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Let $\mathcal{M}$ be a type I von Neumann algebra. Are there necessary and sufficient conditions for $\mathcal{M}$ to admit a faithful normal state?

If I think of $\mathcal{M}$ as the von Neumann algebra of bounded linear operators on some separable Hilbert space $\mathcal{H}$, then I know that there always exists an infinite number of faithful normal states. The way in which I am able to build such states is to consider an orthonormal basis $\{|j\rangle\}_{j=1,\dotsc,\dim(\mathcal{H})}$ in $\mathcal{H}$, a sequence $\{p^{j}\}_{j=1,\dotsc,\dim(\mathcal{H})}$ of strictly positive real numbers summing to $1$, and set:

$$ \rho=\sum_{j=1}^{\dim(\mathcal{H})}\,p^{j}\:|j\rangle\langle j|\,. $$ Then, the normal state $\omega_{\rho}$ given by:

$$ \omega_{\rho}(\mathbf{A})\,:=\,\operatorname{tr}(\rho\,\mathbf{A}) \quad \forall \mathbf{A}\in\mathcal{M}=\mathcal{B}(\mathcal{H}) $$ is faithful. As it is clear, this construction depends on the fact that $\mathcal{H}$ is separable, and thus also the predual $\mathcal{M}_{*}=K(\mathcal{H})$ of $\mathcal{M}$ is separable. Is this true in general?

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If $H$ is not separable, there is no faithful normal state on $B(H)$. That's (one of) the reason(s) one considers weights.

Indeed, if $f$ is such a normal state, you take the sequence $\{q_j\}$ of projections as you did, and you have $\sum_jq_j=I$. Then, using normality, $$ 1=f(I)=\sum_j f(q_j). $$ As $f$ is faithful, $f(q_j)>0$ for all $j$. But you cannot have uncountably many positive numbers with finite sum.

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  • $\begingroup$ According to my question and your answer, the separability of $\mathcal{H}$ is a necessary and sufficient condition for $\mathcal{B}(\mathcal{H})$ to admit a faithful normal state. What happens then for a general type I von Neumann algebra? For instance if we consider the algebra $L^{\infty}(X,\mu)$? $\endgroup$
    – F.M.C.
    Sep 7, 2018 at 15:16
  • $\begingroup$ Now a state is a finite measure $\nu$ such that $\mu$ is absolutely continuous with respect to $\nu$. If I'm not wrong, that can only happen when $X$ is $\sigma$-finite (with respect to $\mu$). $\endgroup$ Sep 7, 2018 at 15:33
  • $\begingroup$ Yes, so a faithful normal state on $L^{\infty}(X,\mu)$ (with $X$ $\sigma$-finite w.r.t. $\mu$) would be identified with an element $h\in L^{1}(X,\mu)$ such that, given $f\in L^{\infty}(X,\mu)$ with $f\geq 0$, it is $\int_{X} f h\,\mathrm{d}\mu=0$ implies $f=0$. How can I be sure that such an $h$ exists? $\endgroup$
    – F.M.C.
    Sep 7, 2018 at 15:55
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    $\begingroup$ If $X$ is $\sigma$-finite, $X=\bigcup_{j=1}^\infty X_j$ with $\mu(X_j)<\infty$. Then take $$h=\sum_j\frac1{2^j\mu(X_j)}\,1_{X_j}.$$ $\endgroup$ Sep 7, 2018 at 16:52
  • $\begingroup$ I just found a proposition (II.3.19 at page 78) in Takesaki's book "Theory of operator algebra I" stating that for a von Neumann algebra acting on some complex Hilbert space to admit a normal, faithful, positive linear functional is equivalent to being $\sigma$-finite. The proof "essentially" generalizes the construction given by Martin Agerami in the previous comment. $\endgroup$
    – F.M.C.
    Sep 11, 2018 at 8:38

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