Can we symmetrize connections on tangent bundles? 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?
 A: Fix a symmetric connection $\partial$ to use as an origin for $\mathcal A$. Any connection $\nabla \in \mathcal A$ can be written $\nabla = \partial + \Gamma$ for some $(1,2)$-tensor $\Gamma$, which can be naturally decomposed as $$\Gamma = \mathrm{Sym}(\Gamma) + \frac 1 2 \tau^\nabla$$ where $\tau^\nabla = 
2 \mathrm{Alt}(\Gamma)$ is the torsion of $\nabla$. Since the sets of symmetric and antisymmetric tensors are vector spaces giving a direct sum decomposition of $T^1_2$, we can take the linear projection on to the symmetric tensors along the antisymmetric ones; i.e. "taking the symmetric part". 
Since the torsion  of a connection is well-defined (i.e. does not depend on our choice of $\partial$), the antisymmetric subspace also does not depend on $\partial$; so this gives us an honest affine projection $P:\mathcal A \twoheadrightarrow \mathcal S$ which we can write as $$P(\nabla) = \nabla - \frac 1 2 \tau^\nabla,$$ or more explicitly $$P(\nabla)_X Y = \nabla_X Y - \frac 1 2\tau^\nabla(X,Y) = \frac 1 2 \left( \nabla_X Y + \nabla_Y X + [X,Y] \right).$$ There will of course be many, many different projections: this is just a fact of linear algebra. I would argue that this one is the most geometrically natural, however: it's the unique symmetric connection having the same geodesics as $\nabla$. 
It's probably also natural/canonical in some rigorous sense: I can't think of any other projection $\mathcal A \twoheadrightarrow \mathcal S$ we could write down without making some arbitrary choice.
A: 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. 
