Data geometry 6. Star graph

This is the final article in the series on di- and bi-coordinates. Consider the graph of the simplest structure and use it for a little research. As data we use a set of integers - this is a convenient field for the demonstration of ideas.

Minimal connectivity graph

Suppose we have a set of elements that look independent of each other, and can serve as vertices (reper) of a basis of a certain space. In order for a metric to be defined on this basis, the elements must be somehow interconnected. What should such a relationship look like so that all elements remain equivalent?

Formulate the problem. It is necessary to determine the topology of the graph for which: 1) all vertices are the same and 2) the total connectivity is minimal. The first condition is satisfied, for example, by a complete graph in which all vertices are connected with all. But such a connection is rare in nature, such a graph does not satisfy the 2nd condition.

On the other hand, the minimum connection (2nd condition) has a chain — each node is connected only with its neighbors, but the 1st condition is not satisfied here. With which neighbors exactly should nodes be connected, if they are all equivalent?
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The answer satisfying the conditions of the problem is the graph-star. In such a graph, all given vertices are not connected with each other, but are connected with a certain central node. Which node is selected as central? If such a node in the set is not visible, then it should be added to the set as a virtual node in order to ensure minimal connectivity of the specified vertices. We can say that the central node is the set itself, in which the elements are included. Star graph satisfies the conditions. Its connectivity is minimal, all vertices (except the central one) are equivalent.

The basis of the space of integers

As you know, any composite number decomposes into a product of simple ones. Then the set of prime numbers will determine the base set. Prime numbers are not related.

Among integers, one stands out - all numbers (even simple ones) are divided by one. We emphasize the special role of the unit by placing it in the center of the star graph (see KPDV). Thus, all prime numbers are associated with the unit.

Gramian Basis Stars

The structure of the graph-stars actually determined the metric of space. If we put the distance between the prime number and the center of the star (one) equal to 1, then the square of the distance between two prime numbers will be equal to 2.

The norms of the basic elements (here - prime numbers) are zero (the elements are local). That is, the Gramian values ​​are already defined, because:

$$$g (A, B) = -q (A, B) / 2 \ quad (6.1)$

For the full humane, add normal to the base set $$$\ mathbf {z}$ . In our case, the basis consists of elements, therefore the products of elements and the normal are equal to 1. As a result, we obtain the form of the Gramian of the star graph for the basis of prime numbers:
\ begin {array} {c | ccc with ccc}
Gm & * & 1 & 2 & 3 & 5 & 7 & 11 & ... \\
\ hline
* & & 1 & 1 & 1 & 1 & 1 & 1 & ... \\
1 & 1 & & -0.5 & -0.5 & -0.5 & -0.5 & -0.5 & ... \\
2 & 1 & -0.5 & & -1 & -1 & -1 & -1 & ... \\
3 & 1 & -0.5 & -1 & & -1 & -1 & -1 & ... \\
5 & ​​1 & -0.5 & -1 & -1 & & -1 & -1 & ... \\
7 & 1 & -0.5 & -1 & -1 & -1 & & -1 & ... \\
11 & 1 & -0.5 & -1 & -1 & -1 & -1 & & ... \\
... & ... & ... & ... & ... & ... & ... & ... & \\
\ end {array}

Laplacian stars

Turning the Gramian, we get the Laplacian: $$$Gm \ cdot Lm = I$
\ begin {array} {c | ccc with ccc}
Lm & * & 1 & 2 & 3 & 5 & 7 & 11 & ... \\
\ hline
* & m / 4 & 1-m / 2 & 0.5 & 0.5 & 0.5 & 0.5 & 0.5 & ...
1 & 1-m / 2 & m & -1 & -1 & -1 & -1 & -1 & ... \\
2 & 0.5 & -1 & 1 &&&&& \\
3 & 0.5 & -1 && 1 &&&& \\
5 & ​​0.5 & -1 &&& 1 &&& \\
7 & 0.5 & -1 &&&& 1 && \\
11 & 0.5 & -1 &&&&& 1 & \\
... & ... & ... &&&&&& ... \\
\ end {array}
In the corner - the norm of the basis sphere $$$rs$ depends on the dimension of space $$$m$ which here is equal to the number of rays of a star: $$$rs = m / 4$ .

Laplacian minor $$$L$ describes the structure of connectivity of a basis of space. We see that all vertices are associated only with unity, that is, the topology is indeed a star .

The Laplacian and the Gramian of a basis are metric tensors of space. Now we turn to the coordinates of the elements.

Element Coordinates

Composite number $$$X$ can be expressed as a linear combination of basis elements $$$A_i$ :

$$$X = bm ^ i (X) \ A_i \ quad (6.2)$

Decomposition coefficients $$$bm (x)$ - This is the bi-coordinates of the element.
Decomposition coefficients for prime numbers are known. For example, the number 12 is two twos and one triple ( $$$12 = 2 ^ 2 \ cdot 3 ^ 1$ ). I.e $$$b ^ 2 (e12) = 2, b ^ 3 (e12) = 1$ .

We use a prefix before numbers $$$e$ , to emphasize that these are not scalars, but elements of space.

In addition to primes, the unit also contains a unit and a normal. We select them explicitly from the decomposition (6.2):

$$$X = w (X) \ \ mathbf {z} + b ^ 1 (X) e1 + b ^ i (X) \ P_i \ quad (6.3)$

Here $$$P_i$ means a set of primes of a basis.

To determine two unknown coefficients (orbitals of $$$w (x)$ and bi components at unit $$$b ^ 1 (x)$ ) use two requirements.

The first follows from the fact that all numbers of a space are elements, that is, the sum of their components (or, equivalently, the product with a normal) must be equal to one. From here we get:

$$$b ^ 1 (X) = 1 - b ^ p \ 1_p = 1 - s1 (X) \ quad (6.4)$

Here $$$s1 (X) = b ^ p \ 1_p$ - the sum of the coefficients for prime numbers.
Since for integers the coefficients $$$b ^ p$ are positive, then the coefficient at one will be either zero or negative.

The second condition is the requirement of locality. All numbers must have a zero rate. Then

$$$n (X) = (w (X) \ \ mathbf {z} + b ^ i (X) \ A_i) ^ 2 = 2 w (X) + (b ^ i (X) \ A_i) ^ 2 = 0$

From where after disclosure of the amount $$$(b ^ i (X) \ A_i) ^ 2$ taking into account the properties of scalar products of basis elements $$$Gm$ we get a simple ratio:

$$$w (X) = (s1 (X) - s2 (X)) / 2 \ quad (6.5)$

$$$s2 (X) = (b ^ p) ^ 2 \ 1_p$ - the sum of squares of coefficients for prime numbers. $$$s1 (x)$ , $$$s2 (x)$ - these are aggregates of number.

It follows that if a number consists of only single components, then its orbital is zero (belongs to the sphere of the basis). Therefore, the orbital of all primoryals is zero.

Examples of decomposition of numbers into components

Knowing the coefficients of the expansion of a number into simple factors, we can determine its bi-coordinates in space - the coefficients of expansion (6.3).

As an example, calculate the bi-coordinates of 6-ki.
Its decomposition into prime factors has the form: $$$6 = 2 \ cdot 3$ . That is, there are two unit coefficients with two and a triple. Then the sum of the coefficients is $$$s1 (e6) = 2$ and the sum of squares is $$$s2 (e6) = 2$ .
The remaining bi-components can now be obtained from formulas (6.4), (6.5).
$$$b ^ 1 (e6) = 1 - s1 (e6) = -1, \ quad w (e6) = (2 - 2) / 2 = 0$ .
We see that the 6-k orbital is zero. Decomposition type:

$$$e6 = -e1 + e2 + e3$

The values ​​of the units and their differences for the first 17 numbers:
\ begin {array} {c | c}
a & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\
\ hline
s1 & 0 & 1 & 1 & 2 & 1 & 2 & 1 & 3 & 2 & 2 & 1 & 3 & 1 & 2 & 2 & 4 & 1 \\
s2 & 0 & 1 & 1 & 4 & 1 & 2 & 1 & 9 & 4 & 2 & 1 & 5 & 1 & 2 & 2 & 16 & 1 \\
s2 - s1 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 6 & 2 & 0 & 0 & 2 & 0 & 0 & 0 & 12 & 0 \\
\ end {array}
B-coordinates of the first (composite) numbers (table columns):
\ begin {array} {c | c}
Bm & 4 & 6 & 8 & 9 & 10 & 12 & 14 & 15 & 16 & 18 & 20 & 21 & 22 & 24 & 25 \\
\ hline
\ mathbf {z} & -1 & 0 & -3 & -1 & 0 & -1 & 0 & 0 & -6 & -1 & -1 & 0 & 0 & -3 & -1 \\
1 & -1 & -1 & -2 & -1 & -1 & -2 & -1 & -1 & -3 & -2 & -2 & -1 & -1 & -3 & -1 \\
\ hline
2 & 2 & 1 & 3 & & 1 & 2 & 1 & & 4 & 1 & 2 & & 1 & 3 & \\
3 & & 1 & & 2 & & 1 & & 1 & & 2 & & 1 & & 1 & \\
5 & ​​& & & & 1 & & & 1 & & & 1 & & & & 2 \\
7 & & & & & & & & 1 & & & & & & 1 & & & \
\ end {array}
In the row names there are elements of the basis, in the column names there are composite numbers.
The coefficient of decomposition at normal $$$\ mathbf {z}$ Is the orbital of the number $$$w$ . You do not need to be an expert in number theory to understand that the orbital should play a significant role in describing the properties of numbers.

Di-coordinates of numbers

Di-coordinates of a number, according to the definition, are a set of scalar products of a number and elements of a basis; here, sets of primes, supplemented by a normal and a unit.

Thus, to obtain the value of the component of the di-coordinates, it is sufficient (scalar) to multiply the linear decomposition of the number (6.3) into the corresponding element of the basis. Multiplication by the normal of space is not very interesting - it gives one, that is, the number is an element of space.

Find the value of the di-component at one. Multiplying (6.3) by $$$e1$ , we obtain in view of (6.4) and (6.5):

$$$dm_1 (x) = -s2 (x) / 2 \ quad (6.6)$

That is, the di-component corresponding to the unit (here - the center of the graph-star) reflects the sum of the squares of the bi-component corresponding to prime numbers (here - the rays of the star).

Similarly, the remaining di-components (corresponding to prime numbers) can be expressed:

$$$dm_p (X) = b ^ p (X) - (s2 (X) + 1) / 2 \ quad (6.7)$

We see that the di- and bi-components in the basis of the graph-star differ only by a scalar, which can be expressed through the di-component of the unit:

$$$b ^ p (X) - dm_p (X) = 1/2 - dm_1 (X) \ quad (6.7.1)$

Distances between numbers

The norm of the difference of elements corresponds to the distance between them. If coordinates are known, then the distance between the numbers can be calculated through the convolution of coordinates, since the norms of numbers are zero:

$$$q (X, Y) = - 2 g (X, Y) = -2 bm ^ i (X) \ dm_i (Y) \ quad (6.8)$

In this basis, the mutual coordinates can be based on the identity (6.7), that is, without using metric tensors.

Let us calculate the distance between 9th and 8th as an example:

$$$q (e8, e9) = - 2 g (e8, e9) = -2 bm ^ i (e8) \ dm_i (e9) = [-3, -2, 3] \ [-2, 4, 5] = 6 - 8 + 15 = $ 1$


Elements of the matrix reflect how far the numbers are from each other in our space.

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The series is complete. We told everything we had planned (and even a bit more). Given the basic concepts and relationships related to the coordinate systems on a basis of elements. The obtained expressions are applicable to any linear spaces, to any data.