Is there an integral condition for zero torsion?

Is there an integral condition for zero torsion? January 27, 2011

Posted by David Speyer in Uncategorized.
trackback

Sooner or later, I’m going to want to mention the Levi-Cevita connection in my course, so I want to understand the meaning of the fact that it has no torsion. I think I have enough understanding for teaching purposes. See these Mathoverflow threads 1 2 and these John Baez essays 1 2 3 (scroll down) for some of the ideas I might present.

However, I am left with a question. I haven’t found any integral characterizations of zero-torsion, only infinitesimal ones. Let me explain what I mean by some examples:

Let $X$ be a smooth manifold.

Let $\omega$ be a differential $k$ -form on $X$ . Then $d \omega$ is zero if and only if, for all contractible $k$ -spheres $S$ in $M$ , we have $\int_S \omega =0$ .

Let $E$ be a vector bundle on $X$ , and let $\nabla$ be a connection on $E$ . Then the curvature of $E$ vanishes if and only if, for every contractible loop $\gamma$ in $M$ , the holonomy around $E$ is trivial.

Both of these characterizations involve quantifying over contractible spheres, and say that some quantity is identically zero.

In contrast, let $\nabla$ be a connection on $T_* X$ . Every characterization of torsion vanishing which I can find involves looking at sufficiently small discs/geodesics/etcetera and involves the order of vanishing of some quantity being small (usually $O(r^3)$ instead of $O(r^2)$ ), rather than being identically zero.

I find this aesthetically unpleasing. Any ideas?

Like this:

Be the first to like this post.

Comments»

1. Ben Webster - January 27, 2011: What about the one mentioned by Deane Yang on MO: torsion-free=a parametrized curve is a constant speed geodesic iff the velocity field is parallel?
2. David Speyer - January 27, 2011: So, I might be missing something, but it looks to me like this is implied by, but not equivalent to, torsion free. Consider a connection on $\mathbb{R}^3$ where, when you travel along any straight line, your coordinate frame spins around your direction of travel, at a rate proportional to your speed. In that case, doesn’t Yang’s condition hold?
3. Greg Kuperberg - January 29, 2011: I have come to think about it this way: A Levi-Civita connection is more naturally a structure on vector bundles over a manifold in general, than on the tangent bundle $TM$ specifically. Even if the bundle $E$ happens to be $TM$ , it’s mixing oil and water in the sense that you make no use of the isomorphism $E \cong TM$ . The connection gives you a way to differentiate an $E$ -valued function (a section), or more generally, the exterior derivative of an $E$ -valued form. The curvature of the connection is the square of the exterior derivative, i.e., the failure of the exterior derivative to be exact.

Let’s say that there is an isomorphism between $E$ and $TM$ ; what of it? Read as an isomorphism from $TM$ to $E$ , it is also an $E$ -valued 1-form $\alpha$ whose matrix (in some coordinates) happens to be square and non-singular. Then the torsion-free condition says that $d\alpha = 0$ . You can also write this as an integral condition if you like.
4. Four Organs - January 30, 2011: 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 (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 neighborhood 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 $R^3$ where straight lines are geodesics and transport of a frame along a line spins the frame is given by $\nabla_X Y=\nabla_Y X = Z$ , etc. This is again a constant curvature connection with no torsion.
5. David Speyer - January 31, 2011: @Greg It took me while to understand what you are saying. I think that I now understand, and disagree.

Let $E$ be a vector bundle with connection $\nabla$ . Let $\alpha$ be an $E$ -valued $1$ -form. As you say, the condition that $\alpha$ pulls $\nabla$ back to a torsion free connection on $TM$ is the same as the condition that $d_{\nabla} \alpha=0$ .

Let’s unpack that subscript there. If $E$ is just a vector bundle, and $\alpha$ an $E$ valued $1$ -form, then there is no notion of $d \alpha$ , because we can’t differentiate sections of $E$ . What we are doing is using the connection $\nabla$ in order to define the exterior derivative of $\alpha$ . We need to remember that $d_{\nabla}$ may not have all the properties we are used to with our old friend $d$ . For example, if $\nabla$ has curvature, then $d_{\nabla}^2$ is not zero.

So, what does it mean to integrate an $E$ -valued $k$ -form? Without $\nabla$ , it means nothing: The different fibers of $E$ lie in different vector spaces, so we can’t add them together. With a connection, I can make sense of integrating an $E$ -valued $1$ -form $\alpha$ along a path $\gamma$ . Let point $\gamma(t)$ , take the tangent vector $\gamma'(t)$ and evaluate $E$ on it to get a vector in $E_{\gamma(t)}$ . Use parallel transport along $\gamma$ to make this into a vector at $E_{\gamma(0)}$ . As we go along $\gamma$ , we get a family of vectors in $E_{\gamma(0)}$ ; take the integral.

I would imagine that you think I can turn $d \alpha=0$ into the statement $\int_{\gamma} \alpha=0$ for any closed contractible loop $\gamma$ . This is wrong.

If I follow the discussion above, this suggests that I should define $\int_{\gamma} \alpha$ as $\int_{t=0}^1 g_t^{-1} \gamma'(t) dt$ , where $g_t$ is the parallel transport along $\gamma$ . This is the same definition I was working with back in this MO question:

And I return to the point I made there: This integral is not zero. One way to think of this is that “Stokes theorem” fails for $E$ -valued $1$ -forms when $\nabla$ has curvature.
6. David Speyer - January 31, 2011: @Four Organs: First of all, I’m sorry that your post got stuck in the spam filter.

I don’t follow everything you are saying (some questions further on). But I think you are wrong about this: “the connection has vanishing torsion if and only if every contractible closed loop in $M$ lifts to a closed loop in $AM$ ?”

I can’t say for sure that you rae wrong, because you don’t say what compatibility condition you are imposing on the loop in $AM$ . But I suspect that you are imposing that this connection be “rolling without slipping”.

This is the thing that everyone seems to believe, and simply doesn’t seem to be true. Take the sphere $S^2$ , with the standard connection. This is torsion free. Run around the equator, and compute the “rolling without slipping” connection. I get that you wind up displaced by $2 \pi$ when you get back where you started. Physical interpretation: If I roll a ball bearing along the floor, it moves!

Maybe I’m being dumb, but it seems to me like lots of people are telling me different ways to think about this integral being zero, and I keep saying “but it isn’t zero!”
7. some guy on the street - February 1, 2011: Hmm… No Torsion means $\nabla_X Y - \nabla_Y X = [X,Y]$ ; it means that the Hessian $X\circ Y - \nabla_X Y$ is a symmetric $\mathcal{C}^\infty$ -bilinear form on $TM$ . Trivially you can rewrite this as $\int g H_{X,Y} f = \int g H_{Y,X} f$ , but I don’t think there are any good ways to use integration by parts or otherwise reduce the domain of the integral.

The book by Misner, Thorne, and Wheeler said this meant that parallel transport around a loop was $O(A)$ , with $A$ the area of a spanning surface, rather than $O(l)$ , with $l$ the length of the loop. As such, *no*, No Torsion isn’t a vanishing integral axiom, it’s an integral *estimate* axiom; it’s also the precondition for a useful synthetic definition of curvature. The curvature is the obstruction to a parallel frame; I don’t know what to call the obstruction to defining an obstruction.

But I can’t be sure there isn’t a clever way around it, either! I hope I’m not duplicating too much?
8. Greg Kuperberg - February 4, 2011: Yeah, you’re right, I made a mistake at the last step.
9. Four Organs - February 6, 2011: @David, don’t worry about the spam filter. I don’t mind.

The rolling of a sphere on the floor is, I believe, what Cartan calls the development of a curve into its conjugate euclidean space. The basic idea is to measure the invariant parameters of the curve and then construct the corresponding curve in euclidean space. When the connection has torsion and curvature the developed curve will exhibit extra translation and twisting not visible in the original curve. One can think of this translation and twisting to have been cancelled in the original curve by the torsion and curvature of the connection. The development construction gives an easy characterization of a geodesic: A curve is a geodesic if and only if it is developed into a straight line.

The invariant parameters of the curve are the usual curvature, torsion and higher order parameters and these are measured by constructing the moving Frenet frame along the curve. To see how this is done consider $R^3$ with the torsion-only connection $\nabla_XY=Z/2, \nabla_YX=-Z/2$ , etc. The torsion two-form, which is vector valued, is $\mathbf{\Theta}(\mathbf{u},\mathbf{v})=\mathbf{u}\times\mathbf{v}$ . The curve to be developed is taken to be the unit circle in the x-y-plane, $\gamma(s)=(\cos s, \sin s, 0)$ . The developed curve should be expected to exhibit extra translation and should therefore be a helix. Thus, the original curve should actually have torsion, yet it is contained in a plane.

This actually makes sense when one constructs the Frenet frame. The curve has velocity $\mathbf{v}=(-\sin s, \cos s, 0)$ and acceleration $\mathbf{a}=(-\cos s,-\sin s,0)$ . Now, the Frenet $\mathbf{t}$ -vector is not $\mathbf{t}=\mathbf{v}$ . Instead the t-vector is defined to be the absolute derivate of the curve. This makes the torsion form show up as a correction factor in the formula for the t-vector: $\mathbf{t}\tilde\mathbf{v}+\mathbf{\Theta}(\mathbf{v},\mathbf{a})$ . The resulting Frenet frame is:

$\mathbf{t}=1/\sqrt{2}\, (-\sin s, \cos s,1)$

$\mathbf{n}=\quad (-\cos s,-\sin s, 0)$

$\mathbf{b}=1/\sqrt{2}\, (\sin s, -\cos s, 1)$

and the Frenet equations become

$\mathbf{t}' = 1/\sqrt{2}\, \mathbf{n}$

$\mathbf{n}' = -1/\sqrt{2}\, \mathbf{t}+1/\sqrt{2}\, \mathbf{b}$

$\mathbf{b}' = -1/\sqrt{2}\, \mathbf{n}$

Thus, the unit circle has constant curvature and torsion and does indeed develope into a helix.

Strangely enough, the Frenet frame does not line up with the original curve. This is similar to how a boat crossing a river has to steer against the current to be able to maintain its route.
10. Anonymous - February 6, 2011: The condition Greg mentioned is pretty standard in the general relativity literature. See, for example, eq 2.14 of gr-qc/9305011v1.

	Mathblogging.org Wee… on The recent difficulties with…
	David Speyer on A proof length challenge
	miforbes on A proof length challenge
	Bruce Bartlett on What would it take to make a n…
	Bruce Bartlett on Journals and the arXiv