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?
Why are There No "Triernions" (3-dimensional analogue of complex numbers / quaternions)? [duplicate]
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25$\begingroup$ For a perfect analogy with the relevant Latin roots, the word you want is trinions. (Singuli, bini, trini, quaterni, quini, seni …) $\endgroup$– Anton SherwoodMay 14, 2016 at 2:54
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7$\begingroup$ 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 DickmanMay 14, 2016 at 3:46
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$\begingroup$ 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 HoffmanMay 15, 2016 at 14:43
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$\begingroup$ Here you can see three examples of commutative 3-dimensional "numbers". math.stackexchange.com/a/4453131/2513 If you want to include non-commutative ones, like the quaternions, the set will be even greater. $\endgroup$– AnixxJun 6, 2022 at 10:46
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7 Answers
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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.
answered May 13, 2016 at 19:35
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3$\begingroup$ 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!) $\endgroup$– pjs36May 13, 2016 at 20:20
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3$\begingroup$ @pjs36 Whoops. :P I just meant order. $\endgroup$ May 13, 2016 at 20:54
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6$\begingroup$ @ArtB It means an order compatible with the field structure, see Ordered field. $\endgroup$– Jack MMay 13, 2016 at 21:52
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4$\begingroup$ @LukaHorvat: There are weaker forms of associativity such as power associativity that disappear from the Caley-Dickson sequence later than the full associativity law. $\endgroup$ May 15, 2016 at 1:15
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5$\begingroup$ 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. $\endgroup$ May 15, 2016 at 3:04
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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.
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2$\begingroup$ +1, although I think a simpler example for your last paragraph is $\mathbb{R}^3$. $\endgroup$ May 13, 2016 at 22:48
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1$\begingroup$ 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? $\endgroup$ May 13, 2016 at 23:40
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3$\begingroup$ The Frobenius theorem only classifies finite division algebras over $\mathbb R$. There are infinite-dimensional algebras too. $\endgroup$ May 15, 2016 at 1:11
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1$\begingroup$ @egreg : the formal Laurent series is explicitly an infinite dimensional algebra over $\mathbb{C}$ (or $\mathbb{R}$ by constraining the coefficients) $\endgroup$– reunsMay 16, 2016 at 11:56
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1$\begingroup$ @egreg: Your list of 3-dimensional associative algebras over $\Bbb{R}$ only contains the commutative ones. Upper triangular real 2x2 matrices form a non-commutative associative one. Possibly there are other (non-isomorphic) ones. $\endgroup$ May 17, 2016 at 5:40
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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.
http://arxiv.org/abs/math/0411428
Nothing like that is known for dimension 3 sitting between 2 and 4.
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$\begingroup$ "Nothing like that is known for dimension 3 sitting between 2 and 4." - what do you mean? Here is an example of 3 3-dimensional commutative associative algebras: math.stackexchange.com/a/4453131/2513 (and all three can be made into 6-dimensional by adding the complex unity). $\endgroup$– AnixxJun 6, 2022 at 10:58
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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, but $[a \times (b \times c)] + [b \times (c \times a)] + [c \times (a \times b)] = 0$
answered May 15, 2016 at 7:41
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1$\begingroup$ Maybe worth pointing out that some people know this structure as the simple Lie algebra $\mathfrak{su}_2$. Also, if one turns the quaternions into a Lie algebra via the commutator bracket, this is the derived subalgebra, the space of "pure quaternions", and isomorphic to the quotient of the full quaternions modulo their centre $\mathbb R$. $\endgroup$ Dec 13, 2021 at 3:46
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$\begingroup$ It amuses me to see "simple" and "Lie algebra" in the same sentence. $\endgroup$ Feb 7, 2022 at 2:22
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$\begingroup$ Would it enhance or reduce your amusement if I said "semisimple" instead? $\endgroup$ Feb 8, 2022 at 17:59
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$\begingroup$ @TorstenSchoeneberg : To paraphrase... Math is not only stranger than you imagined, it is stranger than you can imagine. $\endgroup$ Feb 12, 2022 at 18:25
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$\begingroup$ Could we imagine a computer algebra in dimension 3 by considering $\mathbb{F}_{p^3}$ for p becoming big prime, and mapping the elements to some floating point numbers ? $\endgroup$– Cretin2Feb 22, 2022 at 19:04
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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.
answered May 13, 2016 at 19:34
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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”.
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$\begingroup$ $\mathbb R^2$ is not $\mathbb C$. $\mathbb R^2$ is isomorphic to split-complex numbers. $\endgroup$– AnixxJun 6, 2022 at 11:01
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$\begingroup$ See at the first line where appears $\mathbb R^2\space ( \approx \mathbb C)$ in other words $\mathbb R^2$ is not equal to $\mathbb C$ as you say and me too. $\endgroup$– PiquitoJun 7, 2022 at 15:00
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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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6$\begingroup$ 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? $\endgroup$ May 14, 2016 at 4:57
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$\begingroup$ @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. $\endgroup$– scottJun 20, 2016 at 18:13