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Let $A$ and $B$ be two C$^{*}$-algebras and suppose we have a non-zero projection $p\in A\otimes B$. (We can assume $A$ is nuclear, so that there is only one possible tensor product.)

Does there exist a choice of elements $a_{1},\ldots,a_{n}\in A$ and $b_{1},\ldots,b_{n}\in B$ such that:

  1. $\left\|\left(\sum_{i=1}^{n}a_{i}\otimes b_{i}\right)-p\right\|<\frac{1}{2}$;
  2. If $\pi$ is a non-zero irreducible representation of $A$ and $(\pi\otimes\operatorname{id})(p)=0$, then $\pi\left(\sum_{i=1}^{n}|a_{i}|\right)=0$?

All I can deduce is that for a given choice of $a_{i}$'s and $b_{i}$'s, and and irreducible representation $\pi$ of $A$, we have $$ \left\|\sum_{i=1}^{n}\pi(a_{i})\otimes b_{i}\right\|=\left\|(\pi\otimes \operatorname{id})\left(\left(\sum_{i=1}^{n}a_{i}\otimes b_{i}\right)-p\right)\right\|\leq \left\|\left(\sum_{i=1}^{n}a_{i}\otimes b_{i}\right)-p\right\|<\frac{1}{2}, $$ provided that $(\pi\otimes\operatorname{id})(p)=0$.

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  • $\begingroup$ Are you sure you've stated the problem correctly? What you have written is clearly false, because you can always throw $x\otimes y - x\otimes y$ into the sum. So you would have to have $\pi(|x|) = 0$ for all $x$. $\endgroup$ – Nik Weaver Apr 13 '20 at 16:23
  • $\begingroup$ I am not picky about what the $a_{i}$'s and $b_{i}$'s are so long that the approximation for $p$ holds. I guess I am wondering if there is a way to choose them so that the quoted result is always true. $\endgroup$ – ervx Apr 13 '20 at 16:59
  • $\begingroup$ Given your "I guess I am wondering" could you please edit your question to give a more precise statement of what hypotheses you are assuming and what conclusion you want? $\endgroup$ – Yemon Choi Apr 13 '20 at 17:34
  • $\begingroup$ I have updated the question. $\endgroup$ – ervx Apr 13 '20 at 17:42
  • $\begingroup$ Could you change to a more specific title? $\endgroup$ – YCor Apr 13 '20 at 17:48
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Yes, this is possible, assuming that $A$ (or $B$) is nuclear. The same argument below (using that exact $C^\ast$-algebras are locally reflexive) also works if $A$ or $B$ is exact and the tensor product is the spatial (aka minimal) tensor product.

The role of nuclearity is that any two-sided closed ideal $J \subseteq A\otimes B$ is the closed linear span of all tensor products $I_A \otimes I_B$ of two-sided closed contained in $J$, see Corollary 9.4.6 in Brown and Ozawa's book. Now, let $J_A \subseteq A$ be the intersection of all two-sided, closed ideals $J \subseteq A$ such that $p \in J \otimes B$ and let $I:= \bigcap J \otimes B$ where the intersection is indexed by such $J$. Clearly $J_A \otimes B \subseteq I$. For the converse (which is not as trivial as it a priori looks), take $I_A \otimes I_B \subseteq I$ where $I_A$ and $I_B$ are two-sided closed ideals in $A$ and $B$ respectively. Then $I_A \subseteq J_A$ and $I_B \subseteq B$, so $I_A \otimes I_B \subseteq J_A \otimes B$. By nuclearity of $A$, $I$ is the closed linear span of all such $I_A\otimes I_B$, so $I = J_A \otimes B$. In particular, $p\in J_A \otimes B$.

Now, let $a_1,\dots, a_n \in J_A$ and $b_1,\dots, b_n \in B$ such that \begin{equation} \| \sum_{i=1}^n a_i \otimes b_i - p \| < 1/2. \end{equation} Let $\pi \colon A \to \mathcal B(\mathcal H)$ be an irreducible representation such that $(\pi \otimes \mathrm{id}_B)(p) = 0$. The image of $(\pi \otimes \mathrm{id}_B)$ is canonically $\pi(A) \otimes B$, so by nuclearity of $A$ (or $B$) and exactness of maximal tensor products, it follows that \begin{equation} 0 \to (\ker \pi)\otimes B \to A \otimes B \to \pi(A) \otimes B \to 0 \end{equation} is exact, so the kernel of $(\pi \otimes \mathrm{id}_B)$ is $(\ker \pi )\otimes B$. Hence $p \in (\ker \pi)\otimes B$ so $a_1,\dots, a_n \in J_A \subseteq \ker \pi$ by construction of $J_A$.

If $A$ or $B$ is exact, we get $\ker (\pi \otimes \mathrm{id}_B)$ either by using that $A$ is locally reflexive or $B$ is exact to get that \begin{equation} 0 \to (\ker \pi) \otimes_{\min{}} B \to A \otimes_{\min{}} B \to \pi(A) \otimes_{\min{}} B \to 0 \end{equation} is exact.

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  • $\begingroup$ Excellent answer! $\endgroup$ – Nik Weaver Apr 14 '20 at 4:33

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