Yemon's interests

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Tehran Grand Bazaar, Largest Mall on the Planet, Age Unknown.

http://yomadic.com/tehran-grand-bazaar/

http://yomadic.com/tehran-grand-bazaar/

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Via +Joerg Fliege

Why international students choose not to study in the UK.

Surprise, surprise: because they know they will get thrown out right after they get their degree by a xenophobic government.

Surprise, surprise: because they know they will get thrown out right after they get their degree by a xenophobic government.

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Evening sunshine here on Bailrigg. You may need to live in Lancs to appreciate the joy this can bring

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Hear hear.

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Here we go again ... the latest announcement of a solution to the question "Is Thompson's group F amenable"?

I should really write more about what F is, why the question seems interesting, who has claimed to prove F is amenable, and who has claimed to prove F is not amenable, but to be honest I neither have time nor energy right now. So: knock yersels out, as it were.

#spnetwork arXiv:1310.4395 #amenability #geometricGroupTheory #cursedProblems

I should really write more about what F is, why the question seems interesting, who has claimed to prove F is amenable, and who has claimed to prove F is not amenable, but to be honest I neither have time nor energy right now. So: knock yersels out, as it were.

#spnetwork arXiv:1310.4395 #amenability #geometricGroupTheory #cursedProblems

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Take 2: a public repost of something already shared with a limited circle, to whom I apologize for the repetition.

#spnetwork arXiv:1309.2145 #amenability #Banachalgebras #operatoralgebras

(Not sure if I've correctly understood this Selected Papers Network tagging malarkey -- a concept which surely demands a compound noun in German. Experts, and fans of

Amenable Banach algebras can be thought about in many different ways (or indeed written about without thinking, sad to say). My own preference is the homological one: they are those which have a bounded approximate identity, and are flat as bimodules over themselves in a suitable sense.

This seems recondite but it has the following consequence: if you represent an amenable Banach algebra A on a Hilbert space H, and H has a closed A-invariant subspace V, then one can find a closed A-invariant subspace W with H=V\oplus W (a direct sum, but not necessarily an

Which closed subalgebras of B(H) are amenable? If one restricts attention to subalgebras that are closed under the involution of B(H), the so-called C^\ast-algebras, then there is a deep result (using hard results of Connes/Popa together with details of various other people) that an amenable C^\ast-algebra is nuclear. Nuclear C^\ast-algebras, in this context, should be thought of as those which can be approximated linearly by finite-dimensional matrix algebras in a particularly tractable way. Conversely, Haagerup showed that every nuclear C^\ast-algebra is amenable as a Banach algebra.

If one want amenable subalgebras of B(H) that are not closed under the involution of B(H), one way to do it is to take a nuclear C^\ast-subalgebra A in B(H), and conjugate it with some suitable invertible operator, which will interact badly with the involution of B(H), but result in an algebra that is isomorphic to A and hence is also amenable. The question naturally arises:

The paper http://arxiv.org/abs/1309.2145 finally answers this question, with a negative answer. The counterexample is as an extension by the compact operators of a carefully chosen, commutative, non-separable subalgebra of the Calkin algebra. (In fact, this subalgebra of the Calkin is itself isomorphic to a commutative C^\ast-algebra.) Remarkably, the proof that one can obtain something not isomorphic to a C^\ast-algebra boils down, after a clever use of standard ideas in amenability, to the fact that 2^c > c where c is the cardinality of the continuum!

The construction has other features which run somewhat counter to one's naive intuition/guesses, but I think it will take some time for people in this area to absorb what this all means in the bigger picture. One thing to note is that the original question remains open for separable amenable operator algebras, and for commutative amenable operator algebras...

**Amenable operator algebras**#spnetwork arXiv:1309.2145 #amenability #Banachalgebras #operatoralgebras

(Not sure if I've correctly understood this Selected Papers Network tagging malarkey -- a concept which surely demands a compound noun in German. Experts, and fans of

**Spaced**, may wish to "skip to the end".)Amenable Banach algebras can be thought about in many different ways (or indeed written about without thinking, sad to say). My own preference is the homological one: they are those which have a bounded approximate identity, and are flat as bimodules over themselves in a suitable sense.

This seems recondite but it has the following consequence: if you represent an amenable Banach algebra A on a Hilbert space H, and H has a closed A-invariant subspace V, then one can find a closed A-invariant subspace W with H=V\oplus W (a direct sum, but not necessarily an

**orthogonal**one). So in finite-dimensional settings, we are just talking about the complex associative algebras that are**completely reducible**, which (Wedderburn theorem) means such examples are isomorphic to sums of full matrix algebras.Which closed subalgebras of B(H) are amenable? If one restricts attention to subalgebras that are closed under the involution of B(H), the so-called C^\ast-algebras, then there is a deep result (using hard results of Connes/Popa together with details of various other people) that an amenable C^\ast-algebra is nuclear. Nuclear C^\ast-algebras, in this context, should be thought of as those which can be approximated linearly by finite-dimensional matrix algebras in a particularly tractable way. Conversely, Haagerup showed that every nuclear C^\ast-algebra is amenable as a Banach algebra.

If one want amenable subalgebras of B(H) that are not closed under the involution of B(H), one way to do it is to take a nuclear C^\ast-subalgebra A in B(H), and conjugate it with some suitable invertible operator, which will interact badly with the involution of B(H), but result in an algebra that is isomorphic to A and hence is also amenable. The question naturally arises:

**do all amenable subalgebras of B(H) appear this way?**The paper http://arxiv.org/abs/1309.2145 finally answers this question, with a negative answer. The counterexample is as an extension by the compact operators of a carefully chosen, commutative, non-separable subalgebra of the Calkin algebra. (In fact, this subalgebra of the Calkin is itself isomorphic to a commutative C^\ast-algebra.) Remarkably, the proof that one can obtain something not isomorphic to a C^\ast-algebra boils down, after a clever use of standard ideas in amenability, to the fact that 2^c > c where c is the cardinality of the continuum!

The construction has other features which run somewhat counter to one's naive intuition/guesses, but I think it will take some time for people in this area to absorb what this all means in the bigger picture. One thing to note is that the original question remains open for separable amenable operator algebras, and for commutative amenable operator algebras...

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Someone pointed me to this, linked somewhere in one of the Crooked Timber comment threads.

Not having heard Glass's piece before, I think I prefer this version purely on musical grounds, even without the firehats and Segways (which put me in mind of a Russell T Davies Doctor Who episode made on MDMA)

Not having heard Glass's piece before, I think I prefer this version purely on musical grounds, even without the firehats and Segways (which put me in mind of a Russell T Davies Doctor Who episode made on MDMA)

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The word is that William Thurston has died of cancer. Here he is in 2010 lecturing on Perelman's proof of geometrization.

http://andrewgreene.tumblr.com/post/29943948285/the-mystery-of-3-manifolds-william-thurston-by

http://andrewgreene.tumblr.com/post/29943948285/the-mystery-of-3-manifolds-william-thurston-by

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