Every time I read your essay I seem to understand, it more and more.
I have a couple of questions about Equation 1
(aВІ + bВІ + cВІ + dВІ)uВІ = fВІuВІ
A quote page 3
"The meaning of Equation 1 is that in a 4-D geometry, if a right triangle is constructed from an integer number of basis lengths in each of the four dimensions (a, b, c, and d), then the hypotenuse (f) that traverses through the 4-D space will also have an integer number of the basis lengths."
In Equation 1
Clearly it is the area u2 that is common to both sides. Since its area's four squares when summed gives a transcendent "number" to both (a2 + b2 + c2 +d2) and the area f2. So if we have a 5-d hypotenuse cut from area f2 within our 4-d space-time based on a well understood four squares geometry with an invariant length "the square root of s2". How do you avoid this "cut" being s and not the area s2=(a2 + b2 + c2 +d2) which what equation 1 is saying. That the total area of (a2 + b2 + c2 +d2) times the common area u2 equals the common of area of u2 times the area f2. And ever body knows that (the sign of s2) times (the sign of area u2) equals (the sign of area u2) times (the sign of the area f2).
"Yes, I am treating an octonion as a bi-quaternion. That is what makes the multiplication table work.
The matrix multiplication is interesting. If the complex i commutes normally with the unit vectors, the coefficient matrix uses B. But if the complex i anti-commutes with the unit vectors, the coefficient matrix uses B*."
Bi-quaternions are just directed areas, that is, an area with a + and - sign. Clearly the matrix works because we have the invariant area ijk which then allows us to use octonian logic "based on + and - signs" which are attached to the bi-quaternions' areas. Hence in equation 1 the need of the 5-d hypotenuse cut from the area f2 in our 4-d world which is based on an invariant four squares space-time summation.
Your 5-d area's four squares summation gives us the length of 4-d hypotenuse "the invariant length of the square root s2" not the total invariant area summation. You have 4-d areas with a 5-d hypotenuse length of the four squares for the area f2. We have literally have a 5-d hypotenuse length within our 4-d space-time that any four square summation must obey. Since the area of u2 is the one common transcendental number that bridges both sides of Equation 1, while the 5-d hypotenuse is an invariant 4-d length that any summation must have available to have closure for the geometry of the area of f2.
A number (which is a perfect square) is the summation of four squares. If the area of f2 is n square metres d2ct, then the physical manifestation of that area is a n invariant unit lengths of dct in our 4-d space-time. Not an area. We have an area f2 on the right RHS, then on the LHS, equation 1 has a 5-d hypotenuse cut -- length c(metre) -- an invariant length that, by the 4-S theorem and equation 1 - each and every, any and, all - four square invariant summations must obey within our space-time.
Of course your multiplication matrices Eq 5.4 and Eq 5.5, clearly ties "i" with c(metre), clearly via the common area u2 which is on both sides, where we have units of the summation of transcendental i if we use the 4-S on both sides at once but using your multiplication rules A,B*,A,B* for - and + sign matrix Eq 5.3, which is, after all, a + and - sign summation using "octonian" logic directed bi-quaternion areas i.e. the column [C,D], using Eq 4.1 about a stationary "ijk" invariant the area f2, using f a length "the square root of the area of fВІ" to transverse the equal sign, Equation 1 uses a 5-d length, so cannot be associated 1-1 with a summation of four square labelled A,B,C,D thought of as a "a perfect number as an area" . It is the area uВІ that is, the common "four square summation" i.e. the perfect square, that spans the equal sign using the 4-S theorem on both sides of Equation 1. A number (which is a perfect square) is the summation of four squares). Your Eq 5.3 is a dance using A,B,C,D where A,B,C,D do integral steps on directed areas ALL on the geometry of the area of ijk. More simply the dance is with the directed areas which have a + or - sign, that is, i and * are not moving, i.e. they don't lead! It is --- i and * --- that are stationary and it is Eq 5.3 that moves areas that equal + or - throughout a basic multiplication table page 6, clearly Eq 5.3 only gives the square root of sВІ, a length not an area for how the multiplication table works in your matrices Eq 5.4 and Eq 5.5.
The full 4-S multiplication "of the areas on both sides of Equation 1" is:-
(the sign of the area (aВІ + bВІ + cВІ+dВІ)) times (the sign of the area uВІ on the LHS) = (the sign of the area uВІ on the RHS) times (the sign of the area fВІ).
You will find Eq 5.3 octonian area + and - logic uses only the "square roots for the area uВІ" on the LHS for the bi-quaternions areas plus and minus signs attachment. That is, it is the common area of the transcendent "number" (a summation of four squares) which transverses the equal sign in Eq 1. as perfect numbers). Not your A,B*,A,B*,-,+ matrix dance Eq 5.3. which is after all + and - sign summation using "octonian" logic directed bi-quaternion areas i.e. the column [C,D]; clearly uses Eq 4.1 a stationary "ijk" invariant the area f2.
More simply the area of f2 is ijk equals -1 and then we take the square root of the area of ijk that is в€љ-1 the imaginary unit. Clearly the full 4-S multiplication table for the "equal sign" invariant + and - unit count across the equal sign for Equation 1 is a transcendent dimensional process with "a unit of the square root of the area uВІ (see below)"; we will call the invariant unit of the times table a "sec"" for the area of the total summation of the area of the four squares of space-time. Then the 5-d hypotenuse cut would have a pure number a "transcendental" 5-d number c=i and it's "4-d length" of i(sec). The full 4-S sign multiplication times table used for how the LHS and RHS signs of the area u2 common area behave across the equal sign, are; same signs on the LHS and RHS give +ve while different signs on the RHS and LHS give -ve. Or the appearance of the bridge (common area) across the equal sign is in units of -- +i and -i -- that is, how we cross the equal sign using the area of u2 on the LHS and using the area of u2 on the RHS.
You said in my comments
You have some interesting ideas but they are very speculative. Essay contests such as this are a good place to present such ideas:-) yes you are right about that
I don't think you can set i=c or i=h but I do think you can construct something similar to the following:
PSI = exp(omega) = sqrt[1 - (v/c)^2] + (v/c)i
Then for v=c, PSI=i. I looked at your work instead, to see how you bridged with a common 5-d length (of the square root of f2) the areas on both sides of of the equal sign. Your method mixes lengths with areas across the equal sign. While in the full 4-S, it is the four sums of +i and -i that are the "invariant count" lengths of the area u2. The hypotenuse of the area geometry of f2 is an invariant 5-d length "f" which isn't an area on the LHS.
