8
$\begingroup$

Suppose I have a manifold and a vector bundle over it, but not a connection or a metric. Can I always find a connection on it that has a Riemann curvature tensor that is identically zero? If so, can I always find a connection that has both Riemann curvature and torsion tensors identically zero?

I've attempted to simply for the Christoffel symbols, but couldn't make headway in the equations.

$\endgroup$
  • 5
    $\begingroup$ Your question needs some edition because it is very imprecise. (E.g.a connection is defined on a vector bundle, not on a manifold.) In any case, if the manifold is compact, oriented and the Euler characteristic is $\neq 0$, then there cannot exist any metric on the tangent bundle and connection compatible with it whose curvature is zero. This is a consequence of the Gauss-Bonnet-Chern theorem. $\endgroup$ – Liviu Nicolaescu Oct 2 '13 at 19:56
  • $\begingroup$ Since you are working without a metric, you can take the zero-connection on any vector bundle over your manifold. Its curvature is certainly zero (ie. the connection is flat). $\endgroup$ – Peter Crooks Oct 2 '13 at 19:57
  • 5
    $\begingroup$ @Peter, there isn't a zero connection. Unless you mean something perculiar? Either way, your conclusion isn't right. $\endgroup$ – Paul Reynolds Oct 2 '13 at 20:11
  • 1
    $\begingroup$ Also, Riemann curvature tensor makes no sense for arbitrary vector bundles. $\endgroup$ – Misha Oct 2 '13 at 20:35
  • 1
    $\begingroup$ @PaulReynolds: Yes, there is a notion of connections and curvature for vector bundles, but it is no longer Riemann's (and is no longer a tensor) and should not be referred to by this name, as Riemann curvature tensor is something much more specific. $\endgroup$ – Misha Oct 3 '13 at 2:35
15
$\begingroup$

Milnor proved in [On the existence of a connection with curvature zero, Comm. Math. Helv. v 32] that bundles over a surface of genus g has flat connections iff its Euler class is less than g by an absolute value (see also Wood, Bundles with totally disconnected structure group). Sullivan in "A generalization of Milnor's inequality ...Comm. Math. Helv. v. 51" find a finite upper bound for the Euler class of a R^n-bundle with the affine connection over manifold M^n (the number of n-simplices in the triangulation of M^n). Hope, this might help.

$\endgroup$
8
$\begingroup$

By Chern-Weil theory, the real Pontryagin classes $p_k \in H^{4k}(X, \mathbb{R})$ of a real vector bundle $V$ on a smooth manifold $X$ are determined by the curvature form of any connection on that bundle; in particular, if the curvature vanishes, then so do all of the $p_k$. Hence if any of the $p_k$ don't vanish, then $V$ does not admit a flat connection. (Note that all of the $p_k$ vanish if $\dim X \le 3$; Milnor's result regarding the case $\dim X = 2$ requires more difficult tools.)

If $V$ is taken to be the tangent bundle of $X$, then the first case where this happens is when $\dim X = 4$, where $p_1 \in H^4(X, \mathbb{R})$. If $X$ is closed and orientable then $p_1$ is nonzero iff $X$ has nonzero signature, by the Hirzebruch signature theorem. The simplest example of a $4$-manifold with nonzero signature is $\mathbb{CP}^2$; it follows that the tangent bundle of $\mathbb{CP}^2$ does not admit a flat connection.

$\endgroup$
  • $\begingroup$ You have 666 answers... $\endgroup$ – JP McCarthy Aug 27 '15 at 19:42

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.