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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
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
3 votes
0 answers
31 views

Compactness of the minimal ideal of a compact Hausdorff polytopological semigroup

A semigroup $X$ endowed with a topology is called $\bullet$ a topological semigroup if the semigroup operation $X\times X\to X$ is continuous; $\bullet$ a semitopological semigroup if for every $a,b\...
Taras Banakh's user avatar
  • 41.8k
1 vote
0 answers
139 views

Terminology for an kind-of principal fibration

My interest is in topological monoids, but I think the question may make sense (in some fashion) for monoids of sets. Let $M$ be a topological monoid, and let $X$ be a pointed space that $M$ acts on, ...
Jeff Strom's user avatar
  • 12.5k
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
  • 5,407
6 votes
0 answers
47 views

Special monomorphism to encode the inclusion of topological submonoids

Consider the category $\mathrm{TopMon}$ of topological monoids and continuous monoid homomorphisms. Consider the inclusion $i:\Bbb{R}_{\ge 0}\hookrightarrow \Bbb{R}$, where the spaces are taken with ...
geodude's user avatar
  • 2,129
10 votes
1 answer
578 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
3 votes
1 answer
167 views

When are all the convolution roots of an infinitely divisible probability measure infinitely divisible?

Let $G$ be a topological group. Let us say that a probability measure $\mu$ on $G$ is strongly infinitely divisible (SID) if $\mu$ is infinitely divisible and any probability measure $\nu$ on $G$ ...
Iosif Pinelis's user avatar
2 votes
0 answers
79 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
9 votes
2 answers
667 views

Semi group of polynomials which all roots lie on the unit circle

Let $X=\{f\in \mathbb{C}[z]\mid |z| \neq 1 \implies f(z) \neq 0\} $. The motivation for consideration of such an $X$ is the the concept of Lee-Yang polynomials. With the standard multiplication, $X$...
Ali Taghavi's user avatar
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