The Spin Chronicles (Part 22): Total recall

 We continue our exploration from Part 20. But before diving in, let’s pause to reorient ourselves in the intellectual landscape we’ve been cultivating. It’s a landscape shaped as much by precision as by repetition, a principle philosophers and mathematicians alike have lauded. Aristotle reminds us, “We are what we repeatedly do. Excellence, then, is not an act, but a habit.” Schopenhauer, taking a more hedonistic turn, observes that “Happiness consists in frequent repetition of pleasure.”

Repetition, then, is both the sculptor and the rhythm of mastery. Think of the pianist perfecting a sonata, the athlete refining their form, or the mathematician—us—rewriting the foundations of Minkowski space to craft an associative algebra with unity. Each iteration brings us closer to elegance.

Happiness consists in frequent repetition of pleasure.

Speaking of pleasure, our joy in this series has been the art of complexification. We’ve taken the familiar terrain of Minkowski space and imbued it with richer structure, transforming it into a four-dimensional complex associative algebra. Each layer added is not merely an act of calculation but one of creative fulfillment—a reminder that repetition, when combined with curiosity, is the essence of discovery.

We write a general element of the algebra as p = (p₀,p), where p₀ is a complex scalar, and p is a complex vector with three complex components p₁,p₂,p₃. The multiplication formula is simple to remember:

(p₀,p)(q₀,q) = (p₀q₀ + p_·q,  p₀q + q₀p ₊ i p⨯q).                  (1)

On the other hand we can interpret this rule as coming from the real 8-dimensional Clifford algebra with a basis e₀=1, e₁, e₂, e₃, e₂e₃, e₃e₁, e₁e₂, e₁e₂e₃. When treated as complex space, we set i = e₁e₂e₃, and the complex basis has only four elements 1,e₁,e₂,e₃, since, for instance, e₁e₂=ie₃.

When thinking of Cl(V) as a complex algebra it is useful to remember two formulas. For u,v in V we have

uv + vu = 2u·v

uv - vu = 2i u⨯v.

Adding these two equation together we obtain:

uv = i u⨯v + u·v.

On the other hand, if Cl(V) is being considered as an 8-dimensional real vector space, we have the wedge product and the identity:

uv =  u∧v + u·v .

In calculations it is always necessary to decide in advance whether we perform the calculation in real 8 dimensions or complex 4 dimensions, otherwise contradictions may easily arise. 

These are useful formulas that are worth of being memorized. In the algebra we also have the anti-automorphism ν, which in the above notation, for x = (x₀,x), takes the simple form

ν(x)  = (x₀,-x).                                                                 (2)

It is clear from this last formula that  ν is complex linear i.e. ν(ix) = i ν(x). It readily follows from (1) and (2) that for any element x of the algebra we have that xν(x) is a scalar:

xν(x) = ν(x)x =  ((x₀)² - x², 0).

We have defined the group G as consisting of those elements of the algebra for which this scalar is equal to one:

G = {u: uν(u) = 1} ={u: (u₀)² - u² = 1}.

From the defining formula, uν(u) = 1, we see that for all u in G we have 

ν(u) = u^-1.

Now, after this recall of the main formulas, we can return to our task at hand. We want to see how G acts on the null cone in Minkowski space, and how this action translates to its action on the (x,y) plane of stereographic projection. Since we want real x to be transformed into real x, we choose the action

x' = gxτ(g),

the option already studied as 1) in Part 17. Here we will do it again, this time using the complex notation. But we will do it in the next post.

P.S. 07-12-24 12:04 Here are additional comments in response to the apparent contradiction noted by Bjab in the comments.

We have real algebra Cl(V) and we have the algebra of complex quaternions, let us call it here A. We have a real linear isomorphism of vector spaces, let us call it c from Cl(V) onto A:

c(1)=1

c(e1) = e1, c(e2)=e2, c(e3) = e3,

c(e1e2e3) =i,

c(e1e2) = ie3, c(e2e3)=ie1, c(e3e1)=ie2.

In the above we can write e1∧e2 instead of e1e2 etc. This is the same for orthogonal vectors

The elements of A we can represent in terms of components (u0,<b>u</b>). Then, for any u,v in Cl(V)  c(uv) is given by (1) in terms of components c(u) and c(v). But it is not true that in general  c(u∧v) = i c(u)⨯c(v), even if both sides are well defined. Example: u=e1, v=e2e3.

P.S. 07-12-24 14:27 8 years after I posted this video I am getting a question "Camera setting pls , is this from video mode ?". Like I remember!