# August 3, 1995 {#week59}

> _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_, Springer Verlag, 1994.

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
["Week 61"](#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, $\mathrm{SU}(2)$, rather than the group of
rotations in 3d space, $\mathrm{SO}(3)$, 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!

This game of getting new algebras from old is called the
"Cayley-Dickson" construction. Of course, the fun we've had so far
should make you want to keep playing this 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.
There is something out there called "Pfister forms", which I think
might be related.

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

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

For more about the inner meaning of these rules, see
["Week 5"](#week5). Similarly, we can unravel the above definition
of octonions by writing the octonion $(a+bi+cj+dk,e+fi+gj+hk)$ as
$$a + b e_1 + c e_2 + d e_3 + e e_4 + f e_5 + g e_6 + h e_7.$$
Note: since mathematicians are very impersonal, they usually call these
seven dwarves $e_1,\ldots,e_7$ 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, $e_3 e_7 = -e_7 e_3$. 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 looks like this:
$$\includegraphics[max width=0.65\linewidth]{../images/fano.jpg}$$

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 $e_1$, $e_2$, and $e_3$. 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 $e_1 e_2 = e_4$, $e_2 e_4 = e_1$, and $e_4 e_1 = e_2$. The lines
that are edges of the big triangle also point clockwise, so for example
$e_5 e_2 = e_3$, and cyclic permutations thereof, and $e_6 e_3 = e_4$.
The lines that go through the center point from the vertex to the
midpoint of the opposite edge, so for example $e_3 e_7 = e_1$. I hope
that made sense; you can work it out yourself, of course.

My convention for numbering the seven dwarves in the picture above is
*completely arbitrary*, so don't bother remembering it --- make up your
own if you prefer! The convention I chose looks sort of weird at first,
but it has a couple of endearing features:

-  Index cycling: if $e_i e_j = e_k$, then $e_{i+1} e_{j+1} = e_{k+1}$.
-  Index doubling: if $e_i e_j = e_k$, then $e_{2i} e_{2j} = e_{2k}$.

Here we add and multiply $\mod 7$. Index doubling corresponds to rotating
the Fano plane.

So those are the octonions in a nutshell. I should say a bit about how
they relate to triality for $\mathrm{SO}(8)$, the exceptional Lie group $\mathrm{G}_2$, the
group $\mathrm{SU}(3)$ which is so important in the study of the strong force, and
to lattices like $\mathrm{E}_8$, $\Lambda 16$ and the Leech lattice. But I will postpone
that; for now you can consult Fulton and Harris, and also various papers
by Dixon:

3) Geoffrey Dixon, "Octonion X-product orbits", preprint available as [`hep-th/9410202`](https://arxiv.org/abs/hep-th/9410202).

    "Octonion X-product and $\mathrm{E}_8$ lattices", preprint available as [`hep-th/9411063`](https://arxiv.org/abs/hep-th/9411063).

    "Octonions: $\mathrm{E}_8$ lattice to $\Lambda 16$", preprint available as [`hep-th/9501007`](https://arxiv.org/abs/hep-th/9501007).

    "Octonions: invariant representation of the Leech lattice", preprint available as [`hep-th/9504040`](https://arxiv.org/abs/hep-th/9504040).

    "Octonions: invariant Leech lattice exposed", preprint available as [`hep-th/9506080`](https://arxiv.org/abs/hep-th/9506080).

I am not presently in a position to assess these papers or Dixon's work
relating division algebras and the Standard Model, but hopefully
sometime I will be able to say a bit more.

Let me wrap up by saying a bit about the Leech lattice. As described in
my review of Conway and Sloane's book (["Week 20"](#week20),
there is a wonderful branch of mathematics that studies the densest ways
of packing spheres in n dimensions. Most of the results so far concern
lattice packings, packings in which the centers of the spheres form a
subset of $\mathbb{R}^n$ closed under addition and scalar multiplication by
integers. When $n = 8$, the densest known packing is given by the
so-called $\mathrm{E}_8$ lattice. In ["Week 20"](#week20) I described how to
get this lattice using the quaternions and the icosahedron. Briefly, it
goes as follows. The group of rotational symmetries of the icosahedron
(not counting reflections) is a subgroup of the rotation group $\mathrm{SO}(3)$
containing 60 elements. As mentioned above, $\mathrm{SO}(3)$ has as a double cover
the group $\mathrm{SU}(2)$ of unit quaternions. So there is a 120-element subgroup
of $\mathrm{SU}(2)$ consisting of elements that map to elements of $\mathrm{SO}(3)$ that are
symmetries of the icosahedron. Now form all integer linear combinations
of these 120 special elements of $\mathrm{SU}(2)$. We get a subring of the
quaternions known as the "icosians''.

We can think of icosians as special quaternions, but we can also think
of them as special vectors in $\mathbb{R}^8$, as follows. Every icosian is of the
form
$$(a + \sqrt{5} b) + (c + \sqrt{5} d)i + (e + \sqrt{5} f)j + (g + \sqrt{5} h)k$$
with $a,b,c,d,e,f,g,h$ rational --- but not all rational values of $a,\ldots,h$
give icosians. The set of all vectors $x = (a,b,c,d,e,f,g,h)$ in $\mathbb{R}^8$ that
correspond to icosians in this way is the $\mathrm{E}_8$ lattice!

The Leech lattice is the densest known packing in 24 dimensions. It has
all sorts of remarkable properties. Here is an easy way to get ones
hands on it. First consider triples of icosians $(x,y,z)$. Let $L$ be the
set of such triples with
$$x = y = z \mod h$$
and
$$x + y + z = 0 \mod h^*$$
where $h$ is the quaternion $(-\sqrt{5} + i + j + k)/2$. Since we can think of an
icosian as a vector in $\mathbb{R}^8$, we can think of $L$ as a subset of $\mathbb{R}^{24}$. It
is a lattice, and in fact, it's the Leech lattice! I have a bit more to
say about the Leech lattice in ["Week 20"](#week20), but the real
place to go for information on this beast is Conway and Sloane's book.
It turns out to be related to all sorts of other "exceptional''
algebraic structures. People have found uses for many of these in string
theory, so if string theory is right, maybe they are important in
physics. Personally, I want to understand them more deeply as pure
mathematics before worrying too much about their applications to
physics.

------------------------------------------------------------------------

Here is what Toby Bartels wrote:

> From: Toby Bartels
> Subject: Re: why hexadecanions have no inverses
> To: John Baez
> Date: Sun, 20 Aug 1995 

> 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,
> $$
>   \begin{gathered}
>     (a (c + e) - (d + f)^* b, (d + f) a + b (c + e)^*)
>   \\= (a c + a e - d^* b - f^* b, d a + f a + b c^* + b e^*)
>   \\= (a c - d^* b, d a + b c^*) + (a e - f^* b, f a + b e^*)
>   \\= (a, b) (c, d) + (a, b) (e, f).
>   \end{gathered}
> $$
> Also,
> $$
>   \begin{gathered}
>     ((a, b) + (c, d)) (e, f)
>   \\= (a + c, b + d) (e, f)
>   \\= ((a + c) e - f^* (b + d), f (a + c) + (b + d) e^*).
>   \end{gathered}
> $$
> By the induction hypothesis again,
> $$
>   \begin{gathered}
>     ((a + c) e - f^* (b + d), f (a + c) + (b + d) e^*)
>   \\= (a e + c e - f^* b - f^* d, f a + f c + b e^* + d e^*)
>   \\= (a e - f^* b, f a + b e^*) + (c e - f^* d, f c + d e^*)
>   \\= (a, b) (e, f) + (c, d) (e, f).
>   \end{gathered}
> $$
> 
> $$((a, b) (c, d))^* = (a c - d^* b, d a + b c^*)^* = ((a c - d^* b)^*, -(d a + b c^*)).$$
> Using the induction hypothesis and each of the above identities,
> $$
>   \begin{gathered}
>     ((a c - d^* b)^*, -(d a + b c^*))
>   \\= (c^* a^* - (-b)^* (-d), -d a + (-b) c^*)
>   \\= (c^*, -d) (a^*, -b)
>   \\= (c, d)^* (a, b)^*.
>   \end{gathered}
> $$
> 
> 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.
> $$
>   \begin{gathered}
>     ((a, b) (c, d)) (e, f)
>   \\= (a c - d^* b, d a + b c^*) (e, f)
>   \\= ((a c - d^* b) e - f^* (d a + b c^*), f (a c - d^* b) + (d a + b c^*) e^*)
>   \\=((ac)e - (d^* b)e - f^* (da) - f^* (b c^*), f(ac) - f(d^* b) + (da) e^* + (b c^*) e^*).
>   \end{gathered}
> $$
> On the other hand,
> $$
>   \begin{gathered}
>     (a, b) ((c, d) (e, f))
>   \\= (a, b) (c e - f^* d, f c + d e^*)
>   \\= (a (c e - f^* d) - (f c + d e^*)^* b, (f c + d e^*) a + b (c e - f^* d)^*)
>   \\= (a(ce) - a(f^* d) - (c^* f^*)b - (e d^*)b, (fc)a + (d e^*)a + b(e^* c^*) - b(d^* f)).
>   \end{gathered}
> $$
> 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.
> $$
>   \begin{gathered}
>     (a, b) ((a, b) (c, d))
>   \\= (a, b) (a c - d^* b, d a + b c^*)
>   \\= (a (a c - d^* b) - (d a + b c^*)^* b, (d a + b c^*) a + b (a c - d^* b)^*)
>   \\= (a(ac) - a(d^* b) - (a^* d^*)b - (c b^*)b, (da)a + (b c^*)a + b(c^* a^*) - b(b^* d)).
>   \end{gathered}
> $$
> Meanwhile,
> $$
>   \begin{gathered}
>     ((a, b) (a, b)) (c, d)
>   \\= (a a - b^* b, b a + b a^*) (c, d)
>   \\= ((aa)c - (b^* b)c - d^* (ba) - d^* (b a^*),d(aa) - d(b^* b) + (ba) c^* + (b a^*) c^*).
>   \end{gathered}
> $$
> 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 non-$0$ $x$ and $y$ such that $x y = 0$.
> 
> Just playing around, I found
> $$
>   \begin{gathered}
>     (e_1, e_4) (-1, e_5)
>   \\= (e_1 (-1) - (e_5)* e_4, e_5 e_1 + e_4 (-1)*)
>   \\= (-e_1 + e_5 e_4, e_5 e_1 - e_4).
>   \end{gathered}
> $$
> 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

By the way, in a post to `sci.physics.research` on November 2, 1999, Ralph
Hartley pointed out that even if we start with a field of characteristic
2, repeatedly applying the Cayley-Dickson construction will *not* lead
to an infinite sequence of division algebras, because it's not true in
this case that if $x$ is nonzero, $xx^*$ is nonzero. The problem is that a
field of characteristic 2 can't be an ordered field.