# Approximating a projection by a sum of elementary tensors with a certain property

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 $$(\pi\otimes\operatorname{id})(p)=0$$.

• 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 $x$. – Nik Weaver Apr 13 '20 at 16:23
• 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. – ervx Apr 13 '20 at 16:59
• 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? – Yemon Choi Apr 13 '20 at 17:34
• I have updated the question. – ervx Apr 13 '20 at 17:42
• Could you change to a more specific title? – YCor Apr 13 '20 at 17:48

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 $$$$\| \sum_{i=1}^n a_i \otimes b_i - p \| < 1/2.$$$$ Let $$\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$$) and exactness of maximal tensor products, it follows that $$$$0 \to (\ker \pi)\otimes B \to A \otimes B \to \pi(A) \otimes B \to 0$$$$ is exact, so the kernel of $$(\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$$ is exact to get that $$$$0 \to (\ker \pi) \otimes_{\min{}} B \to A \otimes_{\min{}} B \to \pi(A) \otimes_{\min{}} B \to 0$$$$ is exact.