User four organs - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:24:09Z http://mathoverflow.net/feeds/user/12635 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53387/rolling-without-slipping-interpretation-of-torsion/53868#53868 Answer by Four Organs for Rolling without slipping interpretation of torsion Four Organs 2011-01-31T08:39:34Z 2011-01-31T08:39:34Z <p>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.</p> <p>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) &lt;-> 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.</p> <p>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?</p> <p>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\$?</p> <p>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.</p>