Help wanted. Triyng to define the numbers ²R
#1
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?


Attached Files
.pdf   fracvectors.pdf (Size: 539.49 KB / Downloads: 24)
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#2
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.
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