# The biquaternions and friends

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1. Dec 12, 2015

### EinsteinKreuz

So upon reading the wikipedia entry about the biquaternions I noticed that this algebra has several interesting subalgebras:

1. The split-complex numbers of the form {σ = x+y(hi)| ∀(x,y)∈ ℝ} which have the norm σ⋅σ* = (x2-y2).

2. The tessarines which can be written as {α + βj | ∀(α,β,)∈ℂ1 & j2 = -1}

3. The coquaternions whose bases form the dihedral group D4 and are define as the Span{1, i, (hj), (hk)}

But there is a 4th subalgebra that is somewhat similar to the coquaternions. And its elements can be defined as {g = a + bi + ch + dhi | (a,b,c,d)∈ℝ}. Now of course i2 = h2 = -1 and hi2 =
h2i2 = (-1)2 = +1.

But since I'm not sure how to add a grid for the Cayley table I'll also wrote down the other relational equations:

h⋅i = +(hi)
i⋅h = -(hi)
(hi)⋅i = -h
i⋅(hi) = +h
h
⋅(hi) = -i
(hi)⋅i = +i

Using these rules it can be shown that { g | g ∈ Span[1,i,h,(hi)]} is closed under products and if we define g* = a - bi - ch - dhi, then g⋅g* = a2+b2+c2-d2 = -ds2 where ds2 is the Minkowski metric.

So does this subalgebra have an official name and could it's elements be used as operators to describe the Lorentz transformation?

2. Dec 13, 2015

### Samy_A

A question, as I'm not familiar with Biquaternions:
How did you get i⋅h = -(hi)?

3. Dec 13, 2015

### EinsteinKreuz

Ahhhh...good point! The article states that h⋅i = i⋅h.
Now after tinkering with my original rule I realize that if the product of h with i,j, or k is associative(like the quaternions) then i⋅h = -(hi) → (hi)2 = -1.
So to generate an algebra where i⋅h = -(hi) & (hi)2 = h2i2 = +1, then the product of h with i,j, or k must be non-associative.

Last edited: Dec 13, 2015