Are you sure that such a metric exists? If it does, consider the induced metric on $S^{n-1}$. Then the group $SO(n)$ acts on this Riemannian manifold in a way that topological distance is preserved. But if this action were by isometries in the Riemannian sense, then the dimension of the isometry group implies that you actually have a space-form, so it should be isometric to the standard round metric. So something seems to be strange ...

Thanks for your answer, Could you please more explain on the last part "Then the dimension of isometry group implies that ........".

If a Riemannian manifold of dimension $n$ has isometry group of dimension $n(n+1)/2$ (which is the dimension of the Euclidean group and of $SO(n+1)$), then it has to have constant sectional curvature. This follows since the isomrety group is a Lie group and isometries fixing a point form a closed subgroup. Since an isometry fixing a point is determined by its derivative in that point, this subgroup can be realized as a subgroup of $O(T_xM)$ and it has to have dimension $n(n-1)/2$. Hence it has to contain $SO(T_xM)$. But by construction, the value of the curvature in this point has to be invariant under the action of this group, which implies that the Weyl-curvature vanishes and the Ricci is proportional to the metric. Since this holds in each point, the metric has constant sectional curvature.

Thank you Robert J. Low for your answer. Am I mistaken to think that you implicitly assume that classical geodesics of sphere are geodesic in this new geometry? But such assumption is not legal.

Thanks  again Andreas  Cap for your interesting answer. Is not $n=2$(S^{1}) an exception, since we loose the concept of sectional curvature? I mean that :"Is there a metric in R^2 with this property?

Andreas Cap, I apologize for this second message. after some more thinking on your argument, I think i have some difficulties to understand the final conclusion. my question: What is the contradiction if we assume that, in the sense of metric space, there is  an isometry with a constant distortion,  between (sphere d_{1})  and  (Sphere , d_{2})?d_{1] is riemanian and d_{2} is euclidean?

Dear Ali Taghavi, sorry for the confusion, things are hard to express because of the unclear meaning of "isometry". My last statement was that high enough dimension of the Riemannian isometry group implies isometry in the Riemannian sese to the round sphere and hance also isometry in the sense of metric spaces to the round sphere. So the conclusion is that if a Riemannian metric as asked for in your question existed, then the action of $SO(n)$ would be by diffeomorphsims which are isometries in the sense of metric spaces without being isometries in the Riemannian sense, and it is hard for me to imagine how this could happen. So there is no formal contradiction, just a remark that the outcome looks quite strange to me. Therefore, there is the question, why such a metric should exist at all (which sounds like an interseting question to me). 

Do I understand the question properly: you want a Riemannian metric on Rn

which restricts to a Riemannian metric on Sn-1 giving this Euclidean 'chord distance'?  I think the answer is that the 'chord distance' does not arise from a Riemannian metric on the sphere and for that reason there can be no Riemannian metric on Rn which defines it in this way.  Indeed, its limit as points approach each other is the same as the round spherical metric (up to a constant factor). That is, infinitesimally, on the tangent bundle the two coincide - but as metrics (in the sense of metric spaces) they do not.

in deed there are two reasons why such a metric cannot exist. one due to global geometry (pointed out by robert j. low) and one due to the local one.

the first : euclidean geodesics are always unique, on the sphere they are not always, since the exponential mapping cannot be everywhere injective.

the second : if you consider a plane orthogonal to a radius, it intersects the sphere in a circle S. if you push forward the euclidean metric of the plane locally in a neighborhood of the small disk D bounded by S) on the sphere you obtain a locally defined metric d. this metric defines the good distance only for points on S. for points in the interior of D are strictly shorter than they should be. if you want to remedy this, you will have to change the metric in the interior, but then the distance of points on S will be greater.

Dear Robert J Low thanks for your new answer. I understand from your interesting point that We can summarize the answer in the following way: let we have a  SHORT geodesic on the sphere which arise from this Riemannian metric: the geodesic start from p, pass on q and end at r(p, q, r ) are very close to each other. then d(p,r)=d(p,q)+d(q,r). So we have a triangle equality in euclidian norm. This implies that three points on sphere lie  on the same line , a contradiction. But this situation is  a  motivation for the following question: Let (M,g) be  a  Riemannian manifold and N is  a  submanifold of M such that the metric induced from restriction g to N coincide to the original metric induced from (M,g). Does this imply that N is  a  totally geodesic submanifold of M? 

Dear James Montaldi, Could you please more explain on your argument that two metric coincide on the tangent space Up to a constant.

Dear Andreas Cap, thanks again for your information on dimension group. Could you please give a reference(book) about dimension group?

Dear Robert J. Low, should I understand from your 2 previous answer that the following general statement true: Let  M be  a Riemannian manifold of dimension 2 and $\gamma$ is  a  curve such that all points of gamma are fixed point of  certain isometry. Then gamma is  a  geodesic?

Dear Ali Taghavi

1) If we take two points on the unit sphere, then the Riemannian distance is equal to the angle theta (I'll use t instead) between them.  On the other hand, the chord distance is 2sin(t/2). The Taylor series of this is t - t^3/24 + ... .  The linear term agrees with the Riemannian distance.  So in the limit the ratio chord/Riemannian is 1. Does this answer your question?

2) Regarding isometries: it is easy to show that if S is an isometry then Fix(S) is a totally geodesic submanifold.  In particular, if the fixed point space is 1-dimensional then it is a geodesic. (I have avoided making more careful statements: strictly speaking the fixed point subset could be a disjoint union of smooth submanifolds of different dimensions: so I am really talking about each connected component.)

Dear James  Montaldi. Is it obvious that the fixed point of  a particular isometry is  a  totally geodesic submanifold?may be  a refgerence? regarding your first point (1) in your recent answer:  Why the ratio of chor/riemanian is equal to the ratio of two norms of a given vector in the tangent space?I think that may be some thing is missing in your computation. Forgive me if my question is elementary.

Dear Ali Taghavi

Results of this type (symmetries) often rely on uniqueness.  In this case, let x \in Fix(S), and v be a tangent vector at x which is tangent to FIx(S). Now consider the (unique) geodesic through x tangent to v.  Call it \gamma(t). If it does not lie in Fix(S) then S(\gamma(t)) will be a different geodesic, with the same initial condition (x,v) (because (x,v) is fixed by S).  Hence S(\gamma(t)) = \gamma(t) - by uniqueness.  And I think that is the definition of a totally geodesic submanifold.

For the other question, the issue is not really well-defined.  It is only Riemannian metrics that are defined by norms on tangent vectors (I think).  My point is that if the chord metric were defined by a Riemannian metric (hence by a norm on the tangent space) it would have to be the same as the usual Riemannian metric since infinitesimally they are equal.  Is that clearer?

Dear James Montaldi

Thanks for your recent answer.

I underestand your argument on fixed point of isometries.  But I do not underestand some thing on infinitesimaly computation. I denote by \phi,the flow of geodesic flow.

For a vector V in the tangent space T_{p}M we have |V|=d(\phi_{s}(P),P)/s. where d is the distance

Now  we have two riemannian metric g1 and g2 with geodesic flow \phi1 and \phi2. and norms |. |1 and |. |2.

When we compute the ratio of |V|1 by |V|2 we are not sure that \phi1(s) is located on (or has any relation with) \phi2(s). However you compute the ratio of riemannian and chord distance. 

I realy do not underestand  some relations between this ratio and the ratio of |V|1 and |V|2.

Am I mistaken?Where is my mistake?

 Well, you could use your formula |V|=d(\phi_{s}(P),P)/s to show that the chord-metric is incompatible with being a Riemannian metric, since the right-hand-side is not independent of s:  you get 2sin(s/2)/s, which is not constant.

In my earlier discussions, I was thinking of the limit of this expression (as s-> 0). But your formula works fine.

Dear James Montaldi   this is motivation to ask a more general question: Let g_1 and g_2 be two Riemannian metric on a manifold M such that for every p, d1(x,p)/d2(x,p) approaches 1 when x-->p. Does this implies that g1=g2?  If not what can be said about the comparison of g1 and g2?

I'm sure the answer is yes, g1=g2.  I don't have a proof off hand though.

Dear Robert J.  Low,  Can you give  a precise  argument or  a reference? It is obvious for one dim manifold. But what about higher dimension?

Dear James Montaldi, a few hours ago, I realized that I have still problem to understand your argument(Except for one dim. case). I re investigate your computation: I think we lead to th|V|1/|V|2=|V|1/|V|2    ! 

I am interested to understand your answer, in order to generalize it to arbitrary  Manifold (that is:  My recent question)

Dear Ali

Which argument is there a problem with?  I don't have an argument that g1=g2, though I'm sure it's true.  It's the "surely this implies" part of Robert's answer that I don't see how to do (though I agree with him). 

This looks OK to me too. 

Dear Ali:

The metric  d(x,y) = |x-y|  for  x,y \in S^{n-1}  does not come from any Riemannian metric on S^{n-1}; in particular it is not induced by any Riemannian metric on R^n. The reason is that S^{m-1} with that metric is not a length space. A metric space (X,d) is called length space if for any x,y\in X the distance d(x,y) is the infimum of the length of all rectifiable curves from x to y. It means that the distance  d(x,y)  adds up as the sum of distances of "neighbouring points" between x and y. Aparently, d(x,y) = |x-y| does not have this property: Adding up the distances of very close points on a great circle between x and y one obtains the euclidean length of the great circle segment which is bigger than |x-y|.

Concerning your later question: "Let (M,g) be  a  Riemannian manifold and N is  a  submanifold of M such that the metric induced from restriction g to N coincide to the original metric induced from (M,g). Does this imply that N is  a  totally geodesic submanifold of M?" The answer is yes (the Riemannian metric is just the infinitesimal version of the distance function), but the converse statement does not hold in general: A totally geodesic submanifold need not have the same induced distance function as the ambient manifold since the shortest geodesic in M between two points in N need not be contained in N. If this stronger property holds, the submanifold  N is called convex in M; convex is stronger than totally geodesid. Here is an example of a submanifold which is totally geodesic but not convex: Let M be a flat 2-torus whose fundamental domain is a rhombus with a short and a long diagonal, and let N be the long diagonal. Let p be the center of the rhombus (the intersection of the two diagonals) and q the vertices of the rhombus which are identified to one point in M. The distance between p and q in M is the half length of the short diagonal, but the distance in N is the half length of the long diagonal.

Best regards

Jost

What  about Finslerian norm?

Is there an smooth norm structure F on TS^{n} which corresponding distance equal the chord distance?  

Dear Ali,

no, there is no such Finsler metric. A Finsler space, like a Riemannian space, is a length space, that is the distance between two points is the infimum of the lengths of all curves between the points, or in other words, the distance between farther points is the sum of the distances between intermediate points which are close together. The chord metric |x-y| does not have this property: Infinitesimally (for y --> x) it would give the induced Riemannian metric, but this does not add up to |x-y| in the large. No chance for Finsler.

Best regards

Jost

Dear  Prof  Eschenburg

Thank you very  much for  your help and your interesting answer.

I understand from your answer that there  are a lot of common points in Riemannian and  Finslerian case.  

Is the  concept of (closed) geodesic a common concept? In particular is there a  compact  Finslerian metric  without  closed geodesics?

Best regards

Ali 

Dear Ali,

yes, the concept of geodesics is common to Riemannian and Finsler metrics. A Finsler metric, like a Riemannian one, is a way of assigning a length |v| to each tangent vector v with |tv| =t|v| for any t>0 and such that the "unit ball" {v\in T_pM : |v| \leq 1} is convex. This concept is needed to define the length of a curve, L(c) = \int |c'|  where c' is the derivative vector of the curve c. A geodesic is a curve of (locally) minimizing length. Of course, to ensure its existence you need some sort of completeness which is true when the manifold is compact. Then the most elementary methods to construct closed geodesics are still applicable, e.g. Birkhoff's method: Try to slide a loop of fixed length over a stone (whose surface represents the manifold). If it is too narrow, it will not fit. Make it longer and longer until it fits the first time. During the process of sliding you will reach a position where the loop is attached to the stone along its full length,without leaving any space between loop and surface. This position marks a closed geodesic on the surface. I think by this method one can find always n+1 closed injective geodesic loops on a compact n-manifold with Finsler (or Riemannian) metric. There is a famous example of a 2-sphere with a non-symmetric (|-v| \neq |v|) Finsler metric and exactly 3 closed geodesics. (But I am not expert about this).

Kindest regards

Jost

Dear  Prof. Eschenburg 

Thank you very  much for your answer and  the idea of n+1 closed geodesics on a  Finslerian or Riemannian compact  n-manifold.

Best regards

Ali

You are very welcome!

Best regards

Jost

I found a related MSE post as follows

https://math.stackexchange.com/questions/106508/existence-of-a-riemannian-metric-inducing-a-given-distance/106532#106532