# Spectral measure for a finite set of mutually commuting normal operators

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 $$E$$?

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 $$\mathbb C^p$$.
• Thank you for your answer! May I ask how do you know that the joint spectrum is equal to the set $\Big\{ \Big(\, (\gamma(N_1), \gamma(N_2), \cdots, \gamma(N_p) \,\Big): \gamma\in X \Big\}$? Also, could you elaborate a bit on what you meant by the joint spectrum is like a "curve"? Do you mean there could exist a continuously changing tuple of $(U_1, U_2, \cdots, U_p)$, and $(\lambda_1, \lambda_2, \cdots, \lambda_p)$ such that $\sum_{i\leq p}U_i(\lambda_i - N_i)$ is equal to (or distant away from) the identity when $\{U_i\}, \{\lambda_i\}$ is continuously changing? Commented Sep 23, 2022 at 0:40
• Now I realize my last question is not accurate. It is inspired by a theorem in later chapters in the textbook, which states that when $H$ is separable, I can find a self-adjoint $A\in B(H)$ such that $C^*(A)$ contains each $N_i$. I first wonder if I could find a normal operator whose spectral measure coincides with the one defined on $\bigcup_{i\leq p}\sigma(N_i)$. I wonder will this happen when $\{\sigma(N_i)\}_i$ is mutually disjoint, so that the sum of their spectral measures, which is another spectral measure on $\bigcup_{i\leq p}\sigma(N_i)$, can return a new normal operator. Commented Sep 23, 2022 at 0:50
• I'm defining the joint spectrum that way. It is a way in which it is straightforward to see that it is homeomorphic to the character space of $C^*(N_1,\ldots,N_p)$, so one gets the expected isomorphism $C^*(N_1,\ldots,N_p)\simeq C(\sigma(N_1,\ldots,N_p)$. And it is a direct generalization of what one does in one variable. The definition you used I have to admit I don't understand it (I mean, I understand the terminology and even the analogy with the one-operator case, but it says nothing to me). Commented Sep 23, 2022 at 1:06
• For the $N_j$ to act separately what you need is not that the spectra is dsjoint, but that the spectral measures act on pairwise orthogonal subspaces. That is, you need $E_j(\Delta)E_k(\Delta')=0$ whenever $k\ne j$ and for all $\Delta,\Delta'$. In that case the operators behave as if they were the components of a direct sum. Commented Sep 23, 2022 at 1:10
• I noticed I didn't answer the "curve" question. What I mean is that the Cartesian product of two intervals is a rectangle; but if you have two operators each with spectrum an interval, the joint spectrum is more likely to look like a curve within that rectangle. For instance, suppose that $f,g\in C[0,1]$ satisfy $fg=0$ and both have range $[0,1]$. The characters are the point evaluations. So the joint spectrum is $$\sigma(f,g)=\{(f(x),g(x)):\ x\in[0,1]\}=([0,1]\times\{0\})\cup(\{0\}\times[0,1]).$$ Commented Sep 23, 2022 at 15:47