Spectral measure for a finite set of mutually commuting normal operators

Question

The following question is from Exercise $\S 11.11$ in A Course in Operator Theory written by John B. Conway:

Suppose $\{N_1, \cdots, N_p\}$ is a finite set of mutually commuting normal operators in $B(H)$ such that for each $1\leq j, k\leq p$, $N_j N_k^* = N_k^* N_j$. Show that there is a unique spectral measure $E$ defined on a compact subset $X$ of $\mathbb{C}^p$ such that for each $1\leq k \leq p$, we have $$N_k = \int_X z_k\,d E(z)$$ where, for each $z\in\mathbb{C}^p$, $z_k$ is the $k$-the coordinate of $z$

My attempt is to apply the definition of joint spectrum introduced in this post. The definition is that, given a $n$-tuple of commuting normal operators $\{N_1, \cdots, N_p\}$, we say a point $z\in\mathbb{C}^p$ is in their joint spectrum iff for any $U_1, \cdots, U_p, V_1, \cdots, V_p\in B(H)$, we have:

$$ \sum_{i\leq p}U_i (z_i - N_i)\neq I_H,\hspace{1cm} \sum_{i \leq p}(z_i - N_i)V_i \neq I_H $$

If we use $\Sigma$ to denote the joint spectrum of $\{N_1, \cdots, N_p\}$, whenever $\lambda-N_k$ is invertible, we then have:

$$ (0, 0, \cdots, \lambda, 0, \cdots, 0)\in \mathbb{C}\backslash\Sigma \hspace{0.5cm}\implies\hspace{0.5cm} \lambda\in P_k(\mathbb{C}\backslash\Sigma) \hspace{0.5cm}\implies\hspace{0.5cm} P_k(\Sigma)\subseteq\sigma(N_k) \hspace{0.5cm}\implies\hspace{0.5cm} \Sigma\subseteq\times_{k=1}^p\sigma(N_k) $$

Then, my questions are:

1. How to show $\Sigma$ is closed? (p.s. later in this paper I found a proof).
2. How to define the desired spectral measure $E$ on $\Sigma$ such that $N_k = \int_{\Sigma}z_k\,d E(z)$ (since possibly $P_k(\Sigma)$ might be properly contained in $\sigma(N_k)$)? When will $\Sigma$ be equal to $\times_{k=1}^p\sigma(N_k)$?
3. Is there a normal operator $M$ defined on another Hilbert space such that the spectral measure of $M$ coincides with the $E$?

Answer 1

I would suggest using the following way more straightforward definition of joint spectrum. Consider $\mathcal A=C^*(N_1,\ldots,N_p)$. Then $\mathcal A$ is an abelian C$^*$-algebra, so $\mathcal A\simeq C(X)$ where $X$ is the character space of $\mathcal A$. The joint spectrum is then the set $$ \sigma(N_1,\ldots,N_p)=\{((\gamma(N_1),\ldots,\gamma(N_p)):\ \gamma\in X\}\subset\prod_{k=1}^p\sigma(N_k). $$ It follows from the definitions that $\sigma(N_1,\ldots,N_p)$ is homeomorphic to $X$ (as in the single variable case), which gives $\mathcal A\simeq C(\sigma(N_1,\ldots,N_p))$. One can then consider the representation $\rho:C(\sigma(N_1,\ldots,N_p))\to B(H)$ induced by $\rho(z_k)=N_k$. By Theorem IX.1.14 in Conway's A Course in Functional Analysis there exists a spectral measure $E$ on the Borel sets of $\sigma(N_1,\ldots,N_p)$ such that $$ \rho(f)=\int_{\sigma(N_1,\ldots,N_p)}f\,dE. $$

I think it's almost impossible for the joint spectrum to be equal to the Cartesian product of the spectra. The joint spectrum is more akin as a "curve" (in the $p=2$ case).

Finally, I cannot make sense of your last question. A single operator will have a a spectral measure on a compact subset of $\mathbb C$, whereas the one constructed here is defined over a compact subset of $\mathbb C^p$.