Both definitions are equivalent in Riemannian and hence Euclidean case.  But in pseudo-Riemannian they are not.

The first definition considers vectors as elements of a vector space and the second considers vectors as geometric objects that have length and angle in between.

I  think the main definition is the first one as they are elements of a vector space and in Euclidean sense there are many equivalent definitions like the second one.

I have no good defense to choose the first, but it does make sense.

Dear Sameh,

You are not right. The second definition is more correct, because it is coordinateless. A use of coordinates in definition makes this definition unreliable. In application to the space-time of Minkowski the fist definition leads to smooth world lines of tachyons. As far as tachyons have not be discovered experimentally, it means that tachyons do not exist. According to the second definition spacelike world lines of tachyons wobble with infinite amplitude. Tachyons cannot be discovered experimentally because of this wobbling. But it does not mean, that tachyons do not exist. A single tachyon cannot be discovered experimentally only. However the tachyon gas can be discovered by its gravitational field. The tachyon gas explains freely the dark matter in astrophysics. See "Dynamic equations for tachyon gas", Int. J. Theor. Phys. 52, 133( 10), 3683- 3695, (2013),. doi:10.1007/s10773-013-1674-4 see also electronic version

http://gasdyn-ipm.ipmnet.ru/~rylov/detg3e.pdf

Dear Yuri Rylov,

According to my paper " The Quantization of General Relativity: Photon Mediates Gravitation" http://dx.doi.org/10.14299/ijser.2014.10.002 I found there is no space-time continuum, it is only time! Reinterpretation of Lorentz transformation according to Copenhagen school will lead the Lorentz transformation is vacuum energy dependent and not depend on the relative velocity. Do you know why the discreteness and discontinuity, indeterminism, uncausality which existed in quantum can't be interpreted according to relativity, because of the concept of Einstein of the space-time continuum which is adopted in his interpretation to the Lorentz transformation. Because of the space-time continuum, it is resulted determinism, causality and continuity in SRT. Because of that Einstein could not interpret the Spooky action and quantum entanglement according to his space-time continuum in SRT.  The discreteness, indeterminism, and the uncausality  exist in quantum depend on observation only but not locally. Because of that the speed of light is locally constant and equals to c speed of light in vacuum. According to my transformation it is possible measuring faster than light without violation of Lorentz transformation or causality, but that depends on observation but not locally. Because of that quantum theory built basis on observation and experiment. My transformation will lead to the Heisenberg uncertainty principle, and the wave-particle duality, which they are one thing.

Dear Azzam,

It seems to me, that your answer has no relation to my question. It was necessary to deside, which of two definitions of equality is valid. Instead of answer you consider quantization of General relativity. I believe that it is incorrect to explain simple problems of a geometry via complicated problems (such as quantization)

If you need a definition independent of a basis, but you allow to use the scalar product, then I suggest you the following:

Let $(V,g)$ be a linear space, $g$ being a metric on $V$ whose radical is zero: $rad(g)=\{ 0 \}$. 

Then, two vectors $u, v\in V$ are equal iff $g(u,x)=g(v,x)$ for any other $x\in V$.

Dear Aleksei,

In the proper Euclidean space two vectors AB and CD are equivalent, iff

(AB.CD)=|AB| |CD| and |AB|= |CD| . This definition does not use the linear vector space. The vector AB is the ordered set of two points A and B  AB={A,B} (but not an element of the vector space, where linear operations over vectors are determined). This coordinateless definition is expanded to any geometry, which is described completely by only metric. The scalar product is defined in terms of only metic in the form

(AB.CD)=1/2(|AD|2+|BC|2-|AC|2-|BD|2) This relation is valid in the proper Euclidean geometry. Any reference to the linear vector space are redundant. Thus, the second definition of the vectors equality is true.

Dear Miguel,

Your suggestion contains a reference to a linear space, where the vectors are given. In the geometry, which is described by metric only, the linear space may be not exist. Coordinateless definition of  the vector equivalence must not contain a reference to a vector space.

Dear Aleksei,

You are right in the relation, that my definition of the scalar product differs from usual definition in terms of linear vector space. My definition of the equality of two vectors is intransitive in the geometry with arbitrary metric. But in the Euclidean it is transitive. Geometry with intransitive equality relation is nonaxiomatizable. Mathematicians consider that nonaxiomatizable geometries do not exist, because they know only one method of the geometry construction: deducing all geometric statements from axioms.

I am physicist. I am investigating the space-time, and this position of mathematicians has no relation for me. It indifferent for me, whether or not the space-time geometry is axiomatizable, because I do not deduce geometrical statements and concepts from axioms. I use a more simple way of the geometry construction. I consider the proper Eucledean geometry GE and express all statements of GE in terms of Euclidean metric dE. It is always possible. To construct a geometry G, which is described by metric completely, I replace the Euclidean metric dE by the metric d of geometry G in all statements of Euclidean geometry GE. As a result I obtain all statements of geometry G expressed via metric d of geometry G. It is a very simple method of the geometry construction. In this construction one does not use proofs of theorems. All necessary theorems are proved in GE. If you have time, look my paper (it is a short paper about 5 pages) and write me your objections, if they will appear.  My paper:

“ Metrical conception of the space-time geometry” Int. J. Theor, Phys.54, iss.1, 334-339, (2014), Electronic version on my site:

http://gasdyn-ipm.ipmnet.ru/~rylov/mcstg2e.pdf

   Offhand, I would give the following definition (which may coincide with one that some of you have given): two vectors in a vector space are equal if, for any basis of the space, component i of one vector equals component i of the second vector.

   This clearly applies to Minkowski space, among other spaces. In Minkowski space, since 4-vectors are made up of a 3-vector and a scalar, equality in the above definition becomes, separately,  equality of 3-vectors and equality of the respective scalars.

The Minkowski spacetime should be regarded as an affine space (as introduced in linear algebra), with a difference vector space that carries a Minkowski metric. To every ordered pair of spacetime points there belongs a unique difference vector, and two such vectors are equal if they are equal as vectors in the difference space. Historically, this was clearly presented in H. Weyls "Space-Time-Matter", in particular on p. 163.

Dear Norbert,

Why do you think that space-time is an affine space? If you postulate this, it will be clear, that one can introduce coordinate system, and the first version of definition is valid.

In reality, one can introduce in the space-time only space-time interval (distance). Whether or not one can introduce an affine space and the coordinate system, it is a question, especially the space-time is curved. Euclid described the Euclidean geometry without introducing coordinate system. If one knows geometry, one must be able to describe a geometry by coordinateless way.

It is especially valid for curved space-time. If one can do this, one can describe a geometry without any coordinate system. In this case the second version of coordinateless definition of equality is valid. How to describe the space-time geometry without introducing coordinate system, see my paper: "Metrical conception of the space-time geometry". Int. J. Theor, Phys.54, iss.1, 334-339, (2014). Electronic version: http://gasdyn-ipm.ipmnet.ru/~rylov/mcstg2e.pdf

Dear Norbert:

   I read Weyl's book a long time ago; I will look up page 163. Thanks.

Dear Raul,

Since you read Weyl's great book, I should correct the page number: The one I gave refers to the last german edition; in the english edition of 1952 in Dover this is page 173. Weyl elaborates on his statement on the next few pages of section 21.

To Yuri I should add that I consider only special relativity (no gravity). On GR I wrote a book on more than 700 pages; the last edition appeared in 2013 in the Graduate Texts in Physics of Springer. 

Dear all,

Let us stick to special relativity, as Norbert does, and consider the Minkowski space-time as being first a 4-dimensional vector space (an affine space is but a vector space with a given origin). In any vector space, two vectors are equal if and only if their difference is the null vector. So in a  4-dimensional vector space, v = (x, y, z, t) is null iff x=y=z=t=0. This is not much of an answer ! To go further, we must not forget that the Minkowski space-time comes equipped with an extra structure, which is the Lorentz quadratic form q(v) = x^2 + y^2 + z^2 - t^2 (take c = 1). But this quadratic form does not give an "euclidian metric", i.e. q(v) can be null without v being null (such vectors constitute the "light cone"). I  think that the key property to be taken into account is that the associated bilinear form f(v, v') = xx' + yy' + zz' - tt' is "non degenerate", which gives exactly what we want:  v is null iff f(v, v') = 0 for all v'. One doesn't need to check this for all v', only for 4 vectors v' which form a basis of our space. Of course, taking this basis to be the "canonical basis" (1,0,0,0) etc. will only yield x=y=z=t=0, but this actually shows that the so called canonical basis is not canonical at all : when we wrote v = (x, y, z, t), we took any basis ! A "canonical" choice of a basis should come from the physical problem we deal with, and I guess that was the original motivation of the question.

Dear Charles,

If your critics concern my answer, may I ask you to re- read  my very last sentence ? Of course the equality of two vectors is independent of any choice of coordinates, but since you calculate on coordinates, in order to ease your calculations you must choose a system (basis) which is best adapted to your physical problem . Besides, what "scalar product" a.c do you consider ? Coming back  to Yuri's original question, he gives an equivalence between   (1) equality of coordinates  and   (2) equality of length and nullity of angle (as I understand it). This implies that he thinks of the usual euclidian scalar product. This seems rather irrelevant (I dare not say "silly") because first it would apply to any vector space, second in the Minkowski space, the Lorentz scalar product is basic, so you cannot avoid using it

Dear Charles,

Euclid formulated geometry without a use of term “coordinate system”. I learnt geometry in the middle school, using textbook, where the term “coordinate” was absent at all. It means that a person, who knows geometry properly, can presented geometry without a use of terms “coordinate”, “coordinate system”, “coordinate transformation”, because the term "coordinate" concerns to description of a geometry, but not to the geometry in itself. The geometry description, without using coordinate systems, contains many advantages. For instance, one does not need to consider coordinate transformations. Besides, one cannot introduce coordinate systems in non-Riemannian geometries, which cannot be described in terms of coordinates and where one cannot introduce linear vector field. One cannot introduce coordinate systems in the discrete space-time geometry, where discreteness arises on the Minkowski manifold as a result of a choice of metric, but not as a result of transformation of the Minkowski point set into lattice. Unfortunately, the mathematical community knows slightly geometry, considered as a science on shape of geometrical objects and their mutual disposition. Mathematical community considered geometry as a logical construction, although not all space-time geometries can be considered as a logical construction.

Metrical approach to geometry admits one to construct non-Riemannian geometries, using coordinateless description.. See details in my paper “Metrical conception of the space-time geometry” Int. J. Theor, Phys.54, iss.1, 334-339, (2014), Electronic version in

http://gasdyn-ipm.ipmnet.ru/~rylov/mcstg2e.pdf .The paper is short (about 5 pages)

Dear Charles,

You wrote: “I also learned Euclid before I learned coordinate geometry”. However, there is the third representation of Euclidean geometry, which is not known to you, apparently. It exists about twenty years. I name this representation as \sigma-representation. The \sigma-representation is a monistic representation, where all geometric statements are expressed via one quantity world function \sigma =1/2d2, where d is the Euclidean metric.

Ususlly, constructing a generalized geometry, one modify the Euclidean axioms and construct the generalized geometry, using the method of the Euclidean geometry construction. This method contains proofs of numerous theorems. Learning Euclidean geometry in the middle school, one proves theorems, and most persons think, that any geometry consists of theorems and of their proofs. In reality, proof of theorems is only a method of the Euclidean geometry construction, but not the geometry itself. Any geometry is a set of statements on properties of geometrical objects. These statements describe, how to construct geometrical objects.

Constructing a generalized geometry, there is no necessity to repeat proofs of Euclidean theorems at modified axioms. It is sufficient to transform existing Euclidean rules of the geometrical objects construction. If. one uses monistic \sigma-representation, when these rules are expressed in terms of one quantity (world function), it is sufficient to replace the Euclidean world function by the world function of the generalized geometry G. One obtains the rules of the geometrical object in the generalized geometry  G. No theorems are not to be proved. All necessary theorems have been proved in the Euclidean geometry. It is very important, that the geometry representation be monistic. If the representation of the Euclidean geometry is not monistic and there are several fundamental quantities, then modifcation these quantities must be concerted, in order that the modified geometry G be consistent. In the monistic representation of the Euclidean geometry the problem of agreement between the modified basic quantities (axioms) is absent. One can use arbitrary world functions.

For instance, a segment of straight line T[AB] between two point A and B is described bythe relation

T[AB]={R|dE(A,R)+ dE(B,R) - dE(A,B)=0}

The straight line has the same form in any other generalized geometry G. One needs only to replace the Euclidean metric dE by the metric d of G. Note, that the straight line segment is described by one equation. This equation describes, generally speaking, (n-1)-dimensional surface in n-dimensional space. The Euclidean segment T[AB] is always one-dimensional. This one-dimensionality is a corollary of the specific form of the Euclidean metric, which is not characteristic for other generalized geometries. Even in the Minkowski geometry the segment T[AB]is three-dimensional surface, if interval between points A and B is spacelike (d2(A,B)

Dear Charles,

You suggest, that description in terms of world function is not a geometry. I am not going to discuss the problem of terminology. I am physicist, and I am interesting in science, which describes properties of the space-time and properties of physical bodies in the space-time. Geometrical images of physical objects are geometrical objects. I consider this science as a geometry (more exactly, as “physical geometry”). Namely, this science is interesting for physicists. Mathematicians may treat the symplectic geometry as a geometry, although it has no relation to description of geometric objects. I consider the symplectic geometry as a mathematical geometry. I am not interesting in mathematical geometries.

 In general, I see a situation with geometry as follows. There are three representations of the Euclidean geometry: (1) Euclidean representation (E-representation), (2) representation on the basis of linear vector space (V-representation), and (3) \sigma-representation, when geometry is described completely in terms and only in terms of the world function. The first two representation are known to mathematicians, but the \sigma-representation is unknown, apparently. See details in “Different representations of Euclidean geo\smetry and their application to the space-time geometry” arXiv:0709.2755. The \sigma-representation admits one to construct maximal number of generalized geometries. They are constructed as a result of deformation of the Euclidean geometry (replacement of Euclidean world function by the world function of the generalized geometry). This deformation generates intransitivity of equality relation of vectors. As a result the generalized geometry becomes multivariant. It means, that vectors AB1, AB2,… are equivalent to some vector CD, but vectors AB1, AB2,…are not equivalent between themselves. In the \sigma-representation of Euclidean geometry all geometric definitions may be of two kinds: (1) general geometric, which are expressed in terms of world function, and (2) special Euclidean relations, which disintegrate at the Euclidean geometry deformation. In E-representation and in V-representation one cannot distinguish between the general geometric relations and special geometric relations.

In particular, in E-representation and in V-representation the dimension of a geometry is considered as a fundamental quantity, which is given in the beginning of the generalized geometry construction. In the \sigma-representation the dimension is a derivative quantity, which is constructed on the basis of special property of the Euclidean geometry. In many generalized geometries the dimension cannot be constructed, because the special properties of the geometry, which are needed for construction of dimension are not fulfilled. Nevertheless the generalized geometry exists without a definite dimension, because all geometrical objects can be constructed on the basis of the world function. Mathematicians, which know only E-representation and V-representation cannot imagine geometry without a definite dimension.

Dear Charles,

You say, that I am using language you simply do not understand. However, the question is very simple. What of two definition of the vector equality is true? My post was not devoted to the problem of terminology. Solution of the important cosmological problem (the problem of dark matter) depends on the answer to my question. Experiment shows, that one cannot detect a single tachyon. Then we have two different situations: (1) Let two vectors be equal, if their components are equal, then tachyons do not exist. (2) Let two vectors be equal, if their scalar product is equal to the product of their lengths, and their lengths are equal. Then tachyons may exist, and tachyon gas form the dark matter, which has been observed. See for details my paper: "Dynamic equations for tachyon gas", Int. J. Theor. Phys. 52, 133( 10), 3683- 3695, (2013) ,

Electronic version http://gasdyn-ipm.ipmnet.ru/~rylov/detg3e.pdf . Thus, my question is not an academic question. It is very important physical question, which can be tested experimentally. Problems in answer to my question arise, if we know Euclidean geometry insufficiently.

Dear Yuri,

Now I understand better the motivation of your original question. Allow me to recap the setting in order to make things clearer in my own head . At the risk at looking pedantic when recalling "silly" details of common knowledge, let me borrow (oh, very briefly) the mathematical language of "categories" , which encompasses your V- and E- representations language. To be concrete, fix the scalars to always be the real numbers, and consider all the vector spaces over these scalars.  Since these vector spaces are not mere sets but possess a "linear" structure, we must require the maps inside and between them to be linear maps, i.e. to respect the addition of vectors  and multiplication of vectors by scalars : h(v+v') = h(v) + h(v') and h(rv) = rh(v). This is the category of real vector spaces, which seems to be what you call V- representations. A map which respects a given structure is a "morphism". So the category of V-representations is made not only with objects (vector spaces), but also with morphisms (linear maps). Now you can put additional structures on your vector spaces, such as a scalar product . Call these new  categories E-representations and L- representations if you want.  

Coming back to your original question, you asked for a criterion for the nullity of a given vector in the L-category. Consider the following  criteria in a vector space V:

(V1) The vector v has null coordinates (this is your property (1))                           

(E1) e(v, v') = 0 for all v' in V

(E2) e(v,v) = 0    (note that e(v,v) is the square of the usual "length" of v)

It is known (and easily shown) that (E1) and (E2) are equivalent. It follows that both criteria are equivalent to your condition (2)

(L1) l(v,v') = 0 for all v'    

(L2) l(v,v) = 0

Note right away that (L1) and (L2) are not equivalent, as shown by the existence of the light cone. This non-equivalence also implies that the E- and L-categories are not "the same". 

 The characterization of the nullity of v is different according to the category where you consider it. As the V-category is the largest one, no wonder that the nullity of v is equivalent to (V1) in all the 3 categories. It is equivalent to (E1) and (E2) in the E-category, equivalent to (L1), but not to (L2), in the L-category. This is why I thought that, in your question as it was asked, the appearance of condition (2) was rather irrelevant (I didn't say "silly"). But now it seems that you have something more in mind, which I interpret as follows: starting from a "monistic" description of the E-category from a single "world function" sigma, you expect to extend this monistic approach to other geometries (*) (such as L ?) in view of applications to cosmology. But I have difficulties understanding  the properties (1) and (2) in your last post . As I said, the nullity of v is equivalent to criterion (V1) in the largest category considered, namely the V-representations. And criteria (E1) and (V1) are not equivalent in the Minkowski space-time. It follows logically that tachyons do not exist ??? Unless you use define differently what a vector is according to the representation/category in which you work ?

(*) It is true that etymologically, the term "geometry" contains a notion of "measure", which implies the existence of a "metric", or a "distance". But one of the axioms (or even "common sense" properties) of a distance  d  between two points A, B is that      d(A, B) = 0 if and only if A = B. This excludes the Minkowski space-time from geometry since the only natural distance there would be the square root of l(vector AB), vector AB). "Silly". Reasonably, one should abandon the etymological sense of the word  (which dates back to times when the only known geometry was the euclidian one) and define geometry as the "study of forms up to continuous deformations"(Poincaré's definition of "analysis situs", or "topology" as it is called nowadays)

Equality of two vectors on any manifold can be established if they can be brought together and compared. I understand that there is no ambiguity on how this is done if the manifold has a flat metric structure, as is the case here. But even in the flat case (or no metric at all)  there is an alternative method of transport: Lie transport. For this one needs a vector field, V, which I like to think of as a flow field,  such that if an integral curve joins the location of the two vectors. Then the vector field, V, can be used to transport one vector to the other and so compare.  Is this what you were asking?

Dear Thong nguyen quang,

I could not follow exactly your consideration, but I am to note, that you use the concept of linear vector space in your considerations. You may introduce the vector space in the Euclidean geometry, but, strictly speaking, you cannot do this in the Minkovskian geometry GM, because some vectors in GM have imaginary length (negative values of the world function \sigmaM). In reality, there is no necessity for introduction of linear vector space in GM. Geometric vector (g-vector) AB is defined as the ordered set of two points A and B, (but not as an element of vector space). The scalar product (AB.CD) of two g-vectors AB and CD is defined directly via the world function

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

This definition coincides with conventional definition in the Euclidean geometry (It can be tested easily). This definition does not contain a reference to the vector space. It can be expanded to any space-time geometry (in particular to GM). It is very important, because linear vector space can be introduced not in any space-time geometry. In particular, strictly speaking it cannot be introduced in GM. Your considerations are based on a possibility of the vector space introduction, and they cannot be applied to solution of the question of the two vectors equality.

Dear Peter,

It is useless to define equality of two vectors via more complicated concepts. It is useless to define the rules of arithmetic via rules of differential calculus. If equality of two vectors is defined outside the linear vector space, it is useless to define it via linear vector space, because in the linear vector space the linear operations on vectors are defined. Equality of vectors is one of these linear operations. It is useless to define the vector equality via transport of the vector, because this transport is defined in the linear vector space, or some other operation over vectors. If we want to use a monistic conception of a geometry, where all geometrical quantities are expressed in terms of the world function \sigma=1/2d2, where d is metric, it is the only way of definition is the definition of the scalar product (AB.CD) of two vectors AB and CD directly via the world function

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

In this case the coordinateless definition of two vector equality is defined only via world function (or via metric). This definition may be extended to any space-time geometry, even if one cannot introduce linear vector space in this geometry.

Dear Marcel,

I agree, that exact definition must be known in order to discuss problems with mathematicians.

But I am physicist. I needs only that the space-time geometry describes properties of the space-time correctly. I know, that all properties of the space-time are described by world function and only by world function. All concepts, that you have enumerated are derivative concepts. It is important, how they are expressed via world function (but not, which terms are used by mathematicians for them). Besides, mathematicians are interested in logical connection between all these concepts. However, they are not interested, how these concepts describe properties of the real spacetime. The logical connection between geometrical concepts is important, if you construct any geometry deducing it from axioms. These logical connection exists in the Euclidean geometry. The logical connection is violated after deformation, which is used to obtain space-time geometry from the Euclidean one. So I ignore logical connection between different concepts of the space-time geometry. The real space-time geometry is not known, and it is important to have a simple method for transition from one space-time geometry to another space-time geometry. Mathematicians prefer the difficult and complicated method of the geometry construction by deduction of a geometry from axioms. They are not interested in question to what extent their geometrical concepts corresponds to real properties of the space-time. Unfortunately, creating the universe, the God did not think about mathematicians and about their problems.

Dear Yuri,

Thank you for your answer. But it raised even more questions, which I list below. I apologize in advance for the points which I possibly misunderstood. 

1) About the « use of the concept of linear vector spaces », you say that « strictly speaking, there are no vectors in Minkowski geometry  », and even more generally, « linear vector spaces cannot be introduced in any space-time geometry ». I disagree with such a statement. Although we have not defined precisely what« geometry » means, let us stick to the physical example of the theory of (special or general) relativity and patiently (and pedantically) recap the setting :                                                                                                                          - in special relativity, surely you agree  (not only with myself, but also with Norbert Straumann, Raul Simon and all the textbooks) that the (flat) Minkowski space-time is actually a 4-dimensional real vector space, or an affine space if you fix an origin. The linear transformations (changes of referentials) are required to respect the Lorentz quadratic form , and you derive from this and from F = m.Gamma all the results of the einsteinian mechanics, just as you did for the newtonian mechanics when requiring only the linear transformations to respect the euclidian quadratic form  (the time cordinate  being then independent from the space coordinates). Speaking of the mathematical axioms defining a given geometry (flat here), you add that « the logical connection [between them] is violated after deformation, which is used to obtain space-time geometry from the euclidian one ». I cannot agree, unless « deformation » means that, at non relativistic speeds, time can be considered as independent from space and the newtonian mechanics are a good approximation of the einsteinian ones. Nevertheless the use of the word « deformation » is unfortunate, and besides, I don’t think that the axioms of the two geometries are « violated » in any way : the models are consistent, it is only the description they give of the physical world which is more or less efficient                                                                                                                                                - - in general relativity, because of the curvature due to gravitation, the modellized universe can no longer be a mere vector space (although v.s. are still lurking behind the scene, see below), it is a 4-dimensional differentiable « variety » or « manifold ». To compare vectors, Peter Bollada suggests a « Lie transport », viewed as a flow on the manifold to bring the vectors together, but you discard this, saying that « it is useless to define the vector equality via transport of the vectors, because this transport is defined in the linear vector space, or some other operation over vectors ». The only way I can understand this statement is that you think of the manifold as contained in an ambient vector (or affine) space, for example just as the Earth is contained in a real 3-dimensional v.s. But this is not necessarily so. Actually, a differentiable variety of dimension is defined as a collection of « local charts », more precisely local « diffeomorphisms », which means that the local neighbourhood of any point of the variety (i.e. the local surroundings of any « observer ») can be mapped in a bi-univocal (and differentiable) way onto a -dimensional vector space (i.e. the observer’s measures can be translated fully faithfully into coordinates). Of course the structural properties of the variety are translated into structural properties of the local charts. For example, in the universe of general relativity (probably what you call the Minkowski geometry ), the 4-dimensional v.s. which are the targets of  the local diffeomorphisms are just the lorentzian v.s. we introduced above. But one must insist on the unavoidable local point of view, because e.g.  the speed of light is a limit. There is no theoretical internal need for the-universe to be embedded in an ambient v.s. Besides, this global point of view would raise problems of logic (if the universe is everything, then it would be equal to the ambient v.s., which is impossible because of curvature) and  metaphysics (since you speak of God, only God can take a global look at the universe, and from the outside at that!)   

2) The point of view of differential geometry introduced above allows also to speak of vectors, not on the variety itself, but in the v.s. which come with the local charts. Recall that, locally speaking, on the variety or in the v.s., this « amounts to the same » because of the definition of the diffeomorphisms. I think this answers to your objection that « [my considerations on categories] are based on a possibility of the introduction of vector spaces, and they cannot be applied to solution of the question of the two vectors equality ». From your « monistic » point of view, you propose to introduce in any space-time geometry the world function sigma = (1/2) d^2  without saying what the function d is, but you call it a « metric », so I suppose it verifies the usual properties of a distance, namely d(A,B) positive, and null if and only if A = B, and d(A,B) + d(B,C) greater than d(A,C) (the so-called triagular inequality). You even think of the usual euclidian distance when you give your expression of the scalar product of two vectors. Here, if I understand well, you define a vector just to be a bi-point (A,B). I have two immediate objections :                                                                                                                                                                 - - if a vector is just a bi-point, this notion is valid for any set without any structure (say a Xmas pudding), so it can’t be interesting. Surely you must add supplementary structural conditions, as e.g.  the usual axioms of an affine space when the structure under study is a v.s. Or on a variety, but only locally, in using the local charts. But then we come back to my « categorical » considerations                                              - if  d is absolutely any distance, how can it be related to the specific space-time geometry (e.g. the Minkowski ) under study ? Somehow, there must be a « canonical » choice of d related to the structure which is considered. But again, my categorical considerations imply that this is not always possible

I have been lavishly too long. Please point out my possible misunderstandings/misinterpretations.

Dear Charles,

It is a misunderstanding, that I restrict myself by the Euclidean geometry GE. Monistic representation of GE, when all relations and geometric objects are expressed via the only fundamental quantity (world function \sigma), is only a starting point for construction of a space-time geometry. Replacing the Euclidean world function \sigmaE by the world function \sigma of a geometry G in all relations of GE , one obtains all relations of the space-time geometry G. Any new \sigma corresponds to a new space-time geometry G. Such a replacement is more simpler, than a deduction of geometric relations from a new system of axioms. Besides such a replacement admits one to construct multivariant and nonaxiomatizable geometries, which cannot be deduced from a system of axioms.

Dear Thong nguyen quang do,

In your post you consider different sides of deduction of Euclidean geometry GE from Euclidean axiomatics. However, deduction of GE from axiomatics has no relation to my question: “How does one define equality of two vectors in the space-time geometry of Minkowski?” If you know, how to deduce GE from the Euclidean axiomatics, it does not mean that you know the Euclidean geometry. GE can be deduced from different equivalent axiomatics, but GE is a set of many statements on properties of geometrical objects. Different axiomatics are only some subsets of these statements. In GE there are logical connections between all statements of GE. These logical connections describe the fact that any geometrical object of GE can be constructed of blocks (segments of the straight line). In other space-time geometries this logical connection is violated, and they cannot be constructed as a deduction from a some axiomatics. Such geometries are constructed as a deformation of GE. All statements of GE (but not only axioms!) are presented in terms of the Euclidean world function \sigmaE. Thereafter \sigmE is replaced by other world function \sigma of a geometry G. As a result one obtains all statements of the geometry G. The process of replacement \sigmaE by \sigma means a change of distances in GE. It means a deformation of GE. For a use of the deformation the geometry GE must be presented in a monistic form, when all geometric relations are presented in terms of only world function \sigmaE. Thus, by a deformation of GE one may construct such space-time.

geometries G, which cannot be constructed as a deduction from an axiomatics. Space-time geometries constructed by a deformation of GE are multivariant and noaxiomatizable. The geometry of Minkowski GM is constructed by means of a deformation of GE. Logical connection between timelike vectors preserves, whereas it violates for spacelike vectors. It is a reason, why one should use coordinateless equality, written in terms of world function \sigmaM.

In general, GE in terms of \sigmaE is perceived rather hardly. It is better to see original paper

“Metrical conception of the space-time geometry” Int. J. Theor, Phys.54, iss.1, 334-339, (2014), Electronic version: http://gasdyn-ipm.ipmnet.ru/~rylov/mcstg2e.pdf

The problem with Minkowski Space is that it contains an error. You have the future light cone and you have the past light cone. The difference between the two is a measurement. If it is a measurement, then it can only be certain to planck's constant divided by 2. So you have a present rectangle, not a present plane. There is a worse problem than that, but that is part of a paper the I am currently writing. I will get back to you on that, as soon as I finish the idea in print. The good news is that one can successfully wed Quantum Mechanics to Minkowski Space, but I can't tell you how, at least until I get the print out.