How do we compute the geodesic between two points on a flat manifold?

[mnb96]

Hello,
I read that when a manifold has a flat metric, the geodesics are always straight lines in the parameter space. I have two questions:

(1)
If we are given a Clifford torus S^1 \times S^1 (which is flat), how do we compute the geodesic between two points? Is the following correct?

\sqrt{\min\left\{ \left|\theta_{1}-\theta_{2}\right|,\,2\pi-\left|\theta_{1}-\theta_{2}\right|\right\}^2 +\min\left\{ \left|\phi_{1}-\phi_{2}\right|,\,2\pi-\left|\phi_{1}-\phi_{2}\right|\right\}^2}(2)
Let's consider a torus in \mathbb{R}^3 with R=r=1, where R and r are respectively the distance from the center, and the cross-section radius.
Now the geodesics are not straight lines in the parameter space, however can we say that the above formula defines a distance between points on the torus?
If the answer is yes, shall we conclude that, when simply looking for a 'distance', knowing whether a manifold is flat or not is irrelevant?

Thanks!

 

[arkajad]

Hello,
I read that when a manifold has a flat metric, the geodesics are always straight lines in the parameter space.

The correct statement is: if the manifold is flat that one locally introduce coordinates in which the geodesics are given by straight lines. In other coordinates, that are not adapted to the flat geometry of the manifold, geodesics will not be straight lines.

 

[mnb96]

I see...
but does that mean that the formula I wrote to compute the length of the shortest path is wrong?
That formula would at least work in the case of a S^1 sphere, as it gives the length of the shortest arc, so I thought it might work for S^1 \times S^1.

As we can "unwrap" the unit circle S^1 into a straight line, I thought I was allowed to unwrap S^1 \times S^1 onto a xy-plane.

Anyways, were you basically saying that what I wrote is simply a distance, and I didn't prove that it is also the length of the geodesic shortest path between two points on the Clifford torus?

 

[arkajad]

Now, you need to specify which geometry you want to put on your torus. The flat geometry induced by the parameters \theta,\phi? Then the geodesic segments for this geometry are straight lines - induced from the flat square. What you wrote is the distance along the geodesics of this geometry.

As fror your question:

If the answer is yes, shall we conclude that, when simply looking for a 'distance', knowing whether a manifold is flat or not is irrelevant?

the answer is "sometimes". When you see that in some coordinates the metric coefficients are constant, then your manifold is flat. When you them non-constant, then you still don't know you need to calculate the curvature tensor.

 

[mnb96]

Now, you need to specify which geometry you want to put on your torus. The flat geometry induced by the parameters \theta,\phi?

Yes! ... did I actually have other choices?


... should I also expect that the formula I wrote will give me the same results as the formula D(q,q') you wrote some time ago https://www.physicsforums.com/showpost.php?p=2911801&postcount=6" (applied to the Clifford Torus)?

 

[arkajad]

The distance between two quaternions is measured using the geometry of S^3. Geodesics there are great circles. I don't think that lines of constant \theta and \phi that you are using are great circles of S^3?

 

[mnb96]

I guess in general they are not. However using Hopf coordinates (\psi,\theta,\phi) the Clifford torus is given by the "slice" (\frac{\sqrt{2}}{2},\theta,\phi). I will to figure out if the two geodesic distances coincide, in this case. Intuitively they should, the same way the geodesic distance for two points on the equator of the S^2 sphere reduces to the shortest-arc on the circle, but I am probably wrong, as my intuition is not yet well developed.

Thanks for your help!

 

[mnb96]

...Intuitively they should, the same way the geodesic distance for two points on the equator of the S^2 sphere reduces to the shortest-arc on the circle, but I am probably wrong, as my intuition is not yet well developed...

Indeed, I was wrong but at least now I clearly understand why. Brutally using the same metric for the sphere on a submanifold of it won't ensure that the measured length will be the length of a geodesic sitting on that submanifold.

This also happens for the S^2 sphere. If we constraint ourselves to move along a parallel close to the north-pole, using the geodesic distance on the whole sphere will give us the length of the shortest path which might even pass through the north pole! and these are not a geodesics on our submanifold (the parallel).

Now I see more clearly, why we had to compute the pullback metric of the S^3 sphere on the Clifford torus.