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12 votes
1 answer
624 views

Stone–Čech compactification as a semigroup

Let $G$ be a topological group (we can assume that $G$ is countable and discrete) and let $\beta(G)$ be the Stone–Čech compactification of $G$. It is known that $\beta(G)$ can be turned into a left ...
Serge the Toaster's user avatar
10 votes
1 answer
579 views

Group completion of topological monoids

Let $M$ be an abelian monoid. For sake of simplicity we shall assume that in $M$ the cancellation law holds true. With this last assumption we define the group completion $G$ of $M$ as $$G:=M\times M/\...
Vincenzo Zaccaro's user avatar
9 votes
1 answer
743 views

Why is choice needed in Ellis' Lemma?

Ellis Lemma on idempotent elements asserts that: Lemma (Ellis). Every compact semigroup has an idempotent. The proof below is excerpted from Todorcevic's Introduction to Ramsey Spaces, Lemma 2.1. ...
Clement Yung's user avatar
  • 1,442
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
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
157 views

Closure of the inverse image under the projection map

Let $S$ be a subsemigroup of a semitopological semigroup $(T,+)$, let $e$ be an idempotent in $T\setminus S$ such that $e\in cl_T(S)$, let $\mathcal{E}$ be a subsemigroup of $S\times S$ such that $(e,...
John's user avatar
  • 85
2 votes
1 answer
194 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
2 votes
0 answers
73 views

Nonzero idempotents in compact semitopological semigroups with zero

Let $S$ be a compact semitopological semigroup. Then, $S$ contains minimal idempotents by Ellis' theorem. Ellis' Theorem: In a compact left-topological (resp. right-topological) semigroup, every ...
Onur Oktay's user avatar
  • 2,605
1 vote
0 answers
48 views

Neighborhoods of idempotents in topological inverse semigroups

In a topological group, for any neighborhood $U$ of the origin, there is another such neighborhood with the property that $V.V\subseteq U.$ I conjecture a similar property for topological inverse ...
Bumblebee's user avatar
  • 1,093
1 vote
0 answers
43 views

Interleaving in Viennot's Heaps models?

I am looking for past results on interleaving of heaps (in the sense of Viennot). For a very simplified example, suppose I have two pieces, (b1 a1 b1), and (b2 c2 b2), where the letter represents a ...
holloway's user avatar
1 vote
0 answers
121 views

Why $\beta S$ is not a semigroup when $S$ is a (directed) partial semigroup?

Given a semigroup $(S, *)$ we extend the semigroup operation $*$ of $S$ to a operation $*$ on $\beta S$ (the set of ultrafilters on $S$) defined as $$ \mathcal{U} * \mathcal{V} = \left\{ A \...
andpe's user avatar
  • 59