Are the ideals in two $C^*$-algebras the same?

Question

Let $V_{1}, V_{2}$ be the commuting isometries. By Wold decomposition theorem, we know that $V_{i}$ admits decomposition $$V_i \cong V^s_{i}\oplus V^{u}_{i},$$ where $V^{s}_{i}$ is the shift and $V^{u}_{i}$ is the unitary for $i=1,2$.

Let $$I_{1}:=\left\langle 1-V_{2}\right\rangle \text{ be the ideal in } C^*(V_{1}, V_{2}),$$ and $$I_{2}:=\left\langle 1-V_{2}\right\rangle \text{ be the ideal in } C^*(V^{s}_{1}, V_{2}).$$ And assume that $V^{s}_{i}\neq 0$ for $i=1,2.$

Does there exist a isomorphism $$\pi: I_{1}\to I_{2}$$ such that

$$ \pi(I-V_{2})= I-V_{2}? $$

PS. What I understand that $I_{1}$ and $I_{2}$ both contain the $C^*$-algebra $C^*(I-V_{2}).$

Any thoughts would be greatly appreciated.

Answer 1

Counterexample. Let $H = L^2(\mathbb{T}) \oplus l^2(\mathbb{N})$ and define $V_1 = M_{e^{2\pi it}} \oplus S$ and $V_2 = -I_1 \oplus I_2$. Here $M_{e^{2\pi it}}$ is a multiplication operator, $S$ is the unilateral shift, and $I_1$ and $I_2$ are the identity operators on the two summands of $H$.

We have $I - V_2 = 2I_1 \oplus 0$. The ideal it generates in $C^*(V_1, V_2)$ is $C(\mathbb{T}) \oplus 0$ (with the $C(\mathbb{T})$ realized as multiplication operators), and $V_1^s = 0 \oplus S$, so the ideal $I - V_2$ generates in $C^*(V_1^s, V_2)$ is just $\mathbb{C} \oplus 0$.

Edit: a modified version of the question is posed in the comments which also asks that $V_2$ have a nonzero shift part. That can be achieved by the following modification: $H = (\mathbb{C}^2\otimes l^2(\mathbb{N})) \oplus l^2(\mathbb{N})$, $V_1 = (M_{(1, -1)}\otimes I) \oplus S$ and $V_2 = (I\otimes S) \oplus I$ (now using ``$I$'' generically for the identity on any Hilbert space). Then the ideal generated by $I - V_2$ in $C^*(V_1, V_2)$ is isomophic to $\mathcal{T} \oplus \mathcal{T}$ (where $\mathcal{T}$ resembles, actually is an ideal of, the Toeplitz algebra) and the ideal it generates in $C^*(V_1^s, V_2)$ is isomorphic to $\mathcal{T}$. These two ideals are not isomorphic as C${}^*$-algebras because factoring out the commutator ideal yields $C_0(\mathbb{R}) \oplus C_0(\mathbb{R})$ and $C_0(\mathbb{R})$ in the respective cases.