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?


marked as duplicate by MJD, Jyrki Lahtonen abstract-algebra May 16 '16 at 18:19

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    $\begingroup$ For a perfect analogy with the relevant Latin roots, the word you want is trinions. (Singuli, bini, trini, quaterni, quini, seni …) $\endgroup$ – Anton Sherwood May 14 '16 at 2:54
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    $\begingroup$ 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. $\endgroup$ – Benjamin Dickman May 14 '16 at 3:46
  • $\begingroup$ 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. $\endgroup$ – Darrel Hoffman May 15 '16 at 14:43

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.

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    $\begingroup$ I'm glad you mentioned Hamilton! But I think you, ironically, lost a portion of the the last paragraph: What do we lose, going from real to complex? (I assume order, but maybe even more!) $\endgroup$ – pjs36 May 13 '16 at 20:20
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    $\begingroup$ @pjs36 Whoops. :P I just meant order. $\endgroup$ – Noah Schweber May 13 '16 at 20:54
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    $\begingroup$ @ArtB It means an order compatible with the field structure, see Ordered field. $\endgroup$ – Jack M May 13 '16 at 21:52
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    $\begingroup$ @LukaHorvat: There are weaker forms of associativity such as power associativity that disappear from the Caley-Dickson sequence later than the full associativity law. $\endgroup$ – Henning Makholm May 15 '16 at 1:15
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    $\begingroup$ So it's really only a matter of what niceness properties you want to give up. $\mathbb{C}$ is really, really, really, really, really damn nice. $\endgroup$ – The_Sympathizer May 15 '16 at 3:04

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.

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    $\begingroup$ +1, although I think a simpler example for your last paragraph is $\mathbb{R}^3$. $\endgroup$ – Noah Schweber May 13 '16 at 22:48
  • $\begingroup$ By the way, are all $2$- and $3$-dimensional $\mathbb{R}$-algebras known, associative or not? If so, do you know where to find a list of non-isomorphic such algebras? $\endgroup$ – Batominovski May 13 '16 at 23:40
  • $\begingroup$ @Batominovski I guess that three dimensional (associative) algebras are $\mathbb{R}\times\mathbb{C}$ and $\mathbb{R}^3$. $\endgroup$ – egreg May 14 '16 at 8:32
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    $\begingroup$ The Frobenius theorem only classifies finite division algebras over $\mathbb R$. There are infinite-dimensional algebras too. $\endgroup$ – Henning Makholm May 15 '16 at 1:11
  • $\begingroup$ @HenningMakholm I was assuming finite dimension. $\endgroup$ – egreg May 15 '16 at 8:18

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.

  • $\begingroup$ Neat! I didn't know about this. $\endgroup$ – Noah Schweber May 13 '16 at 23:17

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


Have you heard of the Frobenius theorem?


Triernions would not be an associative division algebra.


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”.


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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    $\begingroup$ You seem to be saying that quaternions are to complex numbers what complex numbers are to reals. I'm not sure if that's true (but I think it is), but what's got to do with the (non-)existence of something else that aren't complex numbers? $\endgroup$ – immibis May 14 '16 at 4:57
  • $\begingroup$ @immibis Once I asked a friend, who was a doctoral candidate in mathematics at the time, how you'd extend fractals (that is, Mandelbrot and its cousins) into 3 dimensions. He said, "Quaternions." That explains to me why there are no "triernions." To answer your question: Since they don't exist, they aren't complex numbers, but if they did, they would have to be complex, or at least based on complex numbers. The OP seemed to imply that, and I tend to agree. $\endgroup$ – scott Jun 20 '16 at 18:13

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