# What specific algebraic properties are broken at each Cayley-Dickson stage beyond octonions?

I'm starting to come around to an understanding of hypercomplex numbers, and I'm particularly fascinated by the fact that certain algebraic properties are broken as we move through each of the $2^n$ dimensions. I think I understand the first $n<4$ instances:

• As we move from $\mathbb{R}$ to $\mathbb{C}$ we lose ordering
• From $\mathbb{C}$ to $\mathbb{H}$ we lose the commutative property
• From $\mathbb{H}$ to $\mathbb{O}$ we lose the associative property (in the form of $(xy)z \neq x(yz)$, but apparently it's still alternative and $(xx)y = x(xy)$. Is that right?)
• The move from $\mathbb{O}$ to $\mathbb{S}$ is where I start to get fuzzy. From what I've read, the alternative property is broken now, such that even $(xx)y \neq x(xy)$ but that also zero divisors come into play, thus making sedenion algebra non-division.

My first major question is: Does the loss of the alternative property cause the emergence of zero divisors (or vice versa) or are these unrelated breakages?

My bigger question is: What specific algebraic properties break as we move into 32 dimensions, then into 64, 128, 256? I've "read" the de Marrais/Smith paper where they coin the terms pathions, chingons, routons and voudons. At my low level, any initial "reading" of such a paper is mostly just intent skimming, but I'm fairly certain they don't address my question and are focused on the nature and patterns of zero divisors in these higher dimensions. If the breakages are too complicated to simply explicate in an answer here, I'm happy to do the work and read journal articles that might help me understand, but I'd appreciate a pointer to specific papers that, given enough study, will actually address the point of my specific interest--something I can't necessarily tell with an initial glance, and might need a proper mathematician to point me in the right direction.

Thank you!

UPDATE: If the consensus is that this is a repeat, then ok, but I don't see how the answers to the other question about why algebraic properties break answers my questions about what algebraic properties break. Actually, the response marked as an answer in that other question doesn't actually answer that question either. It provides a helpful description of how to construct a multiplication table for higher dimension Cayley-Dickson structures, but explicitly doesn't answer the question as to why the properties break.

The Baez article many people suggest in responses to all hyper-complex number questions like mine is truly excellent, but is mostly restricted to octonions, and, in the few mentions it makes of higher dimension Cayley-Dickson algebras, does not refer to what properties are broken.

Perhaps the question isn't answerable, but in any case it hasn't been answered in this forum.

UPDATE 2: I should add that the sub question in this post about whether the loss of the alternative property specifically leads to the presence of zero divisors in sedenion algebra is definitely unique to my question. However, perhaps I should pose that as a separate question? Sorry, I'm not sure about that aspect of forum etiquette here.

• possible duplicate of Why are properties lost in the the Cayley-Dickson construction?, which links to anothe relevant version of this question math.stackexchange.com/questions/86434/… . The Baez link in the comments of both questions is very helpful for this topic. – rschwieb Jan 17 '14 at 17:13
• The Baez link is pretty helpful, as well as this other one math.ucr.edu/home/baez/week59.html, although it doesn't tackle the full tower mentioned here. – rschwieb Jan 17 '14 at 17:23
• It's also interesting to ask what does manage to stick. If I remember right, they retain power associativity and I thought also the flexible identity, but I can't find a source for this latter claim. – rschwieb Jan 17 '14 at 17:30
• I don't think this is a duplication; that question is asking asking why properties get lost, but doesn't reference what properties are lost. The Baez link is indeed fascinating, but I've read through large portions of it, and haven't seen any significant discussion beyond octonions. There are a few brief mentions of sedenions, but only to point out that the alternative property fails and zero divisors emerge as I already stated in the question. Baez then moves on to Clifford Algebras, which, as I understand, do not replicate 2^n-ions beyond octonions. – Pat Muchmore Jan 17 '14 at 17:47
• Yes, Clifford algebras are not really the same. They are close though! They are all simple rings or else squares of a simple ring, which is kind of in the spirit of "doubling" in the Cayley-Dickson construction. However, they're all associative, and they have zero divisors at lower dimensions. It is interesting how authors find relevance for both classes of algebras. – rschwieb Jan 17 '14 at 17:52

What specific algebraic properties break as we move into 32 dimensions, then into 64, 128, 256?

There are always trivial or uninteresting algebraic properties that break when passing from one algebra to the next (for example, that the sedenions have 16 dimensions as a real algebra). To make the question have an interesting answer, then, we have to restrict ourselves to some particular group of properties which share similar features with commutativity and associativity. We will see that this can be done.

First, some definitions. For the purposes of this answer, a *-algebra is a structure with associative and commutative addition, a (not necessarily associative or commutative) multiplication operation which distributes over addition, identities $0, 1$, additive inverses, and a conjugation $*$ satisfying the conditions $1^*=1$, $(x+y)^*=x^*+y^*$, $(xy)^*=y^*x^*$ and $(x^*)^*=x$. The Cayley-Dickson process acts on any *-algebra $\mathbb{A}$ producing an algebra $CD(\mathbb{A}) \simeq \mathbb{A} \oplus \mathbb{A}$ of twice the dimension, where the new product and conjugation are given by

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

Now, in any *-algebra we can have the following algebraic properties (where $x,y,z$ stand for arbitrary elements):

• Characteristic 2: $1+1 = 0$ (for an example, see the finite field $\mathbb{F}_2$ equipped with trivial conjugation).
• Hermiticity: conjugation is trivial, every element is equal to its conjugate.
• Commutativity: $xy = yx$.
• Associativity: $(xy)z = x(yz)$.

The Cayley-Dickson process is closely related to these four properties. We have the following facts, whose proof can be seen e.g. in Toby Bartels's discussion from Baez' TWF59 here: iff a *-algebra $\mathbb{A}$ is Hermitian and has characteristic 2, its Cayley-Dickson double $CD(\mathbb{A})$ is Hermitian. Iff $\mathbb{A}$ is commutative and Hermitian, $CD(\mathbb{A})$ is commutative. Finally, iff $\mathbb{A}$ is associative and commutative, $CD(\mathbb{A})$ is associative. In the most familiar case, we start with $\mathbb{R}$, a Hermitian, commutative and associative algebra. The previous facts make clear which properties break at the first three steps of the construction and why.

To make the relationship between these properties a bit clearer, we can express them in a more suggestive way. For any *-algebra we define the following maps (a nullary map is the same as a constant):

\begin{align*} F_0: [\:] &= 1 - (-1),\\ F_1: [x] &= x - x^*,\\ F_2: [x,y] &= xy - yx,\\ F_3: [x,y,z] &= (xy)z - x(yz). \end{align*}

These maps are, respectively, the number two, the imaginary part of an element (save for a factor of 2), the commutator of two elements and the associator of three elements. The previous observations can be reformulated in terms of these maps: for any *-algebra $\mathbb{A}$ and $0<k\le 3$,

$$F_k\equiv 0 \quad \mathrm{in} \quad CD(\mathbb{A}) \quad \Longleftrightarrow \quad F_k\equiv F_{k-1}\equiv 0 \quad \mathrm{in} \quad \mathbb{A}.$$

Seeing the form of the maps above, we are drawn to restrict ourselves to a certain subtype of algebraic properties: the properties expressed by a single identity $f(x,y,z,\ldots)=g(x,y,z,\ldots)$ whose two terms $f$ and $g$ are linear on all its arguments. Linearity basically implies that $f$ and $g$ must be sums of $n$-ary products where each term appears no more than once per product, such that this generalized distributive law holds:

$$f(\ldots, s+t, \ldots) = f(\ldots, s, \ldots)+f(\ldots, t, \ldots),$$

and the same for $g$. Each property has then an associated map $[x,y,z,\ldots] = f(x,y,z,\ldots)-g(x,y,z,\ldots)$, which vanishes identically whenever that property is fulfilled. Note that the requirement that the maps be multlinear excludes alternativity, since in its defining equation one of the variables appears twice.

After this long rambling, the question then becomes: is there any map of this type which vanishes over the octonions and not over the sedenions? My initial suspicions were in the negative, but the answer is seemingly yes! In section 5 of this article, the author defines the "commu-associator"

$$F_4: [x,y,z,w] = ( (x(yz))w+(w(yz))x+(wz)(yx)+(xz)(yw) )-( w((zy)x)+x((zy)w)+(xy)(zw)+(wy)(zx) ),$$

which is always zero when $x,y,z,w$ are octonions, but not when they are sedenions. The corresponding property could be called Moufangness, since it is a linearized form of the Moufang identities, which hold in the octonions (and imply alternativity). This is what breaks when passing from $\mathbb{O}$ to $\mathbb{S}$.

Moreover, later in that section the claim is made that further results (related to projective geometry over $\mathbb{F}_2$) suggest the existence of similar multilinear maps $F_{n+1}$ of this type, which vanish in the $n$th Cayley-Dickson algebra over the reals, $\mathbb{A}_n$, but not in $\mathbb{A}_{n+1}$. If this is true, there is indeed an infinite sequence of nameless properties which get broken at each step of the process.

Does the loss of the alternative property cause the emergence of zero divisors (or vice versa) or are these unrelated breakages?

In the discussion from TWF59 I linked above, it is shown that $CD(\mathbb{A})$ is a division algebra if $\mathbb{A}$ is an associative division algebra where $x^*x = xx^*$ commute with everything and $x^*x+y^*y = 0$ implies $x=y=0$. It is also shown that $CD(\mathbb{A})$ is alternative iff $\mathbb{A}$ is associative and both $x^*x = xx^*$ and $x+x^*$ associate and commute with everything. All of this is automatic if we start our process from a Hermitian, commutative, associative and ordered algebra such as $\mathbb{R}$, as we can then show that the terms $x+x^*$ always belong to this algebra and that the terms $x^*x = xx^*$ are positive elements.

Thus, the division algebra property is preserved by three steps of the Cayley-Dickson process, if we start at a Hermitian, commutative, associative and ordered algebra (if one tries to start with a characteristic-2 algebra, one gets zero divisors already at the first step: consider $(1,1)\cdot(1,1)=(1+1,1-1)=(2,0)=(0,0)$). Similarly, the ordering property is necessarily lost at the first step, since the existence of $(0,1)$, which is a square root of -1, contradicts the axioms of an ordered ring.

But note that while the ordering and division algebra properties are interrelated in this way, both of them lie outside the main sequence discussed above: they aren't defined by an equation nor have a corresponding linear map. We could consider them "accidental" properties of the starting algebra $\mathbb{R}$, rather than properties related to the Cayley-Dickson procedure per se.

To summarize, ordering is not really the important property we lose when passing from the reals to the complex numbers, but another more subtle property called hermiticity. That property, together with commutativity, associativity, and a stronger form of alternativity, is seemingly part of an infinite sequence of properties which break consecutively at each step of the Cayley-Dickson construction.

Alternativity is indeed indirectly related to the division algebra property in this context, in that both of them are implied by associativity of the previous algebra in the Cayley-Dickson sequence together with some other conditions. However, like ordering, the division algebra property does not "fit into" the mentioned infinite sequence.

• Very interesting! I'm wondering if there's some formula for generating $F_n$ – RothX Mar 6 '18 at 14:44
• Nothing you said here is wrong, but there is something that is a bit misleading. A reader could infer that each $F_n$ property is lost when applying $CD$ to an algebra with that property, but this isn't the case. Applying $CD$ to an algebra with characteristic 2 results in an algebra that is characteristic 2. If $a, b \in \mathbb A$ where $\mathbb A$ has characteristic 2, then $(a, b) + (a, b) = (a+a, b+b) = (0, 0)$. – RothX Apr 30 at 17:56
• @RothX Thanks, you're right, I see how my wording can be misleading there. I've been wanting to revisit my entire answer anyways since I think some things are poorly explained, I'll see if I can do it next month when I have more time (you're also free to make an edit now if you want). – pregunton Apr 30 at 20:05

The breakdown actually occurs at the octonions when associativity is lost. Because of the loss of associativity, the octonions cannot be represented as matrices under normal matrix multiplication. Matrix multiplication is an associative operation.

They can be represented as matrices given special rules for multiplication of those matrices. The way in which octonions multiply can be represented by the metric tensor of Riemannian multiplication.

It is because of these special multiplication rules that you can end up with zero divisors.

Here is an example: In special relativity, the length of a space-time position vector $\vec x$ is

$$|\vec x|^2 = c^2 t^2 - x^2 - y^2 - z^2 = \begin{bmatrix}ct&x&y&z \end{bmatrix} \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} \begin{bmatrix}ct\\x\\y\\z \end{bmatrix}$$

So, you can see that the length of a position vector in space-time can equal zero for an infinite number of positions. This is due to that metric tensor in the middle, there.

You'll find more information on zero divisors and the hypercomplex numbers in this paper: http://arxiv.org/pdf/q-alg/9710013v1.pdf

Check out Corrolary 2.12 - It says that zero divisors of the octonions are a special sort of zero divisor.

• Could whoever is downvoting my answer please explain why they are doing so? – Jade196 Sep 16 '14 at 2:19
• Thanks for this info. I'm still trying to parse all of it, and I'm looking into the paper, so thanks for that link. I think I'm misunderstanding something, because it seems to me that Corollary 2.12 (and, in fact, all of the zero divisors discussed in the paper) belong to the Sedenions and above. The corollary is defined in $\mathbb A_4$. The definition at the start of the paper is: $$\mathbb A_n = \mathbb R^{2^n}$$ so I think that this corollary only applies to $\mathbb R^{16} = \mathbb S$. Am I missing something? – Pat Muchmore Sep 19 '14 at 20:21
• I think you are right. I misread that corollary as applying to all systems up to order 4, not only those of order 4. – Jade196 Sep 20 '14 at 4:00
• Also, I was doing some reading on the Cayley-Dickson construction method. Multiplication is defined differently in the Cayley-Dickson construction than normal matrix multiplication. This new multiplication is $$(AB)_{ij} = \sum_k a_{ik} b_{kj} + ((-1)^{i+j+k} + 1)[b_{kj},a_{kj}]/2$$ reu.dimacs.rutgers.edu/~cskalit/cayley_dickson.html If you follow out these new multiplication rules, you should be able to determine at what order zero divisors crop up. I'm pretty sure it all comes down to how Cayley-Dickson construction defines multiplication, & how those rules nest at higher order. – Jade196 Sep 20 '14 at 4:01