Sunday, November 10, 2024

The Spin Chronicles (Part 11) - Geometric algebra in human language

This post is a continuation of The Spin Chronicles (Part 10) - Dressing up the three involutions

The three Graces—Aglaea, Euphrosyne, and Thalia—are gathered once more under the olive trees, where sunlight filters through like language weaving through a complex idea. Today, they’ve found themselves discussing something a bit unusual: the nuances of translation and language, specifically in the realms of human language and mathematical symbols.


Human language and mathematical symbols

Thalia sips thoughtfully, setting her cup aside. “Language, my dear sisters, is so… slippery. We think we understand each other, and yet, we don’t always share the same concepts. Consider translation. We can translate the same information from Greek to Latin or English, but is it truly the same message?”

Aglaea nods, looking intrigued. “That’s true, Thalia. Translators now have AI to assist them in bridging language divides, which is incredible. But we’ve all seen how things can get lost in translation. A good translator knows one method to verify accuracy: by performing a backward translation, starting, say, in Greek, then moving to English, and then translating that English back into Greek. When the meaning holds up—well, we’ve done our job.”

Euphrosyne raises an eyebrow, a bit skeptical. “And yet, sisters, isn’t it still true that each language carries unique concepts, shades of meaning? You can’t always find exact matches. Think of the Bible translations! The original Scriptures were in Hebrew or Greek, and then they went to Latin, and then to English… translators often go back to the roots to get closer to the original meaning. How many nuances must slip through each transition!”

Aglaea gestures thoughtfully. “This is why it becomes even more fascinating when we look at mathematics. We expect math to be different, don’t we? Since mathematics operates with precisely defined concepts, we hope its ideas should be universal. A mathematical notion expressed in German or English—or purely in symbols—should retain its exact meaning. Or so we believe…”

Thalia leans forward, eyes sparkling with interest. “Oh, but even math is like language in its own way! Just consider the geometric Clifford algebra we’ve discussed recently. We’ve explored it from so many angles, and yet each approach—plain language, logical symbols, matrix representation—feels a bit… what’s the word? Artificial?”

Euphrosyne laughs, nodding. “Yes! It feels as though we’re moving through a maze of mirrors. We understand what we’re doing when we work with matrices, for instance, but are we truly seeing the essence of ‘space’ through them? When we use these matrices, it feels a little like those quantum theorists who say, ‘Shut up and calculate!’ As though we’re so focused on the tools, we lose sight of what they’re trying to reveal.”

Aglaea sighs, smiling. “Exactly. Today, however, we’ll try a new approach. We’ll use complex vectors and their familiar operations to represent the geometric algebra. And maybe, just maybe, this simpler language will bring us closer to feeling ‘at home.’ Imagine, Thalia—no more foreign constructs!”

Thalia claps her hands in delight. “Oh, finally! I love a new perspective! Mathematics, when we translate it into familiar terms, starts to feel less like a maze and more like, well, a map. Perhaps this language of complex vectors will reveal something we’ve missed. If it brings clarity and warmth, then we might finally feel that we understand, that we’re truly ‘seeing’ the space it describes.”

Euphrosyne sighs, content. “So we go from Greek to Latin, to symbols, to matrices, to vectors… I suppose, like in language, there’s no ‘perfect’ translation, but perhaps there is a way to feel at home in it. That’s what we’re aiming for today.”

The three sisters raise their goblets, exchanging warm glances, each in tune with the others’ anticipation. They’re ready to explore this new representation, hoping that in its simplicity and familiarity, they’ll find a language that lets them see the essence of the world they’re describing.


And so, with minds open and spirits light, the three Graces prepare to dive into their next exploration.


From the previous posts we already know that each element u of the Clifford algebra Cl(V) of the real, Euclidean, orinted 3D space V can be written (uniquely) as

u = a + b1e1 + b2 e2 + b3 e3 + c1 e2e3 +c2 e3e1 + c3 e1e2 + d e1e2e3,

where a,b,c,d are real numbers. We have denoted e1e2e3 = ι, and realized that ι2 = -1, and that ι is in the center of Cl(V), i.e. that it commutes with all the elements of Cl(V). In other words: it introduces a complex structure in Cl(V). That is Cl(V) can be considered as a 4D complex space C4, instead of the original 8D real R8. Moreover, we notice that e2e3 = ι e1, etc. Therefore u can be rewritten as

u = (a+ι d) + (b + ι c)·e,

where bold letters are being used for vectors with three components. We set

p0 = a + ι d,

p = b + ι c.

In what follows we will use the symbol "i" instead of "ι" - as it is more familiar, and it has the same properties. Then p0 is a complex scalar, and p is a complex 3D vector. And we know the Clifford (and Grassmann) algebra interpretation of its real and imaginary parts. Thus u can be written as a pair

 u = (p0,p).

 What we need now is the expression of the Clifford product in this language. It is a simple exercise, left for the reader (I hope I am not making mistakes here...):

Exercise 1. Show that: If u = (p0,p), v = (q0,q), then the Clifford algebra product uv is given by

uv = (p0q0 + p·q,  p0q + q0p + i pq).

We can now express our three involutions in this new language.

Exercise 2. Show that

τ(u) = (p0,p)*,

ν(u) = (p0,-p),

π(u) = (p0,-p)*,

where "*" denotes the complex conjugation.

It is so simple now! In fact these operations are well known to the bi-quaternion community. The first one is the complex conjugation, the second one is the quaternionic conjugation in the bi-quatenion algebra. But, coming from the geometric algebra, we have an additional perspective on these operations.

Bilinear forms

For u = (p0,p) we set tn(u) = p0. ("tn" is a short for "trace normalized). We just take the scalar part. This operation corresponds to taking the sum  a + ι d in the Clifford algebra language, and to taking the 1/2 of the trace of the matrix in the matrix representation discussed in the previous post. From the multiplication formula of Exercise 1 we instantly have that tn(uv) = tn(vu) - the standard property of the trace.

We can now define our first natural bilinear form B0(u,v):

B0(u,v) =tn(uv).

We instantly get

B0(u,v) = p0q0 + p·q,

which is the standard complex-valued product in C4. It has symmetric and non-degenerate real and imaginary parts, both of neutral signature (++++,----) if C4 is considered as real R8.

But then we have also the second natural "bilinear" form:

Bτ(u,v) =tn(τ(u)v),

In fact, this form is, when considered as a form on a complex vectors space,  Hermitian: it is complex linear in v, but complex anti-linear in u. Its real part is symmetric positive definite, signature (++++++++) in R8. This is our scalar product (u|v) that we have already met before. Its imaginary part is anti-symmetric ("symplectic").

We leave the other two natural forms for the next post.

P.S. 11-11-24. In reply to Bjab's commentDetails of B0.

Let us introduce a real basis EA (A=0,1,...,7) in Cl(V) as follows:

E0 = I, E1 = e1, E2 = e2, E3 = e3, E4 = ie1, E5 = ie2, E6 = ie3E7 = iI.

The matrix of B0 in this basis is given by

(B0)AB = tn( EAEB ).

Explicitly, after calculations (I am using Mathematica): we get the following results:

B0:





Re(B0):





Im(B0)





Eigenvalues of Im(B0): (-1,-1,-1,-1,1,1,1,1)

Eigenvectors (unnormalized) of Im(B0)





Which should be read as: the first eigenvector of Im(B0) is E7-E0, (or E0-E7, if you prefer) the second is E6-E3, etc.

P.S. 11-11-24 12:15 This kind of scalar products are not usually considered by the Clifford algebraists. Once the Clifford algebra is over the real space, they consider only real-valued forms. We are considering forms with values in the center of the algebra, which happens to be a field, isomorphic to the field of complex numbers. I am not aware of any books or papers discussing this kind of constructions. But they have been probably discussed within the bi-quaternion (or hypercomplex numbers) formulations. They should come naturally there.

P.S. 11-11-24 16:10 Something to think about: 

Fifty Shades of Grey [Aliens]

Beneath the Surface: We May Learn More about UAP by Looking in the Ocean, Rear Admiral Tim Gallaudet, PhD, US Navy (ret.), SOL Vol. 1 No. 1, March 2024.

20 comments:

  1. Now (in part 10) the formula for u is different from formula for u in part 9. (They differ at the end.)
    I'm surprised that this could happen.

    ReplyDelete
  2. Replies
    1. We have denoted e1e2e2 = ι ->
      We have denoted e1e2e3 = ι

      Delete
    2. we notice that e2e3 += ι e1 ->
      += is a sequence from some programming languages.

      Delete
    3. then the Clifford algebra product uv is given by

      uv = (p0q0 + p·q, i p⨯q).


      I don't see in that formula a (from p) times b (from q).

      Delete
    4. "and to taking the trace of the matrix in the matrix representation discussed in the previous post."

      Strictly speaking to the half of the trace?

      Delete
    5. standard propert ->
      standard property

      Delete
  3. "For u = (p0,p) we set tr(u) = p0"

    Ark, could you consider using the abbreviation htr (half of the trace) instead of tr (in several places)?
    tr reminds me so much of trace.

    ReplyDelete
    Replies
    1. Good idea. I replaced tr by t_n, so that it will be ok also for more than 2 dimensions.

      Delete
  4. "It has symmetric and non-degenerate real and imaginary parts, both of neutral signature (++++,----)"

    I don't quite understand that. Is this some kind of shortcut?

    I understand that real part of this form is:
    a_u a_v + b1_u b1_v + ... - c1_u c1_v - ... - d_u d_v

    but what about imaginary part?


    ReplyDelete
    Replies
    1. Thanks for asking. Will add P.S. with an explanation.

      Delete
    2. Indeed, what I wrote is confusing. I have to clarify it, first in my head! Working on it.

      Delete
    3. I think that B0 matrix is bad.
      For example E1E4 should not be zero (and so on).

      Delete
    4. OK. I noticed that in my calculations I used different numeration for E_A Will change it and post the corrected tables. Thank you!

      Delete

Thank you for your comment..

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