Why are so few operations with arity bigger than 2? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T01:11:09Z http://mathoverflow.net/feeds/question/49437 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2 Why are so few operations with arity bigger than 2? Carlos Sáez 2010-12-14T21:39:18Z 2012-02-06T20:32:40Z <p>In usual algebraic structures, like groups, rings, monoids, etc, or in algebras coming from logics like Boolean algebras, Heyting algebras and the like the operations are usually of arity 0 (constants), 1 or 2. My question is two-fold:</p> <ol> <li><p>Provide examples of algebras arising naturally in some field (I'm mainly interested in algebras coming from logics, but I'm open to any field) with operations of arity 3 or bigger.</p></li> <li><p>Is there any reason (more or less profound) for being so few algebras with operations of arity bigger than 2?</p></li> </ol> <p>Thank you in advance.</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49441#49441 Answer by Alberto García-Raboso for Why are so few operations with arity bigger than 2? Alberto García-Raboso 2010-12-14T21:50:12Z 2010-12-14T21:50:12Z <p>Higher arity operations appear quite naturally when homotopy theory enters the stage; e.g., <a href="http://ncatlab.org/nlab/show/A-infinity-algebra" rel="nofollow">$A_\infty$-algebras</a>, <a href="http://ncatlab.org/nlab/show/L-infinity-algebra" rel="nofollow">$L_\infty$-algebras</a> and <a href="http://ncatlab.org/nlab/show/E-infinity+algebra" rel="nofollow">$E_\infty$-algebras</a>.</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49442#49442 Answer by Michael Hardy for Why are so few operations with arity bigger than 2? Michael Hardy 2010-12-14T21:59:02Z 2010-12-14T21:59:02Z <p>$\text{average}(x_1,\dots,x_n) = \dfrac{x_1 + \cdots + x_n}{n}.$</p> <p>$\text{cross-ratio}(z_1,z_2;z_3,z_4) = \dfrac{(z_1-z_3)(z_2-z_4)}{(z_2-z_3)(z_1-z_4)}.$</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49443#49443 Answer by José Figueroa-O'Farrill for Why are so few operations with arity bigger than 2? José Figueroa-O'Farrill 2010-12-14T22:00:12Z 2010-12-14T22:00:12Z <p>First a trivial remark: if you have a binary operation you automatically have higher arity operations by nesting. Hence I would not say that there are fewer such algebras. But there is a sense in which that is cheating. Examples of these are some of the triple systems, say Lie triple systems, which are to symmetric spaces what Lie algebras are to Lie groups: namely, the best linear approximation.</p> <p>Starting at least in the 1940s, the Russian algebraist AG Kurosh and his school sought to generalise many of the algebraic structures with a binary operation to an $n$-ary operation. This is explained in the paper/monograph <em>Multioperator rings and algebras</em> from 1969 as well as in work of Baranovic and Burgin from 1975 on <em>Linear $\Omega$-algebras</em>. Perhaps the best known example of this kind of structure are the $n$-Lie algebras introduced by VT Filippov in 1980.</p> <p>3-Lie algebras had previously appeared in work of Nambu trying to generalise Hamiltonian mechanics by replacing the symplectic form by a closed 3-form. This line of work was continued by Takhtajan and collaborators.</p> <p>In the last few years, $n$-ary Leibniz algebras (but mostly $n=3$) have been given lots of attention due to the unexpected rôle they play in the AdS$_4$/CFT$_3$ correspondence for M2-branes. Two years ago I gave some lectures on some of the underlying algebraic story at Nordita (Stockholm) and <a href="http://arxiv.org/abs/0812.2865" rel="nofollow">wrote them up</a>. You may wish to peruse them for the references. </p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49444#49444 Answer by David Harris for Why are so few operations with arity bigger than 2? David Harris 2010-12-14T22:02:25Z 2010-12-14T22:02:25Z <p>Any $k$-ary relation can be expressed in terms of binary relations by means of projection maps, i.e. introduce new objects which correspond to $n$-tuples of the original objects ($n \leq k$), and introduce binary projection relations (i.e. $P(x,y)$ iff x is the first $n-1$ coordinates of $y$). Then $k$-ary relations are equivalent to a unary relation on $k$-tuples, and the $k$-tuples are all expressible in terms of the original objects via the binary projections maps. </p> <p>In brief, 2-ary relations are sufficiently expressive to handle all arities. (And similarly 2-ary functions can express all functions)</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49448#49448 Answer by Gerhard Paseman for Why are so few operations with arity bigger than 2? Gerhard Paseman 2010-12-14T22:14:36Z 2010-12-14T22:14:36Z <p>Perhaps you mean fundamental operations instead of operations. Others have noted that composition, projection, and changing one's point of view allows you to handle operations of higher arity.</p> <p>I imagine that fundamental operations are usually of such low arity because we prefer simplicity. Doing the maximum or the sum of a tuple of numbers can be acheived by iterating the corresponding binary operation on certain parts of the tuple. Anything more gets uncomfortably complicated.</p> <p>Having said that, there are examples like multi-linear functions (especially the determinant) that come up in various fields of analysis, not to mention infinitary operations like integration. Even then, we like to break things down into iterates of simpler terms, or compositions thereof. </p> <p>William DeMeo has been doing many posts in MathOverflow in re universal algebra. He will probably suggest the majority function on the set {0,1}, varieties which have a ternary or 4-place discriminator term, ternary groups, and the like. He may also point to places in the literature where your question has been raised. </p> <p>Gerhard "Memory Not So Good Lately" Paseman, 2010.12.14</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49449#49449 Answer by maxdev for Why are so few operations with arity bigger than 2? maxdev 2010-12-14T22:15:50Z 2010-12-14T22:15:50Z <p>Is this like the question why matrices are more common than multi-matrices? Feel free to flag this as spam because I don't have enough mathoverflow bucks to comment.</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49450#49450 Answer by Qiaochu Yuan for Why are so few operations with arity bigger than 2? Qiaochu Yuan 2010-12-14T22:31:39Z 2010-12-14T22:31:39Z <p>Here's one I learned from Todd Trimble. Giving a set $X$ the structure of a compact Hausdorff space is the same as equipping $X$ with $J$-ary operations $X^J \to X$ for every set $J$, one for each ultrafilter $P$ on $J$, corresponding to the $P$-limit of a $J$-tuple of elements of $X$, with the appropriate compatibility relations. </p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49458#49458 Answer by arsmath for Why are so few operations with arity bigger than 2? arsmath 2010-12-14T23:05:16Z 2010-12-14T23:05:16Z <p>Operations of arity 3 naturally arise in universal algebra. For example, one strand of research is to characterize the properties of the lattice of congruences of a variety by the existence of special terms -- these usually have arity 3. For example, if a variety has a ternary operation m(x, y, z) such that m(x, y, y) = x and m(x, x, y) = y, then the lattice of congruences is modular. (The converse is not true, but there is a weaker statement involving ternary operations that is true.) Examples of this include groups ($m(x, y, z) = x y^{-1} z$) and vector spaces ($m(x, y, z) = x - y + z$).</p> <p>The ternary operation for vector spaces has a natural geometric interpretation as vector addition in affine space, where vectors are not required to be based at the origin. If you draw a vector from y to x and a vector from y to z, then $m(x, y, z)$ is the vector from y to x + z. You can think of addition as defined by drawing a parallelogram $xyzw$. Then $m(x,y,z)=w$.</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49460#49460 Answer by Stefan Geschke for Why are so few operations with arity bigger than 2? Stefan Geschke 2010-12-14T23:25:50Z 2010-12-14T23:25:50Z <p>Someone already mentioned determinants. Here is a related $n$-ary operation, the vector product in dimension $n+1$: Fix a basis $b_1,\dots,b_{n+1}$ of $\mathbb R^{n+1}$. To $n$ elements $v_1,\dots,v_n$ of $\mathbb R^{n+1}$ assign the unique vector $v_{n+1}$ that is orthogonal to $v_1,\dots,v_n$, such that $v_1,\dots,v_{n+1}$ is of the same orientation as $b_1,\dots,b_{n+1}$ and such that the length of $v_{n+1}$ is the $n$-dimensional volume of the parallelopiped spanned by $v_1,\dots,v_n$. </p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49464#49464 Answer by none for Why are so few operations with arity bigger than 2? none 2010-12-15T00:33:18Z 2010-12-15T00:33:18Z <p>I'd have thought it was just notational. When the arity is > 2, we usually end up coding the operands into a vector or tensor or whatever. The determinant mentioned above is an obvious example of that: a unary operation on a matrix, or a tensor acting on n vectors, depending on how we look at it.</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49471#49471 Answer by Theo Johnson-Freyd for Why are so few operations with arity bigger than 2? Theo Johnson-Freyd 2010-12-15T01:00:39Z 2012-02-06T19:16:16Z <ol> <li><p>I'd say that a lot of "higher-dimensional mathematics" concerns spaces with operations of arbitrary finite arity. I'm thinking of things like planar algebras, operads, ...<br><br>Let me mention one area that I like, which are various things I'd call "associative", and rather than trying to give precise definitions I'll mention planar algebras. A <em>planar algebra</em> includes in the data a $k$-ary operation for every way to draw nonintersecting curves (that are either closed or end on a boundary) on a disk minus $k$ subdisks. These operations are required to compose in any way that you can stick a disk-minus-holes into a hole in another disk (with the requirement that any curves ending on the glued-along boundary components match up). Then there's also an <em>associativity</em> requirement that says that everything only depends on the topology of the diagram, not the geometry.<br><br>Anyway, it is possible to write any "planar operation" as a composition of binary operations (although you need infinitely many "basic" binary operations), but this is the wrong way to think about it, I claim. In particular, there's really no canonical choice how to write something as a composition of binaries.</p></li> <li><p>From this point of view, let's now revisit usual associative multiplication. The associativity says nothing more nor less than: <code>ab c = a bc</code>. Drawn this way, it's clear that this is again a statement that "only the topology matters, not the geometry". But the point is that the usual multiplication is "one-dimensional", in that the ambiant space where things like "a", "b", "c" are put is a line. (Compare planar algebras, which are inherently two-dimensional.) It took a while to invent two-dimensional mathematics, because we're used to thinking of "functions" acting consecutively in "time", and our experience is that "time" is one-dimensional. Anyway, the point is that if your mathematics is one-dimensional, then it's much easier to see how to break any one-dimensional picture into "basic" subpictures with only two things going on. I think this is the answer to your question 2, why most of the time we only think about 2-ary operations.</p></li> </ol> <p>Finally, I'll mention that there's another direction you can go, which is to include "coalgebra" along with your algebra. By "algebra" I mean a theory with some "$k$-ary operations" that take in $k$ inputs and spit out one output. But "coalgebra" has operations that have multiple outputs. Coalgebraic operations are very important, especially in computing: you wouldn't want a computer program that only does <em>one</em> thing when you ran it, because then it couldn't also tell you that it had done it!</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49491#49491 Answer by Denis Serre for Why are so few operations with arity bigger than 2? Denis Serre 2010-12-15T07:14:11Z 2010-12-15T07:14:11Z <p>Lie and Jordan <strong>Triple system</strong> have arity <strong>3</strong>. A Jordan triple system is a vector space with additional structure is given by a triplet $$(x,y,z)\rightarrow \{x,y,z\}$$ that satisfies the identities $$\{u,v,w\} = \{u,w,v\}$$ and $$\{u,v,\{w,x,y\}\} = \{w,x,\{u,v,y\}\} + \{w, \{u,v,x\},y\} -\{\{v,u,w\},x,y\}.$$ See <a href="http://en.wikipedia.org/wiki/Jordan_triple_system#Jordan_triple_systems" rel="nofollow">link text</a>. Every Jordan algebra can be embedded in a Jordan triple system but the converse is not true. Any Jordan triple system is a Lie triple system with respect to the product $$[u,v,w] = \{u,v,w,\} − \{v,u,w\}.$$ The structure of a Lie triple system is given by a bracket satisfying the identities $$[u,v,w] = − [v,u,w], \qquad [u,v,w] + [w,u,v] + [v,w,u] = 0$$ and $$[u,v,[w,x,y]] = [[u,v,w],x,y] + [w,[u,v,x],y] + [w,x,[u,v,y]].$$</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49509#49509 Answer by Andrew Stacey for Why are so few operations with arity bigger than 2? Andrew Stacey 2010-12-15T10:28:01Z 2010-12-15T10:28:01Z <p>To add to the list of examples:</p> <ol> <li><p><a href="http://ncatlab.org/nlab/show/heap" rel="nofollow">Heaps</a> have a single ternary operation (identities on linked page). In short, a heap is to a group what an affine space is to a vector space: as soon as you pick an identity then you get a group.</p></li> <li><p><a href="http://ncatlab.org/nlab/show/totally+convex+space" rel="nofollow">Totally convex spaces</a> which are spaces that allow arbitrary convex combinations. Simple examples are the unit balls of normed vector spaces, but others such as $(0,1)$ exist.</p></li> <li><p>Similarly, $C^*$-algebras and there's a theory closely related to Banach algebras. See <a href="http://ncatlab.org/nlab/show/algebraic+theories+in+functional+analysis" rel="nofollow">this page</a> on the nLab where I started gathering together a few details on these.</p></li> </ol> <p>To address the point as to why we often only use operations of arity at most 2, here's a neat little fact. Abstractly, we can consider operations of arbitrary arity with arbitrary identities, but in concrete situations the operations usually have a high level of compatibility. A common one to ask for is commutativity. This is commutativity of operations, which is ever-so-slightly different from what we normally think of as commutativity (though the two are very closely related). If we have a binary operation with a unit, then any operation that commutes with that operation (and its unit) turns out to be formed by iterating the binary operation. This is an easy generalisation of the <a href="http://ncatlab.org/nlab/show/Eckmann-Hilton+argument" rel="nofollow">Eckmann-Hilton argument</a>. Therefore, once we start applying common identities, we find that we can often reduce the arity down to something palatable.</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49510#49510 Answer by Piero D'Ancona for Why are so few operations with arity bigger than 2? Piero D'Ancona 2010-12-15T10:34:54Z 2010-12-15T10:34:54Z <p>I'm not sure to understand the question: don't you consider differential geometry, with all its multilinear algebra, as a good source of examples? e.g. the volume form on a manifold is a quite important and natural multilinear operator.</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/49511#49511 Answer by Max for Why are so few operations with arity bigger than 2? Max 2010-12-15T10:38:40Z 2010-12-15T10:38:40Z <p>In an <a href="http://en.wikipedia.org/wiki/Affine_space" rel="nofollow">affine space</a> $A$, the displacement (difference) between two points is a vector, and one can add a vector to a point, but not two points. However these can be replaced by a ternary operation in terms of points alone: the parallelogram rule $\nearrow : A \times A \times A \to A,\,\nearrow(p, a, b) = p+(a-b)$.</p> <p>You can even add scalar multiplication of the difference into the bargain.</p> <p>Why would you want to do this? Well affine spaces are more primitive than a vector space -- yet we use a vector space in defining them. To me the more natural approach is to define them without it, and watch the vector space (of displacements) drop out.</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/87587#87587 Answer by Sebastian for Why are so few operations with arity bigger than 2? Sebastian 2012-02-05T14:28:31Z 2012-02-05T14:34:39Z <p>I am surprised that nobody came up with median algebras. That is, algebras that are equipped with a single ternary (fundamental) operation (see <a href="http://en.wikipedia.org/wiki/Median_algebra" rel="nofollow">http://en.wikipedia.org/wiki/Median_algebra</a> for a definition and some references). They generalize distributive lattices since the median function $(x \vee y) \wedge (y \vee z) \wedge (z \vee x)$ of any distributive lattice gives rise to a median algebra. Although median algebras still have many of the nice properties of distributive laticces, the concept is more subtle and the category of median algebras is not equivalent to that of distributive lattices. So the idea of having this single ternary fundamental operation really gives you something new and, at least in my oppinion, very interesting to look at.</p> <p>To support my case: One might also be interested in Median algebras since they have beautiful duality with so-called Isbell spaces, first described by, you guessed correclty, John Isbell (the reference is given in the wikipedia article mentioned above). An Isbell space is a bounded Priestley spaces that is also equipped with certain (unary) complement operation.</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/87611#87611 Answer by robot for Why are so few operations with arity bigger than 2? robot 2012-02-05T20:29:30Z 2012-02-05T20:29:30Z <p>More than two-ary operations pop up in algebraic approach to CSP (constraint satisfaction problem). See e.g. <a href="http://www.ams.org/mathscinet-getitem?mr=2470592" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=2470592</a> or <a href="http://www.ams.org/mathscinet-getitem?mr=2137072" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=2137072</a></p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/87614#87614 Answer by Marcos Cossarini for Why are so few operations with arity bigger than 2? Marcos Cossarini 2012-02-05T20:40:25Z 2012-02-06T09:47:58Z <p>The composition law in a monoid is usually represented using a binary operation (multiplication) and a zeroary operation (unit), but I view it more naturally as an operation (say, a bracket) taking any finite list of arguments and being associative in the sense that brackets can be eliminated in any expression, for example we have the identities</p> <p>[a,[],b,[[c,d],e],[f]]=[a,b,c,d,e,f]</p> <p>and </p> <p>[a]=a. </p> <p>Then we can define a zeroary operation 1:=[] and a binary operation a*b:=[a,b], and recover the bracket from them, using identities such as [a,b,c,d]=[a,[b,[c,d]]]. These two operations satisfy the usual axioms of a monoid, and any two operations satisfying them can be extended to an associative finitary bracket. </p> <p>I view the usual representation by a binary and a zeroary operation as an artifact for being able to produce simpler-looking proofs that the structures that we encounter are monoids. </p> <p>My point is that naturally binary operations are not that common either! Perhaps an example is the Lie bracket. </p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/87713#87713 Answer by Terry Tao for Why are so few operations with arity bigger than 2? Terry Tao 2012-02-06T19:54:16Z 2012-02-06T20:05:24Z <p>In higher order Fourier analysis, there are d-dimensional parallelopiped structures which can be viewed as a $2^d-1$-ary relation of the form "given all but one vertex of a parallelopiped as input, return the final vertex as output". (Here "parallelopiped" should be interpreted in a suitably abstract sense, as a family of $2^d$-tuples obeying a certain number of axioms.) This point of view is taken for instance in <a href="http://arxiv.org/abs/1009.3825" rel="nofollow">this paper of Camarena and Szegedy</a>, building on the <a href="http://arxiv.org/abs/math/0606004" rel="nofollow">earlier work of Host and Kra</a>. In the d=2 case, this ternary operation is essentially equivalent to an additive operation once one fixes an origin (in which case the ternary operation becomes $(x,y,z) \mapsto x-y+z$). For d=3, these operations are governed by 2-step nilpotent groups, and more generally d-dimensional parallelopiped structures are governed by d-1-step nilpotent groups.</p> <p>These parallelopiped structures can be viewed as the abstract foundation of the Gowers uniformity norms, and they also share some formal resemblance to cubic complexes, which are constructs that appear mostly in algebraic topology and are discussed for instance <a href="http://www.ams.org/mathscinet-getitem?mr=2002612" rel="nofollow">here</a>. There is a simplicial version of the latter concept known as a <a href="http://ncatlab.org/nlab/show/Kan+complex" rel="nofollow">Kan complex</a>, but I do not know the details of how they are used. But I think Kan complexes come equipped with high arity relations of the general form "given data for all but one face of a simplex, supply the data for the remaining face (and also for the interior) of that simplex". Among other things, such structures can be used to define n-groups; see for instance <a href="http://terrytao.wordpress.com/2009/10/19/grothendiecks-definition-of-a-group/" rel="nofollow">my blog post on this topic</a>.</p> http://mathoverflow.net/questions/49437/why-are-so-few-operations-with-arity-bigger-than-2/87717#87717 Answer by Ryan Budney for Why are so few operations with arity bigger than 2? Ryan Budney 2012-02-06T20:32:40Z 2012-02-06T20:32:40Z <p>In knot theory, <em>splicing</em> generally has more than just one or two inputs. </p> <p>Splicing with one input generates things like Whitehead doubles: <img src="http://upload.wikimedia.org/wikipedia/commons/thumb/a/ad/B_sat2.png/200px-B_sat2.png" alt="Whitehead double of figure-8"></p> <p>and cabelling: <img src="http://upload.wikimedia.org/wikipedia/commons/thumb/5/5d/B_sat3.png/350px-B_sat3.png" alt="cable of a connect-sum of trefoil and figure-8"></p> <p>There are many $n$-ary operations. The first one noticed (historically) is connect-sum: <img src="http://upload.wikimedia.org/wikipedia/commons/thumb/d/dc/Sum_of_knots3.svg/300px-Sum_of_knots3.svg.png" alt="connect-sum of figure-8 and trefoil">. The issue one might have with the $n$-ary connect-sum is its generated by the 2-ary connect sum. So here is an example of a $2$-ary splice that's not a connect sum: <img src="http://upload.wikimedia.org/wikipedia/commons/thumb/a/a3/B_sat1.png/250px-B_sat1.png" alt="alt text"></p> <p>There are a countable infinite collection of $n$-ary splices for any $n \geq 1$, moreover, there's still countably-infinite many primitive ones for any $n \geq 1$, where primitive means "can't be expressed in terms of $j$-ary operations for j less than n" These primitive splicing operations turn out to be specified (uniquely) by hyperbolic $(n+1)$-component links in the 3-sphere $L \subset S^3$, $L=L_0 \sqcup L_1 \sqcup \cdots \sqcup L_n$ such that the sublink $L_1 \sqcup \cdots \sqcup L_n$ is the trivial link. The hyperbolicity means $S^3 \setminus L$ has a complete hyperbolic structure of finite volume. </p> <p>Splicing can be put in an operadic framework and this is the topic of one of my papers. So you can turn it into a purely algebraic formalism as well, by taking the homology of the space of all knots and the splicing operad, respectively. </p> <p>It's not clear to me there's any <em>reason</em> for the seeming prevalence of 2-ary operations in mathematics. It appears to be more of an accident -- two things interacting is simpler, easier to contemplate. </p>