Normal Operators: Meet

Question

Given a Hilbert space $\mathcal{H}$.

Normal Operators: $$\mathcal{N}(\mathcal{H}):=\{N:N^*N=NN^*\}$$

Borel Calculus: $$\mathcal{B}(N):=\{\eta({N}):\eta\in\mathcal{B}(\mathbb{C},\mathbb{C})\}$$

Commutativity: $$N_\pm\in\mathcal{N}(\mathcal{H}):\quad N_+N_-=N_-N_+$$

Borel Calculus: $$\mathcal{B}(N_+)\subseteq\mathcal{B}(N_-)\lor\mathcal{B}(N_-)\subseteq\mathcal{B}(N_+)$$

Meet Operator: $$N_+\wedge N_-\in\mathcal{N}(\mathcal{H}):\quad\mathcal{B}(N_\pm)\subseteq\mathcal{B}(N_+\wedge N_-)$$ (Symbolic Meet!)

Answer 1

Yeaaahh, I got it! :D
(Thanks to David C. Ullrich!)

Borel Calculus

Given Hilbert Space $\mathbb{C}^2$.

Normal Operators: $$N_+:=1\oplus0\quad N_-:=0\oplus1$$

Commutativity: $$N_+N_-=0\oplus0=N_-N_+$$

Borel Calculus: $$\mathcal{B}(N_+)\cap\mathcal{B}(N_-)=\{0\oplus0\}$$ Counterexample!

Meet Operator

Product Measure: $$(E_+\wedge E_-)(A_+\times A_-):=E_+(A_+)E_-(A_-)$$

Borel Isomorphism: $$\Phi:\mathbb{C}\times\mathbb{C}\leftrightarrow\mathbb{C}:\quad\Phi\mathcal{B}(\mathbb{C}\times\mathbb{C})=\mathcal{B}(\mathbb{C})$$

Meet Operator: $$N_+\wedge N_-:=\Phi(E_+\wedge E_-)$$

Meet Calculus:* $$\eta(N_+\wedge N_-)=(\eta\circ\Phi)(E_+\wedge E_-)$$ $$\vartheta(E_+\wedge E_-)=(\vartheta\circ\Phi^{-1})(N_+\wedge N_-)$$

Borel Calculus: $$\mathcal{B}(N_\pm)\leq\mathcal{B}(E_+\wedge E_-)=\mathcal{B}(N_+\wedge N_-)$$

*See thread: Pushforward