2
$\begingroup$

Let $A$ and $B$ be $C^{\ast}$-algebras. It is well known that maximal tensor product of simple $C^{\ast}$-algebras need not be simple. So basically the ideal structure of $A\otimes_{max}B$ does not really depends on ideal structure of $A$ and of $B$.

What is known about ideals of $A\otimes_{max}B$ .

Any references?

Improve this question
$\endgroup$
10
  • 1
    $\begingroup$ You've asked a very similar question before. The question is so broad that it is very hard to think of a meaningful answer. If there is something specific you want to know, ask that. $\endgroup$
    – Nik Weaver
    Commented Jul 15, 2020 at 15:14
  • 1
    $\begingroup$ As it stands it sounds a bit like "somebody prove some things for me, I don't know what, you'll have to figure that out"... $\endgroup$
    – Nik Weaver
    Commented Jul 15, 2020 at 15:15
  • 1
    $\begingroup$ Clearly the kernel $\ker (A \otimes_{\max{}} B \to A\otimes_{\min{}} B)$ contains no non-zero elementary tensors. More generally, a two-sided closed ideal in $A\otimes_{\max{}} B$ contains an elementary tensor if and only if it has non-zero image in $A\otimes_{\min{}} B$. This can be shown by tweaking the proof of Kirchberg's Slice Lemma (Lemma 2.15 in Blanchard, Kirchberg "Non-simple purely infinite $C^\ast$-algebras: the Hausdorff case"). $\endgroup$
    – Jamie Gabe
    Commented Jul 16, 2020 at 2:17
  • 1
    $\begingroup$ @MathLover I did not intend such a statement. Rather, the kernels of irreducible representations are known as primitive ideals. If a C$^\ast$-algebra is type I, meaning that in every representation the bicommutant of its image is a type I von Neumann algebra, then the map from irreducible representations to primitive ideals is a bijection (because it's an injection). For type I C$^*$-algebras the maximal and minimal tensors coincide, and you may get a good answer to your question, if restricted to primitive ideals. Some information on this is in Dixmier's book on C$^\ast$-algebras. $\endgroup$
    – Robert Furber
    Commented Jul 20, 2020 at 5:51
  • 1
    $\begingroup$ @MathLover Of course, I meant "isomorphism classes of irreducible representations" in my previous comment. The statement I was referring to is the theorem at the start of Chapter 9 of Dixmier's C$^\ast$-algebras. $\endgroup$
    – Robert Furber
    Commented Jul 20, 2020 at 6:34

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .