Rolling without slipping interpretation of torsion

Question

This is, in a sense, a follow up to this question.

Hehl and Obukhov try to give an intuitive description of torsion. I am confused about their description. I am looking at the following paragraph on page 5:

How can a local observer at a point $p$ with coordinates $x^i$ tell whether his or her space carries torsion and/or curvature? The local observer defines a small loop (or a circuit) originating from $p$ and leading back to $p$. Then he/she rolls the local reference space without sliding ... As a computation shows, the added up translation is a measure for the torsion and the rotation for the curvature.

So, my input data is a manifold $M$ with a connection $\nabla$ on $T_* M$, and a path $\gamma : [0,1] \to M$. From this data, I am supposed to obtain an affine linear map $v \mapsto Av+b$ from $T_{\gamma(0)} M$ to $T_{\gamma(1)}(M)$. In particular, if $\gamma(0)=\gamma(1)=p$, I obtain an endomorphism of $T_p M$. Then $b$ is a measure of the torsion, and $A$ is a measure of the curvature.

I am happy with the curvature part. This is parallel transport: Given a tangent vector $u \in T_p M$, I am to find the unique local section $\sigma$ of $\gamma^* T_p(M)$ such that $\sigma(0) = u$ and $\nabla \sigma=0$. Then $Au = \sigma(1)$.

I have two confusions about the torsion part:

(1) Hehl and Obukhov cite their definition of rolling without slipping to five sources (hidden by the ellipses above). The most readable of the ones I have access to is Differential Geometry, by Sharpe. But Sharpe (at least in Appendix B) only gives a definition for the Levi-Civita connection, not for an arbitrary connection. I think I have guessed what the definition should be, but could someone please write it down so I can be sure?

(2) In any case, it is my understanding that the Levi-Civita connection should be torsion-free and that, for the Levi-Civita connection, rolling without slipping corresponds to physically rolling my manifold along a plane. So the statement should be, if I take a surface $S$ in $\mathbb{R}^3$, and physically roll it along a table top, when I get back to the same tangency point on $S$, I should be at the same point on the table. Physical experimentation has not made it clear to me whether or not this is true. However, if I roll my surface along a non-small path, this is definitely false: otherwise, ball bearings could not roll!

Technically, this is not a contradiction, because Hehl and Obukhov only speak of a small circuit. But the usual situation in differential geometry is that when some quantity vanishes everywhere on a manifold, then "small" can be replaced by "contractible". And the example of a rolling ball bearing definitely shows that rolling-without-slipping a plane along a contractible loop according to the LC connection can produce a nontrivial translation.

What's going on?

Answer 1

"Rolling without slipping" is a powerful idea, but the phrase doesn't necessarily lead one to the intended mental model. In particular, torsion is something that is at issue only for manifolds of dimension 3 or higher. Perhaps you can imagine taking a 3-manifold, and rolling it along a hyperplane in 4-space --- but the metaphor becomes strained, partly because most Riemannian 3-manifolds cannot be smoothly isometrically embedded in $\mathbb E^4$.

Another way to think of it is this: suppose you have a smooth parametrized curve say in a Riemannian 3-manifold, $\alpha: [0,T] \rightarrow M^3$. Then the claim is that there exists exists a matching curve $\beta: [0,T] \rightarrow \mathbb E^3$ together with a map $\phi$ from a neighborhood of the image of $\beta$ to a neighborhood of the image of $\alpha$ that takes the Euclidean metric to the metric of $M^3$ up to first order along the curve. Furthermore, $\beta$ is uniquely determined up to an isometry of $\mathbb E^3$. Basically, $\beta$ is what you get if you "roll" $M^3$ along $\mathbb E^3$ along $\gamma$ in a way that best maintains contact between the two spaces: that is, "without slipping".

To see that the curve $\beta$ (assuming it exists) is uniquely determined by $\alpha$, we can imagine trying to send a neighborhood of $\beta$ to a neighborhood of another curve $\gamma: [0,T] \rightarrow \mathbb E^3$ in a way that maintains first order contact of the metric. If you look at curves parallel to $\beta$ in $\mathbb E^3$, the first derivative of their arc length is negative in the direction that $\beta$ is curving. The logarithmic derivative of arc length, for curves displaced along a normal vector field that remains as parallel to itself as possible, is the magnitude of the curvature.

If you try to twist the normal coordinate system, this corresponds to the concept of torsion. It's easiest to visualize along a straight line: if you twist a neighborhood of a line in space, you distort the metric on each concentric cylinder, by changing the angles between cross-section circles and generating lines. I.e., threads that wind around a hose at angles $\pm \pi/4$ are effective at preventing twisting (= torsion). The same principle holds for any curve in space: the first order behavior of the metric in a neighborhood of the curve locks in the Frenet characterization of the curve (well, the curvature and torsion as a function of arc length, but these are different from but related to the curvature and torsion of a connection).

Why does the matching curve exist? You can check derviatives etc. but better to just imagine it. Basically, you could reparametrize $\alpha$ by arc length, then project a neighborhood of $\alpha$ back to $\alpha$ by sending each point to the closest point of $\alpha$, and parametrize the lines of projection by their arc length. On each concentric tube, there's a unique unit vector field orthogonal to the preimages of projection to $\alpha$. Scale this vector field so that it commutes with projection, to get a full set of cylindrical coordinates for a neighborhood of $\alpha$. The only first-order invariant for the metric that is free is the first derivative of scaling function. Using that, you can match the first derivative by using the curvature and torsion of a curve in space.

This process defines the affine connection on the tangent bundle. The Levi-Civita connection is the linear part of the affine connection, which is automatically by definition torsion free. The non-torsion-free connections are ones that impart twists on little neighborhoods of curves. This is usually expressed by translating it into a formula about covariant derivatives of two vector fields not being as commutative as it should be.

This really calls for pictures. Any volunteers?

Answer 2

So, I can now explain almost all the material that confused me. I don't know if this will help anyone else, but here is the answer that would have helped me.

Let $M$ be a manifold with a connection on $T_* M$. Let $\gamma: [0,1] \to M$ be a curve in $M$. Then parallel transport gives us maps $g_t: T_{\gamma(0)} M \to T_{\gamma(t)} M$. So we get a vector $g_t^{-1} \gamma'(t)$ in $T_{\gamma(0)} M$ for every $t$. Let $b(u) = \int_0^u g_t^{-1} \gamma'(t) dt$. So $b(u)$ is a path in $T_{\gamma(0)} M$.

The physical meaning of $b(u)$ (when $\nabla$ is the LC conection) is that, if we roll $M$ along a table, so that at time $t$, $M$ is tangent is the table at position $\gamma(t)$, then $b(u)$ is the path traced on the table by the tangency point. The affine linear map in my question is more or less $v \mapsto g_t (v + b(t))$. See Bill Thurston's answer for physical intuition underlying this.

I believe that the statement which Hehl and Obukhov intended was the following:

Let $\gamma$ be a small curve, bounding a "disc of size $r$", for some small $r$. If $\nabla$ is torsion free then $b(1) = O(r^3)$. Otherwise, if $\gamma$ is in "the plane spanned by the vectors $X$ and $Y$", then $b(1) = T(X,Y) r^2 + O(r^3)$, where $T$ is the torsion tensor and I might be missing some factors of $2$ or $\pi$.

Here the phrases in quotes are meant to be nonrigorous: the disc could be a square or an oval rather than a perfect circle, and I don't claim to have a definition of "plane" in a general Riemmannian manifold. The point is that we should be should be working locally enough that our Euclidean intuition for these concepts is adequate.

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Let's do some sanity checks. First, let's work on $\mathbb{R}^n$ with a constant connection, meaning that $$\nabla_{\partial_i} \sigma = \partial_i(\sigma) + A_i \sigma,$$ where $A_i$ is a constant $n \times n$ matrix. (It must be skew-symmetric, in order to have $\nabla$ respect the metric.) Let's go around a square in the $\mathbb{R}^2$ plane; traveling $r$ in direction $e_1$, $r$ in direction $e_2$, $r$ in direction $-e_1$ and $r$ in direction $-e_2$. So the path $b$ consists of four segments of length $r$, in directions $$e_1,\ e^{-r A_1} e_2,\ -e^{-r A_1} e^{-r A_2} e_1 \ \mbox{and} \ -e^{-r A_1} e^{-r A_2} e^{r A_1} e_2.$$

The total displacement is $$r((e_1+e_2-e_1-e_2)-r(A_1 e_2 + A_1 e_1 + A_2 e_1 + A_2 e_2) + O(r^2))$$ $$=r^2(A_1+A_2)(e_1+e_2)+O(r^3).$$

Intuitively, we are adding up four vectors of length $r$. They can be grouped into two nearly antiparallel pairs; in each pair, the angle between a vector and its near-negative is $O(r)$. So the sum in each pair is $O(r^2)$, and there is no reason to expect further cancellation.

For our second sanity check, let's look at the LC connection on $S^2$ (of radius $1$). This has no torsion, so we should be able to see that.

Let's roll the sphere, keeping the contact point on a line of constant latitude $\phi$. (Here $\phi \in (0, \pi)$, with $\pi/2$ meaning the equator.) The path $b$ is an arc of a circle. The arc has radius $\tan \phi$, and sweeps out the angle $2 \pi \cos \phi$. So the distance from one end of the arc to the other is $2 (\tan \phi) \sin\left( \pi \cos \phi \right) = 2 O(\phi) \sin (\pi - O(\phi^2)) = O(\phi^3)$. So we see that the displacement goes to $0$ as $O(\phi^3)$, as desired.

In response to Bill Thurston's request for pictures, here are the paths traced by the ball for $\phi=(0.1 \ \mbox{radians})*k$, with $1 \leq k \leq 7$. Note that the radii of the arcs are shrinking linearly, but the gap between the endpoints of the arc is shrinking cubicly.

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So, let me now try to address my confusion in part (2), using the example of $S^2$. I was confused by two things. The first is that I took the authors to be saying the translation would vanish, when in fact it died off like $O(r^3)$. But the second was that I had a fallacious argument in my head suggesting that, if we had an $O(r^3)$ bound, then it would imply that the displacement $b$ around any contractible loop would be zero. I don't know if this will help anyone else, but I will now spell out the fallacy and expose it.

False Proof: Identify my contractible path $\gamma$ with a square of side length $1$. Subdivide it into $N^2$ little squares $s_1$, $s_2$, ..., $s_{N^2}$. Let $p$ be a corner of the big square. Let $\delta_i$ be a path which goes from $p$ to $s_i$, circles $s_i$, and goes back to $p$. Let $\delta$ be the concatenation of the $\delta_i$'s. If you choose the ordering of the $s_i$'s correctly and choose the right $\delta_i$'s, then $\delta$ is simply $\gamma$ with a whole lot of backtracking put in.

So the path $b_{\delta}$ coming from $\delta$ will simply be $b_{\gamma}$, plus a lot of backtracking segments. So $b_{\gamma}$ is a concatenation of $N^2$ paths, each of length $O(1/N^3)$. We see that the length of $b_{\gamma}$ is $O(1/N)$. Since $N$ was arbitrary, this concludes the fallacy. QFD

This is similar to a (correct) proof that, if $\omega$ is a closed one-form, then $\oint \omega$ around any contractible loop is $0$: You directly compute that, for a closed one-form, $\oint_{r \cdot \gamma} \omega = O(r^3)$ and then run a similar subdivision argument.

So, why doesn't this work? Going around $s_i$ really does only contribute $O(1/N^3)$. But going from $p$ to $s_i$, around $s_i$, and back to $p$ contributes $O(1/N^2)$! The reason is that, because of curvature, the trip around $s_i$ rotates my reference frame. Thus, when I travel back the way I came, the resulting path is rotated and does not backtrack along itself. How large is this effect? The angle of rotation is $2 \pi - O(\mathrm{Area}(s_i))$, which can be treated as $O(\mathrm{Area}(s_i))$ since we only care about the net effect and not the winding number. The area is $O(1/N^2)$. So we travel a path of length $O(1)$, and then backtrack along the rotation of that path through an angle of $O(1/N^2)$. So our net displacement is $O(1/N^2)$, and the fallacy falls apart.

Here are some figures of taking a ball which is resting on its equator, rolling it to latitude $\phi$, once around at latitude $\phi$, and back the the equatorial starting point. The values of $\phi$ are $0.1$, $0.2$ and $0.3$ radians. What you are supposed to see is that the radii of the arcs are dropping linearly, the separation between the endpoints of the arcs is dropping cubically, and the separation between the end points of the spokes is dropping quadratically.

Answer 3

Judging by your point (2), I think your picture is not entirely correct. The concept you are referring to is called a development. The idea is something like this (I "get" it for the torsion-free case, but have no idea how to deal with the torsion case). Take the flat plane to be your model space (local reference space) and it has a canonical connection on it. Imagine it being a table. Let's say your table is a long rectangle, so you can imagine it having nice gridlines running horizontal and vertically.

Now take an orange (or a watermelon, in fact, a watermelon may be easier to use in practice). Put it somewhere on the table. Look from above the orange and locate the top-most point (you should really do this at the point of contact, but it is easier to draw on the antipode). Draw a blue arrow toward along the vertical direction and a green arrow along the horizontal direction, the green one clockwise from the blue one. Now roll the orange around without slipping or twisting, until the original contact point is in contact with the table again (so your chosen antipode is again on top). Now compare the frame defined by the blue and green arrows and the frame defined by the table. This measures the angle defect of your closed loop. If you just roll the orange directly away from you (travel along a geodesic/great circle on the orange), the two frames will line up. But if you roll it along a different loop in general the two frames will not line up. This rotation is the curvature part.

If you look at it this way, it is clear what you are trying to do: the path traced out by the orange on the plane is the path that experiences the same "acceleration" as your path on the orange at every single point. Because the differing geometry the path on the plane will in general not be closed. What's important is the directional differences which "shows" the effect of parallel transport: note that it is only meaningful to consider a full loop on the fruit (when the arrows came back on top) as there is no canonical frame that is taken a priori on the orange...

(I'm sorry if my explanation is still lacking.)

For the case of torsion: I've tried to work out the idea for the Cartan staircase, but I don't think my attempt using a roller-coaster analogy is the best. In any case, since the surface geometry of physical objects tend to be Levi-Civita, I don't think it will be possible to actually demonstrate this using an analogous experiment, nor conduct a thought-experiment without seriously twisting your imagination. One of the problems is that torsion is generally thought of as the rotation of the normal tangent space to the curve about the curve (under certain assumptions). In two dimensional examples that we like to think of, the normal tangent space to the curve is one dimensional and the rotation group for it is rather trivial.

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I am also not completely sure about your interpretation on the parallel transport (or their statement about translation equal to torsion, which I think is a bit fishy). Here's why: the parallel transport of a vector is linear with respect to scalar multiplication. The whole business about a translation suggests an affine-linear transformation as you wrote. You should not have the parallel transport of the zero vector be anything other than the zero vector!

Answer 4

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.