TITLE: Can we symmetrize connections on tangent bundles?
QUESTION [5 upvotes]: Let $M$ be a smooth manifold. Denote by $\mathcal{A}$ the space of all affine connections on $TM$, and by $\mathcal{S}$ the affine subspace of all the symmetric connections.  
Is there a projection $P:\mathcal{A}\to \mathcal{S}$? (i.e $P^2=P$).
If there is more than one, is there some natural choice?
Note that $S$ is convex, so perhaps introducing some metric on $\mathcal{A}$ will be useful?

REPLY [2 votes]: The naturality statement in the end of the answer by @AnthonyCarapetis can indeed be made precise: There is a natural action of local diffemorphisms on affine connections (basically characterized by $(f^*\nabla)_{f^*\xi}(f^*\eta)=f^*(\nabla_\xi\eta)$). In this sense, forming the torsion is natural, i.e. if $\nabla$ has torsion $T$, then the torsion of $f^*\nabla$ is $f^*T$ (pullback of tensor fields). In particular, if $\nabla$ is symmetric, then so is $f^*\nabla$. On the other hand, naturality of forming the torsion also implies that the projection $P$ in Anthony's answer is compatible with the action of local diffeomorphism, i.e. $P(f^*\nabla)=f^*(P(\nabla))$ for any connection $\nabla$. 
The projection $P$ is not quite uniquely determined by this naturality propery (plus maybe some assumptions like locality), since you can play algebraic games with the torsion tensor, compare with the beginning of section 25 of the book MR1202431 by Kolar, Michor and Slovak, which is available online here. It becomes unique if you require that the symmetrized connection has the same geodesics.