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Let suppose that we focus our scientific curiosity on the concept of “division” and that we must explain it to someone else. We would probably start with a simple question like: “What do you find when you divide six by three?” Most of the spontaneous answers will be: 2. But why? And the next answer would certainly come: “It’s natural!”. And, as a matter of facts, 2, 3 and 6 are natural numbers with which we are seemingly customized to count.
Unfortunately for those who are looking for the simplicity, and as one teaches it in schools, there are a lot of other categories of numbers: relative natural: Z = {…, 2, 1, 0, 1, 2, …}, fractional: Q ={…5/17, …, 9/253, …}, real: R, complex: C, etc. Therefore, the next question is: “Why did nobody give the following answers to my question: 6 = ½. 3 + 9/2 or 6 = 2,3. 3  0,9?” These answers would also have been correct because, my fault, I haven’t precise an important parameter in my initial question: “What do you find when you divide six by 2 and your answer must be a fractional number (respectively a real number, etc.)?” So: I should have first given information on the set in which the solution(s) must be.
What do we state next? The answers are, most of the time, far to be unique; and what else? Any correct answer is not only made of one number, but of a pair of numbers which we may generically label (the main result, the remaining part).
The trivial answer accompanying the answer when we stay in N = {0, 1, ….} is misleading our thoughts on the topic because it gives us the sensation that all responses contain only one number. The correct answer to my initial and quasinaïve question when the solutions must be in N should have been the pair: (2, 0). It is just because there is no remaining part that we didn’t realize the duality of the answer.
These basic, but in some way fundamental elements of a discussion about the concept of division, can now be extrapolated. We may ask if we can divide six by a vector, for example? Or we can consider any vector and ask if we can decompose it with a matrix? These interrogations would slowly drive the discussion into the direction of the concept of torsion or an abstract generalization of it.
These first paragraphs, here, are only a short presentation of what is waiting for you on this website if you want to deepen the mathematics illustrating these preliminaries; they are the essence of the theory of the (E) question (alias TEQ).
© by Thierry PERIAT: The Theory of the (E) question – Semantic
Items 
Peculiar terms 
Details Comments 

Products and their extensions 
Tensor product 
See any good book or, as first help: the extern link on Wikipedia – GB The operator is denoted Ä(…, …) 

Projectile 
First argument: Ä(Projectile, …) 

Target 
Second argument: Ä(…, Target) 

Cube 


Symmetric


Antisymmetric


Reduced


Antireduced


Symmetric and reduced 

Antisymmetric and antireduced 

Null 

Hypercube 
A hypercube is a generalization of the concept of cube to a space with a physical dimension greater than three. 

Deformed tensor product 
A deformed tensor product is a classical tensor product that has been deformed by a cube


Deformed exterior product 


Deformed Lie product 
A deformed Lie product is a deformed exterior product built on an antisymmetric cube 

Elements of a decomposition 
Intrinsic ingredients 


Main part 
([P], …) is the main part in a decomposition ([P], z). This is an element of M(D, K) 

Residual part 
(…, z) is the residual part in a decomposition ([P], z); this is an element in E(D, K) 

Trivial 
A decomposition is said to be trivial when: Ä_{A}(a, b) > = [P]. b > + 0 > 

Nontrivial 
A decomposition is nontrivial when its residual part doesn’t vanish. 

Methods of decomposition 
Intrinsic 
An intrinsic method of decomposition is a mathematical method allowing the discovery of one or several pair(s) ([P], z) with the help of intrinsic ingredients only. Up to now, I have only done in in a threedimensional context for deformed Lie products. 

Extrinsic 
An extrinsic method of decomposition is any mathematical method offering an answer to the (E) question with the help of ingredients which are not only intrinsic to the question. 

Russian dolls 
The Russian dolls method is inspired by the wellknown traditional objects and describes any procedure allowing the discovery of decompositions when the (E) question is asked in E(D + 1, K) but has been answered in E(D, K). 
© Thierry PERIAT
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