# abstract algebra – Why are There No “Triernions” (3-dimensional analogue of complex numbers / quaternions)?

## The Question :

108 people think this question is useful

Since there are complex numbers (2 dimensions) and quaternions (4 dimensions), it follows intuitively that there ought to be something in between for 3 dimensions (“triernions”).

Yet no one uses these. Why is this?

• For a perfect analogy with the relevant Latin roots, the word you want is trinions. (Singuli, bini, trini, quaterni, quini, seni …)
• The question reminds me of an analogous one around forming topological spaces by “crossing them” e.g. $\mathbb{R} \times \mathbb{R}$ for $\mathbb{R}^2$. In particular, one may ask whether there exists a topological space $X$ for which $X \times X$ is homeomorphic to $\mathbb{R}^3$. The answer is no: cf. MO 60375.
• There are, however, octonions (8×8) and sedenions (16×16). And by extension, 32×32, 64×64, and basically any $2^n×2^n$ formation, though I’ve never heard names for anything larger than 16×16.

101 people think this answer is useful

It’s because there isn’t one! (Indeed, Hamilton was originally searching for such a thing, and found the quaternions instead; it was only later that people understood why he hadn’t been successful, initially.)

The quaternions – along with the real numbers and the complex numbers – have a number of nice properties: specifically, they form a real division algebra. This is a mouthful, but basically amounts to:

• Addition/multiplication of quaternions satisfy the ring axioms.

• We can divide by quaternions.

• We can multiply a quaternion by a real (and this “scalar multiplication” satisfies the basic properties it should).

It turns out the only finite-dimensional real division algebras are $$\mathbb{R}$$, $$\mathbb{C}$$, and the quaternions; see this. (I include associativity in the definition of algebra: if we allow non-associative algebras, then the octonions also count.)

By the way, there is a way to (sort of) keep going past the quaternions: the Cayley-Dickson construction. This produces things like the octonions and the sedenions, and other delightfully weird algebraic structures. However, it has a couple drawbacks:

• Each time you apply Cayley-Dickson, the dimension of the starting algebra doubles. So this won’t help us get to $$3$$.

• Also, you keep losing nice properties. Passing from the reals to the complex numbers, we lose order; going from the complexes to the quaternions, we lose commutativity of multiplication. If we keep going, we lose associativity of multiplication, in increasing degrees: the sedenions are even less associative than the octonions, etc.

44 people think this answer is useful

Assume $$A$$ is a three-dimensional (associative) algebra over $$\mathbb{R}$$. We can assume $$\mathbb{R}$$ is embedded in $$A$$. If $$a\in A$$, $$a\notin\mathbb{R}$$ the map $$l_a\colon A\to A$$, $$l_a(x)=ax$$, is an endomorphism of $$A$$ as a vector space over $$\mathbb{R}$$.

Let $$\lambda$$ be a real eigenvalue of $$l_a$$, with eigenvector $$b\ne0$$, so $$ab=\lambda b$$. Such an eigenvalue exists, because the characteristic polynomial of $$l_a$$ has degree $$3$$. Then $$(a-\lambda)b=0$$. Note that $$a-\lambda\ne0$$, so $$A$$ has zero divisors, in particular $$A$$ is not a division algebra.

It’s a bit more complicated showing that a finite-dimensional division algebra over $$\mathbb{R}$$ can only have dimension $$1$$, $$2$$ or $$4$$ and it is isomorphic to $$\mathbb{R}$$, $$\mathbb{C}$$ or $$\mathbb{H}$$ (the quaternions); this is known as Frobenius’ theorem.

On the other hand, three-dimensional algebras over $$\mathbb{R}$$ exist (but they have zero divisors, as shown above). A simple example is $$\mathbb{R}[X]/(X^3-1)$$, but they can be non-commutative as well.

20 people think this answer is useful

There is an algebra in dimension 6, halfway between 4 and 8, that is not a division algebra but correctly interpolates various constructions.

http://arxiv.org/abs/math/0411428

Nothing like that is known for dimension 3 sitting between 2 and 4.

10 people think this answer is useful

The closest thing to triterniums would be the structure $[\mathbb R^3, +, \times]$ where $“\times”$ represents the cross product

$$(a_1 \mathbf i + b_1 \mathbf j + c_1 \mathbf k) \times (a_2 \mathbf i + b_2 \mathbf j + c_2 \mathbf k) = \left| \begin{matrix} \mathbf i & \mathbf j & \mathbf k \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ \end{matrix} \right|$$

It distributes over $“+”$,

is anti commutative,

and isn’t associative yet
$[a \times (b \times c)] + [b \times (c \times a)] + = 0$

6 people think this answer is useful

Have you heard of the Frobenius theorem?

https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)

Triernions would not be an associative division algebra.

2 people think this answer is useful

Among the vectorial spaces $$\mathbb R^n$$, only $$\mathbb R$$ and $$\mathbb R^2\space ( \approx \mathbb C)$$ admit a multiplication that gives them a structure of field. For the other values, only for $$\mathbb R^4$$ we can have a multiplication without divisors of zero and associative seeming a field; actually,with this multiplication, $$\mathbb R^4$$ becomes a division ring or, also called, skew field and is named quaternions.

According to a celebrated theorem of Wedderburn all finite division rings are necessarily commutative so quaternions are the first example of a non-commutative skew field. French mathematicians used the terminology “corps” for both “fields and skew fields” so there are for them commutative and non-commutative corps. By the theorem of Wedderburn, quaternions give the first example of a non-commutative “corp”.