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It struck me that I know nothing about the full (universal) free product of the $B(\mathcal H_i)$ amalgamated over $\mathbb C$ for Hilbert spaces $\mathcal H_i$ with identified unit vector $\xi_i$. So my question is:

What is $$B(\mathcal H_1)\ \check *_\mathbb C \cdots \check *_\mathbb C \ B(\mathcal H_n)?$$ Namely, is it obviously not $*$-isomorphic to $B(*_{i=1}^n (\mathcal H_i, \xi_i))$ or any other $B(\mathcal K)$ for some $\mathcal K$?

Edit: Thanks to Caleb Eckhardt for pointing out that $B(*_{i=1}^n (\mathcal H_i, \xi_i))$ is not $*$-isomorphic to $*_{i=1}^n (B(\mathcal H_i),\xi_i)$. Similar to the above question, I suppose one could ask whether the reduced free product is also obviously not $*$-isomorphic to some $B(\mathcal K)$.

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    $\begingroup$ I'm probably missing something, but I don't see why the isomorphism you stated is true in the reduced case. For example if $H_1$ and $H_2$ are 2-dimensional and $n=2$, then it seems that the left hand side will be norm-separable and the right hand side will not be norm-separable. $\endgroup$
    – Caleb Eckhardt
    Commented Jan 26, 2017 at 3:33
  • $\begingroup$ Yes, of course! The identification would only hold if the vector states were faithful, but that's impossible. I'll alter the question accordingly. $\endgroup$
    – Chris Ramsey
    Commented Jan 26, 2017 at 4:15
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    $\begingroup$ I think it is obviously not isomorphic to some $B(K)$ when the $H_i$ are finite dimensional, but not one dimensional. In that case the free product will be infinite dimensional and norm separable which can't happen for any $B(K)$ $\endgroup$
    – Caleb Eckhardt
    Commented Jan 26, 2017 at 13:36
  • $\begingroup$ @CalebEckhardt Any ideas for the infinite dimensional case? $\endgroup$
    – Chris Ramsey
    Commented Jan 26, 2017 at 14:23
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    $\begingroup$ It's probably the same answer. $B(K)$ has a unique closed 2-sided ideal when $K$ is infinite dimensional. The full free product would have more than one such ideal if $H_1$ and $H_2$ are both infinite dimensional--just take representations that alternately kill/preserve the compacts in each factor. If one of the Hilbert spaces is finite dimensional and the other is infinite dimensional, I'm not sure (my knowledge of free products is shallow) but I would guess it is not isomorphic to some $B(K)$ in that case as well $\endgroup$
    – Caleb Eckhardt
    Commented Jan 26, 2017 at 17:45

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