Let $N_1$ and $N_2$ be two distinct commuting normal elements. $C*$-algebra generated by $N_1$ and $N_2$ is a commutative $C*$-algebra. So, it will be of the form $C(X)$ for $X$ some compact set in a topological space.
Is $X$ of the form $\sigma(N_1) \times \sigma(N_2)$?
Thanks in advance for the help.
I have mentioned two distinct normal elements after Polp's comment.
No, this is not the case in general, even if $N_1$ and $N_2$ are distinct and differ from the identity. For example, let $N_1$ be a normal operator on a finite dimensional space with two distinct eigenvalues and let $N_2$ denote the spectral projection associated with one of them. Note that $N_2 \in C^*(N_1)$ and so $$C^*(N_1, N_2) = C^*(N_1) \cong C(\sigma(N_1)).$$ However, $\sigma(N_2) = \{0,1 \}$. Thus, the dimension of $C(\sigma(N_1) \times \sigma(N_2))$ is double that of $C(\sigma(N_1))$.