Isometric domains are isospectral for the Laplacian Let

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*$X$ and $Y$ be separable Hilbert spaces;

*$(e_n)_{n\in\mathbb N}$ and $(f_n)_{n\in\mathbb N}$ be orthonormal bases of $X$ and $Y$, respectively;

*$(\lambda_n)_{n\in\mathbb N},(\mu_n)_{n\in\mathbb N}\subseteq[0,\infty)$

*$$S(t)x:=\sum_{n\in\mathbb N}e^{-\lambda_nt}\langle x,e_n\rangle_Xe_n\;\;\;\text{for }x\in X\text{ and }t\ge0$$ and $$T(t)y:=\sum_{n\in\mathbb N}e^{-\mu_nt}\langle y,f_n\rangle_Hf_n\;\;\;\text{for }y\in Y\text{ and }t\ge0$$
We can easily show that $(S(t))_{t\ge0}$ is a contractive $C^0$-semigroup on $X$ with generator $$Ax:=-\sum_{n\in\mathbb N}\lambda_n\langle x,e_n\rangle_Xe_n\;\;\;\text{for }x\in\mathcal D(A):=\left\{x\in X:\sum_{n\in\mathbb N}\lambda_n^2\lambda_n^2\left|\langle x,e_n\rangle_X\right|^2<\infty\right\}$$ and the same clearly applies to $(T(t))_{t\ge0}$ (let $B$ denote its generator).

I want to know when we can find an unitary $U\in\mathfrak L(X,Y)$ such that $$\forall t>0:US(t)=T(t)U\tag1.$$ Since I'm not sure how specific this question is, the particular situation I'm interested in is $X=L^2(\Omega)$ and $Y=L^2(\tilde\Omega)$, where $\Omega,\tilde\Omega\subseteq\mathbb R^d$ are bounded and open such that there is an isometry $\iota:\Omega\to\tilde\Omega$, and $(e_n)_{n\in\mathbb N}$ and $(f_n)_{n\in\mathbb N}$ are the orthonormal bases consisting of eigenvectors of $-\Delta$ on $\{u\in H_0^1(\Omega):\Delta u\in L^2(\Omega)\}$ and $\{u\in H_0^1(\tilde\Omega):\Delta u\in L^2(\tilde\Omega)\}$, respectively, and the sequences $(\lambda_n)_{n\in\mathbb N}$ and $(\mu_n)_{n\in\mathbb N}$ are the corresponding eigenvalues.

Unless, I'm missing something, we should be always to find an unitary $U\in\mathfrak L(X,Y)$ with $Ue_n=f_n$ for all $n\in\mathbb N$. This should simply follow from the fact that $V:X\to\ell^2(I),x\mapsto(\langle x,e_i\rangle)_{i\in I}$ and $W:Y\to\ell^2(I),y\mapsto(\langle y,f_i\rangle)_{i\in I}$ are unitary isomorphisms so that we can take $U:=W^{-1}\circ V$. However, in generality I'm not able to deduce $(1)$ from that (using this particular $U$.)
In the particular case of the Laplacian, we could consider $Uf:=f\circ\iota^{-1}$ for $f\in L^2(\Omega)$, which is clearly unitary and hence maps ortrhonormal bases to orthonormal bases. But, again, I'm not able to show $(1)$ for this choice of $U$ either ...
 A: The question seems to be motivated by the "Can you here the shape of a drum" question.
First, let me briefly give a general answer to the question and then explain how it is related to the specific setting of Laplace operators.
General situation.
Let $H_1$ and $H_2$ be two Hilbert spaces and let $A_1: H_1 \supseteq D(A_1) \to H_1$ and $A_2: H_2 \supseteq D(A_2) \to H_2$ be self-adjoint operators with compact resolvent which have finite spectral bound and thus generate $C_0$-semigroups $T_1$ and $T_2$.
Proposition 1.
The following assertions are equivalent:
(i) The operators $A_1$ and $A_2$ have the same eigenvalues, including multiplicities - i.e., they are isospectral.
(ii) There exists a unitary operator $U: H_1 \to H_2$ that intertwines $A_1$ and $A_2$ (equivalently, that intertwines the semigroups $T_1$ and $T_2$).
(iii) There exists a bounded linear bijection $U: H_1 \to H_2$ that intertwines $A_1$ and $A_2$ (equivalently, that intertwines the semigroups $T_1$ and $T_2$).
Sketch of proof.
"(i) $\Rightarrow$ (ii)" Let $(\lambda_n)$ denote the sequence of eigenvalues of both $A_1$ and $A_2$, and let $(e_n)$ and $(f_n)$ be corresponding ONBs of eigenvectors.
Define $Ue_n := f_n$ for each $n$. Then $U$ is a unitary that intertwines $A_1$ and $A_2$.
"(ii) $\Rightarrow$ (iii)" Obvious.
"(iii) $\Rightarrow$ (i)" The fact that $U$ intertwines both operators means precisely that $A_1$ and $A_2$ are similar; hence, they have the same eigenvalues. $\square$
The Laplace operator
Now we specialize to the situation where $H_1 = L^2(\Omega_1)$ and $H_2 = L^2(\Omega_2)$ for two bounded domains $\Omega_1$ and $\Omega_2$ in $\mathbb{R}^d$, und where $A_1$ and $A_2$ denote the Dirichlet Laplace operators on both spaces.
Proposition 2. If there is a Euclidean isometric $\iota: \mathbb{R}^d \to \mathbb{R}^d$ which maps $\Omega_2$ to $\Omega_1$, then $A_1$ and $A_2$ are isospectral.
Sketch of proof. Such a $\iota$ is a composition of a translation and a rotation. Hence, the mapping $U: f \mapsto f \circ \iota$ commutes with the distributional Laplace operator. Moreover, the same mapping maps $H^1_0(\Omega_1)$ to $H^1_0(\Omega_2)$, so it actually intertwines the operators $A_1$ and $A_2$. Hence, isospectrality follows from the implication "(ii) $\Rightarrow$ (i)" in Proposition 1. $\square$
Note that the proof of Proposition 2 has nothing to do with the converse implication "(i) $\Rightarrow$ (ii)" in Proposition 1, i.e., the unitary $U$ in the proof of Proposition 2 is constructed by a very different method than the one in the proof of the implication "(i) $\Rightarrow$ (ii)" in Proposition 1. (But, of course, one can show a posteriori that the unitary $U$ from Proposition 2 maps each eigenspace of $A_1$ to the corresponding eigenspace - for the same eigenvalue - of $A_2$).
Hearing the shape of a drum.
This is a famous problem, made popular by Kac, which asks under which circumstances the converse to Proposition 2 holds. I.e., assume that $A_1$ and $A_2$ are isospectral - does it follows that $\Omega_1$ and $\Omega_2$ they are congruent?
In full generality, the answer has been known to be "no" for a long time. In certain special cases, the answer is known to be "yes".
But for large classes of domains, the problem is still open.
Isothermal drums.
If we are interested in the "Can you here the shape of a drum" problem, why should we care about the implication "(i) $\Rightarrow$ (ii)" in Proposition 1?
Here is an intuitive reason: Assume that $\Omega_1$ and $\Omega_2$ are "good" (for instance, have $C^\infty$-boundary) and that $A_1$ and $A_2$ are isospectral - then Proposition 1 tells us that there exists an operator $U$ as in (ii) (or (iii)). If we could prove that $\Omega_1$ and $\Omega_2$ are congruent, we would have a positive answer to Kac's problem - but as mentioned above, this is open.
Now assume, however, that the intertwining operator $U$ does not only exists but can be chosen such that both $U$ and $U^{-1}$ are positive (in the sense of Banach lattice, i.e., $Uf \ge 0$ iff $f \ge 0$). Then it can indeed be shown that $\Omega_1$ and $\Omega_2$ are congruent; see this paper. Hence, we don't have a solution to Kac's problem, but we have a positive result under additional assumptions on $U$.
(Notational remark: If $U$ and $U^{-1}$ can be chosen to be positive, it is tempting to call $A_1$ and $A_2$ isothermal rather than only isospectral - thus the title of this little subsection here.)
