How to define gradient of an affine connection

Question

I heard somewhere (and just read on a physics forum) that the gradient of a smooth function $f$ on a manifold $M$ can be defined when $M$ is equipped with an affine connection on its tangent bundle, i.e. the Riemannian metric is not strictly necessary. Is this true? If so, how can this be done?

Answer 1

No, or at least not if you want to coincide with the usual gradient in the case where the connection is derived from a metric. To show this, just note that whenever $\nabla$ is the metric connection of $g$, it is also that of $2g$ (since $\nabla(2g) = 2\nabla g = 0$); but the latter metric will produce gradients with half the magnitude of the former.

It seems to me that any sensible definition of the gradient of a function will involve a linear 1-1 relationship to the differential; i.e. will have the structure of a vector bundle isomorphism $$\xi: TM \to {TM}^*.$$ This is almost exactly the same data as a metric: if we impose some positivity conditions (which boil down to "if you move for some short time in the direction $\nabla f=\xi^{-1}(df)$ then $f$ increases unless $df=0$") then given either of the two we can get to the other using the equation $g(u,v) = (\xi(u))(v)$.