At metric approach the proper Euclidean geometry can be described by Euclidean metric completely. In particular, linear dependence of n vectors AB1,AB2,...ABn can be formulated in terms of only metric. How? If one cannot answer this question, it means, that the Euclidean geometry is known unsufficiently.
You can't, but I don't see why this means Euclidean geometry is insufficiently known, and one can define a vector space such that it is a metric space using the inner product or orthogonal projections and a distance function d(x,y)= ||x-y|| where || || denotes the norm (alternatively, for any finitely generated inner product space and any two subspaces W, Y of that space, define d(W,Y) by the orthogonal projections ||pw-py|| of the inner product space onto W and Y, respectively).
Dear Andrew, in your answer you use the following geometrical concepts: (1) orthogonal projections, (2) distance, (3) difference x-y, (4) inner product space, (5) two subspaces W, Y. All these concepts can be defined in the Euclidean geometry. But how to determine these concepts in a metric geometry, which does not coincide with the Euclidean geometry? The answer is very simple. One needs to express these concepts via only the Euclidean metric dE. Then transiting to metric geometry G with the metric d, it is sufficient to replace dE by d and one obtains expression of these concepts in the geometry G
Dear Andrew and Robert,
The usual definition of linear dependence is produced via linear operations in the linear vector space. It seems, that linear dependence describes some properties of linear vector space and linear operations (sum of vector, multiplying vectors by a number) in this space. However, it is not so. Note that a n-dimensional plane Sn is described by n-1 vectors P0P1, P0P2,… P0Pn-1 and by the vector P0R, where R is the running point, describing the points of Sn , and all vectors P0P1, P0P2,…
P0Pn--1,P0R are linear dependent, whereas n-1 vectors P0P1, P0P2,… P0Pn-1 are not linear dependent. It is a geometrical definition of such a geometrical object as n-dimensional plane Sn. This definition does not contain a reference to coordinate system, or to linear vector space. It means that the linear dependence is a geometrical property, but not a property of the geometry description in terms of coordinates. A volume of Sn, determined by n vectors P0P1, P0P2,… P0Pn-1,P0R is equal to zero. This volume is proportional to Fn=det||(P0Pk.P0Pl)||, k,l =1,2,…n, P0=R. Here (P0Pk.P0Pl) is the scalar product of vectors P0Pk, P0Pl. This determinant is known as the Gram’s determinant. Thus, n vectors P0P1, P0P2,… P0Pn are linear dependent, if and only if Fn=det||(P0Pk.P0Pl)||=0, k,l =1,2,…n. The scalar product can be expessed via metric in the form
(P0Pk.P0Pl) =1/2( |P0Pl|2 +( |PkP0l|2 -( |P0P0|2 -( |PkPl|2 ).
Thus, the Gram’s determinant can be expressed in terms of only metric. The complete information on a geometry supposed, that one can distinguish between the geometry in itself and the geometry description in terms of coordinates.
Since there are isometries of Euclidean space that do not preserve linear independence this can not be done.
Dear Timothy,
What do you mean, using the term isometry? This term has several meanings.
The vectors AB_1, ... AB_n in R^n are linearly dependent if and only if the volume of the n-simplex AB_1B_2...B_n vanishes. This volume can be expressed in terms of the edge lengths of the simplex only, which gives an equation that you are looking for.
The formula for the volume in terms of edge lengths uses the Cayley-Menger determinant, which can be googled. In the special case of 2 dimensions this turns into the Heron formula for the area of the triangle.
Dear Ivan,
Your answer is rather close to truth. However, the question needs expression of the definition only in term of metric. Your answer contains the metric (lengths of edges), but not only metric. The Cayley-Menger determinant contains a reference to the dimension of the Euclidean space. To answer the question, one needs to express the Euclidean space dimension via metric. It is the problem, which needs a solution.
Dear Peter,
Why do you think, that there is no coordinateless description of a geometry? By the way, I learned the Euclidean geometry in the middle school, using textbook by Kiselev. This book did not contain the term “coordinate” or “coordinate system”. Nevertheless, it was axiomatic presentation of the Euclidean geometry. When I went to university I was surprised, why a geometry is described in some coordinate system. I should like to note, that the coordinateless description of a geometry is such a presentation, which is most suitable for physicists. Unfortunately, most mathematicians do not accept such an approach. My question is a test for mathematicians.
Dear Yuri,
the Cayley-Menger determinant contains no reference to the dimension of the space. With its help one can check whether, say, given 4 points span a non-degenerate 3-dimensional simplex or not, even if the points lie in a higher-dimensional space.
Dear Peter,
As far as I understood from your post, you think, that coordinateless description of a geometry is impossible. However, in the middle school I leaned the Euclidean geometry, using textbook by Kiselev. This textbook did not contain such terms as “coordinate” and “coordinate system”. However, the geometry was presented properly. From the intuitive viewpoint a geometry is described completely, if the distance is given for any pair of points. Unfortunately, this approach has not been realized formally. I succeeded to realize description of a geometry in terms of metric and only of metric. Unfortunately, mathematicians do not accept this approach. Putting my questions, I try to understand, what disliked by mathematicians. In particular, if a geometry (Euclidean) is described completely by metric, then the dimension is to be expressed via metric only. I put the question: How can one express the dimension of the n-dimensional Euclidean space via the Euclidean metric only?. Nobody tried to answer this question.
Then I put a more simple question, which is answered by you in your post: " How can one express the dimension of the n-dimensional Euclidean space via the Euclidean metric only?"
As far as dimension n is the maximal number of linear independent vectors, then answer to this question is an answer to the first half of the main question about dimension n as derivative concept of metric.
Dear Ivan,
You are right the Cayley-Menger determinant does not depend on dimension of a geometry. You can define the linear dependence of vector via corresponding simplex and via the Cayley-Menger determinant. However, physicists-theorists are interested in the space-time geometry, and they do not apply simplex in the space-time geometry. Instead they use the scalar product, which can be defined via triangle (which is a simplest simplex) by means of the relation
(AB.AC) = ½(|AC|2+|AB|2-|BC|2). Instead of the Cayley-Menger determinant one uses the nth order Gram determinant det||(ABi.ABk)||, i,k =1,2,…n. Iff this determinant vanishes, vectors AB1,AB2,…ABn are linear dependent. From physical viewpoint such formalism looks to be more conventional. See also my previous post.
It seems that linear dependance of vectors could be represented as a tensor.
I meant that linear dependence y=Ax of two vectors x and y is generally expressed by the second-rank tensor operator described by matrix A.
This tensor always expresses a linear relationship, but if it is not orthogonal (unitary), then the metric is broken (the distance between two points and the angle between two vectors are distorted). So the linear relationship is a broader concept than the metric.
Dear Gamlet,
It is something new. Look my paper: “Multivariance as immanent property of the space-time geometry”, Int. J. Theor. Phys 52, iss.11, 4074-4082, ( 2013), Electronic version
http://gasdyn-ipm.ipmnet.ru/~rylov/mimst1e.pdf
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