# Does every manifold have a flat connection?

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.

• 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. – Liviu Nicolaescu Oct 2 '13 at 19:56
• 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). – Peter Crooks Oct 2 '13 at 19:57
• @Peter, there isn't a zero connection. Unless you mean something perculiar? Either way, your conclusion isn't right. – Paul Reynolds Oct 2 '13 at 20:11
• Also, Riemann curvature tensor makes no sense for arbitrary vector bundles. – Misha Oct 2 '13 at 20:35
• @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. – Misha Oct 3 '13 at 2:35

## 2 Answers

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.

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.

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