sum $E\oplus E\oplus\cdots\oplus E$ ($m$ times) is a trivial bundle. More precisely, they write on top of page 19, "The vector bundle $V_a$ defined by $a$ is flat so its real Chern classes vanish, hence some multiple $kV_a$ is (unitarily) trivial".

Of course since the Chern character is an isomorphism from $K^\bullet(X)\otimes \mathbb{Q}$ to $H^{2\bullet}(X;\mathbb{Q})$, we know the bundle $E$ (or, more precely its K-theory class) is torsion in topological K-theory but this only tells us that $E\oplus E\oplus\cdots\oplus E$ is stably trivial.

I checked an impressive amount of literature on characteristic classes without finding a clue. Several authors cite directly Atiyah who, however, does not prove the claim.

Other authors say that the above statement cannot be true.

Does someone know the answer ?

domenico fiorenza
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Paolo Antonini
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1 Answers1

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As you already said, if $E$ is flat of rank $r$, then $ch(E)=0$, meaning that the class $[E] \in K^0 (X)$ is torsion and therefore $m[E]=0 \in K^0(X)$ for some $m$. Pick $m$ such that $mr >> dim X$.

Now the stabilization map $BU(n) \to BU$ is roughly $2n$-connected. This means that if $V \to X$ a vector bundle of sufficiently high rank then $V$ is trivial iff it is stably trivial. Apply this observation to $E^{ \oplus m}$. Done.

Johannes Ebert
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  • Sure the stable rank Theorem. Thank you. – Paolo Antonini Feb 23 '12 at 09:17
  • Is there any down-to-earth proof of this fact? It is so innocent-sounding... – diverietti Mar 09 '12 at 16:51
  • That $BU(n) \to BU$ is highly connected follows from the long exact homotopy sequence of $S^{2n-1} \to BU(n-1) \to BU(n)$. The stable cancellation follows by obstruction theory. Alternatively, if you are talking about smooth vector bundles on smooth manifolds, you can use general position arguments, like the following. If $M^m$ is a manifold, $V$ and $W$ vector bundles on $M$ of rank $r$ and $s$. If $s$ is sufficiently large w.r.t. $m$ and $r$, then any bundle map $f:V \to W$ in general position is injective (use transversality). Now if $V^r \oplus R^i \cong R^{r+i}$, you have ... – Johannes Ebert Mar 09 '12 at 18:58
  • two bundle monomorphisms $R^i\to R^{r+i}$, one with $V$ as complement, the other one with trivial complement. Taking any homotopy between them and put it into general position. Hence the two bundle mono's are isotopic; hence their complements, i.e. $V$ and $R^r$ are isomorphic. When done carefully, this argument gives the same bound on the dimensions as the obstruction theory argument. – Johannes Ebert Mar 09 '12 at 19:00
  • Thanks. Indeed is sufficient to take $n$ to be greater tian the stable ranking of the base – Paolo Antonini Jun 25 '14 at 20:15