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.
 A: 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.   
A: 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. 
