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For every group $G$, the reduced group $C^*$-algebra $C^*_{red}G$ is equipped with the inner product $\langle a,b\rangle=tr(ab^*)$ where "$tr$" is the standard trace on group $C^*$-algebras.

For what kind of groups $G$, $C^*_{red}G$ admit a bounded skew-symmetric $2$-linear map $\wedge: C^*_{red}G\times C^*_{red}G \to C^*_{red}G$ with the following properties:

1) For every $a,b \in C^*_{red}G$, $a\wedge b$ is perpendicular to both $a,b$.

2)For every two independent elements $a,b\in C^*_{red} G$, $a\wedge b$ is a nonzero element?

Our question is inspired by the following paper:

W. S. Massey, Cross Products of Vectors in Higher Dimensional Euclidean Spaces. The American Mathematical Monthly, Vol. 90, No. 10 (Dec., 1983), pp. 697-701.

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    $\begingroup$ By bounded you mean for the $L^2$ norm given by the trace ? By 2) do you mean linear independance over $\mathbb C$ ? If so, your question boils down to one for Hilbert spaces, and the answer just depends on the cardinal of $G$ $\endgroup$
    – InfiniteLooper
    Commented Feb 25, 2019 at 10:47
  • $\begingroup$ @Bleuderk By bounded I mean for the original $C^*$ norm.I used the inner product merely for significance of perpendicularity, not any topological consideration. By 2), yes I mean $\mathbb{C}$-independent. $\endgroup$
    – Ali Taghavi
    Commented Feb 25, 2019 at 11:47

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