197 reputation
8
bio website thenestofheliopolis.blogspot.…
location Glasgow, Scotland
age 27
visits member for 3 years, 11 months
seen May 26 at 12:50

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 Space-time (arXiv:1211.7050 [gr-qc]).


May
6
comment Do c.p.c. order zero maps between local C*-algebras map C*-subalgebras to C*-subalgebras?
Ah that's right, I forgot to try and play with this one-to-one correspondence around! Many thanks!
May
5
accepted Do c.p.c. order zero maps between local C*-algebras map C*-subalgebras to C*-subalgebras?
May
5
comment Do c.p.c. order zero maps between local C*-algebras map C*-subalgebras to C*-subalgebras?
Also, about the second part of your answer, is there something already known in that direction for c.p.c. order zero maps between C*-algebras? For example, if $a\in A^+$, then $\phi(C^*(a))\subset C^*(h_\phi,\pi_\phi(a))\cap B$, but are there elements $g_1,\ldots, g_n\in B$ such that $\phi(C^*(a))\subset C^*(g_1,\ldots,g_n)$?
May
4
comment Do c.p.c. order zero maps between local C*-algebras map C*-subalgebras to C*-subalgebras?
Thanks for the clarification, I think I can now see why this is the case (just considering real functions on $X$). However your example shows that there might be a suitable choice of exhaustive families for which the property in the OP holds. Of course, for any c.p.c. order zero map $\phi:A\to B$ between C*-algebras one has that $\phi(A)$ is contained in a sub-C*-algebra of $B$ (which is actually the property I'm trying to get to when $A$ and $B$ are local C*-algebras).
May
4
comment Do c.p.c. order zero maps between local C*-algebras map C*-subalgebras to C*-subalgebras?
Hi Hannes. Many thanks for your answer. In your example I can see how you can get $A$ from the completion of the union of all the finitely generated C*-subalgebras of $A$, but it seems to me that you are somehow implying that one doesn't need to take the norm-completion, and this is not clear to me at the moment. Perhaps what's misleading me is that I'm thinking of $C(X)$ as the infinite tensor product $C(I)^{\otimes\infty}$ with $I=[0,1]$, seen as an inductive limit with the obvious connecting maps.
Apr
28
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 non-trivial 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
comment Do c.p.c. order zero maps between local C*-algebras map C*-subalgebras to C*-subalgebras?
Completely positive contractive
Mar
11
comment 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 pre-C*-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
comment 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 Brown-Ozawa (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
comment 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 vice-versa of course). This should be no surprise I think, as C-algebras 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