bio | website | sci.ccny.cuny.edu/~benjamin |
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location | New York City | |
age | 41 | |
visits | member for | 3 years, 2 months |
seen | 2 days ago | |
stats | profile views | 7,052 |
I am an algebraist interested in a broad range of areas. I've worked on semigroups, geometric group theory, algebraic combinatorics, representation theory, self-similar groups (aka automaton groups), profinite groups and random walks on semigroups and groups. I've been particularly interested in interactions between these areas and Computer Science and am fond of algorithmic questions. I've also dabbled with operator algebras associated to etale groupoids and inverse semigroups. Currently, I am interested in applications of finite semigroup theory to finite state Markov chains.
My blog is http://bensteinberg.wordpress.com/author/bsteinbg/
Aug 1 |
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Can a closure make the index finite?
@HJRW I wrote a similar but more complicated example at the same time. |
Aug 1 |
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Can a closure make the index finite?
added 6 characters in body |
Aug 1 |
revised |
Can a closure make the index finite?
added 6 characters in body |
Aug 1 |
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Can a closure make the index finite?
Which Karrass-Solitar Thm? They have many. |
Aug 1 |
awarded | Favorite Question |
Aug 1 |
answered | Can a closure make the index finite? |
Jul 30 |
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Is every compact monothetic group metrizable?
Basically monothetic compact groups correspond to subgroups of $S^1_\delta$ and the metrizable ones correspond to countable subgroups. |
Jul 30 |
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Is every compact monothetic group metrizable?
Your group G is metrizable iff $\widehat{G}$ is countable. |
Jul 30 |
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How much of homotopy theory can be done using only finite topological spaces?
Do you know the book Barmak, Jonathan, Algebraic Topology of Finite Topological Spaces and Applications, Springer 2011?. |
Jul 30 |
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Is every compact monothetic group metrizable?
The dual of a countable group is metrizable because it embeds in a countable direct product of $S^1$ and a countable product of metric spaces is metric. |
Jul 30 |
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Is every compact monothetic group metrizable?
Did you try using Pontryagin duality? |
Jul 25 |
awarded | Nice Answer |
Jul 24 |
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Examples of famous 'workhorse' theorems
Related mathoverflow.net/questions/99506/blackbox-theorems |
Jul 18 |
answered | a question about semigroups |
Jul 18 |
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a question about semigroups
Are you asking which semigroups have the property that the intersection of any two ideals is there product? |
Jul 18 |
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Reference for subsemigroups of $\mathbb{N}^n$
@LeeMosher, thanks. |
Jul 14 |
answered | Which algebraic theories have the property that $\mid$ is antisymmetric for all free algebras? |
Jul 14 |
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Which algebraic theories have the property that $\mid$ is antisymmetric for all free algebras?
In semigroup theory this preorder is called the $\mathcal J$-order. A semigroup is called $\mathcal J$-trivial if it is a partial order. There are a number of natural varieties of semigroups and monoids where all free algebras are J-trivial. Note J-triviality is given by finitely many quasiidentities. For finitary universal algebras you may need infinitely many. |
Jul 14 |
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Which algebraic theories have the property that $\mid$ is antisymmetric for all free algebras?
But the theory is the theory of monoids and semigroups. In your question you only ask about the relation on free objects. |
Jul 14 |
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Which algebraic theories have the property that $\mid$ is antisymmetric for all free algebras?
This relation is antisymmetric for monoids and semigroups without abelian and for many natural subvarieties. |