Let us take the octonions as having all integer coefficients and the multiplication table at BAEZ

We have a standard conjugation operator with $\bar{1} = 1$ and $\bar{e_i}= - e_i,$ extend by linearity. Then we get a norm, or squared Euclidean length, as $Nx = x \bar{x} = \bar{x} x,$ and the $N(xy) = Nx \; Ny.$ We get the real part $\mathcal Rx = \frac{1}{2} (x + \bar{x}). $ We will call an octonion $o$ pure if its real part is $0,$ that is $\bar{o} = -o.$ We also retrieve the ordinary dot product with $(x,y)= \mathcal R(x\bar{y}).$ Among a million other facts, multiplication on either the left or the right by a fixed octonion preserves angles, and multiplication by any of the $e_i$ rotates everything $90^\circ.$

The main law we need is the "alternative algebra" law, $$ (xy)x = x(yx). $$ I like the quick summary of the laws in Robert Wilson 2008 SEMINAR PDF

LEMMA: $$ (\bar{t} x) t = \bar{t} (x t) $$

Proof: Write $t = r + o $ with $r$ real and $o$ pure. The $\bar{t} = r - o.$ Calculation gives $$ (\bar{t} x) t = r^2 x + r x o - r o x - (ox)o, $$ while $$ \bar{t} (x t) = r^2 x + r x o - r o x - o(xo). $$ Since $ (ox)o = o (xo) $ the lemma follows.

To the question of interest, which was done for the quaternions as the matrix in formula (10) on page 755 of Pall Automorphs (1940) and Theorem 3 on page 176 of Jones and Pall (1939), both available as pdfs at TERNARY

Let $$ o = o_1 e_1 + o_2 e_2 + o_3 e_3 + o_4 e_4 + o_5 e_5 + o_6 e_6 + o_7 e_7 $$ be a "proper" pure quaternion, that is $\gcd(o_1,o_2,o_3,o_4,o_5,o_6,o_7) = 1,$ with square norm $$ No = m^2 = o_1^2 + o_2^2 + o_3^2 + o_4^2 + o_5^2 + o_6^2 + o_7^2 $$ for integer $m.$

QUESTION: is it necessarily the case that we can find some integral octonion $t,$ with $Nt = m,$ such that $$ o = \pm \bar{t} (e_j t) $$ for some $j =1,2,3,4,5,6,7?$

EDDITTT: I got an email from a kind fellow who has a Maple package for playing with octonions, who says that with $o = e_4 + e_5 + e_6 + e_7$ of norm 4 has no expression as $ o = \bar{t} (e_1 t). $ I have asked him if checking all possible $\pm e_j$ makes any difference. It is a big help to be able to experiment, I don't have any computer programs myself this time. I will put in his name if he says that is alright. Oh, his Maple program may be using a slightly different multiplication table from the one I picked, i think there are 240 or 480 possible versions.

Note that the proof by Jones and Pall uses associativity and, essentially, unique factorization (Theorem 2) for a proof by induction on the number of prime factors in out $m.$ That proof cannot work in quite such a pleasant manner. That is, their Theorem 2 is likely false, but perhaps this (Theorem 3) can be recovered as a very special case. Note that I can live with $m$ being odd, Jones and Pall needed to do that. For seven squares I do not think it should be necessary.

Oh, automorphs. It is fairly likely that every rational automorph of the sum of seven squares can be written with the seven rows being $\bar{t} (e_i t),$ and then divide by $m.$ I'm just saying.

EDIT, JUNE 7: note that the question to which I was trying to apply octonions has an affirmative answer due to J. S. Hsia, see Filling in a rational orthogonal matrix given one row while the octonion approach to seven variables was just not very effective. That is, it works for some 7-tuples but the percentage of failures goes up as the coefficients increase in size.


I'm not sure whether you were exploring a particular application of the octonions, but I want to point out that understanding the spaces where $o=\pm \bar{t} ( e_j t )$ exists are interesting. Papers like http://arxiv.org/abs/1002.1497 or http://arxiv.org/abs/0911.2255 use such constructs, and you're wondering about the further restriction of $t$ and $o$ being a 'proper octonion'. Thanks for posting your question, and some investigation. I wonder what actually are the spaces where a $t$ exists for some $e_j$ such that $o=\pm \bar{t} ( e_j t )$ for a given $o$? Quick question: With "integral octonion" did you mean any octonion that is made from integral coefficients? Or did you mean the (arguably) more customary use where "integral octonions" refers to a (reschaled) version of the E$_8$ lattice? Thanks, Jens

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    $\begingroup$ Hi, Jens. I meant integral coefficients, with the multiplication table from Baez. I understand there are several useful sorts of octonions. The reason i picked this one is that someone asked on MSE whether any vector of $n$ integers that also has integral length $m$ can be placed as the first row of a square matrix $P$ with $P P^T = m^2 I,$ that is $\frac{1}{m}P$ is rational and orthogonal. I found it true from $n = 2,3,4,8,$ false for $n=9,$ and computers suggested true for $n=5,6,7$ but I cannot be sure. math.stackexchange.com/questions/261204/… $\endgroup$ – Will Jagy Dec 27 '12 at 6:31

Sometimes nobody knows a question because it is just nonsense.

For distinct indices $i,j,k = 1,2,3,4,5,6,7,$ I have proved this little item by checking 35 triples in the Baez multiplication table: $$ (e_i e_j) e_k = (e_j e_k) e_i = (e_k e_i) e_j. $$ As a result, $$ (e_i e_j) e_k + (e_k e_j) e_i = 0. $$ I was pretty sure I had proved these with some calculations that forced the second version, but I also wanted to check, and it's true.

W. Edwin Clark, emeritus at the University of South Florida, told me the example i edited into the question. A little more is true: if $Nt = 2,$ then the only possibilities I found for $\bar{t} e_i t$ are $\pm 2, \pm 2 e_j.$

Next I went ahead and checked $Nt = 3.$ This is also a complete bust. The possibilities for $\bar{t} e_i t$ are $\pm 3, \pm 3 e_j, \; \pm e_j \pm 2 e_k \pm 2 e_m.$ There is simply not enough variety to create anything with $9 = 4 + 1 + 1 + 1 + 1 + 1 + 0.$

In conclusion, this was stupid, but now I know a bit more than I did.


I agree!

The process of taking Norm or Conjugates of Octonions (as premised by the question) creates a trivial answer to the question.

Just FYI, I too have a Mathematica notebook here that addresses octonions (and its Cayley-Dickson quaternion half). See that here. This is referenced in the mathematica.stackexchange.com forum here.

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    $\begingroup$ It is not clear in what way is this an answer to the question. $\endgroup$ – Mariano Suárez-Álvarez Jun 8 '13 at 1:44

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