# Data Geometry 6. Star Graph

This is the final article in the series on di- and bi-coordinates. Consider a graph of simple structure and use it for a little research. We use many integers as data - this is a convenient field for demonstrating ideas.

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

We formulate the problem. It is necessary to determine the topology of the graph for which: 1) all vertices are the same and 2) the general connection 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 rarely found in nature, such a graph does not satisfy the 2nd condition.

On the other hand, the chain is minimally connected (2nd condition) - each node is connected only with its neighbors, but the 1st condition is not fulfilled here. Which neighbors do the nodes need to be connected with if they are all the same?

The answer satisfying the conditions of the problem is a star graph. In such a graph, all given vertices are not interconnected, but are connected with a certain central node. Which node is selected as the center? If such a node is not visible in the set, then you should add it to the set as a kind of virtual node in order to ensure the minimum connectivity of the given vertices. We can say that the central node is the set itself, into which the elements enter. The star graph satisfies the conditions. Its connectivity is minimal, all vertices (except the central one) are equivalent.

As you know, any composite number is decomposed into a product of primes. Then the set of primes will determine the base set. Prime numbers are not related.

A unit stands out among integers - 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 CPDV). Thus, all primes are associated with a unit.

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

norms of the basic elements (here, prime numbers) are zero (the elements are local). That is, the values of the gramian are already determined, because:

$$

For a full gramian, add the normal to the base set$$. In our case, the basis consists of elements, therefore, the products of the elements and the normals are 1. As a result, we obtain the form of the graph-star gramian for the prime number basis:

\ begin {array} {c | ccc with ccc}

Gm & * & 1 & 2 & 3 & 5 & 7 & 11 & ... \\

\ hline

* & & 1 & 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 & -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

Inverting the Gramian, we get the Laplacian: $$

\ 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 & -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 - norm of the orthocenter basis$$depends on the dimensionality of space $$, which here is equal to the number of rays of the star: $$.

Minor laplacian$$describes the connectivity structure of the basis of space. We see that all the vertices are connected only with unity, that is, the topology is indeed a

The Laplacian and the Gramian of the basis are the metric tensors of space. Now let's turn to the coordinates of the elements.

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

$$

Decomposition coefficients $$Are the bi-coordinates of the element.

The decomposition coefficients for primes are known. For example, the number 12 is two deuces and one triple ($$) I.e$$.

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

In addition to primes, a basis and a normal are also present in the basis. We isolate them explicitly from expansion (6.2):

$$

Here $$means the set of prime numbers of the basis.

To determine two unknown coefficients (orbitals of a number$$ and bi components at unit $$) we will use two requirements.

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

$$

Here $$- the sum of the coefficients for prime numbers.

Since for integers the coefficients$$positive, then the coefficient at unity will be either zero or negative.

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

$$

Where after disclosing the amount $$ taking into account the properties of scalar products of basis elements $$ we get a simple relation:

$$

$$ - the sum of the squared coefficients for primes. $$, $$Are aggregates of a 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 primaries is zero.

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

As an example, we calculate the bi-coordinates of 6-ki.

Its simple factorization has the form:$$. That is, there are two unit coefficients for a two and a three. Then the sum of the coefficients is$$, and the sum of the squares is $$.

The remaining bi-components can now be obtained from formulas (6.4), (6.5).

$$.

We see that the orbit of 6 is zero. Type of decomposition:

$$

The values of the aggregates and their difference 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 & 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}

Bi-coordinates of the first (compound) 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 & -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 - the elements of the basis, in the column names - composite numbers.

Normal decomposition coefficient$$ Is the orbital of a number $$. One does not have to be a specialist in number theory to understand that the orbital must play a significant role in describing the properties of numbers.

The di-coordinates of a number, by definition, are a set of scalar products of a number and elements of the basis, here are the sets of primes, supplemented by a normal and a unit.

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

Find the value of the di-component at unity. Multiplying (6.3) by$$, we obtain, taking into account (6.4) and (6.5):

$$

That is, the di-component corresponding to unity (here, to the center of the star graph) reflects the sum of the squares of the bi-components corresponding to primes (here, to the rays of the star).

Similarly, we can express the remaining di-components (corresponding to primes):

$$

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

$$

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

$$

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

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

$$

Matrix elements reflect how far the numbers are from each other in our space.

Series completed. We told everything that we planned (and even a little more). The basic concepts and relations related to coordinate systems on the basis of elements are given. The resulting expressions are applicable to any linear spaces, to any data.

**Table of contents**

### Minimal connectivity graph

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

We formulate the problem. It is necessary to determine the topology of the graph for which: 1) all vertices are the same and 2) the general connection 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 rarely found in nature, such a graph does not satisfy the 2nd condition.

On the other hand, the chain is minimally connected (2nd condition) - each node is connected only with its neighbors, but the 1st condition is not fulfilled here. Which neighbors do the nodes need to be connected with if they are all the same?

The answer satisfying the conditions of the problem is a star graph. In such a graph, all given vertices are not interconnected, but are connected with a certain central node. Which node is selected as the center? If such a node is not visible in the set, then you should add it to the set as a kind of virtual node in order to ensure the minimum connectivity of the given vertices. We can say that the central node is the set itself, into which the elements enter. The 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 is decomposed into a product of primes. Then the set of primes will determine the base set. Prime numbers are not related.

A unit stands out among integers - 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 CPDV). Thus, all primes are associated with a unit.

### Gramian star base

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

norms of the basic elements (here, prime numbers) are zero (the elements are local). That is, the values of the gramian are already determined, because:

$$

For a full gramian, add the normal to the base set$$. In our case, the basis consists of elements, therefore, the products of the elements and the normals are 1. As a result, we obtain the form of the graph-star gramian for the prime number basis:

\ begin {array} {c | ccc with ccc}

Gm & * & 1 & 2 & 3 & 5 & 7 & 11 & ... \\

\ hline

* & & 1 & 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 & -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

### Laplacian stars

Inverting the Gramian, we get the Laplacian: $$

\ 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 & -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 - norm of the orthocenter basis$$depends on the dimensionality of space $$, which here is equal to the number of rays of the star: $$.

Minor laplacian$$describes the connectivity structure of the basis of space. We see that all the vertices are connected only with unity, that is, the topology is indeed a

*star*.The Laplacian and the Gramian of the basis are the metric tensors of space. Now let's turn to the coordinates of the elements.

### Bi-coordinates of elements

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

$$

Decomposition coefficients $$Are the bi-coordinates of the element.

The decomposition coefficients for primes are known. For example, the number 12 is two deuces and one triple ($$) I.e$$.

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

In addition to primes, a basis and a normal are also present in the basis. We isolate them explicitly from expansion (6.2):

$$

Here $$means the set of prime numbers of the basis.

To determine two unknown coefficients (orbitals of a number$$ and bi components at unit $$) we will use two requirements.

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

$$

Here $$- the sum of the coefficients for prime numbers.

Since for integers the coefficients$$positive, then the coefficient at unity will be either zero or negative.

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

$$

Where after disclosing the amount $$ taking into account the properties of scalar products of basis elements $$ we get a simple relation:

$$

$$ - the sum of the squared coefficients for primes. $$, $$Are aggregates of a 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 primaries is zero.

### Examples of decomposing numbers into components

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

As an example, we calculate the bi-coordinates of 6-ki.

Its simple factorization has the form:$$. That is, there are two unit coefficients for a two and a three. Then the sum of the coefficients is$$, and the sum of the squares is $$.

The remaining bi-components can now be obtained from formulas (6.4), (6.5).

$$.

We see that the orbit of 6 is zero. Type of decomposition:

$$

The values of the aggregates and their difference 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 & 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}

Bi-coordinates of the first (compound) 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 & -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 - the elements of the basis, in the column names - composite numbers.

Normal decomposition coefficient$$ Is the orbital of a number $$. One does not have to be a specialist in number theory to understand that the orbital must play a significant role in describing the properties of numbers.

### Di-coordinates of numbers

The di-coordinates of a number, by definition, are a set of scalar products of a number and elements of the basis, here are the sets of primes, supplemented by a normal and a unit.

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

Find the value of the di-component at unity. Multiplying (6.3) by$$, we obtain, taking into account (6.4) and (6.5):

$$

That is, the di-component corresponding to unity (here, to the center of the star graph) reflects the sum of the squares of the bi-components corresponding to primes (here, to the rays of the star).

Similarly, we can express the remaining di-components (corresponding to primes):

$$

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

$$

### Distances between numbers

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

$$

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

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

$$

**View of the distance matrix for the first 11 numbers**

\ begin {array} {c | ccc with ccccc with c}

Q & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\

\ hline

1 & - & 1 & 1 & 4 & 1 & 2 & 1 & 9 & 4 & 2 & 1 \\

2 & 1 & - & 2 & 1 & 2 & 1 & 2 & 4 & 5 & 1 & 2 \\

3 & 1 & 2 & - & 5 & 2 & 1 & 2 & 10 & 1 & 3 & 2 \\

4 & 4 & 1 & 5 & - & 5 & 2 & 5 & 1 & 8 & 2 & 5 \\

5 & 1 & 2 & 2 & 5 & - & 3 & 2 & 10 & 5 & 1 & 2 \\

6 & 2 & 1 & 1 & 2 & 3 & - & 3 & 5 & 2 & 2 & 3 \\

7 & 1 & 2 & 2 & 5 & 2 &

Q & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\

\ hline

1 & - & 1 & 1 & 4 & 1 & 2 & 1 & 9 & 4 & 2 & 1 \\

2 & 1 & - & 2 & 1 & 2 & 1 & 2 & 4 & 5 & 1 & 2 \\

3 & 1 & 2 & - & 5 & 2 & 1 & 2 & 10 & 1 & 3 & 2 \\

4 & 4 & 1 & 5 & - & 5 & 2 & 5 & 1 & 8 & 2 & 5 \\

5 & 1 & 2 & 2 & 5 & - & 3 & 2 & 10 & 5 & 1 & 2 \\

6 & 2 & 1 & 1 & 2 & 3 & - & 3 & 5 & 2 & 2 & 3 \\

7 & 1 & 2 & 2 & 5 & 2 &

Matrix elements reflect how far the numbers are from each other in our space.

Series completed. We told everything that we planned (and even a little more). The basic concepts and relations related to coordinate systems on the basis of elements are given. The resulting expressions are applicable to any linear spaces, to any data.