From: baez@galaxy.ucr.edu (John Baez)
Subject: Re: This Week's Finds in Mathematical Physics (Week 141)
Date: 29 Oct 1999 19:41:42 -0700
Newsgroups: sci.physics.research,sci.physics,sci.math
Keywords: (oct)onions and other algebras over the reals

In article <941232837.24226.5.nnrp-07.c1ed2859@news.demon.co.uk>,
marijke wrote:

>On 26 Oct 1999 14:52:54 -0700, baez@math.ucr.edu (John Baez) wrote:

>> but now S^7, the unit octonions, doesn't form a group - because the
>> octonions aren't associative.

>What exactly are octonions? 

Ah, what an excellent question!  I've written a lot about octonions, 
but if you just want to know what they *are*, the stuff near the end
of this post should answer your question.  

>I had always unthinkingly assumed the term
>must refer to quaternions with complex number components -- but those
>*are* associative (for multiplication) AFAICS. 

No, those guys are called the "complex quaternions" or "biquaternions",
not "octonions".  And you're right, they're associative.  In fact, they 
form an algebra isomorphic to the algebra of 2x2 complex matrices!  So
when people use 2x2 complex matrices to study "spin" in quantum mechanics,
they're really using complex quaternions - they just don't know it.  :-)

>These critters aren't a group (for multiplication) either BTW, even
>when excluding 0, because you can have nilfactors: A*B can be 0
>without either A or B being 0, and then A and B don't have
>multiplicative inverses.

Right.

>The unit length ones of them are a group OTOH -- if you define length
>a certain way (whereby square modulus can be positive, 0 or negative,
>a bit like in Minkowski space, with nilfactors in a lightlike role). 

Hmm, I only see how to do this if you let the lengths be complex.  Am
I confused?  

>PS intriguing that inconsistent math... thank you for always digging
>out this weird & wonderful stuff for us!

You're welcome!  Glad you like it!  That book on "Inconsistent
Mathematics" just jumped right off the shelf and into my hand -
I couldn't stop it.

......................................................................

Also available at http://math.ucr.edu/home/baez/week59.html

August 3, 1995
This Week's Finds in Mathematical Physics (Week 59)
John Baez

   As you crack your eyes one morning your reason is assaulted by
   a strange sight.  Over your head, humming quietly, there floats
   a monitor, an ethereal otherworldly screen on which is written a
   curious message.  "I am the Screen of ultimate Truth.  I am bulging
   with information and ask nothing better than to be allowed to impart
   it."

It would be nice if more math books started with something attention-
grabbing like this.  In fact, it appears near the beginning of

1) Geoffrey M. Dixon, Division Algebras: Octonions, Quaternions,
Complex Numbers and the Algebraic Design of Physics, Kluwer Press,
ISBN 0-7923-2880-6.

Dixon is convinced that the details of the Standard Model of particle
interactions can be understood better by taking certain mathematical
structures very seriously.  There are very few algebras over the reals
where we can divide by nonzero elements: if we demand associativity
and commutativity, just the reals themselves and the complex numbers.
If we drop the demand for commutativity, we also get a 4-dimensional
algebra called the quaternions, invented by Hamilton.  If in addition
we drop the demand for associativity, and ask only that our algebra be
"alternative", we also get an 8-dimensional algebra called the
octonions, or Cayley numbers.  (I'll say what "alternative" means
in "week61".)  Clearly these are very special structures, and also clearly
they play an important role in physics... or do they?

Well, few people doubt that the real numbers are fundamental to physics
(though some advocates of the discrete might prefer the integers), and
with emergence of quantum theory, if not sooner, the basic role of the
complex numbers also became clear.  Hamilton discovered the quaternions
in the 1800s, and used them to formulate a beautiful theory of rotations
in 3-dimensional space.  They fell out of favor somewhat when the
vectors of Gibbs proved simpler for many purposes, but their deeper
importance became clear when people started studying spin: indeed, the
Pauli matrices so important in physics are closely related to the
quaternions, and it is the group of unit quaternions, SU(2), rather than
the group of rotations in 3d space, SO(3), which turns out to be the
symmetry group whose different representations correspond to particles
of different spin.  But what about the octonions?

Well, there are not too many places in physics yet where the octonions
reach out and grab one with the force the reals, complexes, and
quaternions do.  But they are certainly out there, they have a certain
beauty to them, and they are the natural stopping-point of a certain
finite sequence of structures, so it is natural for people of a certain
temperament to believe that they are there for a reason.  Dixon makes a
passionate case for this in the beginning of his book.

Suppose you were confronted with the Screen of Truth.  What would you
ask it?  Dixon, being a physicist, naturally fantasizes asking it why
the universe is the way it is!  What kind of answer could this possibly
have?  Perhaps there is only one consistent way for things to be, and
mathematics, with its unique and beautiful structures that are pure
expressions of logical necessity, is trying to tell us something about
this?

On the one hand this seems outrageous... especially to the hard-nosed
pragmatist or empiricist in us.  It seems old-fashioned, naive, and too
good to be true.  On the other hand, the universe *is* outrageous!  It's
outrageous that it exists in the first place, and doubly outrageous that
it has the particular physical laws it does and no others.  It has only
been through the old-fashioned, naive belief that we can understand it
using mathematics that we discovered what we have of its physical laws.
So maybe eventually we *will* see that the basic structures of
mathematics determine, in some mysterious sense, all the basic laws of
physics.  Or maybe we won't.  In either case, there is a long way yet to
go.  As Dixon's Screen of Truth comments, before it departs:

  Do you believe that were I to explain as much of what I know
  as you could comprehend that you would recognize it, that you 
  would say, oh yes, this is but an extension of what we have  
  already done, and though the mathematics is broader, the principles
  deeper, I am not surprised?  Do you think you have asked even a
  fraction of the questions you need to ask?

Anyway, it is at least worth considering all the beautiful mathematical
structures one runs into for their potential importance.  For example,
the octonions.  

In order to write this week's Finds, I needed to learn a little about
the octonions.  I wanted some good descriptions of the octonions, that
hopefully would "explain" them or at least make them easy to remember.
So I asked for help on sci.physics.research, and I got some help from
Greg Kuperberg, Ezra Getzler, Matthew Wiener, and Alexander Vlasov.
After a while Geoffrey Dixon got wind of this and referred me to his
work!  I'll probably talk to him later this summer when I go back to
Cambridge Massachusetts, and hopefully I'll learn more about octonions
and the like.

But for now let me just give a quick beginner's introduction to the
octonions.  A lot of this appears in 

2) William Fulton and Joe Harris, Representation Theory --- a First
Course, Springer Verlag, Berlin, 1991.

I should add that this book is a very good place to learn about
Lie groups, Lie algebras, and their representations... I wish I had
taken a course based on this book when I was in grad school!

Let's start with the real numbers.  Then the complex number

                    a+bi

can be thought of as a pair

                    (a,b)

of real numbers.  Addition is done component-wise, and 
multiplication goes like this:

            (a,b)(c,d) = (ac - db,da + bc)

We can also define the conjugate of a complex number by

                (a,b)* = (a,-b).

Now that we have the complex numbers, we can define the
quaternions in a similar way.  A quaternion can be thought of 
as a pair

                        (a,b)

of complex numbers.  Addition is component-wise and 
multiplication goes like this

            (a,b)(c,d) = (ac - d*b, da + bc*)

This is just like how we defined multiplication of
complex numbers, but with a couple of conjugates (*'s)
thrown in.  To emphasize how similar the two multiplications
are, we could have included the conjugates in the first 
formula, since the conjugate of a real number is just itself. 

We can also define the conjugate of a quaternion by

               (a,b)* = (a*,-b).

The game continues!  Now we can define an octonion to be a pair 
of quaternions; as before, we add these component-wise and
multiply them as follows:

           (a,b)(c,d) = (ac - d*b, da + bc*).

One can also define the conjugate of an octonion by

               (a,b)* = (a*,-b).

Why do the real numbers, complex numbers, quaternions
and octonions have multiplicative inverses?  I take it as
obvious for the real numbers.  For the complex numbers, 
you can check that

(a,b)* (a,b) = (a,b) (a,b)* = K (1,0)

where K is a real number called the "norm squared" of (a,b).
The multiplicative identity for the complex numbers is (1,0).
This means that the multiplicative inverse of (a,b) is (a,b)*/K.  
You can check that the same holds for the quaternions and 
octonions!  

Of course, all this should make you want to keep playing the 
game and develop a 16-dimensional algebra, the "hexadecanions,"
consisting of pairs of octonions equipped with the same sort
of multiplication law.  What do you get?  Why aren't there
multiplicative inverses anymore?  I haven't checked, because
this is my summer vacation!  I am learning about octonions 
just for fun, since I just finished writing some rather technical papers, 
and my idea of fun does not presently include multiplying two 
hexadecanions together to see why the norm-squared law (a,b) (a,b)* =
(a,b)* (a,b) = K (1,0) breaks down.  But I'm sure someone out there 
will enjoy doing this... and I'm sure someone else out there has already 
done it!  So they should let me know what happens.  

[Toby Bartels did some nice work on hexadecanions in response to
the above challenge, which appears at the end of this article.]

Now if we unravel the above definition of quaternions, by writing
the quaternion (a+bi,c+di) as a+bi+cj+dk, we see that the multiplication
law is 

i^2 = j^2 = k^2 = -1, 

and 

ij = -ji = k,  jk = -kj = i,  ki = -ik = j.

For more about the inner meaning of these rules, see "week5".
Similarly, we can unravel the above definition of octonions by 
writing the octonion (a+bi+cj+dk,e+fi+gj+hk) as

a + b e1 + c e2 + d e3 + e e4 + f e5 + g e6 + h e7.

Note: since mathematicians are very impersonal, they usually call these
seven dwarves e1,...,e7 instead of Sleepy, Grumpy, etc. as in the Disney
movie.  Any one of these 7 guys times himself is -1.  Also, any two
distinct ones anticommute; for example, e3 e7 = -e7 e3.  There is a nice
way to remember how to multiply them using the "Fano plane".  This is a
projective plane with 7 points, where by a "projective plane" I mean
that any two points determine an abstract sort of "line", which in this
case consists of just 3 points, and any two lines intersect in a point.
It's a bit of a pain to draw this plane using ASCII!  The book by Fulton
and Harris has a nice picture, but the best I can do is:


                      e5


              e2              e3


                      e4 


  e7                  e1                   e6


The "lines" are the 3 edges of the big triangle, the 3 lines going
through a vertex, the center and the midpoint of the opposite edge, and
the circle including e1, e2, and e3.  All the "lines" are cyclically
ordered, and that tells you how to multiply the seven dwarves.  For
example, the line that's actually a circle goes clockwise, so e1 e2 =
e3, e2 e3 = e1, and e3 e1 = e2.  The lines that are edges of the big
triangle also point clockwise, so for example e5 e3 = e6, and cyclic
permutations thereof, and e1 e7 = e6.  The lines that go through the
center point from the vertex to the midpoint of the opposite edge, so
for example e7 e4 = e3.  I hope that made sense; you can work it out
yourself, of course.

So those are the octonions in a nutshell.  

Here is what Toby Bartels wrote:

From  toby@ugcs.caltech.edu Sun Aug 20 23:22:43 1995
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Subject: Re: why hexadecanions have no inverses
To: baez@math.ucr.edu (john baez)
Date: Sun, 20 Aug 1995 23:00:37 -0700 (PDT)
In-Reply-To: <9508172051.AA25548@math.ucr.edu> from "john baez" at Aug 17, 95 01:51:49 pm

I spent a couple days thinking about why hexadecanions have no inverses,
and the first thing I want to say about it is that they do.
However, these inverses are of limited applicability,
because the hexadecanions are not a division algebra.
A division algebra allows you to conclude, given x y = 0, that x or y is 0.
If your algebra has inverses, you might try to multiply this equation
by the inverse of x or y (whichever one isn't 0) to prove the other is 0,
but this only works if the algebra is associative.
Since the octonions and hexadecanions aren't associative,
there's no reason (yet) to think either of these is a division algebra.
It turns out that the octonions are a division algebra,
despite not being associative, but the hexadecanions aren't.

Why aren't the hexadecanions a division algebra?
Because the real numbers aren't of characteristic 2.
Allow me to explain.

I will prove below that the 2^n onions are a division algebra
only if the 2^(n-1) onions are associative.
So, the question becomes: why aren't the octonions associative?
Well, I've found a proof that 2^n onions are associative
only if 2^(n-1) onions are commutative.
So, why aren't the quaternions commutative?
Again, I have a proof that 2^n onions are commutative
only if 2^(n-1) onions equal their own conjugates.
So, why don't the complex numbers equal their own conjugates?
I have a proof that 2^n onions do equal their own conjugates,
but it works only if the 2^(n-1) onions are of characteristic 2.
The real numbers are *not* of characteristic 2,
so the complex numbers don't equal their own conjugates,
so the quaternions aren't commutative,
so the octonions aren't associative,
so the hexadecanions aren't a division algebra.

I require a few identities about conjugates that hold for all 2^n onions:
(x*)* = x, (x + y)* = x* + y*, and (x y)* = y* x*.
(If these identities are reminiscent of identities for transposes of matrices,
it is no coincidence.)
I will prove these by induction.
That is, if an identity holds for 2^(n-1) onions,
I show it holds for 2^n onions.
Since they hold trivially for the reals (n = 0), they hold for all.

((a, b)*)* = (a*, -b)* = ((a*)*, -(-b)).
By the induction hypothesis and the nature of addition (an Abelian group),
((a*)*, -(-b)) = (a, b).

((a, b) + (c, d))* = (a + c, b + d)* = ((a + c)*, -(b + d)).
By the induction hypothesis and addition again, ((a + c)*, -(b + d))
= (a* + c*, -b + -d) = (a*, -b) + (c*, -d) = (a, b)* + (c, d)*.

The next proof needs the distribution of multiplication over addition.
(a, b) ((c, d) + (e, f)) = (a, b) (c + e, d + f)
= (a (c + e) - (d + f)* b, (d + f) a + b (c + e)*).
By the induction hypothesis and the identity immediately above,
(a (c + e) - (d + f)* b, (d + f) a + b (c + e)*)
= (a c + a e - d* b - f* b, d a + f a + b c* + b e*)
= (a c - d* b, d a + b c*) + (a e - f* b, f a + b e*)
= (a, b) (c, d) + (a, b) (e, f).
Also, ((a, b) + (c, d)) (e, f) = (a + c, b + d) (e, f)
= ((a + c) e - f* (b + d), f (a + c) + (b + d) e*).
By the induction hypothesis again,
((a + c) e - f* (b + d), f (a + c) + (b + d) e*)
= (a e + c e - f* b - f* d, f a + f c + b e* + d e*)
= (a e - f* b, f a + b e*) + (c e - f* d, f c + d e*)
= (a, b) (e, f) + (c, d) (e, f).

((a, b) (c, d))* = (a c - d* b, d a + b c*)* = ((a c - d* b)*, -(d a + b c*)).
Using the induction hypothesis and each of the above identities,
((a c - d* b)*, -(d a + b c*)) = (c* a* - (-b)* (-d), -d a + (-b) c*)
= (c*, -d) (a*, -b) = (c, d)* (a, b)*.

In light of the above identities, if I ever come across, say, (x y* + z)*,
I'll simply write y x* + z* without a moment's hesitation.

Since inductive proofs have been so useful,
I'll use one to prove that 2^n onions always have inverses.
First, I'll extend the method in John's article,
beginning with an inductive proof that x x* = x* x is real.
(a, b) (a, b)* = (a, b) (a*, -b) = (a a* + b* b, 0),
and (a, b)* (a, b) = (a*, -b) (a, b) = (a* a + b* b, 0).
The inductive hypothesis states that both a* a = a a* and b* b are real,
so (a, b) (a, b)* = (a, b)* (a, b) is real.
Since the sum of a positive real and a nonnegative real is positive,
I can take this as a proof by induction that x x* = x* x is not only real,
but is also positive unless x = 0 (which will be important).
All you have to do now is check that these things are true of the 2^0 onions,
and they are, quite trivially (since the 2^0 onions are the reals).

Since the 2^n onions are always a vector space over the reals
(as mentioned in John's article), x (x* / (x x*)) = (x x*) / (x x*) = 1.
Since one can always divide by the real x x*,
the inverse of x is x* / (x x*) in any 2^n onion algebra.

To continue with the streak of inductive proofs,
I will now try to prove that the 2^n onions are always a division algebra.
(I will fail.)
Assume 0 = (0, 0) = (a, b) (c, d) = (a c - d* b, d a + b c*).
This gives the system of equations a c - d* b = 0 = d a + b c*.
Multiplying, (a c) c* - (d* b) c* = 0 c* = 0 = d* 0 = d* (d a) + d* (b c*).
If 2^(n-1) onions are associative, I can add the equations to get
a (c c*) + (d* d) a = 0.
Since c c* and d* d are real, they commute with a,
and the division algebra nature of 2^(n-1) onions allows me to conclude that
c c* + d* d = 0 (which implies c = d = 0 in light of positive definiteness)
or that a = 0 (from which the original equation gives b = 0).
Thus, the octonions are a division algebra
(since the quaternions are associative),
but the hexadecanions aren't (since the octonions aren't associative).

(If you're reading carefully, you realize that I haven't really proved
that the hexadecanions aren't a division algebra.
I've failed to prove that they are, but that's not the same thing.
When I first wrote this, I wasn't reading carefully;
I will return to plug this hole later.)

Thus, the 2^n onions are a division algebra
iff the 2^(n-1) onions are a division algebra *and are associative*.
So, let's try to prove associativity of 2^n onions by induction.
((a, b) (c, d)) (e, f) = (a c - d* b, d a + b c*) (e, f)
= ((a c - d* b) e - f* (d a + b c*), f (a c - d* b) + (d a + b c*) e*)
=((ac)e - (d* b)e - f* (da) - f* (b c*), f(ac) - f(d* b) + (da) e* + (b c*) e*).
On the other hand, (a, b) ((c, d) (e, f)) = (a, b) (c e - f* d, f c + d e*)
= (a (c e - f* d) - (f c + d e*)* b, (f c + d e*) a + b (c e - f* d)*)
= (a(ce) - a(f* d) - (c* f*)b - (e d*)b, (fc)a + (d e*)a + b(e* c*) - b(d* f)).
Assuming the induction hypothesis that 2^(n-1) onions are associative,
these are equal in general iff 2^(n-1) onions *also* are commutative.

Thus, 2^n onions are associative
iff 2^(n-1) onions are associative *and are commutative*.
So, let's try to prove commutativity of 2^n onions by induction.
(a, b) (c, d) = (a c - d* b, d a + b c*).
On the other hand, (c, d) (a, b) = (c a - b* d, b c + d a*).
Assuming the induction hypothesis that 2^(n-1) onions are commutative,
these are equal in general iff 2^(n-1) onions *also* equal their own conjugates.

Thus, 2^n onions are commutative
iff 2^(n-1) onions are commutative *and equal their own conjugates*.
So, let's try to prove conjugate equality of 2^n onions by induction.
(a, b) = (a, b).  On the other hand, (a, b)* = (a*, -b).
Assuming the induction hypothesis that 2^(n-1)onions equal their own conjugates,
these are equal in general iff 2^(n-1) onions *also* have characteristic 2.
(b = -b means 0 = b + b = 1 b + 1 b = (1 + 1) b = 2 b;
this is true in general iff 0 = 2, which is what characteristic 2 means.)

Thus, 2^n onions equal their own conjugates
iff 2^(n-1) onions equal their own conjugates *and have characteristic 2*.
Since the reals *don't* have characteristic 2,
there's no point in trying to prove anything about that by induction.
However, it's a general result that any algebra has characteristic 2
if it has a superalgebra of characteristic 2.
Since the 2^n onions are all superalgebras of the reals
(which means the reals are always isomorphic to a subset of the 2^n onions),
none of the 2^n onions can have characteristic 2 if the reals don't.

In summary, the definition of the reals as the complete ordered field,
along with an initial definition that x* = x in the reals,
allows trivial proofs that:
they form a division algebra, they are associative, they are commutative,
and they equal their own conjugates, but they don't have characteristic 2.
(All of these, in fact, are true of any ordered field
with this definition of conjugate, complete or not.)
From  this and the above considerations, the complex numbers form
a division algebra, are associative, and are commutative,
but they neither equal their own conjugates nor have characteristic 2.
From  this, the quaternions form a division algebra and are associative, but they
neither are commutative, equal their own conjugates, nor have characteristic 2.
From  this, the octonions form a division algebra but they neither associative,
are commutative, equal their own conjugates, nor have characteristic 2.
Finally, the hexadecanions neither form a division algebra, are associative,
are commutative, equal their own conjugates, nor have characteristic 2.

At this point, I must return to the logical hole I mentioned earlier.
But I want to work with a different algebraic concept than a division algebra;
instead I will use (inspired by Doug Merrit's post to sci.physics.research)
what I guess is called `alternativity', which says x (x y) = (x x) y.
I don't like putting alternativity into the pattern,
since associativity implies alternativity.
All the other properties (commutativity, conjugate equality, characteristic)
are logically independent in general.
I'd like to prove that every associative 2^n onion algebra is alternative,
just as I proved every commutative one was associative,
without its having been obvious to begin with.
Well, I will be disappointed even more badly later on.

Taking the conjugate of x (x y) = (x x) y, (y* x*) x* = y* (x* x*),
so left alternativity implies right alternativity, for 2^n onions.

I require an additional general identity of 2^n onions.
Earlier, I proved by induction that x x* was real,
but now I need the reality of x + x*.
Like everything else, this is proved by induction.
(a, b) + (a, b)* = (a, b) + (a*, -b) = (a + a*, 0).
Thus, if a + a* is real, (a, b) + (a, b)* is real.
Since x + x* is real when x is real, x + x* is real when x is any 2^n onion.

Now suppose we're working in an alternative 2^n onion algebra.
x (x y) + x* (x y) = (x + x*) (x y).  Since x + x* is real, it associates,
so x (x y) + x* (x y) = ((x + x*) x) y = (x x) y + (x* x) y.
Since x (x y) = (x x) y, x* (x y) = (x* x) y, which will be needed.

Let's attempt to prove by induction that 2^n onions are always alternative.
(a, b) ((a, b) (c, d)) = (a, b) (a c - d* b, d a + b c*)
= (a (a c - d* b) - (d a + b c*)* b, (d a + b c*) a + b (a c - d* b)*)
= (a(ac) - a(d* b) - (a* d*)b - (c b*)b, (da)a + (b c*)a + b(c* a*) - b(b* d)).
Meanwhile, ((a, b) (a, b)) (c, d) = (a a - b* b, b a + b a*) (c, d)
=((aa)c - (b* b)c - d* (ba) - d* (b a*), d(aa) - d(b* b) + (ba) c* + (b a*) c*).
These are indeed equal in general iff 2^(n-1) onions are associative.

The last sentence may not be immediately obvious.
The induction hypothesis and its corollaries leave us with
x (y z) + (x* y) z = y (z x) + y (z x*) as a necessary and sufficient condition.
It may not be clear that associativity implies this, much less vice versa.
However, the reality of x + x* once more enters the picture.
y (z x) + y (z x*) = y (z (x + x*)) = (x + x*) (y z) = x (y z) + x* (y z).
Thus, the condition becomes x (y z) + (x* y) z = x (y z) + x* (y z),
which is equivalent, in the general case, to associativity.

To sum up the findings so far:
For any n, the 2^n onions form a vector space over the reals.
x + x* and x x* are real if x is any 2^n onion; additionally, x x* = x* x.
Every 2^n onion has an inverse, which is a real multiple of its conjugate.
Conjugation is analogous to matrix transposition in that
(x*)* = x, (x + y)* = x* + y*, and (x y)* = y* x*.
Multiplication distributes over addition every time.
For no n do all 2^n onions equal their own negatives.
2^(n+1) onions equal their own conjugates iff
2^n onions equal their own conjugates and their own negatives.
all 2^(n+1) onions commute iff
all 2^n onions commute and equal their own conjugates.
2^(n+1) onions are associative iff 2^n onions are associative and commutative.
2^(n+1) onions are alternative iff 2^n onions are alternative and associative.
The 2^n onions form a division algebra if they are alternative.

I will be satisfied if I can prove the converse of the last statement.
In light of the results about alternativity, my original attempt
to prove that division of 2^n onions requires associativity of 2^(n-1) onions
looks even more convincing, (since alternativity of 2^(n-1) onions
can be included in the induction hypothesis), but it's still not valid.
I still haven't shown that, if 2^(n-1) onions aren't alternative,
there *must* be non0 2^n onions x and y such that x y = 0.
There doesn't seem to be any reason why there shouldn't be,
but there just might happen not to be any.
So, despite the inelegance of it all,
in order to prove that the hexadecanions aren't a division algebra,
I'm forced to exhibit non0 x and y such that x y = 0.

Just playing around, I found (e_1, e_4) (-1, e_5) =
(e_1 (-1) - (e_5)* e_4, e_5 e_1 + e_4 (-1)*) = (-e_1 + e_5 e_4, e_5 e_1 - e_4).
Since e_5 e_4 = (0, i) (0, 1) = (i, 0) = e_1
and e_5 e_1 = (0, i) (i, 0) = (0, i* i) = (0, 1) = e_4,
(e_1, e_4) (-1, e_5) = (0, 0) = 0.

The 2^n onions can't be a division algebra if the 2^(n-1) onions aren't.
If x y = 0 in the 2^(n-1) onions, (x, 0) (y, 0) = (x y, 0) = (0, 0) = 0.
Thus, the octonions and below are the only 2^n onions to be division algebras.
Still, I wish I had a proof of this that didn't require
the ugly brute force use of a specific counterexample.
(This is the interested reader's cue ...)


-- Toby
   toby@ugcs.caltech.edu