- #1

o_parhizkari

- 4

- 0

Is it true to say on a topological manifold one can always find a unique metric such that a set of non intersecting curves become its geodesics? Is there any way to find that metric?

Thanks, Owzhan

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter o_parhizkari
- Start date

- #1

o_parhizkari

- 4

- 0

Is it true to say on a topological manifold one can always find a unique metric such that a set of non intersecting curves become its geodesics? Is there any way to find that metric?

Thanks, Owzhan

- #2

Eynstone

- 336

- 0

Welcome Owzhan.

Is it true to say on a topological manifold one can always find a unique metric such that a set of non intersecting curves become its geodesics? Is there any way to find that metric?

Thanks, Owzhan

A manifold is locally homeomorphic to some n-dimensional space.Just choose the corresponding euclidean metric ;it solves the problem locally.( If the manifold is smooth, we can 'piece together' the patches 7 get a universal metric).

- #3

quasar987

Science Advisor

Homework Helper

Gold Member

- 4,790

- 20

Second of all, given any metric on a smooth manifold and a point p on that manifold, for any tangent vector V at p, there is a geodesic passing through p those tangetn vector at p is V. In particular, in dimension >1, there are infinitely many geodesics passing through any given point, so no metric admits as its set of geodesics a set of nonintersecting curves.

- #4

o_parhizkari

- 4

- 0

Thank you for the replies, I am not a mathematician but I am enjoying my mind being engaged with your interesting ideas.

Let me first correct my question by adding the assumption that the manifold is smooth and metrizable.

Then let me clarify my question further: geodesics of a cylinder are circles and spirals, they are simply curves without requiring to mention their tangent vectors. Yes from each point of this cylinder there passes infinitely many spirals and one circle but any way any spiral or circle is a geodesic of this cylinder in the eyes of its own metric. In the simplest form I want to know if a set of non intersecting spirals and circles, in e.g. a 3D Euclidean space, considered as geodesics can uniquely give a cylindrical metric for this manifold?

Indeed, I am going to work with integral curves of a vector field and that's why I use the phrase "non-intersecting curves".

Also I am not sure how to work locally as is suggested by Eynstone, since my problem is how to reach the metric from geodesics, and geodesics when metric is not known are simply a set of curves, since we still do not know the connection, so we only have topology and not geometry, that is I don't know how to interpret my curves locally in the Euclidean space to obtain a local metric for that, then to patch them together to obtain a universal metric. Although I guess the answer to my initial question is yes (such a metric exists, although I don't know how to find it) at least limited to some conditions, but looking at the question locally makes it difficult for me to believe what I guessed above! Anyway, if such a metric exists globally it must exist locally as well ...

Thanks, Owzhan.

- #5

quasar987

Science Advisor

Homework Helper

Gold Member

- 4,790

- 20

Also, please don't write in green as it is hard to read. Better to stick with black.

- #6

o_parhizkari

- 4

- 0

Also sorry for my lame clarification, put that for me being an engineer not very familiar with the geometrs' language.

I simply imagine some curves in some known geometry (say, in some Euclidean space) and now I am seeking for a new geometry in which the same curves are observable as straight lines. I only want to make sure if such a new geometry always exist and if yes, if it is unique, and if again the answer is yes if there is a way to find it?

Thanks again.

- #7

Eynstone

- 336

- 0

Do you mean to ask if a new metric can be defined on the same space to make the curve a geodesic? If so, the answers depends upon the solution of equation of a geodesic. Since the curve is known, Christoffel symbols are the only unknowns. Thus, we get a set of first order equations in the metric Gij.I simply imagine some curves in some known geometry (say, in some Euclidean space) and now I am seeking for a new geometry in which the same curves are observable as straight lines. I only want to make sure if such a new geometry always exist and if yes, if it is unique, and if again the answer is yes if there is a way to find it?

Thanks again.

A family of curves would be more difficult to tackle (due to more restrictions on the set of equations) & formidable to compute.

(That's what I took 'straight lines' for- geodesics. If you simply had 'linear equations' in mind, the answer should be relatively simple).

- #8

o_parhizkari

- 4

- 0

... the answers depends upon the solution of equation of a geodesic. Since the curve is known, Christoffel symbols are the only unknowns.

Hi,

Thanks for your reply, indeed, that's what I initially tried. Actually I don't have any set of curve in hand (so I said I only imagine them), but only some differential equations. So, as a first step I tried to write my differential equations in the form of geodesic equations to obtain the new Christoffel symbols, however, I became very comfused when I tried to find the connection of the new geometry on the same space. It was its difficulty that made me first try to make sure if such a treatment always has answer, if the answer is unique, and if there is another way to find it more dirctly.

Asking from an expert in unversity he directed me toward discussions on the geodesic sprays, however, I didn't find a document that directly give my answer and so asked my question again in this forum.

Let me add that even in the simplest linear case that I tried to find the christoffel symbol yet it was not clear how to find them and I guess its answer wasn't unique at all, although I already knew what the answer should be!

Thanks again, Owzhan

Share:

- Last Post

- Replies
- 19

- Views
- 602

- Replies
- 3

- Views
- 2K

- Replies
- 21

- Views
- 625

- Last Post

- Replies
- 8

- Views
- 558

- Last Post

- Replies
- 1

- Views
- 1K

- Replies
- 37

- Views
- 4K

- Replies
- 4

- Views
- 729

- Last Post

- Replies
- 12

- Views
- 3K

- Replies
- 9

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 3K