Visualising Torsion Tensor: Is There a Picture?

[chartery]

As I understand it, parallel transport of a vector around a closed loop on a manifold can lead (in the tangent space) to 1) an angular change, given by the Riemann curvature tensor or, 2) a translational defect given by the Torsion tensor.

I can see how the looping on the curvature of a 2D sphere leads to such a changed angle. Is there any equivalent representation of the features of a 2D manifold that would help me visualise pictorially how torsion leads to a translational defect?

Also, is there an easy way to understand how an intrinsic translational defect can be thought of as a generalisation of an extrinsic rotation in flat space?

 

[robphy]

The best I can do
is to point you to possibly interesting reading:

a very long set of discussions at
https://mathoverflow.net/questions/20493/what-is-torsion-in-differential-geometry-intuitively

One answer points to a paper by Hehl
"Elie Cartan's torsion in geometry and in field theory, an essay"
https://arxiv.org/abs/0711.1535
which draws on pg. 3

 

[Ibix]

Isn't the (non)existence of torsion a property of the connection, rather than the manifold? There exists a unique metric-compatible connection that leads to zero torsion. So you can't do anything like the illustration of curved manifolds by embedding a 2-sphere because you can always pick a connection that leads to zero torsion.

But presumably you can also pick a connection that leads to non-zero torsion even in quite trivial cases. Thus (I think!) you could have torsion on a flat 2d plane by picking a non-trivial connection. As far as I understand it, the connection would (via the covariant derivative and hence parallel transport) affect what it means to "travel in a straight line". So you could pick a connection such that the un-accelerated path of a particle on the plane was a parabola (or something) while the shortest distance between two points remained what Euclid would call a straight line.

 

[vanhees71]

You are right, there's a unique metric compatible torsion-free connection on a Riemannian manifold, and it's given by the usual Christoffel symbols wrt. a coordinate basis, and by definition a Riemannian manifold is the one, for which the affine connection is defined as this unique torsion-free connection.

However, there are also metric-compatbile connections that are not torsion-free. Then you get a more general type of manifold, an Einstein-Cartan manifold. If you want to define fields with spin, in general you need these more general manifolds. For a review, see

https://doi.org/10.1103/RevModPhys.48.393

 

[Ibix]

However, there are also metric-compatbile connections that are not torsion-free.

Yes, metric compatibility and torsion free-ness are separate choices. I didn't state that clearly.

 

[Ibix]

a very long set of discussions at
https://mathoverflow.net/questions/20493/what-is-torsion-in-differential-geometry-intuitively

This thread is very interesting, thank you.

 

[chartery]

Thanks all. I guess I should have been clearer about not presuming the manifold was Riemannian. I had seen the mathoverflow link, but find algebra only helps once I have some visual handle on a concept, however simplistic.

On rereading Hehl's paper (https://arxiv.org/abs/0711.1535), the following seems to me to fit the bill:

Figure 2: Schematic view on a two-dimensional Cosserat continuum: Undeformed
initial state.

Figure 11: Deformation of a cubic crystal by edge dislocations of type α121:
The relative orientations of the lattice plains in 2-direction change. A vector
in x2-direction will rotate, if parallelly displaced along the x1-direction. As
a consequence a contortion κ112 emerges and the closure failure occur of the
“infinitesimal” parallelogram.Although I'm not smart/knowledgeable enough to decide which was the more appropriate:

Figure 4: Homogeneous contortion

Figure 6: Conventional rotation ∂[1u2] of the “particles” of a Cosserat continuum
caused by an inhomogeneous strain.