bio  website  thenestofheliopolis.blogspot.… 

location  Glasgow, Scotland  
age  27  
visits  member for  3 years, 10 months 
seen  1 hour ago  
stats  profile views  141 
I'm currently involved in a PhD in Mathematics. My research area is that of Functional Analysis. Specifically I'm working on classification of C*algebras by defining a bivariant version of the Cuntz semigroup, an invariant used to compare positive elements in a C*algebra and hence infer information about its internal structure.
During my MSc I focused on Algebraic Quantum Field Theory and I have worked on DFR models for Quantum Spacetime (arXiv:1211.7050 [grqc]).
1h

accepted  When are countably generated Hilbert modules generated by c.p.c. order zero maps? 
Apr 20 
comment 
When are countably generated Hilbert modules generated by c.p.c. order zero maps?
OK many thanks. That was really helpful! 
Apr 18 
comment 
When are countably generated Hilbert modules generated by c.p.c. order zero maps?
Perhaps I'm overlooking something, but if there are no nontrivial c.p.c. order zero maps between $A\otimes\mathcal K$ and $B$ then the only sequence one can construct out of a countable family of c.p.c. order zero maps is the constant sequence given by the trivial module, which has limit in the trivial module itself. 
Apr 17 
comment 
When are countably generated Hilbert modules generated by c.p.c. order zero maps?
Many thanks for your answer! I was wondering if the case of $A=\mathcal K$ generalises to any stable C*algebra just by considering $E_A:=\overline{\phi(A\otimes e)B}$, where $e\in\mathcal K$ is any minimal projection; and if there is the possibility of explicitly constructing the c.p.c. order zero map associated to the limit. I was thinking along the lines of a representation of $A$ on $\ell^2(\mathbb N)$ tensor with some positive element in $I$, but I'm not sure to what extent this intuition is correct. 
Apr 9 
awarded  Curious 
Apr 8 
asked  When are countably generated Hilbert modules generated by c.p.c. order zero maps? 
Mar 13 
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Do c.p.c. order zero maps between local C*algebras map C*subalgebras to C*subalgebras?
Completely positive contractive 
Mar 11 
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Do c.p.c. order zero maps between local C*algebras map C*subalgebras to C*subalgebras?
@ChrisHeunen The definition of local C*algebra I would like to consider for this question is given in the OP. As a special example you can consider inductive limits of C*algebras (where for simplicity you can take, say, injective connecting maps, although this shouldn't be strictly necessary), but you omit the completion w.r.t to the norm in order to remain with a preC*algebra. In the above example, $M_\infty$ can be constructed this way from the sequence of matrix algebras $M_n(\mathbb C)$ with obvious inclusions, whose C*limit is the C*algebra of compact operators. 
Mar 10 
asked  Do c.p.c. order zero maps between local C*algebras map C*subalgebras to C*subalgebras? 
Jan 7 
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Can the full and reduced group $C^*$algebras be “noncanonically” isomorphic?
@YemonChoi please reread my comment carefully. I've never said they are intuitive, but that it shouldn't be surprising. As for a reference there must be something in BrownOzawa (although just for the discrete case), but I can't check as I don't have a copy of it with me right now. There is a mention to this fact at en.wikipedia.org/wiki/…, although there is no reference cited there 
Jan 7 
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Can the full and reduced group $C^*$algebras be “noncanonically” isomorphic?
It is known that if $C^*(G)$ and $C^*_r(G)$ are isomorphic (through any isomorphism) then $G$ is amenable (and viceversa of course). This should be no surprise I think, as Calgebras are quite rigid because of the C*identity and hence the existence of a unique C*norm on a C*algebra. 
Jul 15 
awarded  Commentator 
Jul 15 
comment 
C*algebras and bounded relations
Thanks for the reference. My question is on the topic, but asks some clarifications about that particular method of proof of said universality. 
Jul 15 
comment 
Universal $C^*$algebra with generators and relations
As an application of this in Quantum Mechanics, one can consider the position and momentum operators $x$ and $p$, which, according to the standard quantization, must satisfy $xp  px \subset 1$ (assuming natural units, where $\hbar = 1$). Then the above argument shows that $x$ and $p$ cannot be bounded operators. To deal with bounded operators one can do the Weil trick to take the exponentials $e^{i\xi p}$ and $e^{i\eta x}$ in order to get unitaries through Borel functional calculus. 
Jun 12 
revised 
C*algebras and bounded relations
added 6 characters in body 
Jun 12 
revised 
Continuous linear functionals in strong operator and $\sigma$strong topologies
added 11 characters in body 
Jun 12 
answered  Continuous linear functionals in strong operator and $\sigma$strong topologies 
Jun 12 
asked  C*algebras and bounded relations 
Apr 5 
comment 
Projective limit construction of a semigroup
Ok so the difference is just in the set $X$, and the restrictions on the universal property account for the different nature of such set $X$. Anyway that is exactly what I was expecting, for this former case actually comes from $F(C_0(\mathbb N),\mathcal K)$, where $C_0(\mathbb N)$ denotes the C*algebra of continuous functions on $\mathcal N$ (which isn't compact) vanishing at $\infty$. So thanks a lot for your help with this matter! 
Apr 5 
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Projective limit construction of a semigroup
Yes I actually meant 0 by "dropping" in this particular example. So the limit would not be a proACom object, but I guess this non compact $X$ is just $\mathbb N$ in this other particular example? 