Help wanted. Triyng to define the numbers ²R

[marracco]

I conjecture that tetration introduces numbers with fractional dimension \( \mathbb{R}^{\mathbb{R}} = ^2\mathbb{R} \).
For that purpose, I'm trying to extend Geometric algebra (Clifford algebra \( \mathcal{C}\ell_{n,0,0} \)), \( n \in \mathbb{N} \) to fractional dimension, defining \( r \)-blades (of grade \( r \in \mathbb{R} \)).
I attached a PDF explaining it. (it is a draft, not a finished text. And I'm not a mathematician).
In Clifford algebra, given two normal vectors (1-blades), their product should be anticommutative:
\[ f \cdot g = -g \cdot f \]
Each vector can be decomposed into \( s \) fracvectors with fractional dimension \( D = \frac{1}{s} \) (\( s \in \mathbb{N} \)).
\[ f = f_0 \cdot f_1 \cdot f_2 \cdot \ldots \cdot f_s \]
\[ g = g_0 \cdot g_1 \cdot g_2 \cdot \ldots \cdot g_k \]
The problem appears when trying to define the dot product. I don't know how to define the anticommutativity of \( f_i \) with \( g_j \).
\[ f \cdot g = -g \cdot f \]
but
\[ f \cdot g = f \cdot g_0 \cdot g_1 \cdot g_2 \cdot \ldots \cdot g_k \]
then \( f \cdot g = -g \cdot f \) should be preserved regardless of how many \( g_i \) there are (independently of the value of \( s \), \( k \), which determine the number of commutations of \( g_i \) with \( f_j \)).
Commuting
\[ f \cdot (g_0 \cdot g_1) = -g \cdot f \]
Should produce the same signs change than
\[ f \cdot (g_0 \cdot g_1 \cdot g_2) = -g \cdot f \]
regardless of if \( g \) is decomposed in 2 or 3 fractional dimensions
I can't figure how to achieve that. Any idea?

[marracco]

I think that my error was to treat fracvectors as vectors, but they should be treated as Basis n-blades. Basis n-blades are anticommutative only when n is odd.

I don't know how the concept of evenness generalizes to fractional numbers.

[MphLee]

Just an algebraic observation. I'm not sure if this can have any geometrical meaning but... have you considered the study of a graded algebraic structures (anticommutative ones graded ring for examples).

In all generality:
Let be a \(G\) group written with additive notation. Assume I have a set of things \(A\) that I want to perform multiplication on. In other words, I assume \(A\) is a monoid, using a multiplicative notation for it.
I also rquire that \(A\) is made by components of various degrees \(\lambda\in G\). This can be said by the two following equivalent conditions:

1) Exists a map \({\rm deg}:A\to G\) assigning to each \(f\in A\) an element of \(G\), its degree;
2) The whole set can be subdivided in subsets of homogeneous degree, \(A=\bigcup_{\lambda \in G}A_d\) where \(A_d\) is the set of elements with the same degree equals \(d\).

On top of this, we want that the operation respects the grading: \[ {\rm deg} (f\cdot g)={\rm deg}(f)+{\rm deg}(g)\] This amounts to asking that multiplying elements \(f\in A_d\) with elements \(f\in A_e\) sends us straight to \(d+e\)-graded elements \(f\cdot g\in A_{d+e}\).

At the end, if I had to define an anticommutation property, I'd like to have a way to generalize the concept of oddness to arbitrary degrees. 
This is usually done via the concept of "signed-group". We have to assign a "sign-rule" to \(G\) where "sign" can be understood as even/odd duality. This means asking an assignement \(\sigma:G\to \mathbb Z/2\mathbb Z\) such that \(\sigma(d+e)=\sigma(d)+\sigma(e) \) where:
 if \(\sigma (d)=0\) we will call thad degree "even", while \(d\) is "odd" if \(\sigma (d)=1\). As you can see, all the rules of the odd/even opposition are respected.

Once we have generalized "oddness" we have to let \( \mathbb Z/2\mathbb Z\) act on \(A\). Roughly as \[f\cdot g= (-1)^{\sigma{\rm deg}(f)\cdot \sigma{\rm deg}(g)}(g\cdot f)\] This can not make sense in full generality because we don't always know what means to evaluate \((-1)f\in A\). Maybe in your particular case we could assume that \(-1\in A_{0_G}\) has the degree corresponding to the zero element (identity) of \(G\).

I believe this can be made more general.