Spatial isomorphism between type I factors I'm reading a paper from Christopher J. Fewster about split property, and I'm trying to catch one of his assertions.
He asserts that every type I factor $M$ (a concrete von Neumann algebra over $\mathcal{H}$) is spatially isomorphic with $B(\mathcal{K})\otimes \mathbb{1}_{\mathcal{K'}}$. Implemented by a isomorphism $U:\mathcal{H}\to \mathcal{K}\otimes \mathcal{K}'$ such that $UMU^{-1}=B(\mathcal{K})\otimes \mathbb{1}_{\mathcal{K'}}$.
Basically I don't underestand who are $\mathcal{K}$, $\mathcal{K}'$ and what is exactly the isomorphism $U$.
In spite of He give us a reference, I failed trying to understand it because it wasn't the explicit construction there.
Many thanks in advance!
 A: We are assuming that $M$ is not degenerate (i.e., it has the same unit as $B(H)$). Because $M$ is type I, it the sot-closed-span of a family $\{E_{kj}\}_{kj\in S}$ of matrix units. In the decomposition you want, $\dim K=|S|$, and $\dim K'=\operatorname{Tr}(E_{11})$. Let $Z$ be any index set with $|Z|=\operatorname{Tr}(E_{11})$.
Explicitly, form an orthonormal basis $\{e_{k\ell}\}_{k,\in S,\ \ell\in Z}$ for $H$, where $E_{kk}H=\operatorname{span}\{e_{k\ell}\}_{\ell\in Z}$, and $E_{kj}e_{j\ell}=e_{k\ell}$. Such a thing exists because $\operatorname{Tr}(E_{kk})=\operatorname{Tr}(E_{11})$ for all $k$ and $\sum_kE_{kk}=1$. Let $K,K'$ be  Hilbert spaces with $\dim K=|S|$, $\dim K'=|Z|$ and orthonormal bases $\{e_{k}\}_{k\in S}$, $\{f_\ell\}_{\ell\in Z}$ respectively. Define $U:H\to K\otimes K'$ by
$$
Ue_{k\ell}=e_k\otimes f_\ell
$$
and extend by linearity. Because $U$ maps an orthonormal basis to an orthonormal basis, it is a unitary.
Let $\{E_{kj}'\}$ be matrix units for $K$, associated with the basis $\{e_k\}$. Then
\begin{align}
UE_{kj}e_{h\ell}&=\delta_{jh}Ue_{k\ell}=\delta_{jh}\,(e_k\otimes f_\ell),
\end{align}
and
$$
(E_{kj}'\otimes 1_{K'})Ue_{h\ell}
=(E_{kj}'\otimes 1_{K'})(e_h\otimes f_\ell)
=\delta_{jk}\,(e_k\otimes f_\ell).
$$
As we get the equality for all elements in the basis,  $UE_{kj}=(E'_{kj}\otimes 1_{K'})U$.  Since $U$ is a unitary, $UE_{kj}U^*=E'_{kj}\otimes 1_{K'}$. As both on $B(H)$ and $B(K)$ the span of the matrix units is sot-dense, this conjugation by $U$ extends to a $*$-isomorphism, and so
$$
UMU^*=B(K)\otimes 1_{K'}. 
$$
If anything of the above is hard to follow, consider the case where
$$
M=\left\{\begin{bmatrix} 
a&0&0&b&0&0\\
0&a&0&0&b&0\\
0&0&a&0&0&b\\
c&0&0&d&0&0\\
0&c&0&0&d&0\\
0&0&c&0&0&d\\
\end{bmatrix}:\ a,b,c,d\in\mathbb C\right\}. 
$$
