Why torsion is only defined for linear connection on TM?

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 ?

• The definition of the torsion tensor $T(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y]$ cannot be extended to arbitrary vector bundles: $\nabla_X Y$ subsumes $X \in \Gamma(TM)$ and $Y \in \Gamma(E)$, and then you exchange $X$ and $Y$... When you speak about "the twist of a moving frame along a curve", don't you mean a different notion of torsion, like in Frenet-Serret formulas? – Ivan Izmestiev Jun 22 '18 at 11:28
• @IvanIzmestiev I know that the usual definition cannot be extended as such. My question is : why there is no more general notion of torsion which applies to any vector bundle? Or is there ? Or to put it anther way : is the curvature the only tensor field which can be constructed out a connection $\nabla$ ? – ychemama Jun 22 '18 at 11:58
• Then a possible approach to this could be by "natural operations", see for example the book "Natural operations in differential geometry" by Kolar, Michor, and Slovak. I would guess that any tensor "constructed out of $\nabla$" depends only on the curvature of $\nabla$. – Ivan Izmestiev Jun 22 '18 at 18:13
• See mathoverflow.net/a/116487/26935 and other answers there for some aspects. – Peter Michor Jun 23 '18 at 17:25

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 $x$.
I believe that no natural'' section of either $E^*\otimes E^*\otimes E$ or $T^*\otimes T^*\otimes E$ or even $T^*\otimes E^*\otimes E$ can be associated with a general connection on a vector bundle $E\to M$, although I would be interested to see a nice proof of this for myself. (I do not recall anything similar in the literature.) Geometrically, torsion is a measure how far is a connection from being symmetric, but there is no reasonable notion of a symmetric connection for a general vector bundle.