20

The tangent bundles of closed hyperbolic surfaces have flat $PSL(2,\mathbb{R})$ connections showing that there can be no integral formula for the Euler class such connections. This contrasts the situation for the Orthogonal group where the Pfaffian applied to curvature integrates to the Euler class. Can anyone suggest a conceptual reason for the absence of a local formula in the case where the connection fails to preserve a metric? Is it related to the fact that the Euler class is unstable?

David Roberts
  • 33,851
michael freedman
  • 201
  • 2
    If this is the Michael Freedman I think it is, welcome to MO! – David Roberts Feb 26 '14 at 00:47
  • Do you mean an $SL(2,\mathbb{R})$ connection? – Ian Agol Feb 26 '14 at 03:59
  • In fact for the Euler class of flat $SL(2,R)$-bundles there is an integral formula. Maybe, the question was about a formula which makes sense for non-flat bundles as well. – Misha Feb 26 '14 at 06:26
  • 1
    There's a very nice discussion of this general phenomenon of flat bundles with nonzero Euler class in William Goldman's paper http://arxiv.org/pdf/1108.0216.pdf. Maybe this is along the lines of what you are looking for. – Robert Bryant Feb 26 '14 at 10:08
  • Regarding Ian's question, technically one should label the connection sl(2,R), the Lie algebra, but I called it PSL(2,R) to alert the reader that this is the structure group for the bundle possessing the flat connection. I'm trying to understand why ( as Milnor points out in his paper from the 50's) there is a valid Chern Weil formula for Pontryagin classes of bundle with a GL(n,R) connection, but no such formula for the Euler class ( which is square root of the top Pontryagin class). Why do some but not all GL(n,R) classes have integral formulas? – michael freedman Feb 26 '14 at 20:40
  • I wanted to thank Johannes Ebert and Chris Gerig for their comments on the general Chern Weil setting. – michael freedman Feb 27 '14 at 21:35
  • 2
    @michaelfreedman and Misha: At the end of Appendix C to Milnor's and Stasheff's Characteristic Classes, they not only construct a flat $\mathrm{PSL}(2,\mathbb{R})$-bundle with nonzero Euler class over a compact Riemann surface $M$ of genus $g>1$, but they show that this can be lifted to a flat $\mathrm{SL}(2,\mathbb{R})$-bundle over $M$, which yields an actual rank 2, oriented, real vector bundle $E$ over $M$ with a flat connection that has Euler class $e(E) = (1-g)[M]\not=0$. In light of this example, I'm not sure how to interpret Misha's comment. – Robert Bryant Feb 28 '14 at 12:54

1 Answers1

8

I think the point is the theorem, due to Cartan, that the universal Chern-Weil homomorphism

$$CW_G:Sym^{k} (\mathfrak{g})^G \to H^{2k}(BG; \mathbb{R})$$

from invariant forms on the Lie algebra to the cohomology of the classifying space is an isomorphism once $G$ is compact. If $G$ is complex reductive (such as $GL_n (\mathbb{C})$), then it follows by the unitary trick that $CW_G$ is an isomorphism as well. But if $G$ is neither compact nor reductive, there is no reason to expect an expression of a general characteristic class to have an expression in terms of the curvature. In fact, Milnor's example shows that $CW_G$ is not surjective for $G=PSL_2 (\mathbb{R})$. I vaguely remember that if $G$ is the $3$-dimensional Heisenberg group (with center $S^1$), then $CW_G$ is the zero map in positive degrees, even though both source and target are nonzero.

Chris Gerig
  • 17,130
Johannes Ebert
  • 20,634