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I am trying to grasp the Riemann curvature tensor, the torsion tensor and their relationship.

In particular, I'm interested in necessary and sufficient conditions for local isometry with Euclidean space (I'm talking about isometry of an open set - not the tangent space - with Euclidean space) - and I'd especially like to grasp these tensors in terms of their measuring the failure of Euclid's parallel postulate in a particular manifold.

In a Riemannian manifold $M$, we can choose the Levi-Civita connexion and null out the torsion. So then the curvature wholly determines whether or not we can find the Euclidean open set: we seek an open set wherein $R(X_p,Y_p)Z_p = 0; \forall X_p,Y_p, Z_p \in T_p M, \forall p \in U \subset M$ and we are done.

Question 1.

However, what happens to the curvature $R$ in the Riemannian case if we use other connexions with the same geodesic sprays but with different, nonzero torsions $T$?

Question 2.

Is there a formula showing how $R$ and $T$ in the case of connexions with the same geodesic sprays?

Question 3.

Can we still test the curvature alone as above to see whether there is a local Euclidean open set, or do we now need to make sure that torsion also vanishes in that open set too?

Question 4.

If we relax the Riemannian condition and talk abstractly about a connexion alone, which of the above answers change?

and lastly:

Question 5.

I'd really like to have an example of a space with zero curvature but nonzero torsion over a whole, open set, not just at a point, if such a thing exists. I think this would help build intuition.

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These are all good questions, but perhaps too many for one post. Let's hope that someone finds time for a decent response. Meanwhile I suggest you to take a look at Steuard Jensen's note "General Relativity with Torsion" available here –  Yuri Vyatkin Aug 12 '13 at 7:39
    
@YuriVyatkin Many thanks. Is your icon a self portrait? –  WetSavannaAnimal aka Rod Vance Aug 12 '13 at 8:50
    
This is a pastel portrait of me made in 1986 by a street artist. I just took a photo of it and used this as an avatar %) –  Yuri Vyatkin Aug 12 '13 at 9:40

1 Answer 1

up vote 10 down vote accepted

Your question is great! Before answering your question, I would like to visualize some concepts for you:

We consider the Riemann tensor first. A crucial observation is that if we parallel transport a vector $u$ at $p$ to $q$ along two different pathes $vw$ and $wv$, the resulting vectors at $q$ are different in general (following figure). If, however, we parallel transport a vector in a Euclidean space, where the parallel transport is defined in our usual sense, the resulting vector does not depend on the path along which it has been parallel transported. We expect that this non-integrability of parallel transport characterizes the intrinsic notion of curvature, which does not depend on the special coordinates chosen.

enter image description here


It is useful to say that in this sense visualization of the first Bianchi identity is very easy:

enter image description here


For visualizing Lie bracket: if $X$ and $Y$ are two vector fields in a neighborhood of $p$, then for sufficiently small $h$ we can

(1) follow the integral curve of $X$ through $p$ for time $h$ ; (2) starting from that point, follow the integral curve of $Y$ for time $h$; (3) then follow the integral curve of $X$ backwards for time $h$ ; (4) then follow the integral curve of $Y$ backwards for time $h$.

When $X$ and $Y$ are (linearly independent) vector fields with $[X, Y]\ne 0$, the parallelogram is not closed.

enter image description here


We next look at the geometrical meaning of the torsion tensor. Let $p \in M$ be a point whose coordinates are $\{x^μ\}$. Let $X = \varepsilon^μ e_μ$ and $Y = \delta^μ e_μ$ be infinitesimal vectors in $T_pM$. If these vectors are regarded as small displacements, they define two points $q$ and $s$ near $p$, whose coordinates are $\{x^μ + ε^μ\}$ and $\{x^μ + δ^μ\}$ respectively (following figure). If we parallel transport $X$ along the line $ps$, we obtain a vector $sr_1$ whose component is $\varepsilon^μ − \varepsilon^{\lambda} \Gamma^{\mu}_{\nu \lambda} \delta^{\nu}$ . The displacement vector connecting $p$ and $r_1$ is

$$pr_1 = ps + sr_1 = δ^μ + ε^μ − \Gamma^{\mu}_{\nu \lambda} \varepsilon^{\lambda} \delta^{\nu} .$$

Similarly, the parallel transport of $δ^μ$ along $pq$ yields a vector

$$pr_2 = ps + sr_2 = ε^μ + δ^μ − \Gamma^{\mu}_{\lambda \nu} \varepsilon^{\lambda} \delta^{\nu} .$$

In general, $r_1$ and $r_2$ do not agree and the difference is

$$r_2r_1=pr_2-pr_1=(\Gamma^{\mu}_{\nu \lambda}− \Gamma^{\mu}_{\lambda \nu})\varepsilon^{\lambda} \delta^{\nu}=T^{\mu}_{\lambda \nu} \varepsilon^{\lambda} \delta^{\nu} \qquad(*)$$

Thus, the torsion tensor measures the failure of the closure of the parallelogram made up of the small displacement vectors and their parallel transports.

enter image description here

Now with respect to the above description it is easy to imagine the Torsion tensor in terms of the Lie bracket and connection:

$$T(u,v)=\nabla_u v −\nabla_v u−[u,v]$$

enter image description here


I hope the above explanations have cleared the matter now I will get to answering your question:

Suppose we are navigating on the surface of the Earth. We define a vector to be parallel transported if the angle between the vector and the latitude is kept fixed during the navigation. [Remarks: This definition of parallel transport is not the usual one. For example, the geodesic is not a great circle but a straight line on Mercator’s projection.] Suppose we navigate along a small quadrilateral $pqrs$ made up of latitudes and longitudes (following figure).

enter image description here

We parallel transport a vector at $p$ along $pqr$ and $psr$, separately. According to our definition of parallel transport, two vectors at $r$ should agree, hence the curvature tensor vanishes. To find the torsion, we parametrize the points $p$, $q$, $r$ and $s$ as in following figure.

enter image description here

We find the torsion by evaluating the difference between $pr_1$ and $pr_2$ as in $(*)$. If we parallel transport the vector $pq$ along $ps$, we obtain a vector $sr_1$, whose length is $R \sin \theta d\varphi$. However, a parallel transport of the vector $ps$ along $pq$ yields a vector $qr_2 = qr$. Since $sr$ has a length $R \sin(\theta − d\theta) d\varphi \simeq R \sin \theta d\varphi − R \cos \theta d\theta d\varphi$, we find that $r_1r_2$ has a length $R \cos \theta d\theta d\varphi$. Since $r_1r_2$ is parallel to $-\frac{\partial}{\partial \varphi}$, the connection has a torsion $T^{\varphi}_{\theta \varphi}$, see $(*)$. From $g_{\varphi \varphi} = R^2 \sin^2 \theta$, we find that $r_1r_2$ has components $(0,−\cot \theta d\theta d\varphi)$. Since the $\varphi$-component of $r_1r_2$ is equal to $T^{\varphi}_{\theta \varphi} d\theta d\varphi$, we obtain $T^{\varphi}_{\theta \varphi} = −\cot \theta$.

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Very pithy and unusual example of the curvatureless but torsioned connexion - I like it. Is the set of geodesics exactly the straight lines on the Mercator projection (are all straight lines on Mercator's projection geodesics and contrawise?) –  WetSavannaAnimal aka Rod Vance Aug 12 '13 at 13:39
    
@WetSavannaAnimal aka Rod Vance- yes! –  Sepideh Bakhoda Aug 12 '13 at 15:29
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Thank you so much for your wonderful diagrams and explanation. The nice thing about your diagrams is that they are almost graph-theoretic, and thus correspond almost 1-1 to steps in rigorous definitions and arguments. I take it you can generalize your example through any smooth invertible mapping (generalizing the Mercator projection) of a 2-surface to a plane: geodesics on the surface are the images of straight lines on the plane, parallel transport is the image of Euclidean parallel transport on the plane: you will always end up with a connexion with a torsion but no curvature? –  WetSavannaAnimal aka Rod Vance Aug 13 '13 at 1:42
    
I think I could almost apply the Bianci identities now with the help from your diagrams! –  WetSavannaAnimal aka Rod Vance Aug 13 '13 at 1:42
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Where did you find these pictures? These are absolutely amazing. Especially that diagram of the torsion tensor. –  Jesse Madnick Sep 13 '13 at 8:59

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