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?
Best Answer
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.