Suggested by YouTube, as an antidote perhaps --- "Feynman's Lost Lecture". https://www.youtube.com/watch?v=xdIjYBtnvZU Pure math-phys conjuring! WFL On 7/28/18, Fred Lunnon <fred.lunnon@gmail.com> wrote:
From the end of the second introductory short at https://www.youtube.com/watch?v=ng1bMsSokgw we are informed complex octonions |C (x) |O "lead to" (via a currently undefined process) a Clifford algebra (also currently undefined). Yet the first is non-associative, the second associative.
Additionally, why insist on stopping the Cayley construction at #4 ? The next iteration has certainly been investigated, and even named --- sedenions, I think.
My BS detector is beginning rumble ominously ...
WFL
On 7/28/18, Mike Stay <metaweta@gmail.com> wrote:
https://www.youtube.com/watch?v=3BZyds_KFWM
On Fri, Jul 27, 2018 at 7:47 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I don't understand any of the physics, but this is an interesting math idea that I'd never encountered before.
I think this refers to the tensor product of the four real division algebras:
R (x) C (x) H (x) O
(where R = reals, C = complexes, H = quaternions, O = octonions, and (x) denotes tensor product over the real numbers.
Tensor products are one of the two basic ways to combine algebras over a ring. But if I recall how I learned about this in a course called 18.26, I think everything was commutative. (Which the quaternions and octonions aren't, but I'm not sure it matters.)
I will attempt to unwind the definition of tensor product of algebras, since I'd like to know what this weird thing is.
*****-----until further notice, R means any commutative ring-----*****
Let's say A and B are algebras over a commutative ring R. Then their tensor product A (x) B will be the quotient of the free algebra over R generated by the elements
{a_j (x) b_j | a_j in A, b_j in B},
i.e.,
A*B = {Sum r_j (a_j (x) b_j), a_j in A, b_j in B, r_j in R, Sum is finite}.
To get the tensor product we need to factor out by the subring Z generated by elements of the form
((ra)(x)b) - (a(x)(rb)), r in R, a in A, b in B
and the like that give the tensor product its bilinearity. -----
*****-----From now on R is just the reals again-----*****
Dimensions of tensor products multiply, so the dimension of
R (x) C (x) H (x) O
where (x) is always understood as tensor with respect to R=reals, is
1 x 2 x 4 x 8 = 64.
Hmm, weird. Anyone care to explain (anything about) the physics?
—Dan
Henry Baker wrote: ----- ... ...
Combined as $latex \mathbb{R}\otimes\mathbb{C}\otimes\mathbb{H}\otimes\mathbb{O}$, the four number systems form a 64-dimensional abstract space.
... ... -----
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://www.math.ucr.edu/~mike http://reperiendi.wordpress.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun