13
$\begingroup$

Chern-Weil theory tells us that the integral Chern classes of a flat bundle over a compact manifold (i.e. a bundle admitting a flat connection) are all torsion. Given a compact manifold $M$ whose integral cohomology contains torsion, one can then ask which (even-dimensional) torsion classes appear as the Chern classes of flat bundles. What is known about this question? I would be interested both in statements about specific manifolds and about general (non)-realizability results.

One specific thing that I know: if $S$ is a non-orientable surface, then there is a flat bundle $E\to S$ whose first Chern class is the generator of $H^2 (S; \mathbb{Z}) = \mathbb{Z}/2$. This shows up, for example, in papers of C.-C. Melissa Liu and Nan-Kuo Ho. As Johannes pointed out in the comments, this also shows that the fundamental class of a product of surfaces can be realized by a flat bundle.

However, I suspect that for a product of 3 Klein bottles, not all the 4-dimensional torsion classes can be realized as second Chern classes of flat bundles. In fact, I think I know a proof of this if one restricts to unitary flat connections: the space of unitary representations has too few connected components.

$\endgroup$
6
  • 1
    $\begingroup$ Take two flat complex line bundles $L_i \to S_i$ whose Chern classes are generators. Then form $L_1 \times L_2$. Doesn't this answer your last question? $\endgroup$ Mar 24, 2011 at 22:40
  • 1
    $\begingroup$ The holonomy of a flat bundle over $M$ factors through $\pi_1(M)$, so the classifying map $M\to BU(n)$ factors through $B(U(n)^d)$ ($U(n)$ with the discrete topology) so your question asks what is the map $H^*(BU(n))\to H^*(B(U(n)^d))$. The algebraic K-theory experts can probably say something, but I suspect the answer isnt completely known for $n>1$. $\endgroup$
    – Paul
    Mar 25, 2011 at 2:40
  • $\begingroup$ @Paul: Maybe $GL(n,\mathbb C)$ instead of $U(n)$? A bundle admitting a flat connection might not admit an inner product compatible with a flat connection. $\endgroup$ Mar 25, 2011 at 2:52
  • $\begingroup$ Johannes - yes, thanks for pointing this out. I think I really meant to say something about classes on a product of 3 non-orientable surfaces. $\endgroup$
    – Dan Ramras
    Mar 25, 2011 at 5:20
  • $\begingroup$ Paul: Morel recently proved the long-standing conjecture of Milnor that the homology of a Lie group with finite coefficients is the same if you use the discrete topology, so this yields no obstruction. To rule out an obstruction, you just need a surjection in cohomology, which is easy to see from the normalizer of a maximal torus. Also, we're probably in the stable range, which was dealt with by Suslin, decades ago. $\endgroup$ Mar 25, 2011 at 6:40

2 Answers 2

9
$\begingroup$

Small correction: For a non-compact manifold $M$, the group $H_{2k-1}(M)$ might not be finitely generated. In this case Chern-Weil does not imply that the $k$th Chern class of a flat bundle on $M$ has finite order. Rather, it just implies that it belongs to the subgroup $Ext(H_{2k-1}(M),\mathbb Z)\subset H^{2k}(M)$.

Positive answer for first Chern class: Use the surjection $Hom(H_1(M),GL_1(\mathbb C))\to Ext(H_1(M),\mathbb Z)$ associated to the exponential exact sequence $0\to \mathbb Z\to \mathbb C\to GL_1(\mathbb C))\to 1$. An element of this $Hom$ group describes a flat complex line bundle on $M$ with prescribed Chern class in the $Ext$ part of $H^2(M)$.

Negative answer in general, for a pretty trivial reason: If $M$ is simply connected, then flat bundles on $M$ are necessarily trivial, but $M$ can still have torsion in $H^{2k}$ if $k>1$.

So a better question is, if $\Gamma$ is a group then can every element of (the $Ext$ part of ?) $H^{2k}(B\Gamma)$ be $c_k$ of a vector bundle arising from a homomorphism $\Gamma\to GL_r(\mathbb C)$ for some $r$? I don't know the answer if $k>1$.

$\endgroup$
3
$\begingroup$

The answer to Tom's formulation is no. It's possible if you restrict to finitely generated groups that my argument falls apart, but I doubt this is essential.

Take a group $\Gamma$ so that $B\Gamma^+=K(Q/Z,2n-1)$, ie, $H^k(\Gamma;Z)=H^k(K(Q/Z,2n-1);Z)$. Since $Ext(Q/Z,Z)=\hat Z$, there lots of interesting classes in $H^{2n}(\Gamma;Z)$. If we could lift them to flat bundles over $B\Gamma$, then after applying the plus construction and profinite completion, we would have split $K(\hat Z;2n)$ off of $BU^{\hat{}}$. But the torsion homology of the Eilenberg-MacLane space cannot be a retract of the torsion-free homology of $BU$.

I wanted to work one prime at a time, but $Ext(Q_p/Z_p,Z)=0$.

$\endgroup$
1
  • 1
    $\begingroup$ I am a naive student and do not know what is the 'plus construction'. Also given any Eilenberg-Maclane space, I do not know how to onstruct a group whose classifying space is related to the given EM space. Could you please advise some references which I could study to learn this stuff ? Thanks a lot $\endgroup$
    – user90041
    Jun 7, 2019 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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