Compare the weight of $p\vee q$ and that of $p+q$

Question

Let $M$ be a von Neumann algebra. If it has a semifinite faithful normal trace $\tau$, then we have $\tau(p\vee q)\le \tau(p)+\tau(q)$. However, for the weight (even a faithful normal state) $\omega$ case, can we compare $\omega(p\vee q)$ and $\omega(p+q ) $? Say, if $p_n\downarrow 0$ and $q_n\downarrow 0$, then can we find some $p_m$ and $q_l$ sucht that $\omega(p_m \vee q_l)$ is very small?

Answer 1

Let $M = M_2$ be the $2\times 2$ complex matrices, acting on $\mathbb{C}^2$. Let $p$ be the orthogonal projection onto ${\rm span}(e_1)$ and for each $n \in \mathbb{N}$ let $q_n$ be the orthogonal projection onto ${\rm span}(e_1 + \frac{1}{n}e_2)$. Both $p$ and $q_n$ are rank one projections, but their join is the identity operator on $\mathbb{C}^2$. Now define $\phi_n(A) = \frac{1}{2^n}\langle Ae_1,e_1\rangle + \frac{2^n-1}{2^n}\langle Ae_2, e_2\rangle$; this is a faithful state with $\phi_n(p) = \frac{1}{2^n}$, $\phi(p \vee q_n) = \phi(I) = 1$ for all $n$, and $\phi(q_n) \to 0$.

For a counterexample to the second question, Let $P_n = 0 \oplus \cdots \oplus 0 \oplus p \oplus p \oplus \cdots \in M_2 \oplus M_2 \oplus \cdots$, with $n$ initial $0$'s, acting on $\mathbb{C}^2 \oplus \mathbb{C}^2 \oplus \cdots$. Let $Q_n = 0 \oplus \cdots \oplus 0 \oplus q_{n+1} \oplus q_{n+2} \oplus \cdots$ and let $\omega$ be the weight $\omega = \phi_1 \oplus \phi_2 \oplus \cdots$. Here $\omega(P_n),\omega(Q_n) \to 0$, and $\omega(P \vee Q_n) = \infty$ for all $n$.