Talk:Sedenion/Archive 1

Archive 1

Duo-tricenians??

This article says:

"Like (Cayley-Dickson) octonions, multiplication of Cayley-Dickson sedenions is neither commutative nor associative. But in contrast to the octonions, the sedenions do not even have the property of being alternative. They do, however, have the property of being power-associative." In turn, is the property of being power-associative an operation that the duo-tricenians no longer have?? Georgia guy 19:31, 11 September 2006 (UTC)

Hi. All Cayley-Dickson construction products remain power associative, including the 32-ions. This does - of course - not apply to modified constructs, like e.g. the split-octonions, which contain nilpotents. Thanks, Jens Koeplinger 12:37, 15 September 2006 (UTC)

Power-associative conic sedenions ?

The article claims that conic sedenions are alternative and flexible. It also claims them to not be power-associative. I assume the conic sedenions are an algebra i.e. multiplication distributes over addition and commutes with scalar multiplication. Doesn't general di-associativity (and hence power-associativity) follow from any two of the alternative laws (left, right, and flexible) in an algebra, giving an alternative algebra? 85.224.17.70 (talk) 19:52, 19 January 2008 (UTC)

Hello - maybe there is a terminology conflict here? I've taken the term "power associative" from the Imaeda/Imaeda publication (personal web space: http://www.geocities.com/zerodivisor/sbasicalgebra.html , formally published: http://dx.doi.org/10.1016/S0096-3003(99)00140-X ), but the definition that is given here is different: In the sense that $x(x(xx))=(xx)(xx)$ you're absolutely correct, of course. The relation used by Imaeda, $S^{n}S^{m}=S^{n+m}$, however does not hold anymore in general for conic sedenions (or complex octonions), due to the presence of nilpotents (e.g. $exp((\pi /2)(i_{1}+\varepsilon _{2}))=1+(\pi /2)(i_{1}+\varepsilon _{2})\neq exp((\pi /2)i_{1})*exp((\pi /2)\varepsilon _{2})$. The stronger $S^{n}S^{m}=S^{n+m}$ implies $x(x(xx))=(xx)(xx)$, so I'm not sure what the correct terminology would be ... help appreciated! Thanks, Jens Koeplinger (talk) 20:50, 19 January 2008 (UTC)

Baez's quote

If octonions are the crazy uncle that no one lets out of the attic, would sedenions be the serial killer maximum-security prison escapee that no one even lets in the house? Phoenix1304 (talk) 16:11, 9 April 2008 (UTC)

Ha! That seems to go in the right direction. Because they contain zero divisors, they're certainly "convicted" of some sort of violation ... :) --- feel free to post any concerns or more detailed questions you may have. Thanks, Jens Koeplinger (talk) 02:53, 10 April 2008 (UTC)

I'm actually less confused about these than I was about the other thngs. Sedenions seem to be "closer" to octonions than octonions are to quaternions, so it's not a huge leap. Phoenix1304 (talk) 15:37, 13 April 2008 (UTC)

No: the leap from octonions to sedenions is much bigger, because zero divisors appear. For example, the only parallelizable spheres S^n are for n=0,1,3,7. (R,C,H,O are the only normed division R-algebras and the only alternative division R-algebras; the only division R-algebras are of dimension 1,2,4,8.) —Preceding unsigned comment added by 66.92.7.54 (talk) 01:09, 10 November 2008 (UTC)

(Non-)alternative subspaces in Cayley-Dickson sedenions

Quick question: Does anyone know about a comprehensive analysis on the (non-)alternative subspaces in Cayley-Dickson sedenions? I've trouble finding such. Maybe I didn't look right ... Thanks, Jens Koeplinger (talk) 17:59, 20 June 2008 (UTC)

PS: It looks as if all non-real basis elements $e_{1}$ through $e_{15}$ taken by themselves are alternative, i.e.

$(xx)y=x(xy)$

for all

$x,y\in b_{S}:=\{e_{1},...,e_{15}\}$.

I can't find references for that. Unless I'm wrong, this means that all linear combinations of these are alternative as well. Is this true? Thanks, Jens Koeplinger (talk) 21:19, 24 June 2008 (UTC)

I guess this only solves part of the question, but say you take four imaginary sedenions $i,j,l,p$ that satisfy the condition that any one of them anti-commutes with any product of the others (e.g. $i=e_{1},j=e_{2},l=e_{4},p=e_{8}$ would satisfy this).

Let's call such a quadruple $i,j,l,p$ an independent quadruple.

Let $x=(ij+lp)$ and $y=(il+jp)$

Then

$xy=(ij+lp)(il+jp)=0$,

so

$x*xy=(ij+lp)*(ij+lp)(il+jp)=(ij+lp)*0=0$.

O.t.o.h.

$xx*y=(ij+lp)(ij+lp)*(il+jp)=-2*(il+jp)=-2il-2jp$.

Actually, for all pairs of numbers $(u,v)=((aij+blp),(cip+djl))$ where $a,b,c,d$ are real, $(a,b)$≠$(0,0)$, $(c,d)$≠$(0,0)$ and either $a$≠$0$≠$b$ or $c$≠$0$≠$d$ (including the case when all of $a,b,c,d$ are non-zero of course), the numbers $u$ and $v$ span a non-alternative algebra (assuming $i,j,l,p$ form an indepent quadruple of course!)

I don't know references for these equations but they're (very) easily checked.

I'm not sure if these spaces ($a,b,c,d$ fixed in the generating $u,v$ in each space) span all non-alternative sub-algebras of the sedenions though.

Sadly I don't know dimensions formulas of these spaces either.

P.S. Sorry about the formatting: I don't know LaTeX so it's formatted by hand.

85.224.19.225 (talk) 11:08, 2 April 2009 (UTC)

Interesting! Thank you very much for detailing this approach. Now, we need to check for completeness :) Thanks again, Jens Koeplinger (talk) 18:19, 16 April 2009 (UTC)

Further extension?

I would assume that there are logical extensions of this form of mathematics into 32-, 64-, and any other 2ⁿ-dimensions. How far has this been taken? And are there sources for this? (I'm still struggling to wrap my brain around quaternions, let alone octonions or sedenions, but it seems logical that the pattern would continue...) Lurlock (talk) 18:01, 18 November 2011 (UTC)

The series continues but beyond 8 the algebraic properties are the same. So if you've studied Sedenions you've studied all the higher dimensions in sense, and saved yourself the trouble of dealing with more dimensions than you've got fingers and toes.--JohnBlackburne^words_deeds 18:09, 18 November 2011 (UTC)

The most comprehensive work for Cayley-Dickson constructs beyond dimension 16 that I know of is done by Robert de Marrais. Specifically, he is investigating the structure of the zero-divisor spaces that occur and gives them funny names like "box-kites" and so on. He chooses to develop his own terminology which makes his work hard to digest at time. The parts that I did review are sound, but I can't claim I reviewed it all. I would start here: http://arxiv.org/abs/math/0207003 Thanks, Jens Koeplinger (talk) 02:05, 19 November 2011 (UTC)

Use and etymology?

Could you describe what are they used for ? --Taw — Preceding unsigned comment added by Taw (talk • contribs) 23:21, 2 December 2001 (UTC)

Also, what is the etymology of the name? Incidentally, I'm [RobertAtFM|http://wiki.fastmail.fm/wiki/index.php/RobertAtFm] on the [FastMail.FM Wiki|http://wiki.fastmail.fm/].

Yes, I agree: this page needs more information. I'd love to learn more about these things, but this page is barely more than a stub. --AlexChurchill 10:46, 27 July 2004 (UTC)

Assessment comment

The comment(s) below were originally left at Talk:Sedenion/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Lacks context for concept, needs cleanup shotwell 06:32, 6 October 2006 (UTC)

Last edited at 22:36, 19 April 2007 (UTC). Substituted at 02:35, 5 May 2016 (UTC)

Is Moreno's result incorrect?

A few days ago I reverted an unsourced edit to section Sedenion#Applications, but the issue that was brought up may have to be taken into account in some way, either by citing the added content or removing the current sentence in this section (see the discussion on my talk page for further details). Regards, Gap9551 (talk) 15:42, 20 April 2016 (UTC)

It is small mistake in Moreno work, not in the proof but in final statement. I saw this statement is repeated in other works. The correct statement should be: the set $\{(a,b):a*b=0;a,b\in S^{15}\}$ is homeomorphic to Lie group $G_{2}$. Thus it is not set of "norm 1 sedenions zero-divisors". It is set of pairs of norm 1 sedenions which product is zero.

The proof is following. The zero divisors in sedenion sphere is $a+b\iota$ where $a,b$ are perpendicular imaginary octonions and $\iota$ is extra element in Cayley-Dickson definition. Thus this is 11-dimensional subset of $S^{15}$ of shape $S^{6}\times \times S^{5}$. The $G_{2}$ we obtain by considering unit octonion $c$ perpendicular to quaternion space generated by $a,b$. Then we have $(a+b\iota )(c+a(bc)\iota )=0$. Regards, Marek Mitros

— Preceding unsigned comment added by 193.41.170.245 (talk) 08:33, 21 April 2016 (UTC)

The correction by Marek Mitros is indeed right; I looked at Moreno's paper and verified this. So, I've edited the article to make this correction. John Baez (talk) 03:48, 28 July 2016 (UTC)