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

The Question :

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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?

The Question Comments :
  • 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.

The Answer 1

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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.

The Answer 2

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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.

The Answer 3

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


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

The Answer 4

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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
$[a \times (b \times c)] + [b \times (c \times a)] + = 0$

The Answer 5

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Have you heard of the Frobenius theorem?


Triernions would not be an associative division algebra.

The Answer 6

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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”.

The Answer 7

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All of these other answers go way over my head, but this is the way I think about it.

The complex numbers can be represented on a 2-dimensional plane, but they are an extension of the one-dimensional real number line. They are not really 2-dimensional, it’s just a convenient way for us to represent them. When we expand the reals from 1 dimension into 2 dimensions, the corresponding complex numbers must double their dimension as well, going from 2 to 4. That is why they appear to “skip” 3.

Perhaps I am wrong, or this argument was folded into the other explanations, and I just didn’t see it.

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