I really don't know what this means, but I've retired from physics and maths, by and large. I don't see why something of order 8 is automatically octonionic. It's also interesting to me how often spacetimes are treated as fundamental. I don't think of them that way. Spinor space is my starting point, and out of this falls spacetime, the groups and particles of the standard model, an explanation for why our universe is constructed of matter (where are the antimatter frogs and asteroids?), and a whole lot more. I don't need to "try" anything, or seek elsewhere for the pieces of the standard model. The mathematics speaks; I listen. Or I used to. I no longer have any faith that it matters. Consent is now being manufactured on an industrial scale.

The best place for the beginning of the answer is Baez's Octonion paper: http://math.ucr.edu/home/baez/octonions/

Reals, complex, quaternions, octonions, sedenions obey the Cayley-Dickson construction http://math.ucr.edu/home/baez/octonions/node5.html

Clifford algebras obey the relation:

vw + wv = -2

http://math.ucr.edu/home/baez/octonions/node6.html

which by the polarization identity:

< x, y> = 1/ (|| x+y||^2 - || x||^2 - || y||^2)

becomes:

v^2 = -||v||^2

Clifford algebras are associative, octonions are not. (in the Cayley-Dickson, after complex numbers, at every step the next algebra loses some property: quaternions are not commutative, octonios are not commutative or associative.

For Cl(3,0) see the appendix in http://arxiv.org/pdf/1109.0535v3.pdf

To compute in it is very easy: use the following simple rules (e1,e2,e3 are not commutative):

e1e1 = e2e2 = e3e3 = 1

and

e1e2 = −e2e1, e1e3 = −e3e1, e2e3 = −e3e2

and remember the basis:

1, - scalar

e1,e2,e3, - vectors

e1e2, e3e1,e2e3 - bivectors

e1e2e3 - pseudo-scalar

The relation with S^7 comes from Hopf fibration remembering that Cayley-Dickson generates the only normed division number systems and this is related with projective planes because of the division property, but this is an entire different large topic.

Cl(3,0) or Cl(0,3)?

I understood from Wikipedia that Cl(0,3) is the basic one coming from geometric algebra and R^3.

Cl(3,0) or Cl(0,3)?

Short answer: It depends on one's convention.

Long answer:

Formally a Clifford algebra is a quotient algebra by an ideal:

Cℓ(V, Q) = T(V)/IQ.

Cl(p,q) comes from R^{p,q} given by the signature of the quadratic form Q. Now there is an ambiguity of the overall sign of Q but this is fixed by stating which Clifford algebra corresponds to complex numbers C: C=Cl(1,0) or C = Cl(0,1) ?

(Joy is using the opposite convention of Wikipedia so I used that in my appendix to match his papers to avoid yet another confusing sigh change).

Clifford algebras (including "the" Clifford algebra Cl_{(0,3)} (R) are constructed in metric spaces, their definition involving a well-defined metric (non-degenerate quadratic form). The constitutive relations are then

$$xy + yx = 2 1$$

where 1 stands for the unit matrix in an appropriate number of dimensions (in fact,

if the dimension of the linear space to which belong the elements x, y,... is N,

then the dimension of the lowest representation of the corresponding Clifford algebra will be $2^{[N/2]}$. Clifford algebras are non-commutative, but they are associative and thus can be represented by matrices.

The octonion algebra in non-associative, and one of its possible definitions is given by

$$ e_i e_k = -\delta_{ik} e_0 + \epsilon_{ijk} e_k$$

The element $e_0$ is neutral (like a group unit), $e_0 e_i = e_i e_0 = e_i$ and $e_0 e_0 = e_0$. The index i runs from 1 to 7. It is easy to see that this algebra is non-associative and cannot have a matrix representation.

I would phrase phrase your question differently:

Is it possible to express the nonassociative octonionic product in terms of associative Clifford algebra product using some clever explicit formula?

I guess it may be possible, perhaps someone already did it, but I do not know the answer.

A geometric algebra is _exactly the same thing_ as a real Clifford algebra.

If it was not already clear from Florin's answer you can use the convention that x^2 = Q(x) or x^2 = -Q(x) in the definition of the Clifford algebra, and both are used (in differential geometry what is convenient is closely related to what you want to call the Laplacian, the analysts choice negative semi define generalising

\frac{\partial^2}{\partial x^2} or the geometers positive definite one which differs by a minus sign)

The relation of S^7 to the octonians is that the action of the octonians parallises the tangent bundle, i.e. the tangent bundle of S^7 \subset \O is trivial and the trivialisation is given by pulling back the map defined by (right) multiplication, just as it does with the tangent bundle of S^3 \subset \H, and S^1 \subset \C. The point is that the multiplication has an inverse because the octonians are a division algebra. One can prove that these are the only parallisable spheres using non trivial algebraic topology. This is described in part in the book by Milnor Stashef on characteristic classes.

The octonians also have to do with the "Hopf fibrations" This is a series of maps:

\R^2 \superset S^1 \to \RP^1 \iso S^1 with fibre S^0 = unit reals

\C^2 \superset S^3 \to \CP^1 \iso S^2 with fibre S^1 = unit complex numbers*

\H^2 \superset S^7 \to \HP^1 \iso S^4 with fibre S^3 = unit quaternions

\O^2 \superset S^15 \to \OP^1 \iso S^8 with fibre S^7 = unit octonians

These are the only spheres fibring over spheres with fibre a sphere if I remember well. Again this is a non trivial theorem in algebraic topology (Adam's Hopf Invariant one theorem).

* this one is usually called "the" Hopf fibration.

No Clifford algebra can be/is isomorphic to the algebra of octonions because of the "non-associativity" of the latter one and associativity of the first one. It is known that Cliff(8=2^3) [complex, i.e., C^8] = End_C( O + O \otimes_R C ) isomorphism of assoc. alg. More similar isoms are in the book Harvey, F. R. (1990): Spinors and Calibration, Academic Press, San Diego. End( O + O) must be understood as the associative algbera of homomorphism of the vector space of (O + O) otimes C. It should be transferable to the reals. Dimension counts 2^8 = 256. 16^2=256. Similarly 8^2 = 2^6. So there might be ""something like End_R(O) = Cliff(R^6)"". I do not know it adn doubt it a bit. In the book of Harvey there are similar isomorphisms. One should have a look. I looked up in a diploma that Cl(C^8)^{even} = End_C(O\otimes C) + End_C(O\otimes C) ( cplx case), 2^8/2=2^7 = 8^2+8^2.

Svatopluk: "No Clifford algebra can be/is isomorphic to the algebra of octonions"

I agree, but I believe no was making such a suggestion. But, for instance, one can define (commutative) Jordan algebra product in terms of Clifford algebra product, even if Jordan algebra is nonassociative. Can something similar be done with noncommutative and nonassociative octonions?

I hope that I am correct; According to the classification of the Clifford algebra in

https://en.wikipedia.org/wiki/Classification_of_Clifford_algebras#Bott_periodicity

The Clifford 8 dimensional algebra you talked about is H\oplus H. These two copies of

H are mutual commute. That is not true for the other algebra. And it is quite clear that they are different.

Also, from the Lie group point of view, the Clifford algebra is used as a foundation to construct Spin(3), which is the universal covering of SO(3). So, it is very important for the Lie group theory. This possibly has nothing to do with S^7. But not the other algebra.

I hope that I answer your question properly.

Dear Joseph,

How do you understand your formula

begin{displaymath}\Cliff (n+8) \iso \Cliff (n) \tensor \R[16] .\end{displaymath}

when n=0?

Cliff(0) = \mathbb{R} is the reel

$n$ $\Cliff (n)$

$0$ $\R$

$1$ $\C$

$2$ $\H$

$3$ $\H \oplus \H$

$4$ $\H[2]$

$5$ $\C[4]$

$6$ $\R[8]$

$7$ $\R[8] \oplus \R[8]$

Dear Daniel,

this table before $n=8$

Starting at dimension 8 we've the relation you asking about which I mentioned. In other words, $\Cliff (n+8)$ consists of $16 \times 16$ matrices with entries in $\Cliff (n)$. This `period-8' is called Bott periodicity, since it has a far-ranging set of applications to topology, some of which were discovered by Bott.

Now, since Clifford algebras are built from matrix algebras over $\R,\C$ and $\H$, it is easy to determine their representations. Every representation is a direct sum of irreducible ones, or irreps.

looking through the table above unless $n$ equals $3$ or $7$ modulo $8$, $\Cliff (n)$ is a real, complex or quaternionic matrix algebra. So we've a unique irreps.

When $n$ is $3$ or $7$ modulo $8$, the algebra $\Cliff (n)$ is a direct sum of two real or quaternionic matrix algebras, so it has two irreps

Ok. Thanks.

I understand it now. R[16] is the is the matrix algebra, not the vector space R^{16}.

Therefore, there is nothing to do with the 8 dimensional division algebra. Or maybe I misunderstand again?

Thanks,

DG

Dear Daniel,

I remind you that any $n$-dimensional normed division algebra gives an $n$-dimensional representation of $\Cliff (n-1)$ and this gave us the table :

$n$ $\Cliff (n)$

$0$ $\R$

$1$ $\C$

$2$ $\H$

$3$ $\H \oplus \H$

$4$ $\H[2]$

$5$ $\C[4]$

$6$ $\R[8]$

$7$ $\R[8] \oplus \R[8]$

notice thta effectively we use $A[n]$ to stand for $n\times n$ matrices with entries in the algebra $A$

Best

Joseph

May be a last word concerning the link with the exceptional group.

The framework is the relation between composition and division algebras and the two approaches for G2 ; indeed, the two conceptions (G2 = Aut (Octonions) and G2 = isotropy group of a generic 3-form in 7 dimensions) have been already noticed as equivalent by Elie Cartan in 1908 and 1914 and others. Specifically, we can construct the octonions starting with the regular 3-form φ, wherefrom the dual role of G2 will be evident: the 3-form determines a regular metric β, which converts the form φ, as (0,3)-tensor, in a (1,2)-tensor: that is, an algebra γ = β(φ), which turns out to be, of course, the octonion algebra: the group G2 becomes automatically the automorphism group of that algebra. Notice that we can elaborate on composition and division algebras and then also extend some remarks on the relations of Spin(8), Spin(7) (both have a 8-dim real representation) and other smaller groups, including G2. Several physical applications are knowning now, including some rôle for SU(3) connected with octonions. So we could think that the gravitation could play a role through the SU(3), etc...

Sorry for the lengthy answer.

Best

You have answered your own question (yes, you are right), but it might help to give further information if you answer the following: What contradiction do you see in the Wikipedia articles?

Ok. Now, if I understand correctly, Joseph just said that there is a relation between

the algebra Cliff(7) and O. Possibly, this answer the original question of the relation between Clifford algebra and O.

Thanks, Joseph.

Some one might look at this:

On Representation Theory and the Cohomology Rings of Irreducible Compact Hyperk\"ahler Manifolds of Complex Dimension Four, Central European Journal of Mathematics, vol 1 (2003), 661--669.

and get a better understanding or progress.

Thanks.

Thanks Daniel for the kind and interesting exchange, and thanks Richard for the opportunity,

Cheers,

Joseph

Many problems occur because there is a confusion between Clifford algebra per se which is a pure algebraic construction, which involves a bilinear symmetric form (on real orcomplex fields) on one hand, and their representtion as matrix algebras. Any Clifford algebra has a faithful representation as an algebra of matrices, but they are different for real and complex Clifford algebras.

Moreover in physics spinors are not representation of elements of a Clifford algebra, but are vectors belonging to a vector space representation of a Clifford algebra. On these subjects you can see my book Mathematics for theoretical physics (2014 edition)

So we can think of O and Cl(0, 3), as sets, as both being sets of pairs of quaternions. In order to consider the set as an algebra we have to define addition and multiplication. The addition rule is the same, and is the natural addition rule. What are the two multiplication rules? Can anyone write them down for me? And I mean: can anyone write them down for me, in terms of quaternionic multiplication?

Richard

 Perhaps this might help to answer your question? 

 1)      From the late Pertti Lounesto’s “Clifford Algebras and Spinors”, p. 302:

 “Octonions can be defined as pairs of quaternions, but … order of multiplication matters

 (p1,q1)○(p2,q2) = (p1p2 - q2*q1 , q2p1 + q1p2*)

 (read * on r.h.s. q2 and r.h.s. p2 as complex conjugation, not as a product; I can't insert Lounesto's over-bars)

 2)   And as you’re probably already well aware, from the wiki link below

"The usefulness of quaternions for geometrical computations can be generalised to other dimensions, by identifying the quaternions as the even part Cℓ+3,0(R) of the Clifford algebra Cℓ3,0(R). This is an associative multivector algebra built up from fundamental basis elements σ1, σ2, σ3 using the product rules.

 σ12 = σ22 = σ32= 1

σi σj = - σj σi         (j ≠ i).

 …In this picture, quaternions correspond not to vectors but to bivectors – quantities with magnitude and orientations associated with particular 2D planes rather than 1D directions."

As you can probably tell, I’m not a mathematician, but I am interested in division algebras through an interest in quantum spin.

 Hope this helps

 Paul

http://en.wikipedia.org/wiki/Quaternion

Dear Richard,

A simple way to see this:

The octonions are not associative, so we cannot hope to obtain them in the form of any

matrix group. Still, we consider an octonion to be a pair of quaternions $(a, b)$, but we set

the product

$$(a, b).(r, s) = (ra - \bar{b}s, sa + b\bar{r})$$   (1)

Another way to create the octonions is to append the symbol $l$ to the quaternionic $i, j, k,$ and enforce $l^2 = -1$ and $l$ anti-commutes with $i, j, k$, to obtain a twisted product on

$$mathbb{O} = \mathbb{H} \oplus \mathbb{H}l $$

For instance, with the product from (1), we obtain $j(il) = kl$ but (ji)l = -kl$. We define

a trace function on $\mathbb{O}$ in the usual way:

if $ x = (o1, o2)$ where $o_1 , o_2 \in \mathbb{H}$, then

$Tr(x) = Tr(o_1)$

This gives us a complex conjugate

$$\bar(x) = Tr(x) - x $$

and a norm

$$N(x) = x\bar{x}$$

Best,

Joseph

I guess that if you really want to consider them as the same vector spaces, then you need to define a 1-1 linear transformation from one space to the other.

Then compare the product table. That should be doable.

Best,

DG

Clifford algebras algebraically represent space-times, for arbitrary numbers of dimensions of space and time. They act on spinor spaces, which is where fermion fields reside - those that represent things like the electron, quarks and neutrinos.

The octonions, O, are initially associated with one of the three parallelizable spheres (as are the quaternions, H, and complex numbers, C). C, H and O are also spinor spaces, so not fundamentally Clifford algebras. But C and H are in fact also Clifford algebras, but in a different guise. O is not. It is a pure spinor space, and the Clifford algebra that acts on it is the algebra of nested actions of O on itself: things like x(y(zO)), where x,y,z are in O also.

It's kind of cool. There are lots of references, including my two books.

Dear Dr Dixon,

Very interesting to see your succinct summary relating division algebras and spinors.

Are you familiar with the work of Merab Gogberashvili? (Not listed in the index of your 2nd book,* though I expect there are numerous works you're familiar with that you’ve not listed there.)

I’m curious to know if you have an opinion on how useful, for physics, is the octonionic interval he uses to describe space-time, incorporating not only c (for time) but also, using h/2π, both energy and momentum too (full copy of reference attached):

"Recently the assumption that our Universe is made of pairs of octonions become an important idea in string theory also. As distinct from string models in present paper we want to apply octonions to describe space-time geometry and not only internal spaces. We want to describe the physical signal s by a 8-dimensional number, the element of octonionic algebra,

       s = ct + xnJn + (h/2 π)λnjn + c(h/2 π)ωI , n = 1, 2, 3                  (eqn 2)

where by the repeated indexes summing is considered as in standard tensor calculus. The eight scalar parameters in (2) we treat as the time t, special coordinates xn, quantities λn with the dimension of the momentum−1 and ω with the dimension of energy−1. In (2) two fundamental constants of physics, the velocity of light c and Plank’s constant h/2 π, are presented also." .

Of course, the combinations in each term of eqn (2) result in 8 terms in the same units, nominally "x", but resulting in a description that incorporates both the x,t and p, E normally handled separately in vector methods concerned with SR.

I wouldn't expect an approach such as this to result directly in any new physics, but its compactness might suggest new ways of thinking and maybe new insights.

Regards - PGE

* I’ve long had your first book, though haven’t worked through it systematically, and after looking on Amazon have been intrigued enough to purchase your second, just now.

I really don't know what this means, but I've retired from physics and maths, by and large. I don't see why something of order 8 is automatically octonionic. It's also interesting to me how often spacetimes are treated as fundamental. I don't think of them that way. Spinor space is my starting point, and out of this falls spacetime, the groups and particles of the standard model, an explanation for why our universe is constructed of matter (where are the antimatter frogs and asteroids?), and a whole lot more. I don't need to "try" anything, or seek elsewhere for the pieces of the standard model. The mathematics speaks; I listen. Or I used to. I no longer have any faith that it matters. Consent is now being manufactured on an industrial scale.

Geoffrey,                  

As someone else with a core interest in spin and spinors, though I've not till now thought much about them in relation to octonions, I am keen to learn whether you mean by:

“Spinor space is my starting point, and out of this falls spacetime, the groups and particles of the standard model…and a whole lot more”

…that you start purely from:

      T = R ⊗ C ⊗ H ⊗ O 

 … an equation featuring in your more recent book (referred to in my preceding post).

Or/and(?) do you see spacetime emerging from something like spin-networks relating to T (e.g. along lines not too distant from [2] Rovelli & Smolin “Spin Networks and Quantum Gravity”, and more recent works by Rovelli), rather than the more conventional considerations of high density fermion fields occurring inside an expanding spacetime (e.g. [3] Bijan Saha “Nonlinear Spinor Fields and its role in Cosmology” and [4])?

Otherwise, is your approach quite different again, in which case I’d be interested to learn what, to you, suggests spinor spaces as a starting point (?)

Regards – PGE

PS And just to round off the earlier post, I think Gogberashvili justifies his order-8 entities as being octonionic via his intro in another paper [1] with the lines:

“In the paper [5] it was assumed that to describe the geometry of real world most convenient is the algebra of split-octonions. With a real physical signal we associate an 8-dimensional number, the element of split octonions,

     s = ct + xnJn + ¯hλnjn + c¯hωI . (n = 1, 2, 3) (1)

Some characteristics of the physical world (such as dimension, causality, maximal velocities, quantum behavior, etc.) can be naturally connected with the structure of the algebra. For example, our imagination about 3-dimensional character of the space can be the result of existing of the three vector like elements Jn in (1).”

https://arxiv.org/pdf/gr-qc/9505006.pdf

https://arxiv.org/pdf/1103.2890.pdf

https://en.wikipedia.org/wiki/Nonlinear_Dirac_equation

https://arxiv.org/abs/hep-th/0212251

Well, fermions are spin 1/2, 3/2, ...

Bosons spin 0,1,2 ...

You cannot combine bosons to make fermions, but you can make bosons from fermions. Fermions are the starting point. They are represented by spinor fields. They transform in a special way wrt the Lorentz group. Bosons transform more like spacetime itself (at least spin 1 bosons).

This is how I think. Spinors are the starting point. And they are associated with spacetimes, although generally there is more than one. Ian Porteous's Topological geometry was my introduction to everything Clifford algebra and division algebra. He uses C and H to represent the Clifford algebras (CAs) and their spinor spaces. Each of the division algebras can be viewed as a spinor space, including O, and including arbitrary tensor products of any of them.

For example, P = R ⊗ C ⊗ H is the spinor space for a CA of 3-space, and T = R ⊗ C ⊗ H ⊗ O is the spinor space of the CA of 9-space. P^2 is a pair of Dirac spinors. The associated spacetime for these spinors is part of the package. I know next to nothing about spin networks. What I do is more pure mathematics, than building things from known pieces of physics theory. The math mirrors physics, and points to solutions to things as yet unexplained. That's all. In my perfect world, we would go back to 1925 (or so) and replace Dirac spinors with spinors in T^2 and start again. We'd have had quarks decades earlier.

Regarding “In the paper [5] it was assumed that to describe the geometry of real world most convenient is the algebra of split-octonions. With a real physical signal we associate an 8-dimensional number", this again is not at all how I use(d) division algebras. The split octonions are half of C ⊗ O, and very much nonassociative. That authors intuition led him to describe the "geometry of the real world" in terms of something that to me is clearly spinorial. How is its algebraic structure used? Is this "geometry" a spacetime? If so, then it is not spinorial. As to the rest of the quote, I can't say if it has merit or not. All I can say is that it does not seem to me to be inevitable. The truth, when and if we find it, will be inevitable. Starting from T many things are inevitable. Not everything, and those things that are not are tantalizing, leading me to think about T^6, and the lattices Λ_2, Λ_8, and Λ_24 (see for example, A Conceptual Breakthrough in Sphere Packing by Henry Cohn). By the way, the spinor space for a CA of 24-space is O^4. (O itself is the spinor space for R^6.)

I don't start with physics, but with mathematics. Anything the maths says about physics I listen to. It's not applied math: it's just math, with an added dollop of "oh, look, isn't that just like physics".

s = ct + xnJn + ¯hλnjn + c¯hωI is applied math. I see that s consists of 8 parts. Why must they be associated with split octonions?

I wish Gogberashvili success with this approach. I don't care if I'm right or wrong (which I can say comfortably, because I know I'm not wrong, mostly). It would just be nice to know what is true. I fear, however, that Truth is beyond our species.

Anyhum, I may bow out of this discussion. Any research involving octonions is good research. It were best, however, not to force the maths to fit. The maths has tons to say on its own. Every theory of quantum gravity of which I'm aware is an effort to force maths to fit a preconception. Ciao.