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33 votes
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Is there a (discrete) monoid M injecting into its group completion G for which BM is not homotopy equivalent to BG?

For a (discrete) monoid $M$, the classifying space $BM$ is the geometric realization of the nerve of the one object category whose hom-set is $M$. (This definition gives the usual classfiying space ...
Omar Antolín-Camarena's user avatar
21 votes
0 answers
869 views

Noncommutative arithmetic mean geometric mean inequality and symmetric polynomials

While analyzing convergence speed of stochastic-gradient methods for convex optimization problems, Recht et al (2011) posed a tantalizing conjecture. It seems quite tricky, so after having struggled a ...
Suvrit's user avatar
  • 28.6k
17 votes
0 answers
536 views

Question about combinatorics on words

Let $\{a_1,a_2,...,a_n\}$ be an alphabet and let $\{u_1,...,u_n\}$ be words in this alphabet, and $a_i\mapsto u_i$ be a substitution $\phi$. Question: Is there an algorithm to check if for some $m,k$...
user avatar
16 votes
0 answers
861 views

Is "being a full ring of quotients" a Morita invariant property?

Definition and context: An (associative, unital, not necessarily commutative) ring $R$ is called classical if every regular element of $R$ is a unit. Equivalently, $R$ is its own classical ring of ...
Torsten Schoeneberg's user avatar
15 votes
0 answers
1k views

Is the category of smooth manifolds equivalent to the opposite category of the category of commutative monoids of some additive symmetric monoidal category?

This is a followup to my previous question, which asked whether the category of commutative or noncommutative C*-algebras or von Neumann algebras is equivalent to the category of commutative or ...
Dmitri Pavlov's user avatar
13 votes
0 answers
251 views

Is every simply connected finite complex the classifying space of a finite monoid

On page 323 of Fiedorowicz, "Classifying Spaces of Topological Monoids and Categories" it was stated that "it seems likely that any finite simply connected complex should [have the same weak homotopy ...
user46652's user avatar
  • 665
12 votes
0 answers
543 views

Does Wedderburn's Little Theorem hold constructively?

Wedderburn's Little Theorem states that every finite division ring is commutative. Perhaps even more surprising, this implies that every finite reduced ring is commutative. The proofs that I am aware ...
Martin Brandenburg's user avatar
12 votes
0 answers
321 views

Combinatorial proof of invertibility of a symmetric matrix associated to the ring of matrices over a finite field

Let $F$ be a finite field of $q$ elements with characteristic $p$. Let $M_n(F)$ be the ring of $n\times n$ matrices over $F$. We define a $q^{n^2}\times q^{n^2}$ symmetric matrix $L$ over the ...
Benjamin Steinberg's user avatar
12 votes
0 answers
185 views

Hopf-Galois extensions where the "extension" is a module?

For $H$ a Hopf-algebra, an $H$-Hopf-Galois extension is a map of rings $\phi\colon\thinspace A\to B$ such that $H$ coacts on $B$ over $A$, $B\otimes_AB\cong B\otimes H$, and the cofixed points, or the ...
Jonathan Beardsley's user avatar
12 votes
0 answers
267 views

Finitely generated skew-fields

There is a well known theorem saying that a commutative field that is finitely generated as a ring has to be finite (Kaplansky). Is the same true for non-commutative "fields" (usually called ...
Miroslav Korbelar's user avatar
12 votes
0 answers
533 views

Does there exist a Noetherian ring of finite injective dimension but higher Krull dimension?

Definition: a (not necessarily commutative) left and right Noetherian ring $R$ is said to be Auslander-Gorenstein if (i) $R$ has finite left and right injective dimension (in which case it turns out ...
Christopher's user avatar
12 votes
0 answers
443 views

Nullstellensatz for quaternionic plane curves?

By a quaternionic plane curve I mean the zero locus of a noncommutative polynomial in two variables, $x$ and $y$ say, over ${\Bbb H}$, Hamilton's quaternions. It is evidently well-known that, after ...
David Feldman's user avatar
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
  • 631
11 votes
0 answers
286 views

Does every finite poset have a rigid endomorphism?

Crossposted on Mathematics. In this post, an order-preserving self-map of a poset $X$ will be called an endomorphism of $X$, and such an endomorphism $f$ will be called rigid if the only automorphism ...
Pierre-Yves Gaillard's user avatar
11 votes
0 answers
265 views

Criteria for a map of rings to induce an equivalence on K-theory?

Algebraic $K$-theory is Morita invariant, but surely it does not detect Morita equivalence. What are some examples of rings (or ring spectra) $R$ and $S$ that are not Morita equivalent, but ...
Reuben Stern's user avatar
11 votes
0 answers
214 views

Is it decidable if a tree-presented semigroup contains an idempotent?

A semigroup presentation $\langle A | R\rangle$ is called tree-like if every relation has the form $ab=c$, $a,b,c$ are in $A$ and if two relations $ab=c, a'b'=c'$ belong to $R$, then $c=c'$ if and ...
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
10 votes
0 answers
367 views

A formula for Frobenius number of certain numerical semigroups

The old formula for the Frobenius number of a numerical semigroup generated by two elements can be stated as follows: assume $\gcd\{a+1,b+1\}=1$, then the Frobenius number of $S= \left<a+1,b+1\...
Hailong Dao's user avatar
  • 30.5k
10 votes
0 answers
314 views

How much do idempotent ultrafilters generate in terms of semigroups?

It is known that the set of ultrafilters on, say, the natural numbers $\mathbb{N}$, can naturally be endowed with the structure of a compact topological left semigroup (which fails to be anything ...
Jakub Konieczny's user avatar
10 votes
0 answers
2k views

Is my definition of a context algebra new?

In my DPhil thesis, I defined what I called a context algebra as a model of meaning in natural language. The idea is to mathematically formalise the notion that meaning is determined by context. It ...
Daoud's user avatar
  • 223
9 votes
0 answers
164 views

Parallelizability of Lie monoids

A Lie monoid is a monoid, together with a structure of a smooth manifold (possibly with a boundary), such that the monoid multiplication is smooth. If all left (or right) translations in a Lie monoid $...
Žan Grad's user avatar
9 votes
0 answers
347 views

What is the precise connection between logarithmic algebraic geometry and the field with one element?

Monoid schemes (a.k.a. $\frak M$-schemes) have been introduced by Deitmar as a possible approach to geometry over the field with one element. These build upon monoids as the basic building blocks for ...
Emily's user avatar
  • 11.8k
9 votes
0 answers
273 views

About an algebraic construction of a sheaf of formal microdifferential operators

While reading these notes by Victor Ginzburg on $D$-modules I found a certain construction of Microlocailzation in the algebraic setting which unfortunately doesn't seem to be elaborated on a lot in ...
Saal Hardali's user avatar
  • 7,789
9 votes
0 answers
373 views

Embedding $\beta\mathbb{N}$ into a product of Cantor sets

Let us consider $\beta\mathbb{N}$, the Stone-Čech compactification of the natural numbers (where we do not take $0$ to be a natural number, so the only idempotent elements are nonprincipal ...
Simon_Peterson'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
354 views

Rough paths, unparametrized path space, and Kontsevich's moduli space of stable maps

Let $X$ be a manifold. Modulo reparametrization, the path space of $X$ is a groupoid $\Pi_X$. In Kapranov's "Free Lie Algebroids and the Space of Paths", Kapranov constructs an associated ...
John Rached's user avatar
8 votes
0 answers
219 views

Differential birational equivalence

Suppose the base field algebraically closed and of zero characteristic. There are two fascinating questions in the intersection of ring theory and algebraic geometry (for which an excellent discussion ...
jg1896's user avatar
  • 3,318
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
8 votes
0 answers
270 views

Does this kind of non-noetherian bimodule exist?

Question: Do there exist simple rings $R$ and $S$ (i.e., rings with no proper nonzero ideals) and an $(R,S)$-bimodule $M$ such that $M$ is finitely generated both as a left $R$-module and a right $...
Manny Reyes's user avatar
  • 5,407
7 votes
0 answers
227 views

On the structure of an algebra as a bimodule

$\DeclareMathOperator\End{End}\DeclareMathOperator\Mod{Mod}\DeclareMathOperator\Ker{Ker}\newcommand{\bi}{\mathrm{bi}}\newcommand{\op}{\mathrm{op}}$Let $K$ be a field (say of characteristic zero), and $...
FPV's user avatar
  • 541
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
221 views

Strange formula for the dimension of a certain space of noncommutative polynomials

Consider a vector space $V_r(n)$ spanned by (noncommutative) monomials in variables $x_1,\ldots,x_r$ $$ x_{1}^{n_1}x_{2}^{n_2}\ldots x_{r}^{n_r} $$ of total degree $n.$ Inside this space consider a ...
Daniil Rudenko's user avatar
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
7 votes
0 answers
579 views

Guises of the noncrossing partitions (NCPs)

From "Noncrossing partitions in surprising locations" by Jon McCammond: Certain mathematical structures make a habit of reoccuring in the most diverse list of settings. Some obvious ...
Tom Copeland's user avatar
  • 10.5k
7 votes
0 answers
540 views

Algebraic-closures of division rings

In what follows, $x$ is always taken to commute with the coefficient ring. This means that for any given polynomial, you can put the coefficients to the right or the left of $x$ as you please. This ...
Jonathan Gleason's user avatar
7 votes
0 answers
291 views

Differentially closed fields

Let F be a field. Recall that an additive map $d: F\rightarrow F$ is said to be a derivation if $d(ab)=ad(b)+d(a)b$. Now let $F$ be a ring and let $d$ be a derivation of $F$. Examples I have in mind ...
Dr. Evil's user avatar
  • 2,751
7 votes
0 answers
438 views

How to prove that a projective module is not free?

Let $A$ be a noncommutative (perhaps $\ast$-) algebra (over $\mathbb{C}$) and let $M$ be a projective module defined via a projector $P\in M_n(A)$; i.e. $M=P(A^n)$. Furthermore, assume that all ...
Joakim Arnlind's user avatar
6 votes
0 answers
79 views

Examples of $\mathbb{N}$-graded algebras whose global dimension is strictly less than the GK dimension

The relationship between the global and GK dimensions of Artin-Schelter regular algebras remains to be mysterious, yet both dimensions are conjectured to be equal. In a more broad setting, are there ...
A. Kabbaj's user avatar
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
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
0 answers
259 views

Usefulness of total algebras and exotic generating series

In his first Algebra volume, Bourbaki [1] defines the structure of a “total algebra” i.e. the space of functions on a monoid $M$ (to a ring $k$) with the convolution product ( a function $f:\ M\to k$ ...
Duchamp Gérard H. E.'s user avatar
6 votes
0 answers
183 views

Examples of groups with a positive homogeneous presentation without the Haagerup property or not of type $F_\infty$

I am looking for groups with a certain presentation that do not have the Haagerup property or are finitely presented but not of type $F_\infty$ (meaninig that for some $n\geq 3$ we cannot find any ...
Arnaud's user avatar
  • 61
6 votes
0 answers
177 views

Is the monoid of all cancellative finitely generated commutative monoids cancellative?

$\DeclareMathOperator\Mon{Mon}\DeclareMathOperator\Grp{Grp}$Let $\Mon'$ be the set of isomorphism classes of (small) commutative, unital, cancellative ($a + t = b + t$ implies $a = b$) monoids. It is ...
Leo Herr's user avatar
  • 1,094
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
0 answers
255 views

Proving the spectrum of the Young-Jucys-Murphy elements by formal computation in the degenerate affine Hecke algebra

This is really a followup to Why are Jucys-Murphy elements' eigenvalues whole numbers? , specifically to Igor Makhlin's beautiful answer. I'm trying to make it even more beautiful by getting rid ...
darij grinberg's user avatar
6 votes
0 answers
584 views

What are the topics in noncommutative algebraic geometry?

Preface: I know very little about noncommutative algebra and noncommutative geometry, so please feel free to make improvement suggestions for my question. Also, to my knowledge there are several ...
6 votes
0 answers
230 views

Gelfand ring in Bourbaki's exercises

In Bourbaki's General Topology, Chapitre III §6 Exercise 11, they define a Gelfand Ring as a topological ring $A$ such that The set $A^*$ ($=A^{-1}$) of invertibles is open. The uniform structure ...
Duchamp Gérard H. E.'s user avatar

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