value decomposition . Furthermore, we have an equality of spectra $$\begin{gathered}
\sigma(H(g))=\sigma(H(g')),\qquad g\in\Pi_r.\end{gathered}$$

Analyticity of ${{\cal I}}(g)$ in $g$ with respect to the HS norm amounts to analyticity of its kernel $\Psi(g;x,y)$ in $g$ in the strong topology of the Hilbert space ${{\cal H}}_K$ (\[cHK\]). From the explicit formula (\[Psi\]) it is readily verified that the kernel is strongly analytic in $g$ in a complex neighborhood of $\Pi_r$, so the first assertion follows.

Next, we fix $g\in\Pi_r$ and note that the vectors $$\begin{gathered}
b_m(g;x)\equiv e_{\tau_g(m)}(g;x),\qquad m\in{{\mathbb N}},\end{gathered}$$ yield an ONB with $H(g)$-eigenvalues $E_m(g)$, cf. (\[perm\]). Recalling (\[key2\]) and (\[cIad\]), we now deduce $$\begin{gathered}
H(g'){{\cal I}}(g')b_m(g)   =  {{\cal I}}(g')H(g)b_m(g)
 =  E_m(g){{\cal I}}(g')b_m(g).$$ In view of (\[KerI\]), this entails that the vectors ${{\cal I}}(g')b_m(g)$ are eigenvectors of $H(g')$ with eigenvalues $E_m(g)$, and that they are total in ${{\cal H}}$. Hence, $$\begin{gathered}
E_m(g)=E_m(g'),\qquad \forall \, m\in{{\mathbb N}},\end{gathered}$$ and so the last assertion follows. Moreover, it follows that we have $$\begin{gathered}
\tau_g=\tau_{g'},\end{gathered}$$ and $$\begin{gathered}
{{\cal I}}(g')b_m(g)=\mu_m(g')b_m(g'),\qquad \mu_m(g')\in{{\mathbb R}}^{*}.\end{gathered}$$

To prove the theorem, it remains to show that all of the numbers $\mu_m(g)$ are positive. Now they do not vanish and are continuous on $\Pi_r$, since their squares yield eigenvalues of the self-adjoint trace class 