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Jul 22, 2020 at 2:43 vote accept truebaran
Feb 8, 2016 at 16:32 comment added Nik Weaver @truebaran: this seems like a very complete answer to your question, why don't you accept it?
Jan 20, 2016 at 3:31 comment added Transcendental @Svinepels: Regarding your latest question, there is no set-theoretical issue. If $ \varphi(x) $ is a formula in the first-order language of set theory, and if $ X $ is a set, then according to the Axiom Schema of Specification, $$ \{ x \in X \mid \varphi(x) \} $$ is a set. In Step 8 of my post, $ X = \Bbb{R}_{\geq 0} $, and $ \varphi(x) $ reads “there exist a $ C^{\ast} $-algebra $ A $ and elements $ s,t $ of $ A $ such that $ (A,s,t) $ is a $ C^{\ast} $-representation of $ \mathcal{A}_{\theta} $ and $ x = \| {\pi_{A,s,t}}(a) \|_{A} $”.
Jan 20, 2016 at 2:53 comment added Transcendental @Svinepels: Hi. I have deleted my answer to your previous question in order to give a more detailed explanation here. A rational non-commutative torus $ A_{\theta} $, where $ \theta \in \Bbb{Q} $, is constructed in exactly the same manner as in my post, but it is not a simple $ C^{*} $-algebra. In fact, we have $ \operatorname{Prim}(A_{\theta}) \cong \Bbb{T}^{2} $ as topological spaces.
Jan 14, 2016 at 16:30 comment added Ulrik Thanks for the previous answer. Another question: Doesn't it lead to size issues considering the supremum taken over all $C^*$-representations?
Nov 23, 2015 at 17:06 comment added Ulrik If one were to construct a rational commutative torus, what would one do differently?
Aug 17, 2015 at 4:23 comment added Transcendental @Branimir: Thanks for the compliment!
Aug 14, 2015 at 22:50 comment added Branimir Ćaćić +1 I'm very late to the party, but this is an excellent answer.
Feb 22, 2015 at 18:59 history answered Transcendental CC BY-SA 3.0