Re: [math-fun] Cohl Furey & the Octonians
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. ... ... -----
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
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
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Sedenions aren't a division algebra. On Sat, Jul 28, 2018 at 8:11 AM, 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
_______________________________________________ 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
https://en.wikipedia.org/wiki/Bott_periodicity_theorem On Sat, Jul 28, 2018 at 12:23 PM, Mike Stay <metaweta@gmail.com> wrote:
Sedenions aren't a division algebra.
On Sat, Jul 28, 2018 at 8:11 AM, 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
_______________________________________________ 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
-- Mike Stay - metaweta@gmail.com http://www.math.ucr.edu/~mike http://reperiendi.wordpress.com
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
For a non-commutative tensor product, you can think of the \otimes as being a 'pipe' which allows elements of R to move between the two algebras. That is to say, A (x) B is the quotient of the vector space generated by the elements of the Cartesian product A x B by an equivalence relation: A (x) B := <A x B> / ~ where (a r , b) ~ (a , r b) for every a in A, b in B, r in R. In other words, left-multiplying b by r is the same as right-multiplying a by r. I haven't done the calculations to confirm whether C (x) H (x) O is isomorphic to H (x) O (x) C or O (x) C (x) H -- I see no reason why that should be the case a priori. (Reversing the chain definitely yields an isomorphic algebra because complex conjugation.) Best wishes, Adam P. Goucher
Sent: Saturday, July 28, 2018 at 2:47 AM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Cohl Furey & the Octonians
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.
... ... -----
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participants (4)
-
Adam P. Goucher -
Dan Asimov -
Fred Lunnon -
Mike Stay