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I've been following one Rădulescu's Theorem proof ($G$ is hyperlinear if and only if $\mathcal{L}_{G}$ can be embedded in a ultrapower of the hyperfinite type-$II_{1}$ factor $\mathcal{R}$, where $\mathcal{L}_{G}$ denotes the von Neumann group algebra). I am, however, stuck in a detail that appears to be trivial, since it is not specified in any article I'm reading. So you'll excuse me if this question is quite trivial, but I've been just learning this by myself, out of interest.

Here it goes : Why is it that given a group monomorphism $\Phi : G\rightarrow \mathcal{R}^{\mathcal{U}}$, where $\mathcal{U}$ is an ultrafilter in $\mathbb{N}$ and such that $\tau(\Phi(g))=0$, for every $g\in G\setminus{1_{G}}$ - where $\tau$ denotes the trace - can be extended to a von Neumann algebra embedding $\mathcal{L}_{G}\rightarrow \mathcal{R}^{\mathcal{U}}$? Thank you for the attention !

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The conditions guarantee that the linear extension of $\Phi$ is an isometry in the $2$-norm. Indeed, $$ \left\|\Phi\left(\sum_k\alpha_k\,g_k\right)\right\|_2^2=\left\|\sum_k\alpha_k\Phi(g_k)\right\|_2^2=\tau_{\mathcal U}\left(\sum_{k,j}\overline{\alpha_j}\alpha_kg_j^{-1}g_k\right) =\sum_{k,j}\overline{\alpha_j}\alpha_j\tau_{\mathcal U}(g_j^{-1}g_k)\\ =\sum_k|\alpha_k|^2=\left\|\sum_k\alpha_k\,g_k\right\|_2^2. $$ Any II$_1$-factor is closed in the 2-norm topology, so the above isometry guarantees the embedding. Interestingly, once you deduce that it is an embedding, it is automatically norm-isometric.

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  • $\begingroup$ Thank you so much for this clear and very instructive answer. $\endgroup$
    – thetruth
    Oct 5 '14 at 9:44
  • $\begingroup$ You are very welcome! $\endgroup$ Oct 5 '14 at 13:44

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