Dear Robert, this is not considered in textbooks on geometry. But this question concerns a connection between the fundamental concepts of a space-time geometry, which can be described in terms of only world function. A use of the coordinate system is not necessary. First of all, one needs to give a definition of the metric dimension. There is more coordinate dimension (the number of coordinates, used at the geometry description). In general, the coordinate dimension and the metric dimension are different concepts, but usually one does not differ between them. We shall consider the metric dimension, using the term dimension. By definition, the dimension is the maximal number of linear independent vectors. Is it possible to determine linear independent vector without introduction of linear vector space? Yes, it is possible.

The vector AB is the ordered set of two points AB = {A,B}. I shall use the term geometrical vector (g-vector) in order to distinguish it from the linear vector (linvector), which is defined as an element of the linear vector space. The mutual orientation of two g-vectors AB and CD is defined by their scalar product

(AB.CD) = \sigma(A,D) + \sigma(B,C) - \sigma(A,C) - \sigma(B,D)              (1)

where \sigma(A,B) is the world function between two points A and B. One can test that this formula is valid in the proper Euclidean geometry. By definition it is valid in any physical geometry, i.e. in a geometry described completely by the world function.

n g-vectors P0P1,P0P2,…P0Pn are linear dependent by definition, if and only if the Gram’s determinant Fn(Pn+1) = det||(P0Pk.P0Pl)||,   k,l =1,2,…n , vanishes

Fn(Pn+1) = 0.                                           (2)

The quantity Pn+1 (skeleton) is Pn+1= {P0,P1,…Pn}

Equality of (2) can be tested in the Euclidean geometry. By definition this equality is true in any physical geometry.

Thus, the condition of linear dependence of n g-vectors is defined in terms of the world function. To determine dimension one needs to determine maximal number of linear independent g-vectors. It means that there exist such a skeleton Pn+1 = {P0,P1,…Pn}, that Fn(Pn+1) is not equal to 0, and for other skeletons Pk+1, k>n Fk(Pk+1) = 0. The set of linear independent g-vectors constructed on the base          Pn+1 can be used for a construction of a coordinate system.

One can see, that there is a lot of conditions, which are to be fulfilled to introduce the metrical dimension in a space-time geometry. If some of conditions are not fulfilled, one cannot introduce dimension, but physical geometry can exist without a dimension, because all geometrical objects, which can be constructed in the Euclidean geometry, can be also constructed in the physical geometry, where the dimension cannot be introduced.

Perception of a geometry from the metric approach distinguishes strongly from the conventional approach, when the most general space-time geometry is a Riemannian geometry. A use of a space-time geometry leads to a necessity of revision of the general relativity, and of space-time geometry in the microcosm, where the space-time geometry is discrete. A use of the conventional method of a geometry presentation shows, that we know only geometry of circumambient world, but we do not know the microcosm geometry and space-time geometry of cosmos.

Dear Yuri, as I see your problem is the following: You have a metric (world function), define a concept of linear dependence on the basis of this concept you define the number "dimension" and ask that the dimension can be or cannot be determined (and calculated) by the given metric. I think that without any other assumption you have not definite dimension for your space. Consider the following example: Let the space the union of an Euclidean line a and a point A, which does not lie on a. Use the Euclidean metric as world function on this set. Your definition of dependence shows that there are independent pairs of vectors (e.g. AB and AC, where B,C are on a, respectively.)  implying that the dimension of this space is 2. If the world function on the examined set determines the euclidean geometry and also its dimension that this model nothing else as the 2-dimensional Euclidean space, namely the Euclidean plane. On the other hand the spanded usually Euclidean plane contains it as a proper subset showing that the concept of Euclidean plane is not well-defined.

Dear A.V.Horath, Frankly speaking, I have not understand your arguments. I understood you as follows. Let we have two-dimensional Euclidean plane  \Omega. Let there are one-dimensional straight line a and the point A. Both a and A belong to \Omega, but A does not belong to a. Let us remove all points of  \Omega except of a and A. What is dimension of the geometry on the set, consisting of A and a? If the world function between the point A and all points of a remains as it were in \Omega, the dimension will be 2. If the world function betwee A and all points of a will distibguish from the world function in \Omega, then one cannot ascribe any dimension to geometry on the set, consisting of A and a.  

Have I presented your viewpoint correctly? If, yes, what is incorrect in such an approch?

Yes, the world function between the points of the line and the point remains Omega. My problem is the following: You have a two-dimensional Euclidean space which does not contains that objects which we know in the Euclidean plane. We can say or not that it is an Euclidean geometry of dimension two? If the answer is not then the world function alone does not determine the concept of Euclidean plane, if the answer is yes that the world function does not give precise description of the concept of Euclidean plane. 

Moreover, observe that in this new geometry all experiments suggest that fact that our geometry is of dimension one.

Dear Roger, you presented a lot of different definitions of the dimension. Such an approach cannot be effective. To obtain effective definition, one needs to present the Euclidean geometry as a monistic conceptions. It means, that all geometrical quantities and definitions must be expressed in terms of ONE fundamental quantity: world function. The definition of dimension in terms of world function as well as other geometrical quantities can be used in all other generalized geometries. One needs only to replace the Euclidean world function by the world function of the generalized geometries in all expressions of other geometrical quantities. In the case of the dimension such a replacement is possible not always, because at definition of the dimension some special properties of the Euclidean world function are used. If these special properties do not take place for the world function of the generalized geometry,  the dimension cannot be introduced in the generalized geometry.

Dear Yuri ~

A Riemannian space is topological space with a metric. I feel that the concept of "dimension" is more primitive than either coordinates or metric. I'm attaching  a page of notes I made many years ago when I was attempting to understand "dimension" as a purely topological concept. It may (or may not) be of interest to you.

Dear Eric,

You cannot say anything pithy about dimension, if you begin from topological space and from topology, because the topology in itself has no relation to the space-time geometry. Let me show this in the example of a discrete space-time geometry.

The discrete geometry is such a geometry, where there are no close points. Mathematically it means that for any points

for any P,Q |\rho(P,Q)| does not belong to (0,\lambda)      (1)

where \lambda is a minimal length of the discrete geometry Gd. The geometry Gd is constructed as a result of modification (generalization) of the Euclidean geometry GE. The distance is the only quantity, which is common for Gd and GE. There are two different method of Gdconstruction.

(1) Conventional method. One fixes the distance function \rho and varies the set of points, where the geometry is given. One removes almost all points, remaining only those points, whose coordinates are multiple to \lambda. One obtains so called geometry on lattice, which is considered as unique possible discrete geometry. This geometry is not uniform and isotropic.

(2) The true method. One fixes the set of points, where Gd is given and varies the form of the distance function \rho in such a way, that to satisfy the relation (1). It is more convenient to use the world function \sigma = \rho2/2. The world function \sigmad can be chosen, for instance, in the form

\sigmad = \sigmaM +\lambda1/2  sgn(\sigmaM)        (2)

where \sigmaM is the world function in the geometry of Minkowski.

It is easy to verify that \rhod=\sqrt(2\sigmad) defined by (2) satisfies the condition (1). The discrete space-time geometry Gd is give on the set of points \OmegaM, where the geometry of Minkowski is given. Gd is isotropic and uniform. Besides it is rather close to the real space-time geometry. Free particles move stochastically in Gd. Statististical description of this stochastic motion leads to quantum description (Klein-Gordon and Schroedinger equation),  if \lambda2 =\hbar/(bc), where b is some universal constant and c is the speed of the light.

As to geometry on lattice, it has no relation to the real space-time geometry and to a geometry at all. It is a mental toy for mathematicians. Playing this toy, mathematician may introduce topological spaces, and other numerous concepts, which has no relation to geometry.

Yuri ~

I appreciate what you are saying. My tentative topological analysis requires the assumption of continuity. It doesn't apply to discrete topologies, as you pointed out. I never imagined that it did - indeed, I suspect that, in general, discrete topologies don't necessarily possess a unique dimensionality.

Riemannian geometry is defined on a differential manifold, the definition of which is, as I recall, "a Hausdorff topological space, every open neighbourhood of which is homeomorphic to an open neighbourhood of Rn".  This structure becomes "Riemannian" by introducing a metric.

My contention is that, subject to certain axioms including a continuity axiom,  the dimensionality of a topological space can be defined prior to any relationship to Rn, and prior to the introduction of a metric.

Dear Eric, Your statement:

My contention is that, subject to certain axioms including a continuity axiom, the dimensionality of a topological space can be defined prior to any relationship to Rn, and prior to the introduction of a metric

is not founded. One neads no axioms. One sets a distance function on any set of points. It is sufficient for construction of the geometry, if one uses the method of deformation of Euclidean geometry. All, what one needs, is concluded in the Euclidean geometry.

Yuri ~

“the method of deforming Euclidean geometry” enables us to define an n-dimensional Riemannian geometry. That is true. The Riemannian space is then n-dimensional because the Euclidean space is n-dimensional. But then what is “dimensionality”? – that question seems to be implicit in your original question “how can one determine the metric dimension of a spacetime geometry, using only metric”. It seems to me that the dimensionality of a topological space can, in certain circumstances, be defined even in the absence of a “metric” – that was my point.

Dear Eric,

In the proper Euclidean geometry the metric dimension is defined as maximal number of linear independent geometricl vectors (g-vectors). Geometrical vector AB is defined as the ordered set of two points A and B. Linear dependence of g-vectors may be defined in terms of world function. n g-vectors AB1, AB2,…ABn are linear dependent if and only if the Gram’s determinant Fn = det||(ABi.ABk)||, i,k =1,2,…n vanishes. The scalar product is expressed via world function

(AB.CD) = \sigma(A,D) + \sigma(B,C) - \sigma(A,C) - \sigma(B,D)

Hence, the Gram’s determinant is expressed via world function. Using Gram’s determinant, one may determine dimension of the Euclidean geometry. Receipt of this determination one can find, for instance, in "Geometry without topology as a new conception of geometry." Int. Jour. Mat. & Mat. Sci. 30, iss. 12, 733-760, (2002),(Available at http://arXiv.org/abs/math.MG/0103002 ). (section 2)

This definition of dimension contains special properties of the Euclidean geometry, which can be not fulfilled for other physical geometries. For instance, for Riemannian geometry one needs to demand infinitesimal vectors for determination of dimension.

Yuri ~

"...for Riemannian geometry one needs to demand infinitesimal vectors for determination of dimension".

What's is wrong with that? "Infinitesimals" are a respectable ingredient of differential calculus - simply a convenient way of speaking of the idea of taking limits.

Dear Eric.

When you  say: "What's is wrong with that? "Infinitesimals" are a respectable ingredient of differential calculus", I have nothing against. I stress only the fact, that the Riemannian geometry is glued from "pieces" of Eucledean geometry and defining a dimension, one needs that both points of the vectors lie inside one piece. In other words, dimension of the Riemannian geometry is determined by dimensions of small (infinitesimal) Euclidean parts of the Riemannian geometry. If you ignore this circumstance, you obtain an indefinite result for value of dimension.