Let $H$ be a separable Hilbert space. Is it true that there is an isomorphism of $C^*$-algebras $$B(H)\hat{\otimes} K(H)\cong B(H)$$ where $B(H)$ is the algebra of bounded operators, $K(H)$ is the ideal of compact operators?
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4$\begingroup$ $A\otimes K(H)$ does not have an identity element. Hence no unital ${\rm C}^*$-algebra can be stable $\endgroup$– Yemon ChoiCommented Oct 7, 2019 at 20:37
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1$\begingroup$ @YemonChoi I am having a slow day. It seems very plausible to me that $A\otimes K(H)$ does not have an identity, but I cannot prove it. What's a simple argument? $\endgroup$– Matthew DawsCommented Oct 8, 2019 at 8:27
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$\begingroup$ @MatthewDaws argh, my bluff has been called :) See below for a rambling version of the idea I had in mind, but it's very much thinking of the cuff, and I might have overlooked something. $\endgroup$– Yemon ChoiCommented Oct 8, 2019 at 14:51
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$\begingroup$ I've just noticed that the $\widehat{\otimes}$ notation is being used for the ${\rm C}^*$-tensor product. This does not seem to be standard notation. I would reserve $A\hat{\otimes} B$ for the Banach-space projective tensor product of $A$ and $B$, which is never isometrically a ${\rm C}^*$-algebra unless either $A$ or $B$ is one-dimensional. $\endgroup$– Yemon ChoiCommented Oct 9, 2019 at 1:31
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In response to Matt "calling my bluff" I realised that my claim takes some work to justify. The following is very much "thinking aloud through a head cold" so is not a polished exposition, and I have probably overlooked simplifications.
View $K(H)$ as generated by the matrix units $\{e_{ij} \colon i,j\geq 1\}$ viewed as rank-one operators on $H=\ell^2({\mathbb N})$.
For each $n\geq 1$ let $A_n$ be the closed subalgebra of $A\otimes K(H)$ generated by $\{a\otimes e_{ij} \colon a \in A, 1\leq i \leq n , 1\leq j\leq n\}$.
Let $\iota: A \to A$ be the identity map and define $\phi_n : K(H) \to {\mathbb M}_n$ by truncating to the top-left $n\times n$ corner. I claim that $E_n := \iota \otimes \phi_n : A\otimes K(H) \to A_n$ has the conditional expectation properties: $E_n(x)=x$ for all $x \in A_n$ and $E_n (xy) = E_n(x)E_n(y)$ if either $x$ or $y$ belongs to $A_n$. (I convinced myself of this with some $2\times 2$ block matrix heuristics, it shouldn't be hard to make the proof fully precise.)
Therefore, if $A\otimes K(H)$ has an identity element $u$, $E_n(u)$ is an identity element for $A_n$, for each $n$.
In particular, $A=A_1$ must have an identity element, which we denote by $1_A$. Then $1_A \otimes I_n$ must be the identity element for $A_n$ (because identity elements in an algebra are unique). We therefore get $E_n(u) = 1_A \otimes I_n$ for all $n$ which should lead to a contradiction with the assumption that $u \in A\otimes K(H)$, by some hacky approximation argument (e.g. find $u_0$ in $A\otimes {\mathbb M}_m= A_m$ for some $m$ such that $\Vert u-u_0\Vert \ll 1$, then $u_0 = E_{m+1}(u_0)$ has to be close to $1_A\otimes I_{m+1}$ which is a contradiction since elements of $A_m$ are annihilated by $E_{m+1}-E_m$ while $1_A\otimes I_{m+1}$ isn't).
answered Oct 8, 2019 at 14:50
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$\begingroup$ Thanks for this! I actually wasn't trying to call your bluff, as it were, just couldn't see for myself which this had to be true... $\endgroup$ Commented Oct 8, 2019 at 15:05
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2$\begingroup$ @MatthewDaws No worries, I just mean that when I wrote the earlier comment I thought "of course $A\otimes K(H)$ can't be unital, that's obvious" and then your comment prompted me to reflect that "it's true because how could it not be true" is not actually a proof (at least, not in our area of mathematics ;-) ) $\endgroup$ Commented Oct 8, 2019 at 15:30
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1$\begingroup$ An argument I thought of was to represent $A\otimes K(H)$ non-degenerately on $H'\otimes H$, so the identity has to be the identity operator on $H'\otimes H$. Then you need to argue that $A\odot K(H)$ cannot approximate this, which seems to basically be your argument, using matrix algebras to approximate, just in a different presentation. $\endgroup$ Commented Oct 8, 2019 at 15:44
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$\begingroup$ @MatthewDaws Indeed, I think this is somehow the non-verbalised POV I had in my head, and the answer I wrote above is just a translation of the "picture" into a certain way of writing things down. It might be possible to carry out the whole proof intrinsically without using a GNS representation, but having a concrete representation does make things intuitively clearer $\endgroup$ Commented Oct 8, 2019 at 16:23
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2$\begingroup$ Alternatively, you can use slice maps to show that if a C*-tensor product $A\otimes_\gamma B$ is unital then both $A$ and $B$ are unital. In fact, if $(a_\lambda)$ and $(b_\lambda)$ are approximate units in $A$ and $B$ (same index), then $(a_\lambda \otimes b_\lambda)$ is an approximate unit in $A\otimes_\gamma B$ (by basic properties of C*-norms), and thus a Cauchy net by unitality. If $f$ is a state on $A$, the induced slice map $A\otimes_\gamma B \to B$ maps $a_\lambda \otimes b_\lambda$ to $f(a_\lambda)b_\lambda$ which becomes Cauchy, but also an approximate unit in $B$, so $B$ is unital. $\endgroup$ Commented Oct 8, 2019 at 23:51