TITLE: Why torsion is only defined for linear connection on TM?
QUESTION [5 upvotes]: The concept of curvature is defined for any linear connection on any vector bundle $E \to M$, but the concept of torsion is only defined for connection on the tangent bundle $TM$ of a manifold $M^n$, or for a connection obtained as the pullback of a connection on a vector bundle $E \to M$ isomorphic to $TM$ via an isomorphism $\theta \colon TM \to E$ equivalent to a solder form.
Why is that so ? If torsion can be interpreted as the twist of a moving frame along a curve, the same phenomena should occur for a connection on any vector bundle.
Is there a way to define a notion of torsion for any vector bundle ?

REPLY [4 votes]: The torsion of a connection on the tangent bundle is a zeroth order invariant tensor. There is no zeroth order invariant tensor associated to a connection on a principal bundle on a smooth real manifold, since in any coordinates, the connection is $A=g^{-1} \, dg + \operatorname{Ad}_g^{-1} \Gamma_i(x) \, dx^i$, and we can then change those coordinates to get $x=0$ at our chosen point, $g=1$ there, and $\Gamma_i(0)=0$, by replacing $g$ by $h(x)g$ so that we replace $A$ by $g^{-1} \, dg + \operatorname{Ad}_g^{-1} \operatorname{Ad}_{h(x)}^{-1}(\Gamma_i(x) \, dx^i+dh \, h^{-1})$. We pick $h(x)$ so that $dh  \, h^{-1}(0)=-\Gamma_i(0) dx^i$. So all connections look the same at a point, to first order. You can only feel that the connection is not flat at second order. This doesn't work on the frame bundle (the principal bundle associated to the tangent bundle) because you can't change $g$ to $h(x)g$ independent of how you change coordinates $x$.