Ring with three binary operations

Question

A rather precocious student studying abstract algebra with me asked the following question: Are there interesting rings where there are not just two but three binary operations along with some appropriate distributivity properties?

Answer 1

The real numbers $\mathbb{R}$ with the following three binary operations:

• The maximum: $(x,y)\mapsto\max\{x,y\}$.

• The sum: $(x,y)\mapsto x+y$.

• The product: $(x,y)\mapsto x\cdot y$.

The maximum is to the sum what the sum is to the product, except for the fact that the maximum does not have inverses, nor a unit, i.e. $(\mathbb{R},\max,+)$ is a semiring, while $(\mathbb{R},+,\cdot)$ is a ring.

Answer 2

An important example is the notion of a Gerstenhaber algebra. It is simultaneously a commutative ring and a Lie algebra, such that the product and bracket satisfy the Poisson identity, except all these things need to be understood in a differential graded sense.

Answer 3

Very much in the spirit of Dan's answer, but more elementary, are Poisson algebras, associative algebras with Lie brackets that act as derivations.

Answer 4

For experimenting you could use alg, a program which computes all finite models of a given theory. The best thing may be to pass the ball back to your student and ask him to use alg to find some interesting structures.

For example, suppse we want a structure $(R, 0, +, -, \times, \&)$ such that $(R, 0, +, -)$ is a commutative group, $\times$ and $\&$ are associative, $\times$ and $\&$ distribute over $+$ and $\&$ distributes over $\times$ (I am making stuff up, the point is to experiment until something interesting is found). In alg the input file would be:

Constant 0.
Unary ~.
Binary + * &.

# 0, + is a commutative group

Axiom plus_commutative: x + y = y + x.
Axiom plus_associative: (x + y) + z = x + (y + z).
Axiom zero_neutral_left: 0 + x = x.
Axiom zero_neutral_right: x + 0 = x.
Axiom negative_inverse: x + ~ x = 0.
Axiom negative_inverse: ~ x + x = 0.
Axiom zero_inverse: ~ 0 = 0.
Axiom inverse_involution: ~ (~ x) = x.

# * and & are associative

Axiom mult_associative: (x * y) * z = x * (y * z).
Axiom and_associative: (x & y) & z = x & (y & z).

# Distributivity laws

Axiom mult_distr_right: (x + y) * z = x * z + y * z.
Axiom mult_distr_left: x * (y + z) = x * y + x * z.

Axiom and_distr_right: (x + y) & z = (x & z) + (y & z).
Axiom and_distr_left: x & (y + z) = (x & y) + (x & z).

Axiom mult_and_distr_right: (x * y) & z = (x & z) * (y & z).
Axiom mult_and_distr_left: x & (y * z) = (x & y) * (x & z).

Let us count how many of these are, up to isomorphism, of given sizes:

$ ./alg.native --size 1-7 --count three.th

size | count
-----|------
   1 | 1
   2 | 4
   3 | 3
   4 | 36
   5 | 3
   6 | 12
   7 | 3

Check the numbers [4, 3, 36, 3, 12, 3](http://oeis.org/search?q=4,3,36,3,12,3)
on-line at oeis.org

We can also look at these structures, but that's the sort of thing a student should do. Here is a random one of size 4 that alg prints out when we omit --count:

~ |  0  a  b  c
--+------------
  |  0  a  b  c


+ |  0  a  b  c
--+------------
0 |  0  a  b  c
a |  a  0  c  b
b |  b  c  0  a
c |  c  b  a  0


* |  0  a  b  c
--+------------
0 |  0  0  0  0
a |  0  a  0  a
b |  0  0  b  b
c |  0  a  b  c


& |  0  a  b  c
--+------------
0 |  0  0  0  0
a |  0  0  0  0
b |  0  a  b  c
c |  0  a  b  c

Up to size 7 I cannot actually see any interesting ones, there are always large blocks of 0's in $\&$. Other things should be tried out.

Answer 5

An interesting non-example is the Eckmann-Hilton theorem, stating that if a set is endowed with two associative unital binary operations that "commute" (i.e., if I get it correctly, each multiplication operator $a\mapsto a*b$ is a homomorphism with respect to the other multiplication) then the two operations are the same.

This would exclude the existence of rings $(R, +,*,\circ)$ with two genuinely different "commuting" multiplications.

Answer 6

An exponential field is a field with an additional unary operation $x\mapsto E(x)$ extending the usual idea of exponentiation. So it satisfies the usual law of exponents $E(a+b)=E(a)\cdot E(b)$ and also has $E(0)=1$. An exponential ring has an underlying ring, rather than field, and the exponentiation function is a homomorphism from the additive group of the ring to the multiplicative group of units.

The example of the real exponential field $\langle\mathbb{R},+,\cdot,e^x\rangle$ has been a principal focus of the research program in model theory that has lead to the theory of o-minimality. Tarski had famously proved that the theory of real-closed fields $\langle\mathbb{R},+,\cdot,0,1,\lt\rangle$ is a decidable theory, and one of the original motivating questions, still open to my knowledge, is whether the first-order theory of the real exponential field is similarly decidable. Meanwhile, the o-minimalists are making huge progress on our understanding of the structure of definable sets in these and many other similar structures.

Answer 7

Not only are there such examples, there is a very natural way of continuing the progression that starts with addition and multiplication.

Begin with sets. Then monoids are just the monoidal objects in the category of sets. That ends the story at level 1. Now consider the objects that ended the story at level 1 which are also commutative, that is, commutative monoids. Then semirings are the monoidal objects in the category of commutative monoids. Here the monoidal operation on the category of commutative monoids is the tensor product, rather than the Cartesian product, but this is as it should be because of the tensor-hom adjunction: $\mathrm{Hom}(A\otimes B,C)=\mathrm{Hom}(B,\mathrm{Hom}(A,C))$. In other words, the composition of the representable functors $\mathrm{Hom}(B,-)$ and $\mathrm{Hom}(A,-)$ is the functor represented by $A\otimes B$. Thus for the present purposes, the tensor product for commutative monoids plays the role of the Cartesian product for sets.

Now what if we try to go one step further? Can we fill in the missing entries in the following table of analogies? (Sets : Cartesian product : monoids) :: (Commutative monoids : tensor product : semirings) :: (Commutative semirings : ?? : ??)

The answer is yes, but there is one more twist to the story, which is that unlike in the categories of sets and commutative monoids, in the categories of commutative semirings, representable functors $\mathrm{Hom}(A,-)$ take values all the way down in the category of sets, not in the category of semirings. This actually occurs already at the tensor product stage above, but we didn't see it because we were working with commutative monoids (i.e. modules over the semiring of natural numbers) instead of modules over more general semirings.

If we let $K$ be any semiring, then $\mathrm{Hom}_K(A,-)$ takes values in abelian group, not in $K$-modules. To make everything above work, we need $A$ to be a $K$-$K$-bimodule. Thus what we really want is to complete the following table of analogies. (Sets : sets : Cartesian product : monoids) :: ($K$-modules : $K$-$K$-bimodules : tensor product $\otimes_K$ : semirings over $K$) :: (Commutative $L$-algebras : ?? : ?? : ??), where $L$ is a fixed commutative semiring.

For the first ??, we want commutative $L$-algebras $A$ such that $\mathrm{Hom}(A,-)$ takes values in $L$-algebras, rather than sets. This extra structure will be the analogue of the right $K$-module structure above. So let us call it an $L$-$L$-bialgebra structure. For instance, $A$ will need to have two co-operations $A\to A\otimes_L A$, which will induce a functorial ring structure on the objects $\mathrm{Hom}(A,B)$. If $L$ is a ring, then this is precisely the structure of a commutative $L$-algebra scheme on $\mathrm{Spec}(A)$.

Then if $A$ and $B$ are $L$-$L$-bialgebras, we can compose the functors $\mathrm{Hom}(B,-)$ and $\mathrm{Hom}(A,-)$. The result is easily seen to be representable, and the representing object is denoted $A\odot B$. (So when $L$ is a ring, we are then taking two affine commutative $L$-algebra schemes over $L$, viewing them as endofunctors on the category of commutative $L$-algebras, and then composing them. This is not such a common thing to do, but the result is another affine commutative $L$-algebra scheme over $L$.) I like to call it the composition product of $A$ and $B$. Then we can define a composition $L$-algebra to be a monoid object in the category of $L$-$L$-bialgebras. Thus the final line of the analogy table above is (Commutative $L$-algebras : $L$-$L$-bialgebras : composition product $\odot_L$ : composition $L$-algebras). The monoidal operation on a composition $L$-algebra is usually denoted $\circ$ and is called composition or plethysm. So the usual hierarchy of operations "(1) addition, (2) multiplication" extends to "(3) plethysm".

There are many examples of composition $L$-algebras (but interestingly not too many when $L$ is a ring). The most basic example is the polynomial algebra $L[x]$, where $\circ$ is given by usual composition of polynomials. The $L$-$L$-bialgebra structure is the one such that $L[x]$ represents the identity functor on commutative $L$-algebras. A more interesting example is the polynomial algebra `A=$\mathbb{C}[\partial^0, \partial^1,\dots]$ in infinitely many indeterminants, which we think of as all algebraic differential operators in one variable. Here $\circ$ is the usual composition of differential operators. (The $L$-$L$-bialgebra structure is determined by requiring each $\partial^i$ to be linear and to satisfy the appropriate Leibniz rule.)

There are also exotic, arithmetic examples when $L$ is not a $\mathbb{Q}$-algebra. These are responsible for concepts like Witt vectors and $\Lambda$-rings. (When $L$ is a field of characteristic $0$, it is conjectured that any composition $L$-algebra can be generated by linear operators, and so all composition $L$-algebras should reduce to more familiar multilinear constructions, as with the differential operators above.) Perhaps the easiest one of these to give has already been mentioned by Darij Grinberg. It is the ring $\Lambda=\mathrm{Symm}$ of symmetric functions in infinitely many variables. Here addition and multiplication are as usual, and $\circ$ is the operation known as plethysm in the theory of symmetric functions. It is the composition algebra whose representations are $\Lambda$-rings and whose co-induction functor is the big Witt functor. (I haven't defined these concepts here.)

This is discussed in only a few places in the literature. In order of appearance: Tall-Wraith, Bergman-Hausknecht (this deals with general cagtegories of a universal-algebraic nature), Borger-Wieland, Stacey-Whitehouse. And it seems that every paper on the subject uses different term for what I called a composition algebra above. On my web page, I have slides from a talk a gave not so long ago on these things.

Answer 8

The claim that there are two binary operations on rings is misleading. Rings are actually equipped with countably many $n$-ary operations, one for each noncommutative polynomial in $n$ variables over $\mathbb{Z}$. These generate the morphisms in a category with finite products, the Lawvere theory of rings $T$, which is a category with the property that finite product-preserving functors $T \to \text{Set}$ are the same thing as rings. It just happens to be the case that as a category with finite products, $T$ is generated by addition and multiplication. The Lawvere theory of commutative rings is similar except that the polynomials are commutative; incidentally, it may also be regarded as the category of affine spaces over $\mathbb{Z}$.

This gives a useful perspective from which to understand other ring-like structures. For example:

• commutative Banach algebras are equipped with an $n$-ary operation for each holomorphic function $\mathbb{C}^n \to \mathbb{C}$.
• smooth algebras like the algebras $C^{\infty}(M)$ of smooth functions on a smooth manifold are equipped with an $n$-ary operation for each smooth function $\mathbb{R}^n \to \mathbb{R}$.

Here is a general procedure for determining what operations are actually available to you when working with some mathematical objects. If $C$ is a concrete category and $F : C \to \text{Set}$ the forgetful functor, then one interpretation of "$n$-ary operation" is "natural transformation $F^n \to F$." If $C$ has finite coproducts and $F$ is representable by an object $a$, then by the Yoneda lemma these are the same thing as elements of $F(a \sqcup ... \sqcup a)$. This reproduces the obvious answers for groups, rings, etc., and when $C$ is the opposite of the category of smooth manifolds and $F : M \mapsto C^{\infty}(M)$ then we get that "$n$-ary operation" means element of $C^{\infty}(\mathbb{R}^n)$ as above.

Answer 9

This isn't an answer to this exact question, but it sounds like the student may also be interested in hearing about Hopf algebras.

Answer 10

The paper "The Natural Chain of Binary Arithmetic Operations and Generalized Derivatives" by M. Carroll (link) is a great paper for undergrads that demonstrates an infinite number of binary operations (defined recursively and in terms of the exponential function) on the reals where the $i$th operation distributes over the $(i-1)$th operation.

Answer 11

Take the ring of polynomials. More generally, any ring of functions (from a ring to itself).

With the functions, you have 3 operations defined:

Pointwise addition: + such that f+g is the function t--> f(t) + g(t) Pointwise multiplication: . such that f.g is the function t --> f(t).g(t) Composition of functions o such that fog is the function t--> f(g(t))

This is defined for functions but in particular for polynomials on some ring R and for matrices with ring coefficients.

Answer 12

A differential-geometric example of such a thing would be differential forms (with operations of addition, wedge product and differentiation). More generally, one considers DGLAs, Differential Graded Lie Algebras (with the usual caveat that the Lie bracket is not quite commutative/associative); the operations are addition, derivation and the bracket (as well as multiplication by scalars as a free bonus). The main example is, of course, differential forms on a manifold with values in a Lie algebra. One uses DGLA's to describe deformations of pretty much anything under the sun, look here.

Answer 13

$L^2$ with addition, multiplication, and convolution, with both multiplication and convolution linear with respect to addition.

Answer 14

Addition and multiplication over $\Bbb{R}$ can actually be thought of as two parts of an infinite chain of binary operations $b_n: \Bbb{R}^2 \rightarrow \Bbb{R}$ where each one distributes over the next-lower-level one: $$b_n(x, b_{n-1}(y, z)) = b_{n-1}(b_n(x, y), b_n(x, z))$$

for all $n \in \Bbb{Z}$. Let $b_0(x, y) = x + y$, and make the definitions

\begin{align*} b_{k+1}(x, y) &:= e^{b_k(\ln(x), \ln(y))} \\ b_{k-1}(x, y) &:= \ln(b_k(e^x, e^y)) \ \end{align*}

Using these definitions, we find, for example:

\begin{align*} b_1(x, y) &= xy \\ b_2(x, y) &= e^{\ln(x) \ln(y)} \\ &... \\ b_{-1}(x, y) &= \ln(e^x + e^y) \\ b_{-2}(x, y) &= \ln(\ln(e^{e^x}+e^{e^y})) \\ &... \ \end{align*}

So for instance, addition $b_0$ distributes over $b_{-1}$: $$b_0(x, b_{-1}(y, z)) = x + \ln(e^y + e^z) = \ln(e^x (e^y + e^z)) = \ln(e^{x+y} + e^{x+z}) = b_{-1}(b_0(x, y), b_0(x, z)),$$

and this can be proven inductively for all $n \in \Bbb{Z}$.

Ignoring domain restrictions on logarithms, I can repeat this construction over any field equipped with an exponential function and its logarithmic inverse. I didn't need to use $e$ and $\ln$ here in particular--the choice of the exponential/log base doesn't actually affect $b_0$ (regular addition) and $b_1$ (regular multiplication), although in general it will affect higher and lower functions in the chain.