Biprojective $C^*$-algebra

Question

Let $A$ be a Banach algebra. Define $\Delta:A\hat{\otimes}A\to A$ with $\Delta(\sum_{n=1}^\infty a_n\otimes b_n)=\sum_{n=1}^\infty a_nb_n$. Now $A$ is called biprojective if there exists a bounded linear map $\rho:A\to A\hat{\otimes}A$ such that $\Delta\circ\rho(a)=a$, for all $a\in A$ and also for every $a,b\in A$ the following conditions hold: $$\rho(ab)=a.\rho(b),~~\rho(ab)=\rho(a).b$$ where $A\hat{\otimes}A$ is projective tensor products of $A$ by itself and the mappings \begin{align} \left\{ \begin{array}{l} .:(A\hat{\otimes}A)\times A \to A\hat{\otimes}A\\ (\sum_{n=1}^\infty a_n\otimes b_n).a\mapsto\sum_{n=1}^\infty a_n\otimes (b_na)\end{array}\right. \end{align} and \begin{align} \left\{ \begin{array}{l} .:A\times(A\hat{\otimes}A)\to A\hat{\otimes}A\\ a.(\sum_{n=1}^\infty a_n\otimes b_n)=\sum_{n=1}^\infty (aa_n)\otimes b_n\end{array}\right. \end{align} are module's products.

Question: We know there is a $C^*$-algebra which is not biprojective.I need an example with direct proof. I mean if we suppose that it is biprojective then we come to a contradiction but without using Hochschild cohomology and amenability.

Answer 1

In general we have the following theorem [1].

A $C^*$-algebra is biprojective iff $A$ is isomorphic to $\bigoplus_0\{\mathcal{B}(H_\lambda):\lambda\in\Lambda\}$ for some family of finite dimensional Hilbert spaces $\{H_\lambda:\lambda\in\Lambda\}$.

As for commutative case we have the following theorem [2].

A commutative $C^*$ algebra $A$ is biprojective iff its spectrum is discrete.

In fact discreteness of spectrum is a necessary condition of biprojectivity for any commutative Banach algebra. See theorem 5.26 in [2], but before that I strongly recommend you to read theorem 3.5.

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1 Банаховы и полинормированные алгебры: Общая теория, представления, гомологии. А. Я. Хелемский, Наука, 1989

2 Гомология в банаховых и топологических алгебрах, А. Я. Хелемский, МГУ, 1986