5

Are there operations/functions that take any other than 2 arguments in abstract algebra? If there are, then why are they not used or shown while teaching the topic? If there are not, then why is the subject constrained to just binary functions?

José Carlos Santos
  • 427,504
aj14
  • 53
  • because they are devoted a whole entire field around it. eg multivariable polynomials is part of algebraic geometry eg. – user29418 Jul 31 '19 at 16:57
  • 3
    Of course there are functions with a single argument as well. Or functions that take tuples as argument and can be interpreted as functins taking arbitrarily many arguments – Hagen von Eitzen Jul 31 '19 at 16:57
  • There is $B(x,y,z)$ to mean $y$ is between $x$ and $z,$ e.g on real line $x<y<z$ or $z<y<x.$ That's also used in projective geometry I believe. – coffeemath Jul 31 '19 at 16:59
  • There is a generalization of the cross product to be an $n$-ary operation. – rschwieb Jul 31 '19 at 17:00
  • There are definitely higher-arity functions in abstract algebra (I feel like unary functions aren't really what you're looking for), but you're quite right that they're not nearly as fundamental as the usual (or even many unusual) binary operations. That said, this mathoverflow question and this one give a few examples of higher-arity operations. – Noah Schweber Jul 31 '19 at 17:00
  • There are things called operads which involve a system of $n$-ary operations for every $n$. – Kyle Miller Jul 31 '19 at 17:01
  • Also $n$-ary operations can be studied as compositions of binary operations in this sense. – rschwieb Jul 31 '19 at 17:01
  • I suppose you can think of the cross-ratio as operating on 4 inputs. – rschwieb Jul 31 '19 at 17:03
  • 5
    Incidentally an interesting argument I once heard runs as follows. We tend (to put it mildly) to write mathematics linearly; this makes infix notation quite tempting, but that can only accommodate binary functions. Certainly writing linearly indicates sequentiality, which is a fundamentally binary thing (see rschwieb's comment above). This suggests that we might get some interesting new perspectives by looking at "two-dimensional mathematics" - taking us directly to operads, particularly planar algebra. I don't know if I buy it but it's neat. – Noah Schweber Jul 31 '19 at 17:03
  • 1
    As an undergrad, I wondered about a certain $3$-ary products that weren't just the composition of two group multiplications: https://www.kylem.net/papers/trinary_groups.pdf – Kyle Miller Jul 31 '19 at 17:04
  • @NoahSchweber your comment is the simplest and makes the most sense, just as binary operations are simplest and easiest to understand and study))) – aj14 Jul 31 '19 at 17:14
  • @KyleMiller thanks for showing that paper, gives me a deeper idea as how complicated things get with more arguments and hence probably are not taught – aj14 Jul 31 '19 at 17:16
  • Look up "brace algebras", "Sabinin algebras", "Jordan triple systems", "$n$-Lie algebras", "Nambu algebras", "$A_\infty$-algebras" and various others (here is a quick overview). – darij grinberg Jul 31 '19 at 17:34
  • It's not constrained to binary operations - you can also discuss ternary operations, etc. But it's the simpler case, so it is better studied and more fundamental. It's the same reason that we study linear mappings more than polynomial mappings, or why we know 2-dimensional geometry better than 7-dimensional geometry, etc. – Jair Taylor Jul 31 '19 at 17:37
  • Sierpinski proved that $n$-ary operations can be reduced to compositions of binary operations. – Bill Dubuque Jul 31 '19 at 17:42

2 Answers2

2

Aside from the usual nullary (constants), unary operations, and binary operations, there are ternary operations also. One example is that of Heaps and semi-heaps which are intended to axiomatize operations such as $\, a,b,c \to ab^{-1}c\,$ in groups. The intent of this operation is analogous to the difference between vector spaces and affine spaces. In a vector space there is a distinguished zero vector, while in an affine space there isn't. Also, in an affine space there is a multi-variable operation of the affine combination of points.

Somos
  • 35,251
  • 3
  • 30
  • 76
1

Inversion is a unary operation. And the cross-product, which is a binary operation, can be generalized to a $n$-ary operation from $\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}\times\cdots\times\mathbb{R}^{n+1}$ ($n$ times) into $\mathbb R^{n+1}$.

José Carlos Santos
  • 427,504
  • But any binary operation can be exteneded to $n$-ary versions, so that's not saying much. And unary (and nullary) operations can also be viewed as binary operations – Bill Dubuque Jul 31 '19 at 17:46
  • 1
    The generalization of the cross-product to a ternary operation from $(\mathbb R^4)^3$ into $\mathbb R^4$ is not obtained from the ordinary cross-product from $(\mathbb R^3)^2$ into $\mathbb R^3$. – José Carlos Santos Jul 31 '19 at 17:59
  • It seems you missed my point, which is that it is trivial to generate $n$-ary operations via said extensions. So you should explain why the case you mentioned is worth special attention (if you think it is). – Bill Dubuque Jul 31 '19 at 18:08
  • I assumed that what you meant was that if $\varphi$ is a binary operation, then you can generate a ternary operation from it by $(a,b,c)\mapsto\varphi(\varphi(a,b),c)$. Is this what you had in mind or not? – José Carlos Santos Jul 31 '19 at 18:14
  • Yes, that is one of many obvious ways to generate $n$-ary operations. – Bill Dubuque Jul 31 '19 at 18:29
  • But the $n$-ary operation that I mentioned is not obtained from the usual cross-product by this process. – José Carlos Santos Jul 31 '19 at 20:55
  • Did someone claim that it was? Again, there are many ways to construct $n$-ary operations. Why is the one you mention of special interest for this question? – Bill Dubuque Jul 31 '19 at 20:58
  • 1
    If that is not your objection, then I don't see what your objection is. I described a non-trivial $n$-ary operation. – José Carlos Santos Jul 31 '19 at 21:03
  • It's not clear what you mean by "non-trivial" and how that relates to the question being asked. This is not an answer to the question, At most it is a comment (and in fact this example is mentioned in a comment on the question). – Bill Dubuque Jul 31 '19 at 21:07