ROUGH DRAFT authorea.com/15541

# Abstract

Investigate what I percieve to be a Hilbert Space of numbers. I use the concept of a number unit, in analogy to length, area, volume etc. Such that a prime has dimensions of $$p$$. Compunds and partitions are visualised.

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# Sum of Primes

Some scope for things such as $2p + 2p = 4p^2 \\ \\ \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \end{bmatrix} \to \begin{bmatrix} \begin{bmatrix} 1 & 0 \end{bmatrix} \\ 0 \end{bmatrix} \to \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$

But this system appears to break down in many circumstances. For this particular example, one could continue to add $$2p$$ and generate $$2^np^n$$ like structures of rank $$n$$ quite comfortably.

Obviously this is the overlap of product and multiplication, as this is still the dyadic product. something else is needed, some other operation.

# Higher Compounds

Now a matrix has two sides, [?]. But when the array is extended to a cube, there will be 6 sides.

If a number has $$3$$ prime factors $$\{A,B,C\}$$, then there are the six selections, as permutations of a pair from this set. Thus, where the simple rules of left and right operations worked before, now there will be more rules. Let us draw up a notation to deal with this formally.

For a number $$N$$, which is an infinite cubic array, $$N_{ijk}$$, whose construction is $$p_1\otimes p_2 \otimes p_3$$. We will have, for matrices $$M_{ij}$$ which represent constructions $$q_1 \otimes q_2$$, the following operations $[A,B]\odot^1[A,B,C]=[C] \\ [A,C]\odot^2[A,B,C]=[B] \\ [B,C]\odot^3[A,B,C]=[A] \\ [B,A]\odot^4[A,B,C]=[C] \\ [C,A]\odot^5[A,B,C]=[B] \\ [C,B]\odot^6[A,B,C]=[A]$

As a test, there can be $$2p\otimes 3p\otimes 5p$$, and the matrix $$2p \otimes 3p$$, Then by summing over two indicies, there will be left the vector representing $$5p$$.

The only element that is one in the cube will be element $$N_{123}$$. The only element that is one in the matrix will be $$M_{12}$$. Thus we must have the product in the form

$\odot^1 := \sum_{i}\sum_{j} M_{ij}N_{ijk}=v_{k}$

and as a result, $$v=[0,0,1]$$.
For the matrix $$2p \otimes 5p$$, the only one in the matrix lies in element $$M_{13}$$. Thus

$\odot^2 := \sum_{i}\sum_{k} M_{ik}N_{ijk}=v_{j}$

and as a result, $$v=[0,1,0]$$.
For the matrix $$3p \otimes 5p$$, the only one in the matrix lies in element $$M_{23}$$. Thus

$\odot^3 := \sum_{j}\sum_{k} M_{jk}N_{ijk}=v_{i}$

and as a result, $$v=[1,0,0]$$.
For all others we take the transpose of the matrix

$\odot^4 := \sum_{i}\sum_{j} M_{ji}N_{ijk}=v_{k} \\ \odot^5 := \sum_{i}\sum_{k} M_{ki}N_{ijk}=v_{j} \\ \odot^6 := \sum_{j}\sum_{k} M_{kj}N_{ijk}=v_{i}$