Let $N$ be a normal operator. When $W^*(N)$ is a MASA?

Question

I am trying to find conditions in which $W^*(N)$ is a MASA, where $N$ is a normal operator acting on some Hilbert space.

1. I know that the multplication algebra $\{M_f| f\in L^{\infty}(X, \mu) \}$ is a MASA, where $(X, \mu)$ is a sigma finite measure space.
   
2. If $X\subseteq \Bbb{C}$ is compact, $f\in C(X)$ separate points (i.e. injective) and $\mu$ is a sigma finite measure on $X$, then $W^*(M_f)$ is the multiplication algebra and thus a MASA.
   
3. If $U$ denoted the bilateral shift, then $W^*(U)$ is a MASA, and actually consists of operators that commute with $U$.
   That's almost immediate from 2, when considering $U$ as $M_z$.
   First, we get from Stone-Weirstrass theorem that $C^*(1,U)=\{M_f| f\in C(\Bbb{T})\}$, because $C^*(1,z)=C(\Bbb{T})$ and $f \to M_f$ is an isomerty.
   Is it connected somehow to the Gelfand isomorphism? From it we can conclude that $C^*(1,U) \cong C(\sigma(U))=C(\Bbb{T})$.
   
4. I somehow got confused from the remark I mentioned in 3. Because then any normal operator N satisfying $C^*(1,N) \cong C(\sigma(N)) \cong \{M_f \in B(L^2(\sigma(N),\mu))| f\in C(\sigma(N))\}$ and the WOT closure of the last is the multiplication algebra (MASA, where $\mu$ is some sigma finite measure...Well, that's nonesense, because not every normal operator generates a MASA.
   
5. Any normal operator is unitarily equivalent to a multiplication operator on some measure space, $M_f$. From the above, if $f$ is injective then $W^*(N)$ is a MASA. That's true?

Answer 1

Your reasoning in point 3 works because you are using an isomorphism induced by the spatial isomorphism $\ell^2(\mathbb Z)\simeq L^2(\mathbb T)$, so masas go back and forth.

In 4 and 5, the isomorphisms you are considering are not spatial (in terms of the original environment where you had $N$). For instance, let $$ N=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}. $$ Then when you say that $C^*(1,N)\simeq C(\sigma(N))$, those two algebras are $\mathbb C^2$. And of course $\mathbb C^2$ (as the diagonal operators) is a masa in $B(L^2(\sigma(N))=B(L^2(\{0,1\})=B(\mathbb C^2)=M_2(\mathbb C)$. But you have no reference to the original environment, where masas as $3$-dimensional.

As for your title question, and as the above example shows, generating a masa is a property of the position of $N$ inside $B(H)$, and not an intrinsic property of $N$. Of course you can easily find some necessary condition; for instance, if the spectrum of $N$ has less cardinality than the dimension of $H,$ then $N$ cannot generate a masa. In finite dimension this condition is also sufficient; but not in infinite dimension (as usual, "infinite=infinite" equalities don't give much information).