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The Cayley-Dickson construction generates higher-dimensional hyper complex numbers from lower-dimensional ones, producing algebras of dimension $2^n$.

I want to generate an algebra of dimension $2^{-1}$, but there is no CD construction to go from higher to lower dimensions.

The question is: how can hypercomplex numbers with dimension $2^{-1}$ be defined?

Here’s what I tried:

Hypercomplex numbers with dimension $D$ are, geometrically, points in a $D$ dimensional space. According to the CD construction, a number $c \in \mathbb{R}$ (with dimension $D=1$) should be expressible as a pair of numbers $(c_0,c_1)$ from the $2^{-1}$ dimensional space.

So as $2^{-1}$ dimensional space, I pick the set of real numbers, which when written in base $b$, have all zeros in odd-placed digits (counting the lowest significant entire digit as $0$ place).

This set of numbers creates a fractal, which, when scaled by $b^2$, forms $b$ copies of the original set. According to the Hausdorff dimension, they have dimension $D=\frac{\log(b)}{\log(b^2)}=\frac{1}{2}$.

I pick the binary base ($b=2$), because it is convenient to do numerical experiments.

Any number $n \in \mathbb{R}$ can be decomposed into its even and odd digits $n_0,n_1 \in \mathbb{R}^\frac{1}{2}$ with a binary $AND$ operation:

$n_0=n AND ...10101010101.0101...$ $n_1=n AND ...01010101010.1010...=(\frac{n}{2}) AND ...10101010101.0101...$

To recover the real number:

$n=(n_0,n_1)=n_0+2n_1$

The product rule in Cayley-Dickson construction is, for $ a \cdot b = c$, $(a, b, c \in \mathbb{R}),$

$$(a_0, a_1) \cdot (b_0, b_1) =(a_0 b_0−b_1 a_1^*, a_0^* b_1+b_0 a_1) = (c_0, c_1)$$

(wikipedia states this next version, but I use the former, as in most books) $$\require{cancel}\cancel{(a_0, a_1) \cdot (b_0, b_1) = (a_0 b_0 - b_1^* a_1, b_1 a_0 + a_1 b_0^*) = (c_0, c_1) }$$

then:

$c_0=a_0 b_0−b_1 a_1^*$

$c_1=a_0^* b_1+b_0 a_1$

The unknown concept is the definition of the conjugates $a_0^*, a_1^*$, so I numerically calculate the conjugates from the former equations:

$a_1^*=\frac{a_0 b_0−c_0}{b_1} $

$a_0^*=\frac{c_1-b_0 a_1}{b_1}$

But this produces inconsistent results, for example, when multiplying $a= [0,1,...,15]$ by $b=15$, the conjugate for $a_0=1$ may be $a_0^*=1,3,9,11...$

Table for the calculation of conjugates

so this doesn't works. Can you give a hint, a different way to define $\mathbb{R}^\frac{1}{2}$, or fix my attempt?

The conjugate of $n_i$ should produce the modulus of $n_i$, by $|n|=n_i n_i^*$, but I have no clue about what is the meaning of length in $\mathbb{R}^\frac{1}{2}$. Does the Pythagorean theorem extend to $\mathbb{R}^\frac{1}{2}$? How?

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    $\begingroup$ How do you even define sums? $\endgroup$
    – Yaakov Baruch
    Commented Jul 18 at 16:02
  • $\begingroup$ @Yaakov Baruch Is not clear to me. But $R$ should lose properties respect to $R^\frac{1}{2}$, and $n+n=2n$ doesn't stay in $R^\frac{1}{2}$, because it shifts bits, but maybe $R^\frac{1}{2}$, requires a pair of numbers in $R^\frac{1}{2}$, (same as complex numbers require a pair of reals). I'm lost. $\endgroup$
    – wepajakeg
    Commented Jul 18 at 16:08
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    $\begingroup$ It is very unclear to me what your goal even is. You also seem to be neglecting any issues arising from nonuniqueness of binary expansions. $\endgroup$
    – Wojowu
    Commented Jul 18 at 16:48
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    $\begingroup$ I'm sympathetic to this sort of question, but I think in its present form it's not answerable. It's not clear what a successful answer would actually constitute: what sort of demands do you place on a candidate $\mathbb{R}^{1\over 2}$? At present I don't think this question is really appropriate for MO. $\endgroup$
    – Noah Schweber
    Commented Jul 18 at 16:49
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    $\begingroup$ This question may be relevant: math.stackexchange.com/questions/1466820/… $\endgroup$
    – Anixx
    Commented Jul 21 at 21:38

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