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Let $G$ be a locally compact group. We denote by $A(G)$ the Fourier algebra of $G$.

An operator algebra is a closed subalgebra of $B(H)$ where $H$ is a Hilbert space.

What are the necessary and sufficient conditions for $G$ to have a Fourier algbra which is isomorphic to an operator algebra?

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Thanks to Ebrahim Samei for this proof.

Assume $G$ is infinnite. Assme $A(G)$ is isomorphic to operator algebra then it is Arens regular [1]. It is shown in [2] and [3] that if $A(G)$ is Arens regular, then $G$ is discrete and non-amenable. By [4], for discrete groups, cb-homomorphism into $B(H)$ is similar to a $*$-representation. But no $*$-representaion of $A(G)$ into $B(H)$ can have a closed range (see the proof of theorem A in [5]). This contradict the assumption that $A(G)$ is cb-isomorphic to an operator algebra. If $G$ is finite the result is true. It is worth to mention that a weighted Fourier algebra can be an operator algebra [6].

[1] Operator Algebras and Their Modules: An Operator Space Approach By David P. Blecher, Christian Le Merdy (corollary 2.5.4)

[2] Forrest, Brian(3-WTRL), Arens regularity and discrete groups. Pacific J. Math. 151 (1991), no. 2, 217–227.

[3] Forrest, Brian(3-WTRL), Arens regularity and the $A_p(G)$ algebras. Proc. Amer. Math. Soc. 119 (1993), no. 2, 595–598.

[4] Brannan, Michael; Samei, Ebrahim The similarity problem for Fourier algebras and corepresentations of group von Neumann algebras. J. Funct. Anal. 259 (2010), no. 8, 2073–2097.

[5] Choi, Yemon; Samei, Ebrahim Quotients of Fourier algebras, and representations which are not completely bounded. Proc. Amer. Math. Soc. 141 (2013), no. 7, 2379–2388.

[6] http://www.icmat.es/NTHA/WS-OSHA/notes/Lee.pdf

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    It seems to me that the multiplication map $A(G)\otimes A(G)\to A(G)$ is only bounded for the projective tensor product. This does not imply that the multiplication map is bounded for the Haagerup tensor product. Then I believe that we cannot apply the BRS theorem.2012-09-26