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8 votes
2 answers
262 views

An operation on binary strings

Consider the “product” $\gamma = \alpha \times \beta$ of two binary strings $\alpha$, $\beta$ $\in \lbrace 0,1\rbrace^+$ which one gets by replacing every 1 in $\beta$ by $\alpha$ and each ...
Hans-Peter Stricker's user avatar
8 votes
2 answers
352 views

Non-density of invertible elements in $\ell_1(\mathbb{N}_0)$

Consider the Banach algebra $\ell_1(\mathbb{N}_0)$ (with convolution / Cauchy product of series). I am looking for an elementary proof of the fact that the group of invertible elements in this algebra ...
Tomasz Kania's user avatar
  • 11.3k
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
8 votes
1 answer
314 views

What algebras are quotients of $\ell_1(\mathbf{N})$?

Every separable Banach space is a linear quotient of $\ell_1$, however not every separable Banach algebra is a Banach-algebra quotient of $\ell_1(G)$ for some group $G$ (these are the so called ...
user512365's user avatar
8 votes
2 answers
483 views

Posets obtained from a semigroup by the definition $x \leq y \iff x \cdot y = x$

A po-groupoid is a groupoid $\langle A,\cdot\rangle $ such that the relation defined by $$ x \leq y \text{ if and only if } x \cdot y = x $$ is a partial order on $A$, the order related to $\langle ...
Pedro Sánchez Terraf's user avatar
8 votes
1 answer
319 views

Über theorem on unavoidable patterns?

Let $A$ be an alphabet of $k$ symbols, and $p$ a pattern. An example of a pattern is $p=XX$, where $X$ is any finite string of symbols from $A^+$. Avoiding $p$ is avoiding any subword repeated twice ...
Joseph O'Rourke's user avatar
8 votes
1 answer
437 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
8 votes
2 answers
826 views

Which semigroups can be linearly ordered?

As usual I consider a semigroup to be a structure $(A, +)$ such that $+$ is an associative binary function over the set $A$. The notion of linearly-ordered semigroup corresponds to structures of the ...
boumol's user avatar
  • 768
8 votes
1 answer
922 views

What's the cokernel of a monoid homomorphism?

Let $f:A\to B$ be a monoid homomorphism. Where can I find an explicit description of the its cokernel? Are there any books on this topic? By the cokernel of $f$, I mean the universal arrow which ...
Arrow's user avatar
  • 10.5k
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
1 answer
261 views

A generalisation of $C_0$-semigroups

A $C_0$-semigroup is a strongly continuous family $\{T(t)\}$ of bounded linear operators on a normed space $X$, indexed in $\mathbb R_+$ and with two additional properties that make it look like an ...
Delio Mugnolo's user avatar
8 votes
1 answer
1k views

Lattice-ordered commutative monoids

By a lattice-ordered monoid, I mean a structure $(A,0,{+},{\vee},{\wedge})$ such that $(A,0,{+})$ is a (not necessarily commutative) monoid, $(A,{\vee},{\wedge})$ is a lattice, and the two ...
François G. Dorais's user avatar
8 votes
2 answers
1k views

Sets of matrices which are irreducible but not strongly irreducible

A set of $d \times d$ real or complex matrices is commonly called irreducible if those matrices do not jointly preserve a linear subspace with dimension strictly between zero and $d$. A stronger ...
Ian Morris's user avatar
  • 6,206
8 votes
1 answer
229 views

Embedding abelian cancellative Hausdorff topological semigroups into abelian Hausdorff topological groups

An abelian cancellative semigroup embeds (via a semigroup monomorphism) into an abelian group. What about an abelian cancellative Hausdorff topological semigroup that does not embed (via a ...
Salvo Tringali's user avatar
8 votes
0 answers
411 views

Semigroups of matrices closed under conjugate transposition

An involution semigroup or $\star$-semigroup is a unary semigroup $\langle S,{\cdot}\,,{}^\star\rangle$ that satisfies the equations $$ (x^\star)^\star = x \quad \text{and} \quad (xy)^\star = y^\star ...
E W H Lee's user avatar
  • 563
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
0 answers
419 views

Are most semigroups nilpotent of degree 3?

A semigroup is nilpotent of degree 3 if every product of 3 elements gives the same result. In 2012, Andreas Distler and James D. Mitchell wrote that: It is part of the folklore of semigroup theory ...
John Baez's user avatar
  • 22.3k
7 votes
4 answers
1k views

Torsors for monoids

Torsors are defined as a special kind of group action. I am wondering whether the analogous notion exists for monoid actions. Some references would be helpful. In general I'm interesting in the ...
supercooldave's user avatar
7 votes
2 answers
587 views

Representation theory of the full linear monoid

The full linear monoid $M_N(k)$ of a field $k$ is the set of $N \times N$ matrices with entries in $k$, made into a monoid with matrix multiplication. A representation of $M_N(k)$ on a vector space $V$...
John Baez's user avatar
  • 22.3k
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
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
2 answers
383 views

Counting nilpotent self-maps of $\{1,\dots,n\}$ with image of a given cardinal

Let $\mathcal{C}_n$ be the monoid of self-maps $\alpha$ of $\{1\dots,n\}$ that are order-preserving ($\forall x,y$, $x\le y$ $\Rightarrow$ $\alpha(x)\le\alpha(y)$ and decreasing ($\forall x$, $\alpha(...
1Spectre1's user avatar
  • 355
7 votes
2 answers
544 views

A linearly orderable monoid which does not embed into a linearly orderable group

It is known (after an example of A.I. Mal'cev) that there exist cancellative semigroups which do not embed into a group. On the other hand, it is not difficult to see that every linearly orderable ...
Salvo Tringali's user avatar
7 votes
1 answer
266 views

Positive cone of a subgroup of $\mathbb{Z}^n$

This question sounds like it should be very well known, but for some reason I failed to find a decent answer anywhere. Let $G\subset\mathbb{Z}^n$ be a subgroup, and $G_+=G\cap\mathbb{Z}_{\ge0}^n$ be a ...
Vladimir Dotsenko's user avatar
7 votes
1 answer
361 views

For a new operation on a finite group of odd order giving a loop structure, when does this also gives a group

For finite groups $G$ of odd order, as $x \mapsto x^2$ is bijection (but no automorphism in general) then, we can define for each $g \in G$ the element $x^{1/2}$ by requiring $(x^{1/2})^2 = x$. Then ...
StefanH's user avatar
  • 798
7 votes
1 answer
271 views

Algebraic proof that the monoid ring of a torsion-free monoid is reduced

In what follows, I say that a monoid $M$ is torsion-free if the $n$-th power map is injective for all $n \geq 1$. I have a proof of the following result: Claim: if $M$ is a torsion-free commutative ...
Moinsdeuxcat'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
7 votes
1 answer
650 views

Cones, monoids, and the space of (very) ample divisors

An interesting and useful tool to study a projective variety is its ample cone. Understanding the structure of this cone reveals information about the variety, and it is an isomorphism-invariant so ...
Noah Giansiracusa's user avatar
7 votes
1 answer
444 views

Which monoids have a faithful irreducible representation?

Let $*$ be a binary operation on a set $M$, with an identity element $e\in M$. A monoid representation of $(M,*,e)$ is a map $\delta:M\to (S\to S)$ for some set $S$, such that $\delta(e)=\mathrm{id}_S$...
Bjørn Kjos-Hanssen's user avatar
7 votes
1 answer
241 views

Lie monoids as monoids internal to the category of smooth manifolds?

This question can be thought as a complement to this one. Lie groups can be defined as groups internal to the category of smooth manifolds. Lie monoids, however, as a particular case of Lie semigroups,...
Dry Bones's user avatar
  • 321
7 votes
1 answer
350 views

Pushouts of injective monoid homomorphisms

Given a pushout square in the category of monoids $$\begin{array}{ccc}A & \rightarrow & M \\ \downarrow && \downarrow \\ N & \rightarrow & P\end{array}$$such that $A \to M$ and ...
HeinrichD's user avatar
  • 5,482
7 votes
2 answers
305 views

Functionals on oriented matroids

Oriented matroids are abstractions of hyperplane arrangements, or equivalently vector configurations. Let me recall the definition in terms of covectors. Let $R=\lbrace 0,+,-\rbrace$ with the monoid ...
Benjamin Steinberg's user avatar
7 votes
2 answers
417 views

Does every commutative monoid admit a translation-invariant measure?

Let $T$ be a commutative monoid, written additively. The set $T$ is equipped with a canonical pre-order, defined by $s \le t$ when there exists $s' \in T$ so that $s + s' = t$. Consequently, $T$ may ...
Tom LaGatta's user avatar
  • 8,512
7 votes
1 answer
167 views

For which abelian groups $G$ does the monoid of zero-sum sequences over $G$ embed into a ring as a divisor-closed subsemigroup?

Let $K$ be a multiplicatively written semigroup (either commutative or not) and $H$ a subsemigroup of $K$. We say that $H$ is divisor-closed (in $K$) if $x \in H$ for all $x, y \in K$ such that $x \...
Salvo Tringali's user avatar
7 votes
3 answers
525 views

Is the class of inverse semigroups globally determined?

This question is a follow-up to this one I asked on math.stackexchange. I've decided to ask here because I believe this is a research-level question. I'm sorry if I'm wrong -- I'm not a researcher ...
Michał Masny'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
7 votes
1 answer
235 views

name for monoids inducing bimonoids in Rel?

Let Rel be the category of sets and relations, which is a (compact closed) symmetric monoidal category under the cartesian product of sets. We write $A \nrightarrow B$ to indicate a relation from $A$ ...
Noam Zeilberger's user avatar
7 votes
2 answers
309 views

Homological questions on monoid algebras

Given a finite monoid G and its group algebra A over a field $K$. I have never really studied such algebras, but I have the following questions (which are probably basic questions on any large class ...
Mare's user avatar
  • 26.5k
7 votes
1 answer
593 views

Question about topological monoid maps

Let Mon be the category of topological monoids. I am happy to work with the model structure mentioned here: Model Structure/Homotopy Pushouts in topological monoids?. I'm looking for a reference ...
John Klein's user avatar
  • 18.8k
7 votes
1 answer
433 views

Powers of maps on finite sets

Let $X_n$ be a set with $n$ elements. Write $F(X_n,X_n)$ for the set of maps from $X_n$ to itself. It is a monoid under the operation of composition. Let $m$ be a positive integer. How many maps in $...
Steven Spallone's user avatar
7 votes
0 answers
295 views

A minimal semigroup generating subset of the additive reals

I asked this on MSE, but I was told to ask it here because it is a difficult question. Consider the additive magma of the real numbers, $(\mathbb{R};+)$. Does there exist a subset $S$ of the reals ...
user107952's user avatar
  • 2,023
7 votes
0 answers
138 views

The smallest cardinality of a cover of a group by algebraic sets

$\DeclareMathOperator\cov{cov}$A subset $A$ of a semigroup $X$ is called algebraic if $$A=\{x\in X: a_0xa_1x...xa_n=b\}$$ for some $b\in X$ and $a_0,a_1,...,a_n \in X^1=X\cup \{1\}$. The smallest ...
Taras Banakh's user avatar
  • 41.9k
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
0 answers
260 views

Generating the monoid of injective endomorphisms of the free group

Let $F$ be the free group of rank $2$ (or any finite rank if this does not matter). The set of injective group endomorphisms $F\to F$ forms a monoid $M$ by compositions. Is there a simple looking set ...
Lvzhou Chen's user avatar
6 votes
2 answers
237 views

Realizing a monoid as $\mathrm{End}(G)$ for some graph $G$

This question relates to Realizing groups as automorphism groups of graphs. Given a monoid $M$, is there a graph $G$ such that the endomorphism monoid $\textrm{End}(G)$ is isomorphic to $M$?
Dominic van der Zypen's user avatar
6 votes
2 answers
416 views

For a monoid with zero $M$, how many additive operations on $M$ can there be making $M$ a ring?

Let $(M,\times)$ be a monoid with zero. Let $\Sigma(M,\times)$ be the set of binary operations $+$ on $M$ such that $(M,+,\times)$ is a ring. Let $\sim$ be an equivalence relation on $\Sigma(M,\times)$...
Michał Masny'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
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
444 views

Homotopy type of a specific discrete monoid

Consider the discrete monoid $M$ of nondecreasing continuous maps from $[0,1]$ to itself preserving the extremities. Note that the monoid is right-cancellative ($x.z=y.z$ implies $x=y$, since $z$ is ...
Philippe Gaucher'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

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