The Spin Chronicles (Part 6): Clifford (or geometric) algebra of space

This post is a continuation of "The Spin Chronicles (Part 5): Exterior algebra of space", and  we will keep the notation introduced therein. But first a little historical interlude.

"I am Symmetric and Non-Degenerate," announced the left shoe...

Once upon a time

Once upon a time, in a tiny village named Space, there lived a girl with a most unusual name: Grassmann Algebra. Grassmann was as kind-hearted as they came, with a mind as sharp as a moonlit blade. But her legs, fragile from birth, made every step she took a challenge, each movement carefully measured and thoughtful. The villagers knew her well for her patient smile and her skill in arranging numbers into beautiful patterns, even though she sometimes longed to dance freely through the fields of Space.

One bright, dewy morning, as Grassmann wandered along the riverbank lost in thought, she spotted something bobbing gently in the water. Curious, she waded in and fished out…a pair of strange, shimmering shoes! They were woven with golden threads that seemed to flicker and vibrate with energy, and to her amazement, they began to speak.

"I am Symmetric and Non-Degenerate," announced the left shoe, with a self-important air.

The right shoe chimed in immediately, a bit more slyly. "And I am also Non-Degenerate, but distinctly Anti-Symmetric."

Grassmann raised an eyebrow. "Symmetric…Anti-Symmetric? I’ve never heard of shoes with such personality!"

"Just put us on," the left shoe insisted with a twinkle in its eye. "Your feet are about to become very well acquainted with mathematics."

Though a bit skeptical, Grassmann couldn’t resist. She slipped her feet into the shoes, and at once, a surge of energy pulsed through her legs. It was as if the entire fabric of Space itself vibrated with her. Grassmann stood, and to her amazement, her legs felt strong—perfectly balanced and coordinated. She took one step, then another, feeling as if she were part of an elegant, invisible dance that echoed through the entire village. She was no longer bound by her limitations but was, at last, fully free.

Unbeknownst to her, these vibrations reached far beyond the village of Space. In a distant, towering castle overlooking the universe, Prince Gravity felt the rhythmic pulse of Grassmann's footsteps reverberate through his bones. The vibrations resonated with him in a way he’d never felt before. Driven by an inexplicable pull, he set off to find their source, crossing dimensions and curvatures until he arrived in the village of Space, drawn to the girl whose steps had shaken his very core.

When Grassmann and Prince Gravity met, it was as if all the hidden forces of Space aligned. Gravity extended a hand to her, and together, they felt balanced, as though neither would fall as long as the other stood by. Grassmann, no longer bound by any limitation, felt a profound happiness blossom within her. And so, they decided to be bound in another way entirely: they would marry.

The marriage was no small event, attracting many peculiar guests from beyond the realms of Space. Lady Consciousness, a regal figure with an all-seeing gaze, sat by Sir Light, who occasionally flickered from one seat to another, always the center of attention. As Grassmann and Gravity exchanged vows, their presence illuminated all of Space, as if new stars were born simply to watch.

In time, Grassmann and Gravity had two children. The first, a curious child they named Matter, was as grounded as Gravity himself. The second, a mischievous twin they called Anti-Matter, delighted in keeping everyone on their toes. Together, they completed the family, their lives a harmonious mix of push and pull, existence and annihilation, symmetry and difference.

And so, Grassmann Algebra and Prince Gravity, their lives woven by fate, lived happily ever after in the village of Space, dancing across the fields, balancing and twirling in perfect harmony, for all of time.

Now that we got an idea of what to expect, we take on the math. The Grassmann algebra Λ(V) of a three-dimensional real vector space V consists of scalars, vectors, bi-vectors , and three-vectors. Its general element is a linear combination of those. We write it as

Λ(V) = Λ⁰(V)⊕Λ¹(V)⊕Λ²(V)⊕Λ³(V),

where "i" in Λⁱ(V) denotes the rank of multi-vectors constituting the subspace Λⁱ(V) of Λ(V).

Suppose now V is endowed with an Euclidean scalar product, denoted u·v. We then endow
Λ(V) itself with scalar product ( | ) as follows:

if f,g are elements of Λ(V) with different ranks, their scalar product (f|g) is, by definition, zero. We set

(1|1) = 1.

If f, g are exterior products of vectors

f = v¹∧..∧vⁱ, g = w¹∧...∧wⁱ,

then (f|g) is the determinant of the matrix of scalar products of the vectors: (vⁱ·w^j). We then extend the scalar product to the whole Λ(V) by linearity.

Danger: It is not at all evident that the above definition (f|g) is mathematically correct i.e. that it does not depend on the representation of f and g as exterior products of vectors. For instance we have

 v¹∧ v²  =  v¹∧ (v¹+v²).

But it can be shown, nevertheless that the definition is correct.

We can easily verify that if eⁱ is an orthonormal basis in V, then

(e¹²|e¹²) = (e²³|e²³) = (e³¹|e³¹) =1,

(e¹²³|e¹²³) = 1,

(e¹²|e²³) = (e²³|e³¹) = (e³¹|e¹²) = 0.

Exercise 1. Verify the above formulas.

Therefore 1, e¹, e², e³, e¹², e²³, e³¹, e¹²³  form an orthonormal basis in the eight-dimensional vector space Λ(V).

We will use this scalar product on Λ(V) to define a new multiplication, the Clifford algebra multiplication, but first let us introduce the "insertion' (or, some would say "annihilation") operators. For each element φ in Λ(V) we define the linear operator i_φ acting on V as the adjoint to the left exterior multiplication by φ:

( i_φ ψ|λ) = (ψ|φ∧λ),

for any φ,ψ,λ in Λ(V).

Since the right hand side of the defining formula is linear in φ, the operator i_φ is also linear in φ. For the right hand side to be non-zero, and for φ,ψ,λ of definite rank, we must have rank(ψ) = rank(φ)+rank(λ). Thus rank(i_φψ) = rank(ψ) - rank(φ), and we see that the action of i_φ  decreases the rank of a multi-vector by the rank of φ. In particular if rank(φ)>0, we must have i_φ 1 = 0.

Exercise 2. What is i₁?

Exercise 3. Show that, for any v,w in V, we have i_v∘i_w + i_w∘i_v = 0, where "∘" denotes the composition of operators.

Exercise 4. Show that, for every v in V, we have (i_v)² = 0.

If u,v are two vectors in V, then

i_uv = i_vu = (u·v)1

Let us find a similar explicit formula for acting with i_u on bi-vectors. If also v,w,z are in V, then, according to the defining formula, we have

(i_u(v∧w)|z) = (v∧w|u∧z) = (u·v)(w·z) - (u·w)(v·z) = ( i_u(v)w - i_u(w)v |z),

thus, since z was an arbitrary vector in V,

i_u(v∧w) = i_u(v)w - i_u(w)v.

We also see that if u is orthogonal to v and to w, then i_u(v∧w) = 0.
Similarly we can obtain

i_u(v∧w∧z) = i_u(v) w∧z - i_u(w)v∧z + i_u(z)v∧w.

We say that the operator i_u is an anti-derivation.

We are now ready to define a new multiplication in Λ(V). For a,b in Λ(V) we will denote the new product simply as ab, and it is defined by

ab = a∧b + i_a(b).

Λ(V) equipped with this multiplication is called the Clifford (or geometric) algebra of the space V endowed with the scalar product u·v. It is denoted Cl(V).

Cl(V)

Thus, in particular, if v,w are vectors in V, we have

vw = v∧w + (v·w)1

Therefore we have

vw + wv = 2 (v·w),


vw - wv = 2 (v∧w).

If v and w are orthogonal, i.e. if v·w = 0, then vw = v∧w = -wv. Orthogonal vectors anti-commute. On the other hand we have that vv = v² = v·v. Similarly, if u,v,w are mutually orthogonal, then

uvw = u∧v∧w.

Exercise 5. Prove this last formula.

It is not evident that so defined multiplication determines an associative algebra. Therefore let us first calculate the products of our basis vectors 1, e¹, e², e³, e¹², e²³, e³¹, e¹²³, assuming eⁱ form an orthonormal basis. Different eⁱ will anti-commute, their squares will be all equal to 1. 

We easily find that the multiplication table is much the same as that of the algebra Λ(V) except that the squares of eⁱ are now 1 instead of 0, as they are in Λ(V). Let us consider, in particular, the basic three-vector e¹²³. We calculate its square

e¹²³ e¹²³ = e¹e²e³e¹e²e³ = (e¹)²e²e³e²e³ = e²e³e²e³  = - (e²)²(e³)² = -1.

Thus our basic three-vectors behaves like the imaginary "i". Notice that it commutes with all basis elements  of the algebra Cl(V), therefore with all its elements. It behaves like a number, but an imaginary number!

Exercise 6. Prove this last statement.

We will denote this three-vector by ι. In the following we will exploit its properties and define a complex structure on our Clifford algebra Cl(V). We will identify Cl(V) with the algebra of complex quaternions (known also as biquaternions)

P.S. 25-10-24 While the comet Atlas is still in the sky:

Comet Atlas (Credits: Damian)

P.S. 26-10-24 16:29 

Renan has said that truth is always rejected when it comes to a man for the first time, its evolution being as follows:

First, we say the thing is rank heresy, and contrary to the Bible.

Second, we say the matter really amounts to nothing, anyway.

Third, we declare that we always believed it.