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8 votes
1 answer
435 views

Function $\phi$ such that $f(\phi(x,y)) = f(x) + f(y)$

I have a continuous function $f:\mathbb{R}^n\to\mathbb{R}$, and I am looking for a continuous (or at least measurable) function $\phi:\mathbb{R}^{2n}\to\mathbb{R}^n$ such that $f(\phi(x,y))=f(x)+f(y)$....
gmvh's user avatar
  • 3,065
11 votes
0 answers
427 views

Is there a theory of completions of semirings similar to $I$-adic completions of rings?

Let $L = \text{Con } (\mathbb{N}, 0, +) \setminus \Delta$ be the lattice of monoid congruences on the naturals, excluding the trivial congruence. As it happens, every $\theta \in L$ is the meet of ...
Keith's user avatar
  • 591
2 votes
0 answers
92 views

Geometric interpretation of flags and the role of the rook monoid and Kazhdan–Lusztig theory in $M_n(\mathbb{C})$

Let $G = GL_n(\mathbb{C})$, $B$ be its Borel subgroup, and $P$ a parabolic subgroup. The space $G/B$ corresponds to complete flags in $ \mathbb{C}^n$, and $G/P$ corresponds to partial flags. The ...
Learner's user avatar
  • 141
0 votes
0 answers
61 views

Defining rank of an abelian subgroup using the second centralizer

I recently posted this on MSE, but didn't receive any feedback; so I'm posting it on MO. I recently came across this article which explored the maximal abelian subgroups of the symmetric group $S_n$. ...
dbossaller's user avatar
4 votes
1 answer
239 views

True or false? Every left or right cancellative, duo semigroup is cancellative

A semigroup $S$ is duo if $aS = Sa$ for all $a \in S$, where $aS := \{ax: x \in S\}$ and similarly for $Sa$; for instance, every commutative semigroup is duo, and so is every group. On the other hand, ...
Salvo Tringali's user avatar
8 votes
1 answer
321 views

Does every cancellative duo semigroup embed into a group?

Prompted by the comments to a recent answer by YCor to a related question (here), I'd like to ask the following: Q. Does every cancellative duo semigroup embed into a group? A (multiplicatively ...
Salvo Tringali's user avatar
6 votes
3 answers
549 views

Conjecture about commutative semigroups

Conjecture: given any commutative semigroup $S$ of order $n \ge 4$, there exist $a, b \in S$ with $a \ne b$, an integer $m \ge \lfloor (n-1)/2 \rfloor$, and two $m$-element subsets $X = \{x_1, \ldots, ...
Fabius Wiesner's user avatar
8 votes
2 answers
596 views

If a semigroup embeds into a group, then is it a subdirect product of groups?

The title has it all: Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups? If yes, then $S$ is a subdirect product of subdirectly irreducible groups,...
Salvo Tringali's user avatar
7 votes
2 answers
487 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
3 votes
0 answers
250 views

Action (of a graded monoid) required

Reference request: Did the construction below appear anywhere before? Any mentions of it or especially any links to something commonly known would be really helpful. I feel that it might be related to ...
Nikita Safonkin's user avatar
3 votes
0 answers
89 views

Ordering the elements of a semigroup by $a \le b$ iff $a=b$ or $b=ab=ba$

Let $S$ be a semigroup, written multiplicatively. The binary relation $\le$ on (the underlying set of) $S$, whose graph consists of all pairs $(a,b) \in S \times S$ such that $a = b$ or $b = ab = ba$, ...
Salvo Tringali's user avatar
0 votes
0 answers
114 views

Clarifications sought on the paper on the semigroup associated with a free polynomial by Ali Abbas and Abdallah Assi

I have three questions regarding the proof of Proposition 4 on page 4 of this paper here. For those interested in addressing these questions, please refer to some definitions in the first two or three ...
Mousa hamieh's user avatar
3 votes
0 answers
161 views

On generators of the multiplicative semigroup $\{r\in\mathbb Q:\ r>1\}$

The set $M=\{r\in\mathbb Q:\ r>1\}$ is a commutative semigroup with respect to the multiplication. For any integers $a>b\ge1$, we clearly have $$\frac ab=\prod_{n=b}^{a-1}\frac{n+1}n.$$ So the ...
Zhi-Wei Sun's user avatar
  • 15.6k
3 votes
0 answers
91 views

Reference for the monoidal category structure $X \otimes Y = X + Y + X \times Y$ on a distributive category

Given a distributive category $\mathscr C$ (more generally a rig category), we can define a (semicocartesian) monoidal category structure on $\mathscr C$ with tensor product given by $X \otimes Y := X ...
varkor's user avatar
  • 10.6k
4 votes
0 answers
147 views

Isomorphism between the reduced C*-algebra of a groupoid and the crossed product of inverse semigroups

In Paterson's book "Groupoids, Inverse Semigroups and their Operator Algebras" he proves that for any r-discrete groupoid $G$ with unit space $G^0$, its full $C^* $-algebra $C^* (G)$ is ...
Tomás Pacheco's user avatar
5 votes
0 answers
191 views

Do most semigroups have a zero?

It is widely believed in finite semigroup theory that asymptotically almost all finite semigroups $S$, up to isomorphism, are 3-nilpotent, i.e., they satisfy $\#\{abc\,:\,a,b,c\in S\} = 1$. My ...
user513093's user avatar
1 vote
1 answer
215 views

Reference about cancellation property for semigroups

Have the semigroups with the following cancellation property been studied? Property: Let $S$ be a semigroup and $x,y\in S$ such that $xz=yz,$ for all $z\in S,$ then $x=y$.
Hector Pinedo's user avatar
2 votes
1 answer
57 views

Are simplicial commutative inverse semigroups fibrant?

Let $X$ be a simplicial object in the category of commutative inverse semigroups (or monoids, if needed). Is the underlying simplicial set of $X$ always a Kan complex? If so, are there some nice ...
Aurélien Djament's user avatar
4 votes
0 answers
174 views

Centers and conjugacy classes of groups relative to a pair of group homomorphisms

$\newcommand{\defeq}{\mathbin{\overset{\mathrm{def}}{=}}}$Given a group $G$, its center $\mathrm{Z}(G)$ and set of conjugacy classes $\mathrm{Cl}(G)$ are defined by \begin{align*} \mathrm{Z}(G) &\...
Emily's user avatar
  • 11.8k
6 votes
0 answers
151 views

On dual notions of morphisms of algebraic structures obtained by replacing equaliser with coequalisers

This question is based on this discussion from the Category Theory Zulip. See also the earlier question Natural cotransformations and "dual" co/limits. Let $G$ and $H$ be groups. We define ...
Emily's user avatar
  • 11.8k
2 votes
1 answer
423 views

Conjecture about semigroups

Let $G$ be a finite semigroup with order $n$ odd. Let $S_i \in G, i=1,\ldots,\binom{n}{(n+1)/2}$ be all the subsets of $G$ of size $(n+1)/2$. Let $E(S_i)$ be the set obtained "expanding" $...
Fabius Wiesner's user avatar
0 votes
0 answers
95 views

Which algebraic structure characterizes the set of non-trivial qudratic residues in a finite field?

I understand this question may be too naive to ask, but I am unable to figure it out. Suppose, $\mathbb{QR^*}$ denotes the set of all quadratic residues in a finite field except the identity element $...
Somudro Gupto'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
8 votes
1 answer
236 views

Quiver and relations for a monoid related to Catalan numbers

Let $C_n$ be the monoid consisting of monotone maps $\{1,...,n\} \rightarrow \{1,...,n\}$ with $f(i) \leq i$ for all $i$. The cardinality of $C_n$ is given by the Catalan numbers. Consider $A_n= \...
Mare's user avatar
  • 26.5k
5 votes
0 answers
107 views

Structure of well-ordered commutative monoids

Let $(M,+)$ be a commutative monoid. Let $<$ be a well-ordering on $M$, where $\forall a\in M,\ 0\leq a$ $\forall a,b,c\in M,\ a<b\Rightarrow a+c<b+c$ The first condition means $M$ will be ...
Pace Nielsen's user avatar
  • 18.7k
2 votes
0 answers
80 views

An alternative definition for finitely generated (and principal) ideals in a semigroup

Let $S$ be a semigroup. An ideal (of $S$) is a subset $I$ of $S$ such that $SI$ and $IS$ are both contained in $I$. The non-empty ideals constitute a subsemigroup, $\mathfrak I(S)$, of the power ...
Salvo Tringali'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
2 votes
0 answers
91 views

A recursive description of the smallest divisor-closed subsemigroup containing a set

Let $S$ be a semigroup and $\widehat{S}$ be its unitization, i.e., the monoid obtained from $S$ by adjoining an identity element if necessary (so that $\widehat{S} = S$ when $S$ is already a monoid). ...
Salvo Tringali's user avatar
3 votes
1 answer
170 views

Every homomorphism between (rational) Puiseux monoids is multiplication by a non-negative rational

Let a (rational) Puiseux monoid be a non-trivial submonoid of the non-negative rational numbers under (the usual operation of) addition. It is not difficult to show that, if $f \colon H \to K$ is a (...
Salvo Tringali's user avatar
6 votes
1 answer
173 views

References on semigroup actions

I posted this question on Math Stack Exchange about 10 days ago, but received no answer (https://math.stackexchange.com/q/4843881/1223994). I would like to ask for references on semigroup actions on ...
Marco Farotti's user avatar
3 votes
0 answers
74 views

Are the automorphisms of the power semigroup of a cancellative semigroup cardinality-preserving?

Let $S$ be a semigroup (written multiplicatively) and $f$ be an automorphism of the power semigroup $\mathcal P(S)$ of $S$, that is, a bijective function on the family of all non-empty subsets of $S$ ...
Salvo Tringali's user avatar
1 vote
1 answer
142 views

Congruences that aren't "finite from above," take 2: semigroups

This is a hopefully less trivial version of this question. Briefly, say that a congruence is parafinite if it is the largest congruence contained in some equivalence relation with finitely many ...
Noah Schweber's user avatar
3 votes
0 answers
103 views

An isomorphism problem for semigroups of ideals

An ideal of a semigroup $S$ (written multiplicatively) is a set $I \subseteq S$ such that $IS$ and $SI$ are both contained in $I$ (here, $XY$ means, for all $X, Y \subseteq S$, the setwise product of $...
Salvo Tringali's user avatar
14 votes
4 answers
742 views

Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for every binary relation $R$ on a set of size $n$

Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for every binary relation $R$ on a set of size $n$. I have verified the statement for $n \leq 4$ with a Mathematica code. I have ...
Geoffrey Critzer's user avatar
0 votes
0 answers
92 views

What is the highest $n\in\Bbb N$ for which a complete classification of inverse semigroups of order up to $n$ is known?

What is the highest $n\in\Bbb N$ for which a complete classification of inverse semigroups of order up to $n$ is known? Given that there are $3{,}684{,}030{,}417$ semigroups of order $8$, I guess $n\...
Shaun's user avatar
  • 379
2 votes
1 answer
193 views

Continuity of Moore-Penrose generalized inversion

Any matrix $A\in\mathbb{C}^{m\times n}$ has a unique generalized inverse $A^{\dagger}\in\mathbb{C}^{n\times m}$ with the properties $$AA^{\dagger}A=A,\qquad A^{\dagger}AA^{\dagger}=A^{\dagger},\qquad (...
Bumblebee's user avatar
  • 1,093
5 votes
0 answers
160 views

$S$ and $T$ globally isomorphic semigroups, with $S$ (commutative and) cancellative, iff $S$ is isomorphic to $T$?

Denote by $\mathcal P(S)$ the semigroup obtained by equipping the non-empty subsets of a "ground semigroup" $S$ (written multiplicatively) with the operation of setwise multiplication ...
Salvo Tringali's user avatar
2 votes
1 answer
271 views

Apropos of two groups being globally isomorphic iff they are isomorphic

Denote by $\mathcal P(S)$ the semigroup obtained by endowing the non-empty subsets of a "ground semigroup" $S$ (written multiplicatively) with the operation of setwise multiplication induced ...
Salvo Tringali's user avatar
0 votes
1 answer
152 views

Name for a monoid on the basis of a vector space?

Is there a name for the structure of a vector space with a monoid defined on its basis? Given a vector space V over a field F, we can choose a basis and define a monoid on it. Now we can use each ...
Spencer Woolfson's user avatar
0 votes
1 answer
327 views

Can we generalise groupoids to monoid-oids? [closed]

Groups correspond to one object categories where every morphism is an isomorphism. Monoids correspond to one object categories. Groupoids correspond to small categories where every morphism is an ...
Diego de la Paz's user avatar
5 votes
2 answers
477 views

Generalization of the concept of a measure

Consider the following generalization of the concept of a measure: Let $L = (X, \lor, \land, \bot)$ be a semi-bounded lattice. Let $M = (Y, \bullet, e)$ be a commutative monoid. An $(L, M)$-measure is ...
user76284's user avatar
  • 2,203
3 votes
0 answers
176 views

The monoid of stably-free modules over integral group rings

Fix a torsion-free group G, let $M_G$ be the monoid of stably-free $\mathbb{Z}G$-modules under operation $\oplus$, the direct sum of modules. In studying objects related to Wall’s D2 problem on CW-...
William Thomas's user avatar
7 votes
1 answer
183 views

Is lambda calculus polymorphism a type of generalized monad?

Let $\mathbf{C}$ be a Cartesian closed category. Then simply typed lambda calculus in $\mathbf{C}$ in one type variable can be interpreted as a category $\mathbf{STLC}_{\mathbf C}$ where the objects ...
Johan Thiborg-Ericson's user avatar
2 votes
0 answers
175 views

On the origin of power semigroups

Let $S$ be a (multiplicatively written) semigroup. Equipped with the (binary) operation of setwise multiplication $(X, Y) \mapsto \{xy \colon x \in X, \, y \in Y\}$, the family of all non-empty ...
Salvo Tringali's user avatar
8 votes
3 answers
1k views

Are all free monoids residually finite?

I cannot manage to prove that a free monoid with operation concatenation, and with at least two generators is residually finite. If there is just one generator, the free monoid $\{a\}^*$ is isomorphic ...
iguessarian's user avatar
1 vote
1 answer
87 views

Semigroup algebras with one dimensional center

Let $S$ be a finite semigroup and $K$ a field of characteristic 0 (we can assume the complex numbers for simplicity). Question: Is there a characterization when the center of the semigroup algebra $...
Mare's user avatar
  • 26.5k
1 vote
1 answer
129 views

Cycles in almost breakable semigroups

Last October, I learned from Benjamin Steinberg's answer to another question of mine that a semigroup $S$ is called breakable if $xy \in \{x, y\}$ for all $x, y \in S$. Let's now say that $S$ is an ...
Salvo Tringali's user avatar
10 votes
0 answers
248 views

What is the tiling semigroup for an einstein "hat" tiling?

My undergraduate dissertation was on inverse semigroups and the key text I used for it was Lawson's, "Inverse Semigroups: The Theory of Partial Symmetries". In said book, Lawson describes ...
Shaun's user avatar
  • 379
0 votes
0 answers
63 views

A construction that sort of merges two semigroups to build a new one

Suppose $H$ and $K$ are semigroups and assume without loss of generality that (the underlying sets of) $H$ and $K$ are disjoint. We can then extend the operations of both $H$ and $K$ to a binary ...
Salvo Tringali's user avatar
0 votes
0 answers
122 views

Is there a name for this condition on a monoid?

Suppose we have a commutative monoid ${\mathcal M}=\langle M,\otimes\rangle$ such that the usual divisibility relation $\leq_\otimes$ given by $a\leq_\otimes b\Leftrightarrow \exists c(a\otimes c=b)$ ...
Steven Stadnicki's user avatar

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