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created hyperring
Corrected the proposition about the structure of $Hom(\mathbb{Z}[X], \mathbf{S})$.
Ah, thanks for fixing that. I had put the statement the way Connes said it in his talk, but evidently that was without some fine print.
Added that we can form many examples of hyperrings by quotienting a ring $R$ by a subgroup $G \subset R^{\times}$ of its multiplicative group.
John says here that the hypersemigroup is a monoid object in the category of sets and multivalued functions. Sounds the way I like, but I do not understand. I mean what is the monoidal product in the category of sets and multivalued functions (I guess a cartesian product, but what is its description in elementwise terms) ? Sounds like a student exercise but I would rather not blunder into possibly wrong attempts now.
I think the monoidal structure is the ordinary cartesian product of sets, which is however not the cartesian product in the category of sets and multivalued functions.
It’s probably high time that all these notions got sussed out properly, because certain aspects of hypergroup and hyperring as so far presented have an ad hoc look to them. These could probably be clarified by placing them in the context of the cartesian bicategory of relations and the cartesian monoidal subbicategory of functions.
But to answer the immediate question: a multivalued function from $A$ to $B$ is just a relation from $A$ to $B$, and we are working in the bicategory of sets and relations. The monoidal structure is given by cartesian product of sets as you surmise, and the tensor product of relations $R: A \to B$, $S: C \to D$ is the relation $R \times S: A \times C \to B \times D$ defined by
$(R \times S)((a, c), (b, d)) = R(a, b) \wedge S(c, d)$This is a monoidal product (in the world of poset-enriched categories, or of 2-categories if you prefer), but not a cartesian monoidal product. Of course, we can define monoids in any monoidal category, so John’s formulation makes sense.
(Apparently the hypergroup experts intend “multivalued” to mean “at least one value”, so here we are talking about total relations. But that’s okay: it’s clear that the monoidal product as defined above restricts to the bicategory of total relations.)
Another way of thinking about all this is that we’re working in the full (poset-enriched) subcategory of the monoidal category of sup-lattices and cocontinuous maps whose objects are power sets. A monoid in the monoidal category of sup-lattices is usually called a (noncommutative) quantale, so John’s definition boils down to a quantale structure on a power object, except that we must restrict to total relations.
The funny thing is: I can’t seem to rummage up many interesting examples of total power quantales off the top of my head! There seems to be a paucity of examples over at hyperring, and it leads me to wonder what can be said about the finite examples. Can these be classified, or if not, can be easily produce a rich supply of them?
Actually, now that I think further, I believe one interesting example would be the Boolean Hecke algebra of a Coxeter group, which Jim Dolan and I were discussing a few years back. This would be worth putting on the Lab.
Hm, there is no extant entry for hypersemigroup or hypermonoid, and it’s not clear we want or need to create one. So I’ll just record the example here for now.
Let $W$ be a finite Coxeter group. A Coxeter group is really a group presentation, with a collection of involutive generators $s_1, \ldots, s_n$ and whose relations are of the form , where $m_{i j}$ is a collection of integers specified by a Coxeter diagram or Coxeter matrix (assumed to be symmetric and with 1’s down the diagonal). We may rewrite the presentation as
$s_{i}^{2} = 1, \qquad s_i s_j \ldots = s_j s_i \ldots$where the words on either side of the second equation alternate between $s_i$ and $s_j$ (for distinct $i$ and $j$) and have length $m_{i j}$.
The Boolean Hecke algebra of a Coxeter group $W$ is a quantale whose underlying sup-lattice is $P W$, and whose quantalic multiplication is uniquely determined by a hypermonoid structure which satisfies
$s_{i}^{2} = \{s_i, 1\}$and described roughly as follows: given two reduced words $w_1$, $w_2$ in the Coxeter group representation, put the word $w_1 w_2$ into reduced form, resolving any occurrences of $s_{i}^2$ that occur according to the equation above. Now group elements $g \in W$ are equivalence classes of reduced words, and the recipe above actually gives a well-defined map $W \times W \to P(W)$ that gives a hypermonoid structure.
The way I just portrayed it makes it seem very combinatorial and dependent on the theory of rewriting systems, but is can be made more conceptual if the Coxeter group $W$ is seen as coming from a BN-pair. How this usually works is that “B” is a Borel subgroup of a simple algebraic group $G$ and “N” is the normalizer of a maximal torus. The basic idea then is that the Boolean Hecke algebra is the algebra consisting of $G$-equivariant Boolean algebra maps
$P(G/B) \to P(G/B)$which as a sup-lattice is isomorphic to $P(B\backslash G/B)$, the power set of the set of double cosets. (Since such equivariant maps compose, this imparts a quantale structure to $P(B\backslash G/B)$.) When I refer to $W$ “coming from” a BN-pair, I mean according to some general theory that these double cosets are in bijection with the elements of some Coxeter group $W$. Hence $P(W) \cong P(B\backslash G/B)$, but the algebra structure on $P(B \backslash G/B)$ can be regarded as a deformation of the group algebra structure.
This example of a hypermonoid deserves to be discussed at much greater length, but not necessarily under hypermonoid.
Great answers! My personal opinion is that long term it would be good to have separate entries hypermonoid, hypersemigroup, hyperfield in addition to the present hyperring. It seems that the subject is becoming important.
This example of a hypermonoid deserves to be discussed at much greater length, but not necessarily under hypermonoid.
Just please don’t forget to put it somewhere on the $n$Lab. Why not at hypermonoid? You can still move it later.
Okay, I just did that Urs.
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