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## torsion in two dimensions?

This is really more of a differential geometry question than a GR question. Is it possible to have torsion in two dimensions? I've seen statements implying that torsion is only of interest in 3 or more dimensions, but I don't see why. My understanding is that in two dimensions, you could not have a totally antisymmetric torsion tensor, and therefore it's not possible to have torsion in which parallel-transporting a tangent vector along a geodesic keeps it tangent to the geodesic. This doesn't seem the same as saying that torsion is impossible or of no interest in 2 dimensions.
 Recognitions: Science Advisor Maybe i m overlooking something, but torsion is just the antisymm. part of the connection. So why can't that exist in 2 dim.?

 Quote by bcrowell This is really more of a differential geometry question than a GR question. Is it possible to have torsion in two dimensions? I've seen statements implying that torsion is only of interest in 3 or more dimensions, but I don't see why.
This is an ambiguous question as worded. What do you mean by torsion in 2D? Certainly it is possible to talk about torsion of 2D surfaces in 3 or n dimensions. A different thing is the possibility of torsion of curves inside a 2D space, there is no such thing.

 Quote by bcrowell My understanding is that in two dimensions, you could not have a totally antisymmetric torsion tensor, and therefore it's not possible to have torsion in which parallel-transporting a tangent vector along a geodesic keeps it tangent to the geodesic.
That is not dependent on the number of dimensions but on the Riemannian connection you are describing. The torsion tensor is a vector-valued two form, when completely anty-symmetrized it vanishes leaving the connection totally symmetric.

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## torsion in two dimensions?

 Quote by haushofer Maybe i m overlooking something, but torsion is just the antisymm. part of the connection. So why can't that exist in 2 dim.?
Yeah, I don't see it either, except that in 2 dimensions you can't have torsion and also have parallel transport of a tangent vector along a geodesic maintain its tangency.
 Didn't you see #3?

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 Quote by bcrowell Yeah, I don't see it either, except that in 2 dimensions you can't have torsion and also have parallel transport of a tangent vector along a geodesic maintain its tangency.
We had a discussion about "geodesics" and torsion earlier. Which definition of "geodesic" do you mean here? A curve whose tangent is parallel-transported along itself, or a curve that locally extremizes the length functional?

It is clear that the first type of curve exists even in the presence of torsion, although it might no longer agree with the second.

 Quote by Ben Niehoff We had a discussion about "geodesics" and torsion earlier. Which definition of "geodesic" do you mean here? A curve whose tangent is parallel-transported along itself, or a curve that locally extremizes the length functional? It is clear that the first type of curve exists even in the presence of torsion, although it might no longer agree with the second.
I took it to mean the latter type of geodesic (the one determined by Levi-civita connection). But as I said that would be independent of the dimensionality of the manifold. If it is the other type, or any other curve my understanding is that there is no torsion in a two manifold, right?
 Recognitions: Science Advisor I see no reason there should be no torsion in 2 dimensions. However, clearly the totally antisymmetric part of the torsion vanishes. This means that, assuming metric-compatibility, any given set of geodesics uniquely determines the connection (because the totally antisymmetric part of torsion is what allows many different metric-compatible connections to have the same geodesics). But this does not mean that the Levi-Civita connection is the only metric-compatible connection. It does mean that any other metric-compatible connection will have different geodesics, as Ben points out.

 Quote by Ben Niehoff I see no reason there should be no torsion in 2 dimensions.
I don't understand this statement. So you are sayin that a curve in a surface can have torsion, can you point me to a reference that shows how to calculate it?

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 Quote by Ben Niehoff We had a discussion about "geodesics" and torsion earlier. Which definition of "geodesic" do you mean here? A curve whose tangent is parallel-transported along itself, or a curve that locally extremizes the length functional?
The point of the stuff you quoted was basically just that I was trying to make the distinction between those two definitions.

 Quote by Ben Niehoff This means that, assuming metric-compatibility, any given set of geodesics uniquely determines the connection (because the totally antisymmetric part of torsion is what allows many different metric-compatible connections to have the same geodesics).
This sounds interesting, but I don't really understand it. Are you a condensing a long, complicated argument that I would need to read in a book to understand it, or is it a simple argument that could be explained here?

 Quote by Ben Niehoff I see no reason there should be no torsion in 2 dimensions.
Isn't a planar curve with non-vanishing curvature defined to have zero torsion at each point?
From the Wikipedia entry for torsion:"A plane curve with non-vanishing curvature has zero torsion at all points."
If you are referring to the torsion of surfaces embedded in higher spaces obviously there is torsion for the surface geodesics. That is why I asked the OP to clarify what he meant by "torsion in 2 dimensions".
 I'm sorry for the intrusion, I'm not understanding how there cannot be torsion in 2 dimensions. Given a connection $\nabla$ on a differentiable manifold the torsion tensor $T$ is defined by $T(X,Y) \equiv \nabla_X Y - \nabla_Y X - [X,Y]$, where $X, Y$ are tangent vector fields. It's antisymmetric by construction and in components it is $T(X,Y)^\mu \partial_\mu = (\Gamma^\mu_{\alpha \beta} - \Gamma^\mu_{\beta \alpha})X^\alpha Y^\beta \partial_\mu$. So why should torsion vanish in two dimensions? Ilm

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 Quote by Ilmrak I'm sorry for the intrusion, I'm not understanding how there cannot be torsion in 2 dimensions. Given a connection $\nabla$ on a differentiable manifold the torsion tensor $T$ is defined by $T(X,Y) \equiv \nabla_X Y - \nabla_Y X - [X,Y]$, where $X, Y$ are tangent vector fields. It's antisymmetric by construction and in components it is $T(X,Y)^\mu \partial_\mu = (\Gamma^\mu_{\alpha \beta} - \Gamma^\mu_{\beta \alpha})X^\alpha Y^\beta \partial_\mu$. So why should torsion vanish in two dimensions?
I'm more comfortable with straight index gymnastics notation, whereas it looks like you're expressing this in a combination of index-gymnastics and index-free notation (?), so I'm not sure how to interpret it.

In the notation I'm more comfortable with, we'd have a three-index tensor $T^c_{ab}$. It's antisymmetric by definition on the indices ab. In $\ge 3$ dimensions, it can also be antisymmetric on all three indices, in which case the connection preserves tangent vectors along geodesics. In two dimensions, it's only possible for a nonzero three-index tensor to be antisymmetric on two of its indices, in which case there are two degrees of freedom, $T^1_{12}$ and $T^2_{12}$. This does *not* imply directly that torsion is impossible in two dimensions.

 Quote by bcrowell I'm more comfortable with straight index gymnastics notation, whereas it looks like you're expressing this in a combination of index-gymnastics and index-free notation (?), so I'm not sure how to interpret it.
Yes, sorry, I should have written in a more standard notation $T(X,Y) = T^\mu_{\alpha \beta}X^\alpha Y^\beta \partial_\mu$.

 Quote by bcrowell In the notation I'm more comfortable with, we'd have a three-index tensor $T^c_{ab}$. It's antisymmetric by definition on the indices ab. In $\ge 3$ dimensions, it can also be antisymmetric on all three indices, in which case the connection preserves tangent vectors along geodesics. In two dimensions, it's only possible for a nonzero three-index tensor to be antisymmetric on two of its indices, in which case there are two degrees of freedom, $T^1_{12}$ and $T^2_{12}$. This does *not* imply directly that torsion is impossible in two dimensions.
So is the point that, in two dimension, there cannot be a connection with non vanishing torsion such that the two definitions of geodesics are equivalent?

Ilm

 Quote by Ilmrak So is the point that, in two dimension, there cannot be a connection with non vanishing torsion such that the two definitions of geodesics are equivalent?
Not exactly, what happens is that in surfaces the torsion tensor is completely determined by the torsion form, that usually is a 2-form but in surfaces is a one-form.
So this is the sense in wich torsion is trivial in 2D, the amount of information it gives is quite limited, basically it gives the orientation of the surface. See relative (geodesic) torsion and the Darboux frame.

 Quote by TrickyDicky Not exactly, what happens is that in surfaces the torsion tensor is completely determined by the torsion form, that usually is a 2-form but in surfaces is a one-form. So this is the sense in wich torsion is trivial in 2D, the amount of information it gives is quite limited, basically it gives the orientation of the surface. See relative (geodesic) torsion and the Darboux frame.
I think I will have to study a bit more differential geometry, I don't know anything about the torsion form