Octonions and Clifford algebra and S^7

[gill1109]

Can someone explain me the difference between Clifford algebra (the usual one - as a vector space over the reals, 8-dimensional) and the octonions?

The Wikipedia articles on these two subjects contradict one another.

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

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

Note: I refer to *the* Clifford algebra Cℓ_{0,3}(R), not to *a* Clifford algebra. The one with starting point R^3, the most common one in Geometric Algebra.


If I understand it correctly they are different. Both contain, as it were, two copies of the quaternions, but the multiplication tables are different.

The Clifford algebra has the right to be called a non-commutative algebra, as the phrase is usually understood, since multiplication is associative.

The algebra of the octonions is not associative but only satisfies a weaker property.

Am I right?

What is the connection of either with S^7 ?

[Joy Christian]

Can someone explain me the difference between Clifford algebra (the usual one - as a vector space over the reals, 8-dimensional) and the octonions?

The Wikipedia articles on these two subjects contradict one another.

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

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

Note: I refer to *the* Clifford algebra Cℓ_{0,3}(R), not to *a* Clifford algebra. The one with starting point R^3, the most common one in Geometric Algebra.


If I understand it correctly they are different. Both contain, as it were, two copies of the quaternions, but the multiplication tables are different.

The Clifford algebra has the right to be called a non-commutative algebra, as the phrase is usually understood, since multiplication is associative.

The algebra of the octonions is not associative but only satisfies a weaker property.

Am I right?

What is the connection of either with S^7 ?

You are right.

S^7 is a set of unit octonions, just as S^3 is a set of unit quaternions.

Two copies of quaternions give you bi-quaternions, not octonions. The former is a torus but the latter is a sphere. Hence the difference in the multiplication tables.

For a relationship between Cl(3, 0) and octonions see Lounesto, Advances in Applied Clifford Algebras 11 No. 2, 191-213 (2001).

[gill1109]

S^7 is a set of unit octonions, just as S^3 is a set of unit quaternions.

Two copies of quaternions give you bi-quaternions, not octonions. The former is a torus but the latter is a sphere. Hence the difference in the multiplication tables.

For a relationship between Cl(3, 0) and octonions see Lounesto, Advances in Applied Clifford Algebras 11 No. 2, 191-213 (2001).

Thanks!

I think you mean Cl(0, 3)?

OK: the notion of *unit* means that the quaternions and octonions also have a norm. And even perhaps an inner-product? I guess the usual Euclidean norm, inner product, when we think of them as elements of R^4 and R^8 respectively. (Similarly with the complex numbers and R^2).

Ah, the Lounesto article has been posted to the fqxi forum:

http://fqxi.org/data/forum-attachments/Lounesto.pdf

[gill1109]

Lounesto writes: "an algebra A with a positive-definite quadratic form N : A -> R ...".

I suppose that we start with the "norm" N, then define a real inner-product by polarisation, now we can define a real quadratic form and demand that that is positive-definite.

A norm is not a quadratic form.

[Joy Christian]

I think you mean Cl(0, 3)?

It is the same thing. There is no standard convention. Some people write Cl(3, 0), like I have done, and some people write Cl(0, 3). What's in the name?

Quaternions and octonions form two of the only four possible normed division algebras, the other two (non-quantum ones) being the reals and the complex.

Norm is not the same thing as a quadratic form. I have discussed some of these notions in this paper: http://arxiv.org/abs/1101.1958.

[gill1109]

No of course norm is not the same thing as quadratic form. The square of the Euclidean norm N gives us the Euclidean quadratic form Q, and from there one gets the bilinear form B(x,y) = Q(x+y) − Q(x) − Q(y) which is twice the Euclidean inner-product (this technique is called polarisation). A norm corresponds to an inner-product if and only if the parallelogram law holds.

I'm just pointing out that Lounesto is good on some things, and sloppy on others.

Now, the paper of yours which you refer to is really interesting, on two counts. One is the statement "Worse still, all such probabilistic counterparts P(A| n, λ) of A(n, λ) surreptitiously presuppose vector-algebraic models of the Euclidean space, which we have shown to be both physically and topologically incomplete". In other words: your model *cannot* be simulated because that requires a "probabilistic counterpart" which surreptitiously puts us back in flatland.

The other is the connection between parallelizability of spheres and quantum correlations. This connection has always fascinated me. I think you should "de-couple" this part of your grand project from the Bell criticism part. It should and can stand alone. The Bell part is a red herring.

[gill1109]

Quaternions and octonions form two of the only four possible normed division algebras, the other two (non-quantum ones) being the reals and the complex.

Yes, there are only four normed division algebras over the reals. This is Hurwitz' theorem on finite-dimensional unital real non-associative algebras endowed with a positive-definite quadratic form. The theorem states that if the quadratic form defines a homomorphism into the positive real numbers on the non-zero part of the algebra, then the algebra must be isomorphic to the real numbers, the complex numbers, the quaternions or the octonions. It can be proved using Clifford algebras!

Now another theorem is that the only parallelizable spheres are S^0, S^1, S^3 and S^7. The parallelizable spheres correspond precisely to elements of unit norm in the normed division algebras of the real numbers, complex numbers, quaternions, and octonions, which allows one to construct a parallelism for each. Proving that other spheres are not parallelizable is more difficult, and requires algebraic topology.

So these are related, but not equivalent theorems.

On these matters, Wikipedia seems to have it right.

[gill1109]

I'd still like to see a clear description of the *difference* between the octonions and "the" Clifford algebra Cl(0,3) (the Geometric Algebra of R^3).

Cl(0,3) aka the split biquaternions is "just" the algebraic direct sum H + H of the algebra of the quaternions, with itself. There is no norm (quadratic form, bilinear form) naturally defined *on* this space, though there is a quadratic form on R^3 associated with the *construction* of the space - that is what the numbers 0, 3 refer to.

Because H is associative we see immediately that the Clifford algebra is associative.

Do the octonions have a simple algebraic relationship with H? Presumably the octonions also contain two copies of H. But they are put together in a different way.

As vector spaces over the reals, both spaces are isomorphic to R^8 and hence to one another. So they can be put in one-to-one correspondence in a way which respects addition and scalar multiplication. But not in such a way that also multiplication is respected. The quadratic form or norm on the octonions is just the Euclidean norm on R^8 but this norm has no special meaning in the Clifford algebra, it seems.

[gill1109]

Here is some more good information, thanks to Florin Moldoveanu

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 constructionhttp://math.ucr.edu/home/baez/octonions/node5.html

Clifford algebras obey the relation:
vw + wv = -2<v,w>
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(0,3) 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.

The theory of parallelizable spheres is related to Hopf fibration which is a large topic. The key idea is like this: in the most general setting, have 2 groups A and C and perform a Cartesian product AxC If you understand C as an equivalence class, this is related with short exact sequences http://mathworld.wolfram.com/ShortExactSequence.html

0->A->AxC->C->0 and C ~ AxC/C

[ An exact sequence is a chain of functions where the image of a function is the ker of the next function and this generalizes the idea that "the boundary of a boundary is zero" in topology. Exact sequences are used in homology and cohomology theory (algebraic topology). A short exact sequence is a short "exact sequence" of 3 spaces sandwiched between null spaces. The null spaces means that the A->AxC function is injective, and the AxC->C function is surjective and sometimes the null spaces are omitted.]

Now in general in a short exact sequence: 0->A->B->C->0 B *IS NOT NECESSARILY ONLY* AxC and given A and C they do not define B uniquely and the problem is to classify all possible B's. Another way to understand B is as a fibre bundle (http://en.wikipedia.org/wiki/Fiber_bundle A=the fiber space, C = the base space, B=the entire space).

A Hopf fibration (http://en.wikipedia.org/wiki/Hopf_fibration) for S3 is a short exact sequence:

0->S1->S3->S2->0

and for S7:

0->S3->S7->S4->0

Now since S3 is not S1xS2, the fibers interlock (when B=/=AxC the fibers twist around one another)

Joy is using the theory of paralellized spheres to get a norm factorization: ||ab||=||a|| ||b|| property which allows him to claim the lack of superposition (in QM Bell states are not separable like this) in his model.