Given a smooth, closed, oriented $n$-manifold $M$ with amenable fundamental group $\pi_1(M)$. Is every $n$-dimensional flat orientable vector bundle $\xi = (p: E \rightarrow M)$ over $M$ trivial?

By an $n$-dim orientable flat vector bundle I mean a vector bundle arising as $ E= \tilde{M} \times \mathbb{R}^n / \pi_1(M)$, where the fundamental group acts on the universal covering $\tilde{M}$ by deck transformations and on $\mathbb{R}^n$ via a homomorphism $\rho: \pi_1(M) \rightarrow GL_n^+(\mathbb{R}) $ and with $p$ induced by the canonical projection.

I don't see why it should be true and I am searching for a counterexample with some manifold and bundle satisfying the assumptions, but since the Euler class always vanishes and so do all characteristic classes in cohomology with real coefficients (e.g. by M. Karlsson in her thesis "characteristic classes and bounded cohomology"), since the real bounded cohomology of $M$ is trivial, I haven't been able to come up with a counterexample myself yet. But my approach would remain to have a bundle some nontrivial characteristic classes that are torsion. I appreciate any suggestions.


2 Answers 2


Like you say, we cannot get real characteristic class obstructions because of the lack of curvature. So we look to the second Stiefel-Whitney class. Googling I find this paper. (I have not verified any claims in there, only extracted the relevant results.) Proposition 4.4 claims to construct a flat oriented non-spin manifold $M$. See section 2 for their notion of "acting diagonally" and the beginning of section 4 for the (complicated-looking) construction of the action they want.

$M$ is given as a quotient of $T^n$ by a free action of $\Bbb Z_2^d$; in particular, its fundamental group fits into the exact sequence $$1 \to \Bbb Z^n \to \pi_1(M) \to \Bbb Z_2^d \to 1.$$ Abelian groups are amenable and extensions of amenable groups are amenable so $\pi_1(M)$ is amenable.

  • 2
    $\begingroup$ Good example. I think, that the right statement is that all flat vector bundles (over compact CW complexes $M$ with amenable fundamental groups) are virtually trivial, i.e. there is a finite cover $M'\to M$ such that the pull-back of $\xi$ to $M'$ is trivial. Note that this is false without amenability assumption already when $M$ is a surface and $\xi$ is of rank 2. $\endgroup$ May 31, 2017 at 2:14
  • $\begingroup$ @MoisheCohen This seems very believable. Do you know a proof? I wasn't able to prove it. $\endgroup$
    – user98602
    Jun 19, 2017 at 19:32
  • $\begingroup$ Yes, I think I do. I will write it when I have more time. $\endgroup$ Jun 19, 2017 at 21:58
  • $\begingroup$ @MoisheCohen Sorry to bug you; I ping you only because I remembered this and will probably forget again. Do you have a brief sketch you can write down? Or is this particularly complicated? $\endgroup$
    – user98602
    Jul 3, 2017 at 20:30
  • $\begingroup$ @MikeMiller: I posted a sketch. $\endgroup$ Jul 4, 2017 at 6:19

As requested, here is a sketch (more precisely, description of steps of the proof).

Theorem. Suppose that $G$ is a discrete amenable group, $G=\pi_1(B)$ and $E\to B$ is a flat vector bundle associated with a linear (finite-dimensional) representation of $G$. Then there exists a finite index subgroup $G_1<G$ such that the lift of $E$ to the covering space of $B$ corresponding to $G_1$ is trivial as a vector bundle.

Sketch of the proof.

I will us the notation $E_\rho$ to denote the vector bundle associated with the linear representation $\rho$.

  1. The first thing to observe is that given a continuous family of representations $\rho_t$, all vector bundles $E_{\rho_t}$ are isomorphic (as vector bundles, not as flat bundles, of course). You should be able to prove this yourself.

  2. Furthermore, assuming that $G$ is a finitely generated amenable group, by Tits' Alternative, $\rho(G)$ is virtually solvable, i.e. contains a solvable subgroup of finite index. Hence, its Zariski closure contains a connected solvable subgroup $H$ of finite index.

  3. Pass to a finite index subgroup $G_1<G$ such that $\rho(G_1)<H$. Now, use the fact that $H$ preserves a full flag in $V^{\mathbb C}$, i.e. is conjugate in $GL(V^{\mathbb C})$ into the Borel subgroup $B$ of upper triangular matrices.

  4. From this, you conclude that $H$ preserves a 2-dimensional subspace $V_2\subset V$. Consider the representation $\rho_2$ of $G_1$ on $V/V_2$ and use the dimension induction. This allows you to deform $\rho_2$ to the trivial representation $\rho'_0$. Lift this to a deformation of $\rho|G_1$ to a representation $\rho_0$ of $G_1$.

  5. The representation $\rho_0$ need not split off $\rho_0'$ as a direct summand as $\rho_0$ is not reductive: it is a representation by block-triangular matrices though. Now, use the "semisimpliciation", i.e. take a suitable family of conjugates of $\rho_0$ by diagonal matrices which limits to a representation $\rho_1$ which has zero upper block in the block-triangular decomposition. (I suggest to work this out first in the case of upper triangular 2-by-2 complex or real matrices: conjugate such a matrix by a family of diagonal matrices until it becomes diagonal itself in the limit.)

Thus, $\rho_1$ is the direct sum of the trivial and a real 2-dimensional representation $\rho_3$. It suffices to deform $\rho_2$ to the trivial representation.

  1. Now $\rho_3(G_1)$ is abelian, conjugate into $R_+\times SO(2)\cong {\mathbb C}^*$. Use the fact that $Hom(G_1, {\mathbb C}^*)$ is connected, which is a pleasant exercise since ${\mathbb C}^*$ is abelian. The result is a deformation of $\rho_3$ to the trivial representation.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.