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8 votes
1 answer
238 views

Functions over monoids which factor in two different ways

This is a follow-up question to this MO question, which was asked by Richard Stanley in a comment to my answer there. Let $S$ be a commutative monoid and $f(x_1, \dots, x_n)$ be a function from $S^n$ ...
Tony Huynh's user avatar
  • 32.1k
8 votes
0 answers
285 views

Matrix decompositions as monoid isomorphisms. Ever considered before?

I've noticed some correspondences between some matrix decompositions and monoid isomorphisms (always to some free commutative monoid), in addition to the one I asked about in a previous question: ...
wlad's user avatar
  • 4,943
8 votes
5 answers
1k views

Examples of left reversible semigroups

I am looking for concrete examples of cancellative, left reversible semigroups. Left reversible semigroups are also called "Ore semigroups". See this wikipedia page for the definition of a left ...
Orr Shalit's user avatar
7 votes
2 answers
370 views

Wants: Polynomial Time Algorithm for Decomposing a Multiset of Rationals into Two Additive Subsets.

First, allow me to say that this problem was posed to me by a professor in the department. It is related to his research in a way that I do not know. However, since I couldn't come up with anything ...
B. Bischof's user avatar
  • 4,842
7 votes
0 answers
194 views

Factoring a function from a finite set to itself

Let $S$ be a finite set and $f: S \to S$ be a function. Let $k = |f(S)|$ and let $\alpha$ be the partition of $S$ into $f$-fibers, i.e. $\alpha = \{ \alpha_t \}_{t \in f(S)}$ where $\alpha_t = f^{-1}(\...
Sophie M's user avatar
  • 695
7 votes
2 answers
488 views

Is every cancellative semigroup a subdirect product of subdirectly irreducible cancellative semigroups?

By a classical result of Birkhoff (that is, Theorem 2 in [G. Birkhoff, Subdirect unions in universal algebra, Bull. AMS, 1944]) and the trivial fact that the class of semigroups is closed under the ...
Salvo Tringali's user avatar
7 votes
1 answer
722 views

How is called a semigroup...

Does anyone know, how is called a semigroup in which every equation $ax=b$ has only a finite set (maybe empty) of solutions?
Boris Novikov's user avatar
6 votes
3 answers
595 views

What's the current state of one-rule semi-Thue system termination problem?

What's the current state of one-rule semi-Thue system termination problem? Search produces a lot of references, but it's hard to find out if decidability of this problem has been proven or not.
TT_ stands with Russia's user avatar
6 votes
0 answers
632 views

Generating functions in countable commutative monoids

Let $f: \mathbb{N}_0 \rightarrow \mathbb{C}$ be a function. The power series of $f$ can be viewed as the function $\mathscr{P}_f : q \mapsto \sum_{n \in \mathbb{N}_0}^{} f(n)q^n$ where $q \in \mathbb{...
Tian Vlašić's user avatar
6 votes
1 answer
622 views

When is the cofibrant replacement of a product the product of the cofibrant replacements?

I'm in a situation where I'd like to prove $Q(E\otimes E) \simeq QE \otimes QE$ for a monoid $E$ in a symmetric monoidal model category. I know it's not true in general that $Q(E\otimes F)\simeq QE \...
David White's user avatar
  • 30.3k
6 votes
0 answers
618 views

Duality between conjugacy classes and irreducible characters for finite monoids?

Qiaochu's answer to this question suggests that the proper way to view the bijection between conjugacy classes and irreducible complex representations of a finite group is via a duality. My question ...
Benjamin Steinberg's user avatar
6 votes
3 answers
393 views

Structure theorem for a class of idempotent monoids (where $xy = x$ or $xy = y$ for all $x, y$)

Question. Is there any structure theorem for the class of monoids $H$ with the property that $xy = x$ or $xy = y$ for all $x, y \in H$? Or does this look hopeless for some good reasons? A monoid with ...
Salvo Tringali's user avatar
6 votes
0 answers
190 views

The highest degree of a polynomial on a finite group

This question is motivated by the comments and the answer to this MO-question. First let us recall some definitions. A function $f:X\to X$ on a group $X$ is called a polynomial if there exists $n\in\...
Taras Banakh's user avatar
  • 41.9k
6 votes
3 answers
472 views

Spaces with unique endomorphism monoids

If $(X,\tau)$ is a topological space, let $\text{End}(X)$ denote the collection of all continuous maps $f: X\to X$. With composition, this becomes the endomorphism monoid $(\text{End}(X), \circ)$. We ...
Dominic van der Zypen's user avatar
6 votes
1 answer
371 views

Embedding a cancellative monoid into another in such a way that a prescribed element becomes left-invertible

Let $\mathbb A = (A, +_A)$ be a cancellative, but possibly non-commutative, monoid with identity $0$, and fix an element $x \in A$. Does there always exist a cancellative monoid $\mathbb B = (B, +)$ ...
Salvo Tringali's user avatar
5 votes
2 answers
236 views

Descent of flatness from algebras to monoids

Consider a morphism of commutative monoids $u\colon M\rightarrow N$. We say that $u$ is flat, if the tensor product functor $\bullet\otimes_MN$ from the category of $M$-modules to the category of $N$-...
Fred Rohrer's user avatar
  • 6,700
5 votes
0 answers
99 views

Zappa-Szép products of the monoid of integers with itself

Question What are all the functions $\alpha , \beta : \mathbb{N} \to \mathbb{N}$ satisfying the following functional equations? $\bullet ~~~~ \alpha(0)=0, \quad \beta(0)=0\\ \bullet ~~~ \...
HeinrichD's user avatar
  • 5,482
5 votes
1 answer
342 views

Can the trivial module be stably free for a monoid ring?

Let $M$ be a non-trivial monoid and $\mathbb ZM$ its monoid ring. All modules are left modules in what follows. Suppose that $M$ contains a zero element (or absorbing element) $z$. That is $mz=z=zm$ ...
Benjamin Steinberg's user avatar
5 votes
0 answers
187 views

Isbell duality for monoids and groups

Isbell Duality $\newcommand{\IsbellSpec}{\mathsf{Spec}}\newcommand{\IsbellO}{\mathsf{O}}\newcommand{\Sets}{\mathsf{Sets}}\newcommand{\rmL}{\mathrm{L}}\newcommand{\rmR}{\mathrm{R}}\newcommand{\B}{\...
Emily's user avatar
  • 11.8k
5 votes
2 answers
402 views

Maximal commuting subsets of $\text{End}(X)$

Let $X$ be a set and let $\text{End}(X)$ be the set of all functions $f:X\to X$. We say that $f, g\in \text{End}(X)$ commute if $g\circ f = f\circ g$, and $S\subseteq \text{End}(X)$ is a commuting ...
Dominic van der Zypen's user avatar
5 votes
2 answers
364 views

Have semigroups with actions on themselves that have a dual to the compatibility axiom ever been studied?

For a semigroup $G$ with a left action on itself, the axiom for compatibility becomes: $$ \forall f,g,h\in G:hg(f)=h(g(f)) $$ Now suppose there is additional axiom, or constraint if you prefer, ...
James Smith's user avatar
5 votes
2 answers
332 views

Questions on weakly symmetric algebras

A finite dimensional algebra $A$ over a field $K$ is called weakly symmetric in case $soc(P)=top(P)$ for every indecomposable projective module $P$ and it is called symmetric in case $D(A) \cong A$ as ...
Mare's user avatar
  • 26.5k
4 votes
1 answer
446 views

What is a "cusp" ("кусок") in relation to Guba's embedding theorem?

I'm confused by the definition of a "cusp" as found in V.S. Guba, Conditions for the embeddability of semigroups into groups, Math. Notes 56 (1994), Nos. 1-2, 763-769 (link). In the words of Mark ...
Salvo Tringali's user avatar
4 votes
1 answer
385 views

Which monoids can be realized as the monoid of ideals of a commutative monoid?

Let $H$ be a commutative monoid (written multiplicatively). We say that a set $I \subseteq H$ is an ideal of $H$ if $IH = I$. The set $\mathcal I(H)$ of all ideals of $H$ is made into a (commutative) ...
Salvo Tringali's user avatar
4 votes
0 answers
158 views

Is the group ring of an amenable group, viewed as multiplicative monoid, amenable?

Motivated by this question, it seems natural to ask the following: Question 1: Is there a [finitely generated discrete] torsion-free virtually Abelian (but not Abelian) group $G$ so that the ...
ARG's user avatar
  • 4,432
4 votes
1 answer
428 views

Cancellable elements of a power semigroup

For a semigroup $S,$ its power semigroup $P(S)$ is the semigroup of all non-empty subsets of $S$ with the operation given by $AB=\{ab\,|\,a\in A,b\in B\}.$ I would like to know about the cancellable ...
Michał Masny's user avatar
3 votes
0 answers
161 views

Making the powerset into a topological monoid

Every monoid $X$ induces a monoid structure $\circledast$ on $\mathcal{P}(X)$ via $$U\circledast V := \{uv\ |\ u\in U,v\in V\}.$$ Moreover, a morphism of monoids $f\colon X\to Y$ induces a morphism of ...
Emily's user avatar
  • 11.8k
3 votes
0 answers
303 views

Pseudomodules, "general coherence theorem"

A pseudomonoid is defined within a monoidal bicategory. It is like a monoid in a monoidal category except that the usual axioms hold up to coherent invertible 2-cells. Pseudomonoid is like a monoidal ...
Dimitri Chikhladze's user avatar
3 votes
1 answer
404 views

How many monoids with $n$ arrows exist?

How many monoids with strictly $n$ arrows exist? Is this known? I ask this only out of curiosity. Looking at $n=1,2,3,4$, this number doesn't appear to be very large relative to $n$.
sergeant jam's user avatar
3 votes
1 answer
285 views

Cancellative semigroup on a distributive lattice

Let $(S,\le)$ be a distributive lattice. Is there a semigroup structure on $S$ such that $S$ is cancellative and always $(x\wedge y)(x\vee y)=xy$?
Minimus Heximus's user avatar
3 votes
0 answers
314 views

Certain conditions on cancellative semigroups

This is extracted from this question following Benjamin Steinberg's suggestion. For a semigroup $S,$ let $P(S)$ denote the power semigroup of $S,$ which is made up of all non-empty subsets of $S$ ...
Michał Masny's user avatar
2 votes
1 answer
182 views

Terminology for the equation $a=a+b$ in commutative semigroups

Let $(S,+)$ be a commutative semigroup. For $a,b\in S$ consider the equation $a=a+b$. Does such a relation between the given $a$ and $b$ have a name? I am currently using such equations quite often ...
Miroslav Korbelar's user avatar
2 votes
1 answer
122 views

If $H$ is essentially equimorphic to $K$, then is $H$ atomic only if so is $K$?

I will first state my question, and then give all the relevant definitions. Q. Let $H$ and $K$ be monoids, and assume $H$ is essentially equimorphic to $K$. Is it true that $H$ is atomic only if so ...
Salvo Tringali's user avatar
2 votes
0 answers
80 views

Are the roots of an infinitely divisible probability infinitely divisible themselves?

Let $\mu$ be an infinitely divisible probability on a topological group $G$. If $\nu ^{* n} = \mu$ for some $n$, is $\nu$ an infinitely divisible probability too? A sufficient criterion would be to ...
Alex M.'s user avatar
  • 5,407
2 votes
1 answer
302 views

Name for this algebraic structure?

I've found myself looking at a structure $\mathbb{M}$ whose important properties are: $\mathbb{M}$ is a discretely ordered additive monoid. $\mathbb{M}$ has a least element, and this least element is ...
Alec Rhea's user avatar
  • 10.1k
2 votes
0 answers
106 views

Descent of flatness from algebras to monoids II

This is a follow-up question to this one. There, Benjamin Steinberg showed that a morphism of commutative monoids $u$ need not be flat if the induced morphism of $R$-algebras $R[u]$ is flat for some ...
Fred Rohrer's user avatar
  • 6,700
2 votes
2 answers
241 views

If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?

Let me start by recalling some basic definitions (just for the sake of avoiding misunderstandings due to the vocabulary of the post). Basically following some ideas of W. Lawvere (but not his ...
Salvo Tringali's user avatar
2 votes
0 answers
122 views

First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric

This is a follow up of Question 163246. For the reader's convenience, let me first copy&paste some basic definitions. We let a semimetric on a set $X$ be a function $d: X \times X \to [0,\infty]$ ...
Salvo Tringali's user avatar
2 votes
3 answers
1k views

on the set of numbers generated by integer linear combination of two real numbers.

Let $b > a > 0$ be two real numbers. I am interested in the set of numbers $X(p,q) = p a + q b$ with $p,q$ positive integers. Basically this is the set $a \mathbb{N} + b \mathbb{N}$. What ...
Skarr's user avatar
  • 29
2 votes
0 answers
119 views

The "matrix direct sum" monoid modulo unitary equivalence

Given a commutative $*$-ring $(R,*)$, let $M(R,*)$ be the monoid whose elements are matrices over $R$ of all possible shapes and entries, including those that have $0$ columns or $0$ rows. Let the ...
wlad's user avatar
  • 4,943
1 vote
0 answers
52 views

Is every cyclic right action of a cancellative invertible-free monoid on a set isomorphic to the set of shifts of some homography?

The terms are defined in a related question. [1] Conjecture 1. Let $A$ be a set, $W$ a cancellative invertible-free monoid, and $\cdot\colon A\times W\rightarrow A$ a cyclic right $W$-action generated ...
David Pokorny's user avatar
1 vote
1 answer
96 views

If $H$ is commutative and unit-cancellative, then so is the monoid of non-empty ideals of $H$

Let $H$ be a (multiplicatively written) commutative monoid with identity $1_H$. Given $X, Y \subseteq H$, we take $$XY := \{xy: x \in X,\, y \in Y\}.$$ We call a set $I \subseteq H$ an ideal of $H$ ...
Salvo Tringali's user avatar
1 vote
2 answers
296 views

Are orbits of an affine algebraic monoid affine?

Let us work over the complex numbers for simplicity. Let $M$ be an affine algebraic monoid and $X$ an affine variety on which $M$ acts regularly, i.e. there is a morphism $\alpha: M\times X\to X$. Let ...
Jesko Hüttenhain's user avatar
1 vote
1 answer
231 views

Continuous semigroup homomorphism of composition to additive structure

Let $G$ be the topological semigroup whose underlying space is $C(\mathbb{R}^d,\mathbb{R}^d)$ equipped with composition as semigroup operation and let $H$ be the topological group whose underlying ...
ABIM's user avatar
  • 5,405
0 votes
1 answer
403 views

When does a power semigroup have a zero, and what can the zero be?

Let $S$ be a semigroup. The power semigroup of $S$ is the set $P(S)=2^S\setminus\lbrace\varnothing\rbrace $ with the operation $$AB=\lbrace ab\ |\ a\in A,\ b\in B\rbrace.$$ This operation is ...
Michał Masny's user avatar
-8 votes
1 answer
351 views

Are there overwhelmingly more finite monoids than finite spaces? [closed]

A function $f:\mathbb{Z}_{\geq 1}\to\mathbb{Z}_{\geq 1}$ overwhelms $g:\mathbb{Z}_{\geq 1}\to\mathbb{Z}_{\geq 1}$ if for any $k\in \mathbb{Z}_{\geq 1}$ the inequality $f(n)\leq g(n+k)$ holds only for ...
firn's user avatar
  • 23

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