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 ?



Why torsion is only defined for linear connection on TM?
Actually, one can define torsion for Lie algebroids, and in particular vector bundles, eg. here: <https://arxiv.org/pdf/math/0105033.pdf>


It is defined as so: given a Lie algebroid $A\to M$ with anchor map $\alpha\,,$ choose a connection $\nabla\,.$ The torsion tensor is then defined as $T(X,Y)=\nabla\_{\alpha(X)}Y-\nabla\_{\alpha(Y)}X-[X,Y]\,.$


When the Lie algebroid is the tangent bundle, you get the usual torsion. When the Lie algebroid is a Lie algebra, you get the commutator (up to a sign). When the Lie algebroid is a vector bundle (by which I mean the anchor map and bracket are zero) the torsion is zero, which is consistent with what the other commentors have said.


