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

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

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

Commuting System: $$\mathcal{S}(\mathcal{H})\subseteq\mathcal{N}(\mathcal{H}):\quad[\mathcal{S}(\mathcal{H}),\mathcal{S}(\mathcal{H})]=0$$ (Spectral Measures!)

Maximal Operator: $$N_\infty\in\mathcal{N}(\mathcal{H}):\quad\mathcal{S}(\mathcal{H})\subseteq\mathcal{B}(N_\infty)$$ (Existence & Uniqueness?)


In application one deals with commuting normal operators.

Reduction from two operators: Meet
Consider commuting normal operators $N$ and $N'$.
As they commute they admit a product spectral measure $E\wedge E'$.
But the situation is even better: They admit a wedge normal operator $N\wedge N'$.
That is its Borel calculus contains both original operators $N$ and $N'$.

Thus one may reduce the case of finitely many commuting normal operators
to the study of only one normal operator and its Borel calculus.

For arbitrarily many the question was still open here...

  • $\begingroup$ @Norbert: I added a motivation. (See edit!) $\endgroup$ – C-Star-W-Star Apr 1 '16 at 11:59

Meanwhile I got it.. :D

Hilbert Space: $$\Omega:=\mathcal{P}^2(\mathbb{C}):\quad\mathcal{H}:=\ell^2(\Omega)$$

Normal Operators: $$N_\omega:\mathcal{D}N_\omega\to\mathcal{H}:\quad N_\omega\varphi:=1_\omega\varphi$$

Commuting System: $$\mathcal{S}(\mathcal{H}):=\{N_\omega:\omega\in\Omega\}$$

Borel Calculus: $$\#\mathcal{B}(N_\infty)\leq\#\mathcal{B}(\mathbb{C},\mathbb{C})\leq\#\mathcal{P}(\mathbb{C})\\ <\#\mathcal{P}^2(\mathbb{C})=\#\Omega=\#\mathcal{S}(\mathcal{H})$$ (Contradiction!)


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