Answer by Four Organs for Rolling without slipping interpretation of torsion

2011-01-31T08:39:34Z

This was meant to be a comment over at Secret Blogging but was probably classified as spam (not that I disagree with that), so I stick it here instead.

An example to consider is $S^3$. The metric from the ordinary embedding into $E^3$ gives a constant curvature connection with no torsion. The geodesics are all great circles and splits into equivalent classes of parallel geodesics (the Hopf fibration). An observer travelling along a geodesic will observe how nearby geodesics twist around him. This is a higher dimensional analogue of how nearby geodesics in two dimensions are observed to do sinusidal oscillations when the curvature is positive. The curvature form is an so(3)-valued two-form which integrated around (the interior of) a closed loop gives the rotation of a frame transported around the loop. Now, given an element of so(3) at a point of the manifold, it can be reinterpreted as an vector in the tangent space. (This is the usual so(3) <-> angular velocity vector isomorphism.) This turns the curvature form into a torsion form, giving a new connection with no curvature but with "constant" (homogeneous) torsion. Integrating the torsion form gives a tangent vector which is the translation of the tangent space when translated around a (not necessarily small) loop.

This absolute parallelism connection has the same geodesics as the constant curvature connection. But in this case, an observer travelling along a geodesic, I believe, does not observe any twisting of nearby geodesics. Instead he sees the nearby geodesics to be completely straight lines, but they lag behind (or run ahead?). In some way a torsion connection introduces an ambiguity in the velocity concept which would be interpreted by an observer that the immediate neighbourhood does not stay in place, it slips?

Now, bundles. Take the tangent bundle $TM$ of some manifold and release each tangent space from its point of contact with the manifold. This turns tangent spaces into affine spaces and the tangent bundle into an affine bundle $AM$. This bundle does not have a distinguished zero section, instead each and every point of an affine fiber have equally right to be considered a point of the underlying manifold. Then, giving a connection on $M$, it should the case that the connection has vanishing torsion if and only if every contractible closed loop in $M$ lifts to a closed loop in $AM$?

The connection in $R^3$ with straight lines as geodesics and when a frame is transported along a line, it spins, is given by $\nabla_XY=\nabla_YX=Z$, etc. This is again a constant curvature connection with no torsion.

User four organs - MathOverflow

Answer by Four Organs for Rolling without slipping interpretation of torsion