I'm searching for a counterexample for $C^*$-algebras $A$ and $B$ and essential ideals (I assume an ideal to be closed and only two-sided ideals) $I\subseteq A$, $J\subseteq B$ , such that the ideal $I\otimes_{max} J$ is not essential in the (maximal tensor product) -$C^*$-algebra $A\otimes_{max} B$. I'm not sure if my idea works:

The algebra of compact operators $K(l^2(\mathbb{N}))$ is an essential ideal in the $C^*$-algebra of bounded linear operators on $l^2(\mathbb{N})$, $B(l^2(\mathbb{N}))$. The reason is that $B(l^2(\mathbb{N}))$ has only the closed ideals $\{0\}$, $K(l^2(\mathbb{N}))$ and $B(l^2(\mathbb{N}))$, thus $K(l^2(\mathbb{N}))\cap M\neq 0$ for all closed nontrivial ideals $M\subseteq B(l^2(\mathbb{N}))$.

But is $K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))$ essential in $B(l^2(\mathbb{N}))\otimes_{max} B(l^2(\mathbb{N}))$?

I don't think so, but I'm stuck to prove that there must be a nontrivial closed ideal $M$ in $B(l^2(\mathbb{N}))\otimes_{max} B(l^2(\mathbb{N}))$ such that $K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))\cap M$ is trivial. For the proof it must be important that one takes the maximal norm-closure of $I\odot J\subseteq A\odot B$ ($\odot$ denotes the tensor product as $*$-algebras), because particularly the spatial tensor product satisfies that if $I$ is essential in $A$, $J$ essential in $B$, then $I\otimes_{min} J$ is essential in $A\otimes_{min} B$ (and here is $B(l^2(\mathbb{N}))\otimes_{min} B(l^2(\mathbb{N}))\neq B(l^2(\mathbb{N}))\otimes_{max} B(l^2(\mathbb{N}))$. If my ideal don't work, what else can I do? I appreciate your help.

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    $\begingroup$ Your example works. Since you seem to have almost all the elements to show this, I'll only suggest using that if $I$ is an essential ideal in a C*-algebra $A$ then $\|a\|=\sup \{\|ax\|\mid x\in I,\|x\|\leq 1\}$ for all $a\in A$. Use also that on $\mathcal K\otimes \mathcal K$ the min and max norms agree. $\endgroup$ May 11, 2016 at 19:57
  • $\begingroup$ thanks! I got it. Your hint to use that $K(l^2(\mathbb{N}))$ is nuclear is very helpful. $\endgroup$ May 11, 2016 at 22:05
  • $\begingroup$ Hey Sabrina :D waving $\endgroup$ Feb 14, 2018 at 5:34
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    $\begingroup$ furthermore, you can check II.9.6.2 in Blackadars book "Theory of $C^*$-Algebras and von Neumann Algebras" $\endgroup$ Feb 16, 2018 at 10:18
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    $\begingroup$ @FreeziiS thank you for having a critical view on my attempts:), in theses cases I always learn something new as well. However, you are right and I was not aware that there this is an issue for general $C^*$-subalgebras ( there is a discussion about this in Brown-Ozawas book, corollary 3.6.4) $\endgroup$ Feb 16, 2018 at 10:28

1 Answer 1


My suggestion (I hope that nothing is wrong):

Let $$F:B(l^2(\mathbb{N}))\otimes_{max} B(l^2(\mathbb{N}))\to B(l^2(\mathbb{N}))\otimes_{min} B(l^2(\mathbb{N}))$$ be the canonical map and define $M:=\ker(F)$. Since $F$ is not injective, $M$ is nontrivial.

Claim: $K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N})) \cap M = 0$.

First note that $K(l^2(\mathbb{N}))$ is nuclear (because it's a inductive limit of nuclear $C^*$-algebras) and simple, therefore $$K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))\cong K(l^2(\mathbb{N}))\otimes_{min} K(l^2(\mathbb{N}))$$ is simple. It follows $K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N})) \cap M=\ker(F_{|K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))})=0$, because $F_{|K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))}$ is nonzero and $\ker(F_{|K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))})$ is an ideal in $K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))$.


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