Talk:Sedenion

Multiplicative inverses and zero divisors
I wonder about the following feature of the sedenions as claimed in the article: They have multiplicative inverses and at the same time zero divisors.

In a matrix algebra, these two features cannot coexist, since a divisor of zero necessarily has determinant zero and thus it is not invertible.

I"d like to see an example of two sedenions with inverses and a product of zero. --J"org Knappen
 * What the article says is:"The sedenions have a multiplicative identity element 1 and multiplicative inverses, but they are not a division algebra. This is because they have zero divisors."Take a look at zero divisors which demonstrates the principle with matrices. A given sedenion might be a zero divisor when paired with one particular other, but nothing stops either having an inverse as far as I can see. HTH --Phil | Talk 11:56, Jan 11, 2005 (UTC)


 * Seen that -- nice demostration of what is possible in infinitely many dimensions. My argument above is for finite matrices (what"s the determinant of an infinite matrix, anyway?) I just like to see an example for sedenions, too --J"org Knappen


 * J"org - Matrices form an associative algebra. Sedenions are non-associative. You are not able to map a sedenion algebra on matrices (as opposed to e.g. octonions or quaternions). For examples of calculations with sedenions see e.g. the two references to articles from K. Carmody. Please note, however, that the sedenions in there are of a different type as the ones displayed in the multiplication table (I'll try to have this corrected). --65.185.222.50 13:19, 12 September 2005 (UTC)


 * Within the current article it says  $$(e_3+e_{10})(e_6-e_{15}) = 0$$, and the inverse of any $$s$$ can be calculated by $$s^*/|s|^2$$, so $$(e_3+e_{10})^{-1} = -1/2 (e_3+e_{10}),  (e_6-e_{15})^{-1}=-1/2 (e_6-e_{15})$$. Generally, the inverses of (nontrivial) zero divisors are again zero divisors. Summsumm2 (talk) 15:19, 5 July 2019 (UTC)

Corrigendum

The multiplication table given in the article applies to sedenions of the type discussed by Imaeda/Imaeda, however, the sedenions discussed by Carmody are of a different type, first proposed by C. Musès. The following reference should be added to establish the correct originatorship of the latter sedenion type: C. Musès, Appl. Math. and Computation, 4, pp 45-66 (1978) --65.185.222.50 13:19, 12 September 2005 (UTC)

History
when were they discovered? are there larger algebras with the same properties? 83.79.181.211 19:05, 29 September 2005 (UTC)


 * About the term "sedenion" and the discovery; as far as I know, the cited articles are the earliest publications I could find that go deeper into arithmetic laws of the sedenions, and thereby solidify the term (so it's quite recent). Surely, as part of the Cayley-Dickson construction, their existence was discovered earlier within that program. Any earlier known uses of the term "sedenion"? Thanks, Jens Koeplinger 12:40, 15 September 2006 (UTC)

There are larger algebras with the same properties, in fact an infinite number of them. One can perform the Cayley-Dickson construction on sedenions to get a 32-element algebra, and again to get a 64-element algebra - in fact, one can get an algebra of 2n for any non-negative n. n=0 gives the reals, n=1 gives complex numbers, n=2 quaternions, n=3 octonions, n=4 sedenions and so forth. --Frank Lofaro Jr. 22:33, 2 March 2006 (UTC)

https://hsm.stackexchange.com/questions/11411/who-pioneered-the-study-of-the-sedenions

Who pioneered the study of the sedenions?

On Bibliography of Quaternions and Allied Mathematics by Alexander Macfarlane I found this:

On page 72; James Byrnie Shaw 1896 Sedenions (title). American Assoc. Proc., 45, 26.

I couldn't find this reference, but the same author wrote this book:

Synopsis of Linear Associative Algebra: A Report on its Natural Development and Results Reached up to the Present Time. 1907.

From its Table of Contents; Part II: Particular Algebras. Section XVIII: Triquaternions and Quadriquaternions. Page 91.

Early on Sedenions were also known as "quadriquaternions".

Section XIX: Sylvester Algebras. Page 93. Covers "Nonions" (9-ions), and "Sedenions" (16-ions). Here the Sedenions are attributed to James Joseph Sylvester.

On page 76; James Joseph Sylvester 1883-4 On quaternions, nonions, sedenions, etc. Johns Hopkins Univ. Circ., 3. Nos. 7 and 9. 4, No. 28.

This second reference can be found among The Collected Mathematical Papers of James Joseph Sylvester, [Volume IV(1882—1897)]:

Sylvester, James Joseph (1973) [1904], Baker, Henry Frederick (ed.), The collected mathematical papers of James Joseph Sylvester, IV, New York: AMS Chelsea Publishing, ISBN 978-0-8218-4238-6

Then, as far as I understood, James Joseph Sylvester appears on the literature as the proponent of two Algebras; the one from the Nonions and the Sedenions. I will leave as an open question for the community to confirm, correct, or debunk that J.J. Sylvester "discovered" or "was the pioneer" of their study in 1883. — Preceding unsigned comment added by 178.48.57.121 (talk) 04:10, 9 February 2020 (UTC)

Is Moreno's result incorrect?
A few days ago I reverted an unsourced edit to section Sedenion, 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)

Double simetry in Sedenion
For a double simetry in a sedenion, resulting in a Quaternion, resultant wave function in the Quantum mechanics, is a matrix that results in the Elementary particle of the standard model (without Higgs).

Please, add the simplification.

--188.171.58.12 (talk) 22:17, 26 December 2019 (UTC)

Multiplication Table
Could the multiplication table be fixed such that it shows -e correctly for all of the elements? In my browser (Chrome) it shows the negative sign above the e for e values with two-digit subscripts, and it took me a moment to figure out what this was. TricksterWolf (talk) 18:55, 29 February 2020 (UTC)

Incorrect number of primitive zero divisors
There are actually 168 primitive zero divisors. The Quaternion triplets mentioned are those created by the Cayley-Dickson doubling process with sequential new basis elements but is not unique, others exist. Please consult my paper: https://vixra.org/abs/2010.0086?ref=11771562 where all 168 primitive zero divisors for the standard Sedenion form mentioned are itemized, the 15 Octonion subalgebra candidates are itemized by Sedenion basis element subsets, an algebraic proof all 15 cannot be oriented as proper Octonion algebras, a method to determine if a set of seven oriented Quaternion triplets otherwise suitable for an Octonion Algebra actually describes one, a method to find all sets of 5 Octonion subalgebra candidates that cannot all be properly oriented and how this shows the maximal number of Octonion Algebra candidates that can be oriented as proper Octonion is eight; any seven sharing a common basis element plus any one of the other eight.

The definition variation for all Cayley-Dickson algebras through Sedenions is discussed in detail in my paper describing all valid Quaternion triplet orientations. A doubling process by variation is provided. There is much information provided that could improve the wiki for Octonion Algebra.

EightOnions (talk) 23:43, 14 January 2021 (UTC)

Flexible alternativity
In it is demonstrated that all Cayley - Dickson algebras are flexible alternative. I feel that only saying 'sedenions are not an alternative algebra', thought it is true, does not describe the essence of sedenions. — Preceding unsigned comment added by Crodrigue1 (talk • contribs) 01:44, 30 January 2021 (UTC)

Zero Divisor correction
I think there are 2 mistakes in the Sedenion Zero divisor list:

"(e1+e13)×(e7+e-14) = 0" is False, but "(e1+e13)×(e2+e-14) == 0" is True. "(e6+e10)×(e7+e-13) = 0" is False, but "(e6+e12)×(e7+e-13) == 0" is True.

Correction would then be:

{e1,e13,e7,e-14} -> changed to -> {e1,e13,e2,e-14} {e6,e10,e7,e-13} -> changed to -> {e6,e12,e7,e-13}

I am not math certified. I wrote a program to locate zero divisors and found these 2 differences. Maybe someone could verify. Peawormsworth (talk) 19:36, 6 August 2022 (UTC)

Names of further algebras
Names have been given to various further algebras, but this appears inconsistent. The 32 unit algebra is referred to in one section as the "pathions" but elsewhere as the "trigintaduonions" (this name presumably courtesy of Raoul E. Cawagas). Both of these are inconsistent with the established naming scheme given by the names of the quaternions, octonions and sedenions. The name of the sedenions shows that the 32 unit algebra following it should be the duotriginions, as the tens prefix (den-) is preceeded by the ones prefix (se-), and since the "ta" of "triginta" would be omitted to fit the established nomenclaturalogical pattern ("...-nions"). This would give:

Real numbers (1)

Complex numbers (2)

Quaternions (4)

Octonions (8)

Sedenions (16)

Duotriginions (32)

Quatersexaginions (64) *

Cenoctoviginions (128)

Duocensequinquaginions (256) *

and as the article says, ad infinitum.

*: The established prefixes "se" and "quater" might be considered only due to ease of pronunciation, and "sex" and "quattuor" (the correct latin prefixes) may in fact be more appropriate. HYPERIAPATH (talk) 22:03, 25 May 2024 (UTC)

Fixed Zero Divisors
I changed...

$$\{e_1,e_{13},e_7,-e_{14}\}$$

$$\{e_6,e_{10},e_7,-e_{13}\}$$

to...

$$\{e_1,e_{13},e_2,-e_{14}\}>$$

$$\{e_6,e_{12},e_7,-e_{13}\}$$

Because this is False...

$$(e_1+e_{13})\times (e_7-e_{14}) = 0$$

$$(e_6+e_{10})\times (e_7-e_{13}) = 0$$

Proof...

$$(e_1+e_{13})\times (e_7-e_{14}) = (e_1\times e_7)+(e_{13}\times e_7)-(e_1\times e_{14})-(e_{13}\times e_{14})$$

$$= (-e_6)+(e_{10})-(e_{15})-(-e_3) = e_3-e_6+e_{10}-e_{15}$$

$$(e_6+e_{10})\times (e_7-e_{13}) = (e_6\times e_7)+(e_{10}\times e_7)-(e_6\times e_{13})-(e_{10}\times e_{13})$$

$$= (e_1)+(-e_{13})-(e_{11})-(e_7) = e_1-e_7-e_{11}-e_{13}$$

And this is True...

$$(e_1+e_{13})\times (e_2-e_{14}) = 0$$

$$(e_6+e_{12})\times (e_7-e_{13}) = 0$$

Proof...

$$(e_1+e_{13})\times (e_2-e_{14}) = (e_1\times e_2)+(e_{13}\times e_2)-(e_1\times e_{14})-(e_{13}\times e_{14})$$

$$= (-e_3)+(e_{15})-(e_{15})-(-e_3) = 0$$

$$(e_6+e_{12})\times (e7-e{13}) = (e_6\times e_7)+(e_{12}\times e_7)-(e_6\times e_{13})-(e_{12}\times e_{13})$$

$$= (e_1)+(e_{11})-(e_{11})-(e_1) = 0$$

I made the change live, because I believe it is true and the last comment was deprecated.

Citation: the multiplication table of this wiki page. Peawormsworth (talk) 11:27, 23 June 2024 (UTC)