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