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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?

thecat
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    For a perfect analogy with the relevant Latin roots, the word you want is *trinions*. (Singuli, bini, trini, quaterni, quini, seni …) – Anton Sherwood May 14 '16 at 02:54
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    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**](http://mathoverflow.net/questions/60375). – Benjamin Dickman May 14 '16 at 03:46
  • There are, however, [octonions](https://en.wikipedia.org/wiki/Octonion) (8×8) and [sedenions](https://en.wikipedia.org/wiki/Sedenion) (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. – Darrel Hoffman May 15 '16 at 14:43

7 Answers7

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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.

J. W. Tanner
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Noah Schweber
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    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!) – pjs36 May 13 '16 at 20:20
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    @pjs36 Whoops. :P I just meant order. – Noah Schweber May 13 '16 at 20:54
  • `Passing from the reals to the complex numbers, we lose order` what is "order" in this context? Is it [partial](https://en.wikipedia.org/wiki/Partially_ordered_set#Formal_definition) vs total ordering? – Sled May 13 '16 at 21:35
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    @ArtB It means an order compatible with the field structure, see [Ordered field](https://en.wikipedia.org/wiki/Ordered_field). – Jack M May 13 '16 at 21:52
  • Dang, I wanted an argument about well-ordering here. – Daniel R. Collins May 13 '16 at 22:26
  • @ArtB In particular, yes, a total ordering. – Noah Schweber May 13 '16 at 22:47
  • Is it worth mentioning Clifford algebras? The geometric algebra with three bases is a bit like what the OP is asking for. – Simon May 14 '16 at 07:38
  • How can you lose associativity to varying degrees? Also, does this suggest some ordering between the properties? Anything interesting about that? – Luka Horvat May 14 '16 at 08:08
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    @LukaHorvat Well of course you could say that either the algebra is associative so there seems to be no degrees. But you can also have things that aren't quite associative but aren't that far. For example for the octonions any subgroup generated by 2 elements is associative since octonions satisfy the moufang identity. – DRF May 14 '16 at 09:00
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    @LukaHorvat: There are weaker forms of associativity such as [power associativity](https://en.wikipedia.org/wiki/Power_associativity) that disappear from the Caley-Dickson sequence later than the full associativity law. – hmakholm left over Monica May 15 '16 at 01:15
  • @mike4ty4 That's a thing one could do; however, that would be going in a very different direction. Nothing wrong with that, it's just a different activity. I suspect the motivation behind wanting division is that we are going to try to interpret "lots" of these numbers as transformations - e.g., rotations - with composition given by multiplication, and the kinds of transformations we're interested in are invertible. Why this is something that we want to do, and why it's promising, are good questions; maybe someone who knows more than I do can answer them. – Noah Schweber May 15 '16 at 02:30
  • Also I think it's just an interesting problem from the point of view of universal algebra: to what extend does demanding divisibility constrain the possible algebras we can build, especially with regard to large (finite) dimension and versions of the associative law? – Noah Schweber May 15 '16 at 02:33
  • I'd want to say that you can actually create, in a number of ways, a ring on $\mathbb{R}^3$, in a manner similar to the complex numbers and so forth. It is even commutative, but it loses division by arbitrary elements. This does eventually happen with the Cayley-Dickson construction once you hit 16 dimensions though and you lose a lot more with that, too. – The_Sympathizer May 15 '16 at 03:03
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    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. – The_Sympathizer May 15 '16 at 03:04
  • Noah Schweber: Yeah, and you lose that interpretation when you hit dimension 8 in the CD construction, because associativity craps out. (And thus it no longer is possible to interpret multiplication as a composition of functions.) – The_Sympathizer May 15 '16 at 03:08
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    @NoahSchweber The reals, complex numbers, and quaternions are the only **finite-dimensional** real **associative** division algebras. I'm not entirely sure finite-dimensional can't be left off but if you leave out associativity then the octonions should also be counted. –  May 15 '16 at 18:17
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    @Bye_World Fixed. Leaving out finite-dimensionality was a mistake. As for associativity, the definition of algebra I learned included associativity, but I've edited for clarity. Thanks! – Noah Schweber May 15 '16 at 18:32
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    There is this [document](http://www.mathe-seiten.de/zahlentheorie.pdf) (german) which explains a hirarchy of properties, where the presence of a property in $\mathbb K^n$ is crucial for the presence of the next higher property in $\mathbb K^{2n}$. Specificially there are characteristic 2 $\rightarrow$ conjugate $\bar x$ equals $x$ $\rightarrow$ commutative $\rightarrow$ associative $\rightarrow$ division algebra. He argues, that the reason that $\mathbb H$ is not commutative is, that in $\mathbb C$ the conjugate is not the identity, and this is because $\mathbb R$ has not characteristic 2. – M. Winter Apr 27 '17 at 09:18
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    About the question of what having a finite dimensional associative division algebra $D$ says about the dimension (as a vector space over the center $K$) we can say the following. The dimension will always be a perfect square. Furthermore, if $K$ is a finite dimensional extension of $\Bbb{Q}$, then all the perfect squares occur. So it is possible to construct 9-dimensional associative division algebras with center $\Bbb{Q}$. In fact there are infinitely many non-isomorphic ones. – Jyrki Lahtonen Feb 27 '19 at 10:31
  • Can you please define "even less associative than"? – Luke Hutchison Aug 11 '20 at 09:42
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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.

J. W. Tanner
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egreg
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    +1, although I think a simpler example for your last paragraph is $\mathbb{R}^3$. – Noah Schweber May 13 '16 at 22:48
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    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? – Batominovski May 13 '16 at 23:40
  • @Batominovski I guess that three dimensional (associative) algebras are $\mathbb{R}\times\mathbb{C}$ and $\mathbb{R}^3$. – egreg May 14 '16 at 08:32
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    The Frobenius theorem only classifies _finite_ division algebras over $\mathbb R$. There are infinite-dimensional algebras too. – hmakholm left over Monica May 15 '16 at 01:11
  • @HenningMakholm I was assuming finite dimension. – egreg May 15 '16 at 08:18
  • @HenningMakholm : the field of formal Laurent series ? – reuns May 16 '16 at 11:19
  • @user1952009: Among others. Concretely I just had in mind using the upward Löwenhiem-Skolem theorem to create some models of the division-algebra axioms with too many elements to have finite dimension -- crude but effective. – hmakholm left over Monica May 16 '16 at 11:25
  • @user1952009 More simply, the rational functions (quotient field of the ring of polynomials) – egreg May 16 '16 at 11:35
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    @egreg : the formal Laurent series is explicitly an infinite dimensional algebra over $\mathbb{C}$ (or $\mathbb{R}$ by constraining the coefficients) – reuns May 16 '16 at 11:56
  • @user1952009 Yes, of course, but it's a bit harder to define. `;-)` – egreg May 16 '16 at 11:59
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    @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. – Jyrki Lahtonen May 17 '16 at 05:40
  • @JyrkiLahtonen Yes, it was incomplete – egreg May 17 '16 at 06:13
  • @NoahSchweber: Just add a constant symbol for each scalar before applying the Löwenheim-Skolem theorem. Since it produces an elementary extension, those symbols will still behave as your original scalars in the extended model. – hmakholm left over Monica May 18 '16 at 06:54
  • As an anecdote there are 3 dimensional algebras over integer modulo a prime. Creates a finite field. I wonder if it could work as an analogy to su2 in certain scenarios, either actually mathematically or as an approximation for applied mathematics. Fun question. – VoronoiPotato Mar 21 '22 at 18:50
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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.

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

Steven Alexis Gregory
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  • 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 $\mathbb R$. – Torsten Schoeneberg Dec 13 '21 at 03:46
  • It amuses me to see "simple" and "Lie algebra" in the same sentence. – Steven Alexis Gregory Feb 07 '22 at 02:22
  • Would it enhance or reduce your amusement if I said "semisimple" instead? – Torsten Schoeneberg Feb 08 '22 at 17:59
  • @TorstenSchoeneberg : To paraphrase... Math is not only stranger than you imagined, it is stranger than you can imagine. – Steven Alexis Gregory Feb 12 '22 at 18:25
  • 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 ? – Cretin2 Feb 22 '22 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.

Chill2Macht
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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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Piquito
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-3

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.

scott
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    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? – user253751 May 14 '16 at 04:57
  • @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. – scott Jun 20 '16 at 18:13