Are there simple examples of Riemannian manifolds with zero curvature and nonzero torsion

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(Xp,Yp)Zp=0;Xp,Yp,ZpTpM,pUM 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.

Answer

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]0, the parallelogram is not closed.

enter image description here


We next look at the geometrical meaning of the torsion tensor. Let pM
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.

Attribution
Source : Link , Question Author : Selene Routley , Answer Author : Sepideh Bakhoda

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