0

What is an example of an infinite dimensional $C^{*}$ algebra with a Hilbert space structure (not merely pre-hilbert structure) such that the orthogonal complement of each closed left ideal is again a closed left ideal?

Ali Taghavi
  • 925
  • 1
    You mean a C*-algebra who's norm is given by a scalar product? – freishahiri Feb 14 '15 at 20:51
  • 1
    Hopefully you don't mean that, because such a thing doesn't exist with dimension greater than 1. What do you mean exactly Ali? – Jonas Meyer Feb 14 '15 at 23:06
  • @JonasMeyer Interesting. Please could you elaborate on your comment? – Rudy the Reindeer Feb 14 '15 at 23:38
  • 1
    @RudytheReindeer:http://math.stackexchange.com/q/112844 – Jonas Meyer Feb 14 '15 at 23:47
  • @JonasMeyer Sorry for my delay. No but I mean a Hilbert space structure whose topology coincide to the original topology of C^* algebra. – Ali Taghavi Feb 15 '15 at 18:33
  • @Freeze_S sorry for my delay. Please read the above comment. My motivation is the finite dimensional algebras. – Ali Taghavi Feb 15 '15 at 18:34
  • 1
    @AliTaghavi: do you have an example of an infinite-dimensional $C^*$-algebra which is isomorphic to a Hilbert space as topological vector space? – Yurii Savchuk Feb 20 '15 at 08:28
  • @YuriiSavchuk No I don't have. What about if we try with that C* algebra generated by diagonal operators on H? – Ali Taghavi Feb 20 '15 at 09:35
  • @AliTaghavi: I have just learned from here http://math.stackexchange.com/questions/112844/c-algebra-which-is-also-a-hilbert-space?rq=1 that every infinite dimensional C-algebra is a strict subset of its second dual. (See the second answer and discussion there). It shows that infinite dimensional $C^$-algebras cannot have the topology of a Hilbert space. – Yurii Savchuk Feb 20 '15 at 09:48
  • @YuriiSavchuk What about if we intersect the later algebra with finite rank or trace class operators? – Ali Taghavi Feb 20 '15 at 09:49
  • 2
    @AliTaghavi: But finite-rank operators and trace-class operators are not closed in $B(H).$ If you take all diagonal operators, it is isometrically isomorphic to $\ell^\infty,$ which is not reflexive. Hence cannot have the topology of a Hilbert space. – Yurii Savchuk Feb 20 '15 at 09:57
  • @YuriiSavchuk thank you for the link. – Ali Taghavi Feb 21 '15 at 17:09

0 Answers0