December 16, 2024
We study the Rellich type theorem (RT) for the Maxwell operator \(\hat{H}^D=\hat{D}\hat{H}_0\) on \({\boldsymbol{Z}}^3\) in a constant anisotropic medium, i.e., the permittivity and permeability of which are constant non-scalar diagonal matrices. We also prove the unique continuation property (UCP) in the exterior of a compact convex set \({\boldsymbol{K}}_{\rm int}\subset {\boldsymbol{Z}}^3\) for the perturbed Maxwell operator \(\hat{H}^{D_p}=\hat{D}_p\hat{H}_0\) on \({\boldsymbol{Z}}^3\) for which the permittivity and permeability are locally perturbed from a constant matrix on a compact subset in \({\boldsymbol{K}}_{\rm int}\).
\(^{\small 1}\) University of Tsukuba, Japan
\(^{\small 2}\) Aix-Marseille Université, France
Primary 35Q61, 47A40, 81Q35.
. Maxwell equations, Scattering theory of linear operators, Quantum mechanics on lattices.
Let us begin with general definitions. Consider an operator \(\hat{Q}\) acting on the space \(({\boldsymbol{C}}^m)^{{\boldsymbol{Z}}^d}\) of sequences \((\hat{u}(n))_{n\in {\boldsymbol{Z}}^d}\) with values in \({\boldsymbol{C}}^m\), where \(m,d\ge 1\). We efine the Besov type space \[\mathcal{B}_0^*({\boldsymbol{Z}}^d;{\boldsymbol{C}}^m) := \Big\{ \hat{u}: \; {\boldsymbol{Z}}^d \to {\boldsymbol{C}}^m \; \mid\; \lim_{R\to\infty} \frac{1}{R}\sum_{|n|<R} |\hat{u}(n)|^2 = 0\Big\}.\] We say that the Rellich type theorem (RT) holds for \(\hat{Q}\), if \(\hat{u}\in\mathcal{B}_0^*({\boldsymbol{Z}}^3;{\boldsymbol{C}}^6)\) satisfies the equation \[\label{e46Qu610} \hat{Q} \hat{u}\, (n) = 0,\quad n\in {\boldsymbol{Z}}^d\tag{1}\] outside a compact set in \({\boldsymbol{Z}}^d\), then \(\hat{u} = 0\) on \(\{n\in{\boldsymbol{Z}}^d\) \(\mid\) \(|n| > R\}\) for some constant \(R > 0\).
(RT) is connected with the unique continuation property (UCP) in exterior domains of \({\boldsymbol{Z}}^d\). Here, by an exterior domain we mean the set of the form \({\boldsymbol{K}}_{\rm ext}={\boldsymbol{Z}}^d\setminus{\boldsymbol{K}}_{\rm int}\), \({\boldsymbol{K}}_{\rm int}\) being compact. We say that (UCP) holds for \(\hat{Q}\) in \({\boldsymbol{K}}_{\rm ext}\) if, all \(\hat{u}\) with compact support in \({\boldsymbol{Z}}^d\) satisfying \(\hat{Q}\hat{u} = 0\) in \({\boldsymbol{K}}_{\rm ext}\) vanishes on \({\boldsymbol{K}}_{\rm ext}\).
We consider (RT) and (UCP) for the discrete Maxwell operator \(\hat{H}^{D}\) defined on the square lattice \({\boldsymbol{Z}}^3\).
Let \(\boldsymbol{\varepsilon}\) and \(\boldsymbol{\mu}\) be the permittivity and the permeability for the ambient space \({\boldsymbol{Z}}^3\), which are the \(3\times 3\) constant diagonal matrices with diagonal elements, \(\varepsilon_1, \varepsilon_2, \varepsilon_3\in (0,\infty)\) for \(\boldsymbol{\varepsilon}\), and \(\mu_1,\mu_2,\mu_3 \in (0,\infty)\) for \(\boldsymbol{\mu}\). Letting \(\hat{H}_0\) be the isotropic discrete Maxwell operator with \(\boldsymbol{\varepsilon}= \boldsymbol{\mu}= I_{3\times 3}\), the anisotropic Maxwell operator is defined by \[\hat{H}^{D}= \hat{D} \hat{H}_0,\] where we set \[\hat{D} = \left(\begin{array}{cc} \boldsymbol{\varepsilon}& 0_{3\times3} \\ 0_{3\times3} &\boldsymbol{\mu} \end{array}\right).\qedhere\] The perturbed Maxwell operator has the form \(\hat{H}^{D_p}=\hat{D}_p \hat{H}_0\) where \(\hat{D}_p\) is a perturbation of \(\hat{D}\), i.e., \[\hat{D}_p := \left(\begin{array}{cc} \tilde{\boldsymbol{\varepsilon}}& 0_{3\times3} \\ 0_{3\times3} &\tilde{\boldsymbol{\mu}} \end{array}\right),\] \(\tilde{\boldsymbol{\varepsilon}}_j(n)>0\), \(\tilde{\boldsymbol{\mu}}_j(n)>0\) for \(1\le j\le 3\) and \(n\in{\boldsymbol{Z}}^3\), and \(\hat{D}_p-\hat{D}\) has compact support.
Our interest in (RT) and (UCP) appears in the application to inverse problems similar to those studied in [1] for the Schrödinger operator on lattices. This aspect will be developed in further publications.
Following the methods in [2], [3], we pass to the Fourier series and consider \(H^D=\hat{D} H_0\), which is a multiplication operator on the real torus \(\mathbb{T}^3\approx (\mathbf{R}/(2\pi {\boldsymbol{Z}}))^3\). The operator \(H^D\) is then represented by the analytically fibered self-adjoint operator \(\mathbb{T}^3 \ni x\mapsto H^D(x)=\hat{D} H_0(x)\) where \(H_0(x)\) is a \(6\times 6\) real symmetric matrix depending on the variable \(y=(y_1,y_2,y_3)\), \(y_i = \sin x_i\) only (see 9 ). The equation \((\hat{H}^{D}-\lambda)\hat{u}\, (n) = 0\) outside a compact set is then equivalent to \[\label{e46Hbu61f} (H^D -\lambda)u = f ,\tag{2}\] where \(f\) is a trigonometric polynomial of the variable \(x\in \mathbb{T}^3\), and where 2 has to be taken in the sense of distributions. Using the correspondence \(\hat{u} \in \mathcal{B}_0^*({\boldsymbol{Z}}^3)\iff u\in \mathcal{B}_0^*(\mathbb{T}^3)\), the problem is reduced to the Besov space in terms of the variable \(x\). Multiplying 2 by the cofactor matrix of \(H^D(x)-\lambda\) leads to the equation \[\label{e46qu61g} q(\cdot;\lambda)u = g \quad {\rm on} \quad \mathbb{T}^3,\tag{3}\] where \(q(x;\lambda):=\det(H^D(x)-\lambda)\) and \(g\) is a trigonometric polynomial. Let us recall the factorization of \(q(x;\lambda)\) and the spectral analysis of \(H^D(x)\) (see [4]). The spectrum \(\sigma(H^D(x))\) of \(H^D(x)\) depends on \(z= (z_1,z_2,z_3)\), \(z_i=\sin^2 x_i\), only, and \(q(x;\lambda)\) is a polynomial of \(z\) and \(\lambda\), which we denote by \(p(z;\lambda)\). We write the diagonal matrices \(\boldsymbol{\varepsilon}= (\varepsilon_1,\varepsilon_2,\varepsilon_3)\), \(\boldsymbol{\mu}= (\mu_1,\mu_2,\mu_3)\) and put: \[\label{def46beta-alpha} \begin{align} \boldsymbol{\beta}&:=& \boldsymbol{\varepsilon}\times \boldsymbol{\mu}=(\beta_1,\beta_2,\beta_3)\\ \alpha_i &:=& (\varepsilon_j \mu_k + \varepsilon_k \mu_j)/2, \\ \gamma_i &:=& \varepsilon_j \varepsilon_k \mu_j\mu_k, \end{align}\tag{4}\] where \((i, j, k)\) ranges over the cyclic permutations of \((1,2,3)\), and \(\boldsymbol{\alpha}:=(\alpha_1,\alpha_2,\alpha_3)\). We put \[\begin{align} {\boldsymbol{B}}_0 &:=& \{(0,0,0)\}, \\ {\boldsymbol{B}}_{3} &:=& \{\boldsymbol{\beta}\in {\boldsymbol{R}}^3\; \mid \; \beta_1\beta_2\beta_3 \neq 0\},\\ {\boldsymbol{B}}_{12} &:=& {\boldsymbol{R}}^3 \setminus ({\boldsymbol{B}}_0\cup {\boldsymbol{B}}_3). \end{align}\]
We have \[\label{val46pz} q(x;\lambda) = p(z;\lambda) = \lambda^2\left(\lambda^4 - 2 \lambda^2\Psi_0(z) + \Psi_0^2(z) - K_0(z)\right),\tag{5}\] where we put \[\label{def46Psi0} \begin{align} K_0(z) &:=& \frac{1}{4}\sum_{i=1}^3(\beta_iz_i)^2 - \frac{1}{2}\sum_{1\leq i<j\leq 3}\beta_i\beta_jz_iz_j,\\ \Psi_0(z) &:=& \alpha\cdot z = \alpha_1z_1 + {\rm c.p.}.\end{align}\tag{6}\] Then, \(p(z;\lambda)\) is rewritten as: \[p(z;\lambda) = \lambda^2(\tau^+(z) - \lambda^2)(\tau^-(z) - \lambda^2),\] where we put: \[\tau^{\pm}(z) := \Psi_0(z) \pm \sqrt{K_0(z)}. \label{taupmzdefine}\tag{7}\] In [4], it is shown that \(\tau^-(z)\ge 0\) for all \(z\in [0,+\infty)^3\), and that \[\sigma(H^D(x)) = \{-\sqrt{\tau^+(z)},-\sqrt{\tau^-(z)},0,\sqrt{\tau^-(z)},\sqrt{\tau^+(z)}\} , \quad x\in\mathbb{T}^3.\] We put \[\lambda_\pm = \max_{[0,1]^3} \sqrt{\tau^\pm(z)}.\] Note that \(\lambda_+ \ge \lambda_- > 0\), and we have \(\lambda_+=\lambda_-\) iff \(\boldsymbol{\beta}\in {\boldsymbol{B}}_{0}\).
Our first main result is concerned with (RT) and is stated in terms of the parameter \(\boldsymbol{\beta}=\boldsymbol{\varepsilon}\times\boldsymbol{\mu}\) and \(\lambda_{\pm}\).
****Theorem** 1**. (RT) for \(\hat{H}^{D}-\lambda\) holds in each of the following cases:
\(\boldsymbol{\beta}\in {\boldsymbol{B}}_0\) and \(0 < |\lambda| < \lambda_+\),
\(\boldsymbol{\beta}\in {\boldsymbol{B}}_3\) and \(0<|\lambda|<\lambda_+\),
\(\boldsymbol{\beta}\in {\boldsymbol{B}}_{12}\) and \(0<|\lambda|<\lambda_-\).
Since \(\lambda_- < \lambda_+\) if \(\boldsymbol{\beta}\in {\boldsymbol{B}}_{12}\), the case (3) in Theorem 1 is critical.
****Theorem** 2**. Assume \(\boldsymbol{\beta}\in {\boldsymbol{B}}_{12}\) and \(|\lambda|>\lambda_-\). Then (RT) for \(\hat{H}^{D}-\lambda\) fails.
****Remark** 1**. Even in the situation of Theorem 2 the following fact holds. Let \(\lambda\in (\lambda_-,\lambda_+)\), and \(u\in\mathcal{B}_0^*(\mathbb{T}^3)\) be such that \((H^D -\lambda)u\) is a trigonometric polynomial of \(x\in \mathbb{T}^3\). Then \(v(x):=(\tau^-(z)-\lambda^2)u(x)\) is a trigonometric polynomial of \(x\in \mathbb{T}^3\), and so, \(u(x)=(\tau^-(z)-\lambda^2)^{-1}v(x) \in L^2(\mathbb{T}^3)\). We consequently obtain the following result.
*****Proposition** 3**. *Assume \(\boldsymbol{\beta}\in {\boldsymbol{B}}_{12}\) and \(|\lambda|>\lambda_-\). Let \(\hat{u} \in \mathcal{B}_0^*({\boldsymbol{Z}}^3)\) such that \((\hat{H}^{D_p}-\lambda)\hat{u} =0\) and \(\hat{u}\neq 0\). Then, \(\hat{u}\) is an eigenvector of \(\hat{H}^{D}\) associated with the eigenvalue \(\lambda\).**
Our second main result is about (UCP) and is stated as follows.
****Theorem** 4**. Let \({\boldsymbol{K}}_{\rm ext}:={\boldsymbol{Z}}^3\setminus{\boldsymbol{K}}_{\rm int}\) where \({\boldsymbol{K}}_{\rm int}\) is a compact convex set of \({\boldsymbol{Z}}^3\). Assume \(\lambda\neq 0\). Then (UCP) holds for the operator \(\hat{H}^{D_p} - \lambda\) in \({\boldsymbol{K}}_{\rm ext}\).
****Remark** 2**. The proof of Theorem 4 shows that (UCP) holds in the exterior of a compact convex set for any scalar operator \(\hat{Q}\) where \(Q\) is the multiplication by a trigonometric polynomial \(Q(x)\) of the form \(\sum_{j=1}^3 c^+_j e^{iax_j} + c^-_j e^{-iax_j} + \sum_{m\in{\boldsymbol{Z}}^3} c(m) e^{im\cdot x}\) with \(a\in \mathbf{N}^*\), \(c_j^\pm\neq 0\), and \(\sum_j |m_j|<a\).
****Remark** 3**. The results of [4] showed that \(\sigma_p(\hat{H}^{D}) = \{0\}\), and so, showed that (UCP) holds for the unperturbed operator \(\hat{H}^{D}-\lambda\) in the whole \({\boldsymbol{Z}}^3\) if \(\lambda\neq 0\). Nevertheless (UCP) does not hold for \(\hat{H}^{D}\) in \({\boldsymbol{Z}}^3\) since the kernel of \(\hat{H}^{D}\) contains trigonometric polynomials.
****Corollary** 5**. Let \(\lambda\neq 0\) and \({\boldsymbol{K}}_{\rm ext}\) be as in Theorem 4. Assume that (RT) for \(\hat{H}^{D}-\lambda\) holds. Let \(\hat{u}\in \mathcal{B}_0^*({\boldsymbol{Z}}^3)\) be a solution of \((\hat{H}^{D_p} -\lambda)\hat{u}=0\) in \({\boldsymbol{K}}_{\rm ext}\). Then \(\hat{u}\) vanishes in \({\boldsymbol{K}}_{\rm ext}\).
****Remark** 4**. *The proof of Theorem 4 shows that Theorem 4 and so Corollary 5 extend to the case where the perturbation \(\hat{D}_p- \hat{D}\) is any operator of multiplication by real or complex valued function with compact support such
that, for all \(n\), \(\hat{D}_p(n)\) is a self-adjoint positive matrix. Nevertheless, it is difficult to extend Corollary 5 to the operator \((\hat{H}^{D}+ \hat{V}-\lambda)\), where \(\hat{V}\) is an operator of multiplication on \({\boldsymbol{Z}}^3\) with compact support, since (UCP) fails for \(\hat{H}_0\) and for \(\hat{H}^{D}\). In fact, the trigonometric polynomial, \(\phi^+(x):=(y,0_{{\boldsymbol{R}}^3})\) (we can choose also \(\phi^-(x):=(0_{{\boldsymbol{R}}^3},y)\)) \(\in {\boldsymbol{R}}^6\) is an eigenmode for \(H_0\) (and so for \(H^D\)) associated with the eigenvalue 0, then (UCP) obviously fails for \(\hat{H}_0\) in the whole space \({\boldsymbol{Z}}^3\). We then have the following general negative result.
Let \(\lambda\in\mathbf{R}\). Then there exists a compactly supported multiplication operator \(\hat{V}\) and a compactly supported eigenmode \(\hat{u}\) of
\(\hat{H}_0+\hat{V}\) associated with the eigenvalue \(\lambda\), i.e., \[(\hat{H}_0+\hat{V}-\lambda)\hat{u}=0.\] In fact, setting \(u=\phi^+\) and \(\hat{V}=\lambda \cdot 1_{{\rm supp}\,\hat{u}}\), the conclusion follows.*
In §2 we give the definition of the discrete Maxwell operator, and recall its main spectral properties. In §3 we prepare Theorem 8 which is a very slight extension of results in [2], [3]. We then give a proof of Theorem 8. In §4 we describe the Fermi surface \(M^{\boldsymbol{C}}(q(\cdot;\lambda))\). The analysis begins with the description of the Fermi surface \(M^{\boldsymbol{C}}(p(\cdot;\lambda))\) which is simpler since the function \(p(\cdot;\lambda)\) is a polynomial of second order of the variable \(z\). The key is Lemma 8. In §5 we give the proofs of Theorems 1, 2, 4, Proposition 3 and Corollary5.
We use the following notations. Letting two sets \(E,F\), we denote by \(\mathcal{A}(E;F)=F^E\) the set of applications from \(E\) into \(F\). For \(n=(n_1,\ldots,n_d)\in{\boldsymbol{Z}}^d\) we set \(|n|=\sum_{j=1}^d |n_j|\). For a map \(f : {\boldsymbol{R}} \to {\boldsymbol{R}}\) (respect., a function from \(\mathbb{T}^1\) into \({\boldsymbol{R}}^1\)), we denote by \(f\) again the mapping \({\boldsymbol{R}}^d\) (respectively, \(\mathbb{T}^d\)) \(\ni x \mapsto (f(x_1),\ldots, f(x_d))\in {\boldsymbol{R}}^d\). If \(E\subset\mathbb{T}^d\) or \(E\subset{\boldsymbol{R}}^d\) we denote by \(f(E)\) the set \(\{f(x)\) \(\mid\) \(x\in E\}\). In particular we set, for \(x\in \mathbb{T}_{\boldsymbol{C}}^d\cup {\boldsymbol{C}}^d\), \[\label{y61sinxformula} \begin{align} y &:= \sin x =(\sin x_1,\dots, \sin x_d), \\ z &:= (\sin^2x_1, \dots, \sin^2x_d). \end{align}\tag{8}\] We use the abreviated expressions "iff" which means "if and only if" and "\(+\) c.p." in a computation to indicate that the value is the addition of the two other similar terms obtained by two successive permutations of \(1\to 2\to 3\to 1\). For example, an expression written as \(p_1+q_{23}+\) c.p. is the sum equal to \(p_1+q_{23}+p_2+q_{31}+p_3+q_{12}\). For \(\tilde{E}\subset \mathbb{T}_{\boldsymbol{C}}^d\cup{\boldsymbol{C}}^d\), we set \(\sin^2(\tilde{E})=\{z\in{\boldsymbol{C}}^d\) \(\mid\) \(x\in \tilde{E}\}\), and letting \(E\subset {\boldsymbol{C}}^d\) we set \((\sin_{\boldsymbol{C}}^2)^{-1}(E)=\{x\in{\boldsymbol{C}}^d\) \(\mid\) \(z\in E\}\) and \((\sin_\mathbb{T}^2)^{-1}(E)=\{x\in\mathbb{T}_{\boldsymbol{C}}^d\) \(\mid\) \(z\in E\}\). We denote by \(\mathcal{C}^\infty_c(E)\) the space of real \(\mathcal{C}^\infty\) functions defined in \(E\) with compact support in \(E\). For \(m\ge 1\) we denote by \((e_1,\ldots,e_m)\) the standard basis of \({\boldsymbol{R}}^m\) or of \({\boldsymbol{C}}^m\). The spaces \({\boldsymbol{R}}^m\), \({\boldsymbol{C}}^m\) are equipped with the inner product \[\begin{align} <f,g>_{{\boldsymbol{R}}^m} & := \sum_{j=1}^m f_jg_j, \quad f=(f_j)_{1\le j \le m},\; g=(g_j)_{1\le j \le m}, \\ <f,g>_{{\boldsymbol{C}}^m} & := \sum_{j=1}^m f_j\overline{g_j}, \quad f=(f_j)_{1\le j \le m},\; g=(g_j)_{1\le j \le m}. \end{align}\] (The formula for \(<f,g>_{{\boldsymbol{R}}^m}\) extends to the case where \(f_j\) and \(g_j\) are complex valued.) If \(E\subset {\boldsymbol{R}}^m\) (respect., \(E\subset {\boldsymbol{C}}^m\)) then we denote \(E^\perp:=\{g\in {\boldsymbol{R}}^m\) \(\mid\) \(<f,g>_{{\boldsymbol{R}}^m}=0\), \(\forall f\in E\}\) (respect., \(E^{\perp_{\boldsymbol{C}}}:=\{g\in {\boldsymbol{C}}^m\) \(\mid\) \(<f,g>_{{\boldsymbol{C}}^m}=0\), \(\forall f\in E\}\)). The Besov space \(\mathcal{B}_0^*(\mathbb{T}^d;{\boldsymbol{C}}^m)\), simply denoted by \(\mathcal{B}_0^*(\mathbb{T}^d)\), is the space of tempered distributions \(u\in\mathcal{S}'(\mathbb{T}^d;{\boldsymbol{C}}^m)\) such that \(\hat{u}\in \mathcal{B}_0^*({\boldsymbol{Z}}^d)\).
If \(\mathcal{A}\) is a \(n\times p\) matrix with real coefficients we denote \({\rm Im}\mathcal{A}:=\mathcal{A}({\boldsymbol{R}}^p) \subset {\boldsymbol{R}}^n\), \({\rm Im}_{\boldsymbol{C}}\mathcal{A}:=\mathcal{A}({\boldsymbol{C}}^p)\subset{\boldsymbol{C}}^n\), \(r= {\rm rank}(\mathcal{A}):= {\rm dim}({\rm Im}\mathcal{A})\).
Let \({\boldsymbol{Z}}^3=\{n=(n_1,n_2,n_3)\) \(\mid\) \(n_j\in {\boldsymbol{Z}}\}\), \(\mathbb{T}^3\approx (\mathbf{R}/(2\pi {\boldsymbol{Z}}))^3\) and \(U\) be the unitary transform between \(L^2(\mathbb{T}^3)\) and \(l^2({\boldsymbol{Z}}^3)\): \[(Uf)(n) := \hat{f} (n) = (2\pi)^{-\frac{3}{2}} \int_{\mathbb{T}^3} e^{inx} f(x) \, {\rm d}x, \quad n\in{\boldsymbol{Z}}^3,\] so that any \(f\in L^2(\mathbb{T}^3)\) can be written \[f(x) = (U^* \hat{f})(x) := (2\pi)^{-\frac{3}{2}}\sum_{n\in{\boldsymbol{Z}}^3} e^{-inx} \hat{f}(n), \quad x\in\mathbb{T}^3.\] The isotropic discrete Maxwell operator is the bounded operator \(H_0\) on \(\mathcal{H}=(L^2(\mathbb{T}^3;{\boldsymbol{C}}))^6\) defined by \[\hat{H}_0 = UH_0U^* ,\] where \(H_0\) is the operator of multiplication by the real symmetric \(6\times 6\) matrix: \[\label{def46matH0} H_0(x) := \left(\begin{array}{cc} 0_{3\times3} & \tilde{M}(y)\\ -\tilde{M}(y) & 0_{3\times3} \end{array}\right) \in {\boldsymbol{R}}^6, \quad x\in\mathbb{T}^3,\tag{9}\] where \(y=\sin x\) (see 8 ) and \(\tilde{M}(y)\) is the real anti-symmetric \(3\times 3\) matrix: \[\tilde{M}(y) := \left(\begin{array}{ccc} 0 & -y_3 & y_2\\ y_3 &0 & -y_1\\ -y_2 & y_1 & 0 \end{array}\right) , \quad y\in {\boldsymbol{R}}^3.\] Then \(H_0\) is a bounded self-adjoint operator on \(\mathcal{H}= L^2(\mathbb{T}^3,{\rm d}x;{\boldsymbol{C}}^6)\) with the inner product \[(u,v) := \int_{\mathbb{T}^3} <u(x),v(x)>_{{\boldsymbol{C}}^6} \,{\rm d}x = \int_{\mathbb{T}^3} \sum_{j=1}^6 u_j(x)\overline{v_j(x)}\, {\rm d}x .\] The anisotropic discrete-Maxwell operator is defined by \[\hat{H}^{D}:= \hat{D} \hat{H}_0,\] with \[\hat{D} := \left(\begin{array}{cc} \boldsymbol{\varepsilon}& 0_{3\times3} \\ 0_{3\times3} &\boldsymbol{\mu} \end{array}\right).\] We set \(H^D :=U^*\hat{H}^{D}U\) so, since \(\hat{D}\) is constant, \[H^D = U^*(\hat{D} \hat{H}_0)U = \hat{D} U^* \hat{H}_0 U = \hat{D} H_0.\] The relation \[(\hat{D}^{-1} H^D u,v)_{\mathcal{H}}= (H_0 u,v)_{\mathcal{H}}\] shows that the operator \(H^D\) is bounded and self-adjoint on the space \(\mathcal{H}^{D}=\mathcal{H}\) equipped with the hilbertian product \[(u,v)_{\mathcal{H}^{D}} := (\hat{D}^{-1}u,v) = \int_{\mathbb{T}^3} <\hat{D}^{-1}u(x),v(x)>_{{\boldsymbol{C}}^6} \,{\rm d}x.\] We write \[H^D = \int_{\mathbb{T}^3}^{\oplus} H^D(x) \,{\rm d}x,\] where \(H^D(x)\) is self-adjoint on \({\boldsymbol{C}}^6\) equipped with the inner product \[<u(x),v(x)>_{{\boldsymbol{C}}^6,D} := <\hat{D}^{-1}u(x),v(x)>_{{\boldsymbol{C}}^6}.\]
Since \(H^D(x)\) depends only on the variable \(y=\sin x\) we write \(H^D(x)=h^D(y)\). In this subsection we consider a more general case \(y\in {\boldsymbol{R}}^3\), and let \(z_j=y_j^2\), \(z\in [0,+\infty)^3\). We recall some results in [4] using the notations \(\boldsymbol{\beta}\), \(\boldsymbol{\alpha}\) in 4 . Since \(\boldsymbol{\beta}\cdot \boldsymbol{\varepsilon}=0\) and \(\varepsilon_i>0\) for all \(1 \leq i \leq 3\) then there exists \(1 \leq j\leq 3\) such that \(\beta_j \beta_i\le 0\) and \(\beta_k \beta_i\ge 0\) for \(i,k\neq j\). If two of the \(\beta_j\)’s vanish then \(\boldsymbol{\beta}\) vanishes. Moreover \(\boldsymbol{\beta}\) is replaced by \(-\boldsymbol{\beta}\) if \(\boldsymbol{\varepsilon}\) and \(\boldsymbol{\mu}\) are exchanged, which involves the same analysis. Hence if \(\boldsymbol{\beta}\neq0\) then we can assume without loss of generality that
(A0): \(\beta_1\ge \beta_2 > 0 > \beta_3\) or \(\beta_1> \beta_2 = 0 > \beta_3\).
Let us observe that \(\tau^{\pm}(z)\) defined by 7 satisfy \[\tau^+(z) \ge \tau^-(z) >0 \quad z\in [0,+\infty)^3\setminus\{0_{{\boldsymbol{R}}^3}\}.\]
****Proposition** 6**. (Spectrum of \(h^D(y)\).) If \(y=0_{{\boldsymbol{R}}^3}\) then \(h^D(y)=0_{6\times 6}\).
Let \(y\in {\boldsymbol{R}}^3\setminus\{0_{{\boldsymbol{R}}^3}\}\). Then \(0\in \sigma(h^D(y))\) has multiplicity two with eigenvectors \((y_1,y_2,y_3,0,0,0)=
y\otimes 0_{{\boldsymbol{C}}^3}\) and \((0,0,0,y_1,y_2,y_3)= 0_{{\boldsymbol{C}}^3}\otimes y\).
1) Assume \(\boldsymbol{\beta}=0\) and \(y\in {\boldsymbol{R}}^3 \setminus\{0_{{\boldsymbol{R}}^3}\}\). Then, \(K_0\equiv 0\) and all the eigenvalues have
multiplicity two. Moreover, the nonzero eigenvalues of \(h^D(y)\) are \[\pm\sqrt{\Psi_0(z)}=\pm \sqrt{\varepsilon_2\mu_3 z_1 + \varepsilon_3\mu_1 z_2+\varepsilon_1\mu_2 z_3},\] where \(z_j:=y_j^2\) hence \(z\in [0,\infty)^3 \setminus\{0_{{\boldsymbol{R}}^3}\}\).
2) Assume \(\boldsymbol{\beta}\neq0\) (so (A0) holds) and \(y\in {\boldsymbol{R}}^3 \setminus\{0_{{\boldsymbol{R}}^3}\}\). Then the nonzero eigenvalues of \(h^D(y)\) are
\(\sqrt{\tau^+(z)}\) and \(-\sqrt{\tau^+(z)}\), simple iff \(K_0(z)\neq 0\),
\(\sqrt{\tau^-(z)}\) and \(-\sqrt{\tau^-(z)}\), simple iff \(K_0(z)\neq 0\).
\(\sqrt{\tau^+(z)}=\sqrt{\tau^-(z)}\) and \(-\sqrt{\tau^+(z)}=-\sqrt{\tau^-(z)}\), double iff \(K_0(z)= 0\).
If, in addition, \(\beta_2=0\), then, \(\tau^+\) and \(\tau^-\) are linear with respect to \(z\): \[\begin{align} \tau^+(z) &=& \varepsilon_2\mu_3 z_1+ \varepsilon_3\mu_1 z_2 + \varepsilon_2\mu_1 z_3,\\ \tau^-(z) &=& \varepsilon_3\mu_2 z_1+ \varepsilon_3\mu_1 z_2 + \varepsilon_1\mu_2 z_3. \end{align}\]
By Proposition 6, we observe that: If \((z_1,z_3)\neq 0_{{\boldsymbol{R}}^2}\), then the nonzero eigenvalues of \(h^D(y)\) are \(\pm\sqrt{\tau^+(z)}\), simple. If \((z_1,z_3)= 0_{{\boldsymbol{R}}^2}\) and \(z_2\neq 0\), then the nonzero eigenvalues of \(h^D(y)\) are \[\pm\sqrt{\tau^+(z)}=\pm\sqrt{\tau^-(z)}=\pm \sqrt{\alpha_2} |y_2| =\pm \sqrt{\varepsilon_3\mu_1} |y_2| ,\]
We set \[\lambda_\pm := \max_{[0,1]^3} \sqrt{\tau^\pm(z)} .\] The following result is proved in [4].
****Proposition** 7**. (1) The spectrum of \(H^D\) is \(\sigma(H^D) = [-\lambda_+,\lambda_+]\).
(2) The pure point spectrum is \(\sigma_{pp}(\hat{H}^{D})=\{0\}\) and the eigenvalue \(0\) of \(\hat{H}^{D}\) has infinite multiplicity.
(3) The singular continuous spectrum of \(\hat{H}^{D}\) is empty.
****Remark** 5**. We have, in addition,
(1) \(\lambda_+ =\sqrt{\tau^+(1,1,1)}\), \(\lambda_- = \max \{\sqrt{\tau^-(1,1,1)},\sqrt{\tau^-(1,1,0)}\}\).
(2) If \(\boldsymbol{\beta}\neq0\) then \(\lambda_+ >\lambda_-\).
We apply to 3 the following general result.
****Theorem** 8**. Let \(m\ge 1\), \(Q\) a scalar trigonometric polynomial of \(x\in \mathbb{T}^d\), \(d\ge 2\)
satisfying \(\overline{Q(x)}=Q(x)\), \(x\in \mathbb{T}^d\). Consider the complex \(d\)-dimensional torus \(\mathbb{T}^d_{\boldsymbol{C}}\), extend analytically \(Q\) to \(\mathbb{T}^d_{\boldsymbol{C}}\), and set \[\begin{align}
M^{\boldsymbol{C}}(Q) &:=& \{x\in \mathbb{T}^d_{\boldsymbol{C}} \; \mid \; Q(x) =0\}, \\ M^{{\boldsymbol{C}}}_{sgn}(Q) &:=& \{x\in \mathbb{T}^d_{\boldsymbol{C}}\; \mid \; Q(x) =0, \; \nabla Q(x)=0\}, \\ M^{{\boldsymbol{C}}}_{reg}(Q)
&:=& M^{{\boldsymbol{C}}}(Q) \setminus M^{{\boldsymbol{C}}}_{sgn}(Q).
\end{align}\] Assume
(A-1-1)\((Q)\): \(M^{{\boldsymbol{C}}}_{sgn}(Q)\) (\(\subset \mathbb{T}_{\boldsymbol{C}}^d \approx(\mathbb{T}\times{\boldsymbol{R}})^d\))
has Hausdorff (\(2d-1\))-measure zero,
(A-1-2)\((Q)\): \(M^{{\boldsymbol{C}}}_{sgn}(Q)\cap \mathbb{T}^d\) is discrete,
(A-2)\((Q)\): Each connected component of \(M^{{\boldsymbol{C}}}_{reg}(Q)\) intersects \(\mathbb{T}^d\) and the intersection is a \((d-1)\)-dimensional real analytic submanifold of \(\mathbb{T}^d\).
Let \(u\in \mathcal{B}_0^*(\mathbb{T}^d)\) be a solution of the equation \[\label{e246Qu61g} Qu = g \quad {\rm on} \quad \mathbb{T}^d ,\qquad{(1)}\] where \(g\) is a trigonometric polynomial of \(x\in \mathbb{T}^d\) with values in \({\boldsymbol{C}}^m\). Then \(u\) is a trigonometric polynomial.
Proof. Lemma 5.2 of [3] shows that \(u\in \mathcal{C}^\infty(\mathbb{T}^d\setminus M^{{\boldsymbol{C}}}_{sgn}(Q))\) and \(g=0\) on \(M^{{\boldsymbol{C}}}_{reg}(Q)\cap \mathbb{T}^d\). Lemma 5.3 of [3] shows that \(g=0\) on \(M^{{\boldsymbol{C}}}_{reg}(Q)\). Let us define the meromorphic function \(v(x):=g(x)/p(x)\) for \(x\in \mathbb{T}^d_{\boldsymbol{C}}\setminus M^{\boldsymbol{C}}(Q)\). The proof of [3]) shows that \(v\) extends continuously to \(\mathbb{T}^d_{\boldsymbol{C}}\). In fact this proof is based on Assumption (A-1)‘(\(Q\)) (i.e., \(M^{{\boldsymbol{C}}}_{sgn}(Q)\) is discrete) which is stronger than (A-1-1)-(A-1-2)(\(Q\)). However [3], which does not use Assumption (A-1)’(\(Q\)), implies that \(v\) is analytic near \(M^{{\boldsymbol{C}}}_{reg}(Q)\) and the singularities of \(v\) are localized in \(M^{{\boldsymbol{C}}}_{sgn}(Q)\). Thanks to (A-1-1), the closed set \(M^{{\boldsymbol{C}}}_{sgn}(Q)\) is negligible in the sense of Shiffman in [5] and [6], which means that \(v\) extends analytically to \(\mathbb{T}^d_{\boldsymbol{C}}\) (see also [7] and [8]).
Let us set \(u'=u-v|_{\mathbb{T}^d}\). Thus, \(u'\) belongs to \(\mathcal{B}_0^*(\mathbb{T}^d)\) and satisfies \({\rm supp}\,u'\subset \mathbb{T}^d\cap M^{{\boldsymbol{C}}}_{sgn}(Q)\). Therefore, thanks to Assumption (A-1-2), \({\rm supp}\,u'\) is discrete. Let us prove \(u'=0\). In fact, if \(u'\neq 0\), then there exists \(x^*\in {\rm supp}\,u'\). Let \(\chi\) be a smooth cut-off function on \(\mathbb{T}^d\) such that \({\rm supp}\,(\chi u')\subset \{x^*\}\) and \(\chi(x^*)\neq 0\). We then use the following equivalent characterization of \(\mathcal{B}_0^*(\mathbb{T}^d)\) (see [2]). Take a \(\mathcal{C}^\infty\)-partition of unity \(\{\chi_l\}_{1\le l\le L}\) on \(\mathbb{T}^d\) where the support of \(\chi_l\) is sufficiently small. Then \[\mathcal{B}_0^*(\mathbb{T}^d;{\boldsymbol{C}}^m) := \{w\in \mathcal{S}'(\mathbb{T}^d;{\boldsymbol{C}}^m)\; \mid \; \chi_l w \in\mathcal{B}_0^*({\boldsymbol{R}}^d;{\boldsymbol{C}}^m) \} ,\] where \[\mathcal{B}_0^*({\boldsymbol{R}}^d;{\boldsymbol{C}}^m) := \{w\in \mathcal{S}'({\boldsymbol{R}}^d;{\boldsymbol{C}}^m)\; \mid \; \lim_{R\to +\infty} \frac{1}{R} \int_{|\xi|<R} |w(\xi)|^2 d\xi =0 \}.\] Let \(l\) be such that \(\chi_l(x^*)\neq 0\). Thus \(u'':=\chi_l\chi u'\) belongs to \(\mathcal{B}_0^*({\boldsymbol{R}}^d;{\boldsymbol{C}}^m)\) and is a finite linear combination of derivatives of Dirac distribution \(\delta_{x^*}\): \[u''= \sum_{|\alpha|\le N} b_\alpha \partial_\alpha \delta_{x^*},\] where \(N\ge 0\) and \(\{b_{\alpha}\}_{|\alpha|=N}\neq \{0\}\). Let \(\tilde{u}''_N(\xi)\) be the Fourier transform of \(u''_N:=\) \(\sum_{|\alpha|= N} b_\alpha \partial^\alpha \delta_{x^*}\). It satisfies \[\tilde{u}''_N(\xi) := (2\pi)^{-d/2}\sum_{|\alpha|= N} b_\alpha \xi^\alpha e^{i\xi x^*}, \quad \xi\in{\boldsymbol{R}}^d .\] Putting \(\xi=|\xi|\omega\) with \(\omega\in {\mathbb{S}}^{d-1}\), we get \[\begin{align} \frac{1}{R} \int_{|\xi|<R} |\tilde{u}''_N(\xi)|^2 \, {\rm d}\xi &=& (2\pi)^{-d/2} \frac{1}{R} \int_{0}^R \Big(\int_{{\mathbb{S}}^{d-1}} |\sum_{|\alpha|= N} b_\alpha \omega^\alpha |^2 \, {\rm d}\omega\Big) t^{2N+d-1} \, {\rm d}t \\ &=& (2\pi)^{-d/2} R^{2N+d-2} \int_{{\mathbb{S}}^{d-1}} |\sum_{|\alpha|= N} b_\alpha \omega^\alpha |^2 \,{\rm d}\omega\\ &=& CR^{2N+d-2} \end{align}\] with \(C>0\), since \(\{b_\alpha\}_{|\alpha|=N}\neq \{0\}\). A similar calculation shows that \[\frac{1}{R} \int_{|\xi|<R} |\tilde{u}''(\xi)- \tilde{u}''_N(\xi)|^2 \, {\rm d}\xi =O(R^{2N+d-4}).\] We thus have \[\frac{1}{R} \int_{|\xi|<R} |\tilde{u}''(\xi)|^2 \, {\rm d}\xi \approx R^{2N+d-2}.\] It contradicts \(u'' \in \mathcal{B}_0^*({\boldsymbol{R}}^d;{\boldsymbol{C}}^m)\). Hence \(u'=0\) and \(u=v|_{\mathbb{T}^d}\) is a trigonometric polynomial. ◻
****Remark** 6**. 1) Theorem 8 is a slight improvement of results in [3] and [2], where Assumption (A-1-1)-(A-1-2) is replaced by: (A-1)’(\(Q\)): \(M^{{\boldsymbol{C}}}_{sgn}(Q)\) is discrete.
2) Theorem 1 for the second case follows from Conditions (A-1-1)-(A-1-2) by taking \(Q\) to be \(q(\cdot;\lambda)\) in Theorem 8 but not from Condition (A-1)’(\(q(\cdot;\lambda)\)) of [3] and [2], which is not fullfield.
We consider \(\lambda\in {\boldsymbol{R}}^*\). Thanks to 5 we write \[\label{form46pz} p(z;\lambda)/\lambda^2 = Az\cdot z + \boldsymbol{b}\cdot z +c \quad z\in{\boldsymbol{C}}^3,\tag{10}\] where we put \[\label{def46A} A := \left( \begin{array}{ccc} \gamma_1 & \varepsilon_3\mu_3 \alpha_3 & \varepsilon_2\mu_2 \alpha_2 \\ \varepsilon_3\mu_3 \alpha_3 & \gamma_2 & \varepsilon_1\mu_1 \alpha_1 \\ \varepsilon_2\mu_2 \alpha_2 & \varepsilon_1\mu_1 \alpha_1 & \gamma_3 \end{array} \right), \quad \boldsymbol{b}:= - 2\lambda^2 \boldsymbol{\alpha}, \quad c := \lambda^4,\tag{11}\] and \(\gamma_i\) is defined by 4 .
We state analytic properties of the variety defined by the polynomial \(p(\cdot ;\lambda)\) in a slightly general form. We consider a polynomial of the form \(P(z)= \mathcal{A}z \cdot z+\boldsymbol{b}\cdot z+c\), \(z\in {\boldsymbol{C}}^d\), where \(\mathcal{A}\) is a real symmetric \(d\times d\) matrix, \(\boldsymbol{b}\in {\boldsymbol{R}}^d\), \(c\in\mathbf{R}\). We set the complex analytic variety \(M^{\boldsymbol{C}}(P) :=\{z\in{\boldsymbol{C}}^d\) \(\mid\) \(P(z)=0\}\), \(M^{\boldsymbol{C}}_{sgn}(P) :=\{z\in M^{\boldsymbol{C}}(P)\) \(\mid\) \(\nabla P(z)=0\}\) the singular part and \(M^{\boldsymbol{C}}_{reg}(P):=M^{\boldsymbol{C}}(P)\setminus M^{\boldsymbol{C}}_{sgn}(P)\) the regular part.
Lemma 1. (1) Assume \(\mathcal{A}=0\), \(\boldsymbol{b}\neq 0\). Then \(M^{\boldsymbol{C}}_{sgn}(P) = \emptyset\) and \(M^{\boldsymbol{C}}_{reg}(P)=M^{\boldsymbol{C}}(P)\) is a \((d-1)\)-dimensional connected analytic submanifold of \({\boldsymbol{C}}^d\). In addition, \(M^{\boldsymbol{C}}_{reg}(P)\cap{\boldsymbol{R}}^d\) is a \((d-1)\)-dimensional real analytic submanifold of \({\boldsymbol{R}}^d\).
(2) Assume \(\mathcal{A}\neq0\). Then:
(2-1) Assume \(\boldsymbol{b}\not\in {\rm Im}\mathcal{A}\). Then, \(M^{\boldsymbol{C}}_{sgn}(P)=\emptyset\), \(M^{\boldsymbol{C}}(P)=M^{\boldsymbol{C}}_{reg}(P)\) is an analytic connected manifold of dimension \(d-1\) and the intersection \(M^{\boldsymbol{C}}_{reg}(P)\cap{\boldsymbol{R}}^d\) is a \((d-1)\)-dimensional real analytic submanifold of \({\boldsymbol{R}}^d\).
(2-2) Assume \(\boldsymbol{b}\in {\rm Im}\mathcal{A}\). Then there exists a unique vector \(z^0\in{\rm Im}_{\boldsymbol{C}} \mathcal{A}\) such that \(\mathcal{A}z^0= -\boldsymbol{b}/2\). In addition, \(z^0\in {\boldsymbol{R}}^d\). Furthermore, setting \(c^*:= \mathcal{A}z^0 \cdot z^0 = -\boldsymbol{b}z^0/2\)
and \(c':=c-c^*\), the following assertions (2-2-1) and (2-2-2) hold.
(2-2-1) Assume \(c \neq c^*\). Then \(M^{\boldsymbol{C}}_{sgn}(P)\) is void and \(M^{\boldsymbol{C}}(P)=M^{\boldsymbol{C}}_{reg}(P)\). Moreover,
Assume \(0<s=r\) and \(c > c^*\), or \(0=s<r\) and \(c < c^*\). Then, \(M^{\boldsymbol{C}}(P)\) does not intersect \({\boldsymbol{R}}^d\).
*Assume \(0\le s<r\) and \(c > c^*\), or \(0<s\le r\) and \(c < c^*\).
(2-2-2) Assume \(c = c^*\). Then, \(M^{\boldsymbol{C}}_{sgn}(P)\) is a \((d-r)\)-dimensional complex analytic submanifold of \({\boldsymbol{C}}^d\) and the intersection \(M^{\boldsymbol{C}}_{sgn}(P)\cap{\boldsymbol{R}}^d\) is a \((d-r)\)-dimensional real analytic submanifold of \({\boldsymbol{R}}^d\). Moreover,
- Assume \(r=d\). Then \(M^{\boldsymbol{C}}_{sgn}(P)=\{0_{{\boldsymbol{C}}^d}\}\) and \(M^{\boldsymbol{C}}_{reg}(P)\) is a connected \((d-1)\)-dimensional submanifold of \({\boldsymbol{C}}^d\).
Assume \(s=0\) (i.e. \(\mathcal{A}<0\)) or \(s=r\) (i.e. \(\mathcal{A}>0\)). Then \(M^{\boldsymbol{C}}_{reg}(P)\cap{\boldsymbol{R}}^d=\emptyset\).
Assume \(1 \leq s \leq d-1\) (and \(r=d\)). Then \(M^{\boldsymbol{C}}_{reg}(P)\cap{\boldsymbol{R}}^d\) is a \((d-1)\)-dimensional real analytic submanifold of \({\boldsymbol{R}}^d\).
- Assume \(r=d-1\).
Assume \(s=0\) (i.e. \(\mathcal{A}\le 0\)) or \(s=r\) (i.e. \(\mathcal{A}\ge0\)). Then \(M^{\boldsymbol{C}}_{reg}(P)\cap{\boldsymbol{R}}^d = \emptyset\).
Assume \(1 \leq s\leq d-1\) (and \(r=d-1\)). If \(d>3\) then \(M^{\boldsymbol{C}}_{reg}(P)\) is a connected \((d-1)\)-dimensional submanifold of \({\boldsymbol{C}}^d\) and \(M^{\boldsymbol{C}}_{reg}(P)\cap{\boldsymbol{R}}^d\) is a \((d-1)\)-dimensional real analytic submanifold of \({\boldsymbol{R}}^d\). If \(d=3\) then \(M^{\boldsymbol{C}}_{reg}(P)\) admits two connected components: \(M^{\boldsymbol{C}}_{reg}(P)=M^{\boldsymbol{C}}(P^+)' \cup M^{\boldsymbol{C}}(P^-)'\), where each \(M^{\boldsymbol{C}}(P^\pm)'\) is a two-dimensional submanifold of \({\boldsymbol{C}}^3\) and each \(M^{\boldsymbol{C}}(P^\pm)'\cap{\boldsymbol{R}}^3\) is a two-dimensional real analytic submanifold of \({\boldsymbol{R}}^3\).
Proof. (1) (Case \(\mathcal{A}=0\), \(\boldsymbol{b}\neq0\).) This is obvious.
(2) (Case \(\mathcal{A}\neq0\).) We observe that if it is not void, then \(M^{\boldsymbol{C}}_{reg}(P)\) is an analytic manifold of dimension \(d-1\). If
\(M^{\boldsymbol{C}}_{sgn}(P)\neq\emptyset\), then \(M^{\boldsymbol{C}}_{sgn}(P)\) is a smooth manifold, since its singular part, \[M^{\boldsymbol{C}}_{sgn,sgn}(P)=\{z\in {\boldsymbol{C}}^d\; \mid \; P(z)=0, \; \nabla P(z)=0, \; D^2 P(z)=0\}\] is necessarily void (we have \(z\in M^{\boldsymbol{C}}_{sgn,sgn}(P)\) iff \(\mathcal{A}=0\), \(b=0\), \(c=0\), which is forbidden). Moreover, \(M^{\boldsymbol{C}}_{sgn}(P)\) is characterized by the
equations \(\mathcal{A}z=-\boldsymbol{b}/2,\quad \boldsymbol{b}\cdot z/2+c=0\). Let \((s,r-s)\) be the signature of the quadratic form \(z\mapsto \mathcal{A}z\cdot
z\). Gauss’s reduction provides a real linear change of coordinates on \(z\), so we can assume that \(\mathcal{A}\) is diagonal and \(\mathcal{A}z
\cdot z=\sum_{j=1}^{s} z_j^2 - \sum_{j=s+1}^{r} z_j^2\). The equation of \(M^{\boldsymbol{C}}(P)\) is then \[\label{e46pz39610} M^{\boldsymbol{C}}(P):
\quad \sum_{j=1}^d a_j z_j^2 + \sum_{j=1}^d b_j z_j + c=0,\tag{12}\] with \(a_j=1\) if \(j\le s\), \(a_j=-1\) if \(s<j\le r\), \(a_j=0\) if \(j>r\). The equations of \(M^{\boldsymbol{C}}_{sgn}\) are: \[M^{\boldsymbol{C}}_{sgn}(P):\quad \sum_{j=1}^d a_j z_j^2 + \sum_{j=1}^d b_j z_j + c=0, \quad {\rm and} \quad 2a_j z_j+ b_j=0 \quad \forall j .\] (2-1) (Case \(\boldsymbol{b}\not\in {\rm
Im}\mathcal{A}\).) Hence, \(r<d\) and there exists \(j_0>r\) such that \(b_{j_0}\neq 0\). Thus \(M^{\boldsymbol{C}}_{sgn}(P)=\emptyset\) and \(M^{\boldsymbol{C}}(P)=M^{\boldsymbol{C}}_{reg}(P)\) is an analytic manifold of dimension \(d-1\). Moreover the
equation of \(M^{\boldsymbol{C}}(P)\) becomes \[z_{j_0} = \frac{-1}{b_{j_0}}(\sum_{j=1}^r z_j^2 + \sum_{j=r+1, \; j\neq j_0}^d b_j z_j + c),\] which shows that \(M^{\boldsymbol{C}}(P)\) is connected and that \(M^{\boldsymbol{C}}(P)\cap{\boldsymbol{R}}^d\) has dimension \(d-1\). This proves the assertion.
(2-2) (Case \(\boldsymbol{b}\in {\rm Im}\mathcal{A}\).) We denote \({\rm Im}_{\boldsymbol{C}} \mathcal{A}=\{\mathcal{A}z\) \(\mid\) \(z\in {\boldsymbol{C}}^d\}\), \(\ker_{\boldsymbol{C}} \mathcal{A}=\{z\in {\boldsymbol{C}}^d\) \(\mid\) \(\mathcal{A}z=0\}\) and \(\ker\mathcal{A}=\ker_{\boldsymbol{C}}\mathcal{A}\cap{\boldsymbol{R}}^d\). We recall that, since \(\mathcal{A}\) is real symmetric, we then have \({\rm Im}\mathcal{A}:=\{\mathcal{A}z\) \(\mid\) \(z\in{\boldsymbol{R}}^d\}=(\ker \mathcal{A})^\perp\), \({\rm Im}_{\boldsymbol{C}}\mathcal{A}=(\ker_{\boldsymbol{C}} \mathcal{A})^{\perp_{\boldsymbol{C}}}\), and \[{\boldsymbol{R}}^d={\rm Im}\mathcal{A}\oplus \ker \mathcal{A}, \quad {\boldsymbol{C}}^d ={\rm Im}_{\boldsymbol{C}} \mathcal{A}\oplus \ker_{\boldsymbol{C}} \mathcal{A}.\] Hence, \(\mathcal{A}\) is an automorphism on \({\rm Im}_{\boldsymbol{C}} \mathcal{A}\). It implies the existence and uniqueness of a vector \(z^0\in{\rm Im}_{\boldsymbol{C}} \mathcal{A}\) such that \(\mathcal{A}z^0= -\boldsymbol{b}/2\). Since \(\boldsymbol{b}\) is a real vector so is \(z^0\). Set \(c^*:= \mathcal{A}z^0 \cdot z^0 = -\boldsymbol{b}z^0/2\), \(c':=c-c^*\), and consider the translation \(z':=z-z^0\). Then 12 becomes \[M^{\boldsymbol{C}}(P):\quad \sum_{j=1}^{s} {z'_j}^2 - \sum_{j=s+1}^{r} {z'_j}^2 + c'=0,\] and the equations of \(M^{\boldsymbol{C}}_{sgn}(P)\) become \[M^{\boldsymbol{C}}_{sgn}(P):\quad \sum_{j=1}^{s} {z'_j}^2 - \sum_{j=s+1}^{r} {z'_j}^2 + c'=0 \quad {\rm and} \;\; z'_j=0 \;\; {\rm for}\;\; 1\le j\le r,\] i.e., \[M^{\boldsymbol{C}}_{sgn}(P):\quad c'=0 \quad {\rm and} \quad z'_j=0 \;\; {\rm for}\;\;1\le j\le r.\] (2-2-1) (Case \(c \neq c^*\), i.e., \(c'\neq0\).) We assume \(c > c^*\), the other case being similar. Then, obviously, \(M^{\boldsymbol{C}}_{sgn}(P)\) is void and \(M^{\boldsymbol{C}}(P)=M^{\boldsymbol{C}}_{reg}(P)\).
Assume \(0<s=r\), i.e., \(\mathcal{A}\ge 0\). Obviously, \(M^{\boldsymbol{C}}(P)\) does not intersect \({\boldsymbol{R}}^d\).
Assume \(0\le s<r\).
(2-2-2) (Case \(c = c^*\), i.e., \(c'= 0\).) Then, \(M^{\boldsymbol{C}}_{sgn}(P)=\{z\in{\boldsymbol{C}}^d\) \(\mid\)
\(z'_1=\ldots=z'_r=0\}\) and \(M^{\boldsymbol{C}}_{sgn}(P)\cap{\boldsymbol{R}}^d=\{z\in{\boldsymbol{R}}^d\) \(\mid\) \(z'_1=\ldots=z'_r=0\}\) are affine spaces with same dimension \(d-r\).
- Assume \(r=d\). Then \(M^{\boldsymbol{C}}_{sgn}(P)=\{0_{{\boldsymbol{C}}^d}\}\). Since \(d\ge 3\) then \(M^{\boldsymbol{C}}_{reg}(P)\) is a irreducible and regular algebric variety and so it is a \((d-1)\)-dimensional connected submanifold of \({\boldsymbol{C}}^d\).
Clearly, \(M^{\boldsymbol{C}}_{reg}(P)\cap{\boldsymbol{R}}^d\) is a regular subset of \({\boldsymbol{R}}^d\), but may be void.
Assume \(s=0\) (i.e., \(\mathcal{A}<0\)) or \(s=r\) (i.e., \(\mathcal{A}>0\)). Then \(M^{\boldsymbol{C}}_{reg}(P)\cap{\boldsymbol{R}}^d=\emptyset\).
Assume \(1\le s\le d-1\) (and \(r=d\)). Then \(M^{\boldsymbol{C}}_{reg}(P)\) intersects \({\boldsymbol{R}}^d\) since it contains the points \(z\in{\boldsymbol{R}}^d\) such that \(z'_d= \pm \sqrt{\sum_{j=1}^{s} {z'_j}^2 - \sum_{j=s+1}^{d-1} {z'_j}^2}\) and \(|z'_1| > \sqrt{\sum_{j=s+1}^{d-1} {z'_j}^2}\). Hence \(M^{\boldsymbol{C}}_{reg}(P)\cap{\boldsymbol{R}}^d\) is a \((d-1)\)-dimensional real analytic submanifold of \({\boldsymbol{R}}^d\).
- Assume \(r=d-1\).
Case \(s=0\) or \(s=r\) (i.e., \(\mathcal{A}\ge0\)): the result is obvious.
Assume \(1\le s\le d-1\) (and \(r=d-1\)).
– Assume \(d>3\). The proof is similar to the proof of case (ii)(B)(\(r=d\)).
– Assume \(d=3\) so \(r=2\), \(s=1\). We have \(M^{\boldsymbol{C}}_{reg}(P)=M^{\boldsymbol{C}}_+(P)'\cup
M^{\boldsymbol{C}}_-(P)'\) with \(M^{\boldsymbol{C}}(P^\pm)' : = M^{\boldsymbol{C}}(P^\pm) \setminus M^{\boldsymbol{C}}_{sgn}(P)\), \(M^{\boldsymbol{C}}(P^\pm) : =
\{z\in{\boldsymbol{C}}^3\) \(\mid\) \(z'_1 = \pm z'_2 \}\). Hence \(M^{\boldsymbol{C}}(P^\pm)' = \{z\in{\boldsymbol{C}}^3\) \(\mid\) \(z'_1 = \pm z'_2\) and \(z'_2\neq 0\}\). Clearly, \(M^{\boldsymbol{C}}(P^+)'\) and \(M^{\boldsymbol{C}}(P^-)'\) are the connected composants of the complex variety \(M^{\boldsymbol{C}}(P)\) and each intersection \(M^{\boldsymbol{C}}(P^\pm)'\) with \({\boldsymbol{R}}^3\) is a two-dimensional open set of the plane \(\{z\in{\boldsymbol{R}}^3\) \(\mid\) \(z'_1 = \pm z'_2\}\).
We have thus completed the proof of Lemma 1. ◻
We put \[\label{def46g} \boldsymbol{g}:=(\varepsilon_1\mu_1\beta_1,\varepsilon_2\mu_2\beta_2,\varepsilon_3\mu_3\beta_3).\tag{13}\]
Lemma 2. We have the following relation: \[\label{r46alphaImA} \boldsymbol{\alpha}\cdot \boldsymbol{g}= \sum_{i=1}^3\varepsilon_i\mu_i\alpha_i\beta_i = \frac{1}{2} \beta_1\beta_2\beta_3.\qquad{(2)}\]
Proof. We easily observe that the following relations and their cyclic changes hold: \[\label{r46alphabeta} \begin{align} \alpha_1^2-\gamma_1 &=& \frac{1}{4} \beta_1^2,\\ \varepsilon_1\mu_1 \alpha_1 - \alpha_2\alpha_3 &=& \frac{1}{4}\beta_2\beta_3,\\ \alpha_3\beta_2 + \alpha_2 \beta_3 &=& -\varepsilon_1\mu_1 \beta_1. \end{align}\tag{14}\] Then, we obtain \[\begin{align} \boldsymbol{\alpha}\cdot \boldsymbol{g}&=& \sum_{i=1}^3\varepsilon_i\mu_i\alpha_i\beta_i = -(\alpha_3\beta_2 + \alpha_2 \beta_3)\alpha_1 + \varepsilon_2\mu_2\alpha_2\beta_2 + \varepsilon_3\mu_3\alpha_3\beta_3 \\ &=& (\varepsilon_2\mu_2\alpha_2-\alpha_1\alpha_3)\beta_2 + (\varepsilon_3\mu_3\alpha_3-\alpha_1\alpha_2) \beta_3 = \frac{1}{4} \beta_1\beta_3 \beta_2 + \frac{1}{4}\beta_1\beta_2\beta_3 \\ &=& \frac{1}{2}\beta_1\beta_2\beta_3. \end{align}\] ◻
Lemma 3. Let \(A\) be defined by 11 and \(\boldsymbol{g}\) by 13 .
(1) Let \(\{a_1,a_2,a_3\}\) be the spectrum of \(A\). Then, we have \[\label{r46vpA} \det A=a_1a_2a_3=0, \quad {\rm tr} A = a_1+a_2+a_3= \gamma_1+\gamma_2+\gamma_3>0, \quad\Pi_{i\neq j} a_ia_j = -\boldsymbol{g}^2/4 \le 0.\qquad{(3)}\] (2) We have \(\boldsymbol{\alpha}\in {\rm Im}A\) iff \(\beta_1\beta_2\beta_3=0\).
(3) Assume \(\boldsymbol{\beta}=0\). Then \(A=\boldsymbol{\alpha}\boldsymbol{\alpha}^T\), \({\rm rank}(A) = 1\), \({\rm Im}A={\rm span}(\boldsymbol{\alpha})\) and \(\ker A\) is the complex plane orthogonal to \(\boldsymbol{\alpha}\).
(4) Assume \(\boldsymbol{\beta}\neq 0\). Then \({\rm rank}(A) = 2\) and \(\ker A= {\rm span}(\boldsymbol{g})\). Moreover, \(A\) has one positive eigenvalue and one negative eigenvalue.
(5) Assume \(\beta_1\beta_2\beta_3=0\) (and \(\boldsymbol{\beta}\neq0\)), so we assume (A0) with \(\beta_1>0=\beta_2>\beta_3\). Then there exists a unique vector \(z^*\in{\rm Im}A\) such that \(Az^*=\boldsymbol{\alpha}\). We have \(z^* = (0,1/\sqrt{\gamma_2},0)\) and \[\label{c46alphaz42} \boldsymbol{\alpha}\cdot z^* = 1.\qquad{(4)}\]
Proof. (1) A simple computation shows that \(A\cdot \boldsymbol{g}=0\) and, if \(\boldsymbol{\beta}=0\), \(A\cdot (1,1,1)=0\). Hence \(\det A=0\). Moreover, we have \[\begin{align} \Pi_{i\neq j} a_ia_j &=& \gamma_2\gamma_3 - \varepsilon_1^2\mu_1^2\alpha_1^2 + {\rm c.p.} =
\varepsilon_1^2\mu_1^2(\varepsilon_2\varepsilon_3\mu_2\mu_3-\alpha_1^2) + {\rm c.p.} \\ &=& \varepsilon_1^2\mu_1^2(\varepsilon_2\varepsilon_3\mu_2\mu_3- (\varepsilon_2\mu_3+\varepsilon_3\mu_2)^2/4) + {\rm c.p.} \\ &=&
-\varepsilon_1^2\mu_1^2 \beta_1^2/4 + {\rm c.p.} = -\boldsymbol{g}^2/4 \le 0.
\end{align}\]
(3) From 6 , 5 and 10 we get \(A=\boldsymbol{\alpha}\boldsymbol{\alpha}^T\).
(2) Assume \(\boldsymbol{\beta}= 0\). Since \(A=\boldsymbol{\alpha}\boldsymbol{\alpha}^T\) (see (3)), then \(\boldsymbol{\alpha}\in{\rm Im}A\).
Assume \(\boldsymbol{\beta}\neq 0\). Since \(A\) is real symmetric then \({\rm Im}A=\ker A^T=\boldsymbol{g}^\perp\), so, thanks to ?? , \(\boldsymbol{\alpha}\in {\rm Im}A\iff \boldsymbol{\alpha}\cdot\boldsymbol{g}=0\iff\Pi_{j=1}^3\beta_j=0\).
(4) (Case \(\boldsymbol{\beta}\neq 0\).) Thanks to ?? and since \(\boldsymbol{g}\neq 0\) then \({\rm rank}(A)=2\) and \(\ker A={\rm span}(\boldsymbol{g})\), and (2) follows from ?? .
(5) The proof of Lemma 1 (see case (2-2)) with \((\mathcal{A},\boldsymbol{b},d,r,s)=(A,-2\boldsymbol{\alpha},3,2,1)\) shows the existence and uniqueness of such a \(z^*\). Put \(z = (0,1/\alpha_2,0)\) and observe that \(\alpha_2=\sqrt{\gamma_2}\). Since \(\boldsymbol{g}=(\varepsilon_1\mu_1\beta_1,0,\varepsilon_3\mu_3\beta_3)\), then \(z\perp \boldsymbol{g}\) and so \(z\in {\rm Im}A\). In addition, we have \((Az)^T= (\varepsilon_3\mu_3\alpha_3/\alpha_2, \alpha_2, \varepsilon_1\mu_1\alpha_1/\alpha_2)\). Thanks to 14 , we obtain \(Az=\boldsymbol{\alpha}\). This shows that \(z^*=z\). Finally, ?? is obvious and (5) is proved. ◻
Let \(\lambda\in{\boldsymbol{R}}^*\). We use the notations of Lemma 1 and apply Lemma 1 to \(P:=\frac{1}{\lambda^2}p(\cdot;\lambda)\), so \((\mathcal{A},\boldsymbol{b},c,d)=(A,-2\lambda^2\boldsymbol{\alpha},\lambda^4,3)\). In view of Lemmas 1 and 3 we obtain
Lemma 4. (1) Assume \(\boldsymbol{\beta}=0\). Then \(b\in{\rm Im}A\), \(z^0= \frac{\lambda^2}{|\boldsymbol{\alpha}|^2}\boldsymbol{\alpha}\), \(c^*=c\), and \(M^{\boldsymbol{C}}_{sgn}(P)\) is a two-dimensional complex submanifold of \({\boldsymbol{C}}^3\).
(2) Assume \(\boldsymbol{\beta}\neq0\).
(2-1) Assume \(\Pi_{j=1}^3 \beta_j\neq 0\). Then \(\boldsymbol{b}\not\in {\rm Im}\mathcal{A}\), \(M^{\boldsymbol{C}}_{sgn}(P)=\emptyset\), \(M^{\boldsymbol{C}}(P)=M^{\boldsymbol{C}}_{reg}(P)\) is an analytic connected manifold of dimension two and its intersection with \({\boldsymbol{R}}^3\) is a real analytic manifold of dimension
two.
(2-2) Assume \(\Pi_{j=1}^3 \beta_j= 0\). Then \(\boldsymbol{b}\in {\rm Im}\mathcal{A}\), \(z^0=\lambda^2 z^*\), \(c^*=c\), \(M^{\boldsymbol{C}}_{sgn}(P)\) is a straight line of \({\boldsymbol{C}}^3\), and the intersection \(M^{\boldsymbol{C}}_{sgn}(P)\cap{\boldsymbol{R}}^3\) is a straight line of \({\boldsymbol{R}}^3\). In addition, \(M^{\boldsymbol{C}}_{reg}(P)\) has two connected components, \(M^{\boldsymbol{C}}(P^+)'\) and \(M^{\boldsymbol{C}}(P^-)'\), and each \(M^{\boldsymbol{C}}(P^\pm)'\cap{\boldsymbol{R}}^3\) is a two-dimensional real analytic submanifold of \({\boldsymbol{R}}^3\).
****Remark** 7**. Let us supplement (5) of Lemma 3 and (2-2) of Lemma 4. We assume \(\beta_1>0=\beta_2>\beta_3\). Then each \(\tau^\pm\) is linear and we have \[A(z-z^0)(z-z^0)=\lambda^{-2} p(z;\lambda)=(\tau^+ -\lambda^2)(\tau^- -\lambda^2),\] where \(z^0=\lambda^2 z^*\) is the corresponding point of Lemmas 1 and 3. Set \(P=p(\cdot;\lambda)\) and the linear functions \(P^\pm:=\tau^\pm-\lambda^2\). Then each \(M^{\boldsymbol{C}}_{sgn}(P^\pm)\) is void and, since \(P=\lambda^2 P^+ P^-\), we have \[\label{val46MCsgnP} M^{\boldsymbol{C}}_{sgn}(P)=M^{\boldsymbol{C}}(P^+)\cap M^{\boldsymbol{C}}(P^-) =\{z\in {\boldsymbol{C}}^3 \; \mid \; K_0(z)=\Psi_0(z)-\lambda^2=0\}\qquad{(5)}\] is a complex analytic variety of dimension one. Then, \[M^{\boldsymbol{C}}_{reg}(P)= M^{\boldsymbol{C}}(P^+)'\cup M^{\boldsymbol{C}}(P^-)' ,\] where \(M^{\boldsymbol{C}}(P^\pm)':=M^{\boldsymbol{C}}(P^\pm)\setminus M^{\boldsymbol{C}}_{sgn}(P)\), as in the proof of Lemma 1. Thanks to Lemma 1, \(M^{\boldsymbol{C}}(P^+)'\) and \(M^{\boldsymbol{C}}(P^-)'\) are the two connected component of \(M^{\boldsymbol{C}}_{reg}(P)\); they are two-dimensional submanifolds of \({\boldsymbol{C}}^3\).
Lemma 4 shows that we can’t apply Theorem 8 directly to prove Rellich’s properties if \(\boldsymbol{\beta}=0\) since Assumptions (A-1-1)(\(p(;\lambda)\)) fails.
Lemma 5. Assume \(\boldsymbol{\beta}=0\). Set \(P=\Psi_0-\lambda^2\). Then \(M^{\boldsymbol{C}}_{reg}(P)=M^{\boldsymbol{C}}(P)\) and
\(M^{\boldsymbol{C}}(P)\cap [0,1]^3\) is a closed convex set. In addition, we have
(1) Let \(|\lambda|\in (0,\lambda_+)\). Then \(M^{\boldsymbol{C}}_{reg}(P)\cap (0,1)^3\) has dimension two.
(2) Let \(|\lambda|=\lambda_+\). Then \(M^{\boldsymbol{C}}_{reg}(P)\cap [0,1]^3=\{(1,1,1)\}\).
Proof. Obviously, \(M^{\boldsymbol{C}}_{sgn}(P)=\emptyset\) and \(M^{\boldsymbol{C}}(P)\cap [0,1]^3\) is a closed convex set as intersection of the plane \(\boldsymbol{\alpha}\cdot z=\lambda^2\) of \({\boldsymbol{R}}^3\) and the cube \([0,1]^3\).
(1) Let \(|\lambda|\in (0,\lambda_+)\). Set \(z_\lambda=\frac{|\lambda|}{\lambda_+}(1,1,1)\in (0,1)^3\), so \(\Psi_0(z_\lambda)=\tau^\pm(z_\lambda)=\lambda^2\), since we have \(\lambda_+=\Psi_0(1,1,1)=\sum_{j=1}^3\alpha_j\). Thus, \(z_\lambda \in M^{\boldsymbol{C}}(P)\cap
(0,1)^3\), and the conclusion follows.
(2) Assume \(|\lambda|=\lambda_+\). Since \(\partial_{z_j}\Psi_0(z)=\alpha_j>0\) for all \(j\) then \(|\Psi_0(z)|<\lambda_+ =
\Psi_0(1,1,1)\) for all \(z\in [0,1]^3\setminus\{(1,1,1)\}\). Hence \(\{(1,1,1)\}=M^{\boldsymbol{C}}(P)\cap [0,1]^3\), which proves (2). ◻
Lemma 6. Under the assumptions in Remark 7 with the same notations (hence we assume \(\boldsymbol{\beta}\neq 0\) and \(\Pi_{j=1}^3\beta_j=0\) and we have \(P=p(\cdot;\lambda)\), \(P^\pm:=\tau^\pm-\lambda^2\)), we have
(1) The intersection \(M^{\boldsymbol{C}}_{sgn}(P)\cap[0,1]^3\) has one point at most.
(2) Let \(|\lambda| \in (0,\lambda_-)\). Each \(M^{\boldsymbol{C}}(P^\pm)'\cap (0,1)^3\) has dimension two.
(3-1) Let \(|\lambda| \in (0,\lambda_+)\). Then \(M^{\boldsymbol{C}}(P^+)'\cap (0,1)^3\) has dimension two.
(3-2) Let \(|\lambda| \in (0,\lambda_-)\). Then \(M^{\boldsymbol{C}}(P^-)'\cap (0,1)^3\) has dimension two.
(3-3) Let \(|\lambda| \ge \lambda_-\). Then \(M^{\boldsymbol{C}}(P^-)'\cap [0,1]^3\) has at most one point.
Proof. 1) Thanks to ?? and since \(\beta_1>0>\beta_2\), we have the equivalence (\(K_0(z)=0\) and \(z\in [0,1]^3\)) iff \(z_1=z_3=0\). Hence \[M^{\boldsymbol{C}}_{sgn}(P) \cap [0,1]^3 =\{z=(0,z_2,0)\; \mid \; z_2\in [0,1] \quad {\rm and} \quad z_2=\lambda^2/\alpha_2\} \subset \{z^0\},\] with \(z^0=(0,\lambda^2/\alpha_2,0)\). This proves (1).
(2), (3-1) and (3-2). It is similar to the proof of Lemma 5.
(3-3) If \(|\lambda|>\lambda_- = \sup_{[0,1]^3} \tau^-\) then, obviously, \(M^{\boldsymbol{C}}(P^-) \cap \mathbb{T}^3\) is void. If \(|\lambda|=\lambda_- =
\sqrt{\tau^-(z)}\) and \(z\in [0,1]^3\), then \(z=(1,1,1)\), so \(M^{\boldsymbol{C}}(P^-) \cap [0,1]^3 = \{(1,1,1)\}\). The conclusion follows. ◻
Lemma 7. Assume \(\Pi_{j=1}^3\beta_j\neq0\). Let \(\lambda\neq 0\) and set \(P=\lambda^{-2} p(\cdot;\lambda)\), so \(M^{\boldsymbol{C}}_{sgn}(P)=\emptyset\).
(1) Let \(|\lambda| \in (0,\lambda_+)\). Then \(M^{\boldsymbol{C}}(P)\cap (0,1)^3\) has dimension two.
(2) Let \(|\lambda|=\lambda_+\). Then \(M^{\boldsymbol{C}}(P)\cap [0,1]^3=\{(1,1,1)\}\).
Proof. We have \(P(z)=(\tau^+(z)-\lambda^2)(\tau^-(z) - \lambda^2)\) so \(M^{\boldsymbol{C}}(P)\cap (0,1)^3=\{z\in (0,1)^3\) \(\mid\) \(\tau^+(z)=\lambda^2\}\cup \{z\in (0,1)^3\) \(\mid\) \(\tau^-(z)=\lambda^2\}\). Since \(\nabla P(z)\neq 0\) for all \(z\in [0,1]^3\) then \(M^{\boldsymbol{C}}(P)\cap (0,1)^3\) has dimension is 2 iff it is non empty.
(1) Let \(|\lambda| \in (0,\lambda_+)\) and set \(z(t)=t(1,1,1)\). Since \(\tau^+(z(0))=0\) and \(\tau^+(z(1))=\lambda_+^2\)
then there exists \(t\in (0,1)\) such that \(\tau^+(z(t))=\lambda^2\). Hence \(z(t)\in M^{\boldsymbol{C}}(P)\cap (0,1)^3\). (2) Let \(|\lambda|\ge \lambda_+\). Since \(\lambda_+<\lambda_-\) then \(\tau-(z)<\lambda^2\) for all \(z\in [0,1]^3\). Assume \(|\lambda|> \lambda_+\). Then \(\tau^+(z)<\lambda^2\) for all \(z\in [0,1]^3\) so \(M^{\boldsymbol{C}}(P)=\emptyset\).
Assume \(|\lambda|= \lambda_+\). Since we have (\(\tau^+(z)= \lambda_+^2\) and \(z\in [0,1]^3\)) iff \(z=(1,1,1)\), the
conclusion then follows. ◻
We denote \(\sin_\mathbb{T}:\: \mathbb{T}_{\boldsymbol{C}}^d\ni x \mapsto z=\sin^2 x \in {\boldsymbol{C}}^d\) and set \[X_{0,1} := (\sin_\mathbb{T}^2)^{-1} (\{0,1\}^3) =\{x\in \mathbb{T}^3 \; \mid \: z\in \{0,1\}^3\}.\]
Lemma 8. Let \(d\ge r\ge 1\) and \(E\subset{\boldsymbol{C}}^d\) a connected set. Assume that \(\tilde{E}:=(\sin_\mathbb{T}^2)^{-1}(E)\) is a \(r\)-dimensional smooth submanifold of \(\mathbb{T}_{\boldsymbol{C}}^d\). Let \(C_x\) a connected component of \(\tilde{E}\). Then \(C_x\) is open and closed and \(\sin^2(C_x)=E\).
Proof. Let us consider the topological set \(E\) with the topology induced by those of \({\boldsymbol{C}}^d\) and the topological set \(\tilde{E}\) with the topology induced by those of \(\mathbb{T}_{\boldsymbol{C}}^d\). Then, it is easy to see that \(\phi_E\): \(\tilde{E} \ni x\mapsto z \in E\) is continuous. Since \(\tilde{E}\) is a \(r\)-dimensional smooth submanifold of \(\mathbb{T}_{\boldsymbol{C}}^d\), near any \(x^0\in \tilde{E}\), there exists an open ball \(B(x^0,r)\subset \mathbb{T}_{\boldsymbol{C}}^d\) and a smooth function
\(f:B(x^0,r)\mapsto {\boldsymbol{C}}\) such that \(\nabla f(x)\neq 0\) for all \(x\in B(x^0,r)\). Then, \(\tilde{E}\cap B(x^0,r)=
f^{-1}(0_{{\boldsymbol{C}}})\). Hence \(\tilde{E}\) is locally connected and each of its connected components is open and closed. We now prove that the map \(\phi_E\) is both open and
closed. It then implies that the set \(\phi_E(C_x)\) is open and closed hence equals to \(E\), since \(E\) is connected. Since \(\sin^2\) is a non-constant holomorphic function from \({\boldsymbol{C}}^d\) into itself then it is an open map and \(\sin_\mathbb{T}^2: \:
\mathbb{T}_{\boldsymbol{C}}^d={\boldsymbol{C}}^d/(2\pi {\boldsymbol{Z}})^d\mapsto {\boldsymbol{C}}^d\) is also open. Let \(V\) an open set of \(\tilde{E}\) so \(V=\tilde{E}\cap V'\) where \(V'\) is an open set of \(\mathbb{T}_{\boldsymbol{C}}^d\). Then \(\phi_E(V)=\phi_E((\sin_\mathbb{T}^2)^{-1}(E)\cap V')=E\cap \sin_\mathbb{T}^2(V')\). Since \(\sin_\mathbb{T}^2\) is an open map then \(\sin_\mathbb{T}^2(V')\) is open. Thus \(\phi_E(V)\) is an open set of \(E\). Hence \(\phi_E\) is an open map. Let us prove
that \(\phi_E\) is a closed map. We observe that \(|\sin^2(\Re x+i\Im x)|=\sinh^2(\Im x)+\sin^2(\Re x)\) so a set of the form \(\sin_\mathbb{T}^2(A)\), \(A\subset\mathbb{T}_{\boldsymbol{C}}^d\), is unbounded iff \(A\) is unbounded. (In fact \(\sin_\mathbb{T}^2\) is proper.) Let \(F\) a closed subset of \(\tilde{E}\). Let \(z_n\) a sequence of values in \(\phi_E(F)\) which tends to some \(z\in E\). Then \(z_n=\phi_E(x_n)\) tends to \(z\), \(x_n\in F\), so, by the above observation, the sequence \((x_n)\subset F^\mathbf{N}\) is bounded. Let \(x'\) a subsequential limit of \(x_n\). Then \(z=\phi_E(x')\) so \(z\in \phi_E(F)\). Hence \(\phi_E(F)\) is closed.
The conclusion then follows. In fact, since \(\phi_E\) is an open and closed map and \(C_x\) is open and closed then \(\phi_E(C_x)\) is both open and closed
in the connected space \(E\), so it coincides with \(E\). ◻
Lemma 9. Assume \(\boldsymbol{\beta}=0\). Set \(Q(x)=\Psi_0(z)-\lambda^2\) with \(z=\sin^2 x\). Then \(M^{\boldsymbol{C}}_{sgn}(Q)\) is discrete. In addition, assume \(|\lambda|\in (0,\lambda_+)\). Then, each connected component of \(M^{\boldsymbol{C}}_{reg}(Q)\) intersects \(\mathbb{T}^3\) and the intersection is a two-dimensional real manifold.
Proof. Since \(\partial_{z_j}\Psi_0(z)\neq 0\) for all \(z\in {\boldsymbol{C}}^3\), and since \(\partial_{x_j}z=\sin(2x_j)e_j\) vanishes iff \(z_j\in \{0,1\}\), then \(\nabla Q(x)=0\) iff \(z\in \{0,1\}^3\). Hence \(M^{\boldsymbol{C}}_{sgn}(Q)= X_{0,1}\) is discrete. Let \(C_x\) be a connected component of \(M^{\boldsymbol{C}}_{reg}(Q)\). Set \(P=\Psi_0-\lambda^2\). We have \(M^{\boldsymbol{C}}_{reg}(Q)=(\sin_\mathbb{T}^2)^{-1}(E)\) with \(E:= M^{\boldsymbol{C}}_{reg}(P)\setminus\{0,1\}^3\). Since \(M^{\boldsymbol{C}}_{reg}(P)\) is a connected two-dimensional complex manifold, then so is \(E\). Since \(E\) is a two-dimensional complex manifold, so is \(C_x\). Thanks to Lemma 8, we obtain \(\sin^2(C_x)=E\). Since \(E\subset M^{\boldsymbol{C}}_{reg}(P)\) and \(M^{\boldsymbol{C}}_{reg}(P)\) intersects \((0,1)^3\), \(\sin^2(C_x)\) intersects \((0,1)^3\). Hence \(C_x\) intersects \(\mathbb{T}^3\). Finally, since the jacobian of \(\sin^2|_{\mathbb{T}^3}\) does not vanish on \((\sin_\mathbb{T}^2)^{-1}((0,1)^3)\) (\(\subset \mathbb{T}^3\)), then \(C_x\cap \mathbb{T}^3\) is a two-dimensional real manifold. ◻
Lemma 10. Assume \(\boldsymbol{\beta}\neq 0\) and \(\Pi_{j=1}^3\beta_j=0\). Set \(Q(x)=q(x;\lambda)\).
(1) The analytic variety \(M^{\boldsymbol{C}}_{sgn}(Q)\) has Hausdorff (\(5\))-measure zero and \(M^{\boldsymbol{C}}_{sgn}(Q)\cap \mathbb{T}^3\) is
finite.
(2) We have \(M^{\boldsymbol{C}}_{reg}(Q)=M_{\boldsymbol{C}}^+ \cup M_{\boldsymbol{C}}^-\) where each \(M_{\boldsymbol{C}}^\pm\), defined by \[M_{\boldsymbol{C}}^\pm := \{x\in \mathbb{T}^3_{\boldsymbol{C}} \; \mid \; z\not\in \{0,1\}^3, \; \tau^\pm(z)=\lambda^2 \neq \tau^\mp(z)\}\] is a two-dimensional submanifold of \(\mathbb{T}_{\boldsymbol{C}}^3\).
(3) If \(|\lambda|\in (0,\lambda_-)\), each connected component of \(M_{\boldsymbol{C}}^\pm\) intersects \(\mathbb{T}^3\) and the intersection is a
two-dimensional real manifold.
(4) If \(|\lambda|\in [\lambda_-,\lambda_+)\), each connected component of \(M_{\boldsymbol{C}}^+\) intersects \(\mathbb{T}^3\) and the intersection is a
two-dimensional real manifold. However, the set \(M_{\boldsymbol{C}}^- \cap \mathbb{T}^3\) is finite.
Proof. We use the notations in Lemma 6, so \(Q(x)=P(z)\), and we set \(Q^\pm(x)=\tau^\pm(z) -
\lambda^2=P^\pm(z)\).
(1) We have \[\begin{align}
\nonumber M^{\boldsymbol{C}}_{sgn}(Q) &=& (M^{\boldsymbol{C}}(Q)\cap X_{0,1}) \cup (M^{\boldsymbol{C}}(Q^+)\cap M^{\boldsymbol{C}}(Q^-)) \\
\label{val346MCsgn} &=& (M^{\boldsymbol{C}}(Q)\cap X_{0,1}) \cup (\sin_\mathbb{T}^2)^{-1}(M^{\boldsymbol{C}}_{sgn}(P)).
\end{align}\tag{15}\] The set \(M^{\boldsymbol{C}}(Q)\cap X_{0,1}\) is finite since \(X_{0,1}\) is finite. In addition, the set \((\sin_\mathbb{T}^2)^{-1}(M^{\boldsymbol{C}}_{sgn}(P))\) has Hausdorff \(k\)-measure zero for all \(k\ge 3\), since \(M^{\boldsymbol{C}}_{sgn}(P)\) is a complex straight line (see Lemmas 4, 6) and \(\sin_\mathbb{T}^2\) is a local smooth diffeomorphism except on a finite set of \({\boldsymbol{C}}^3\). Hence \(M^{\boldsymbol{C}}_{sgn}(Q)\) has Hausdorff \((2d-1=5)\)-measure zero. Since \(M^{\boldsymbol{C}}_{sgn}(P)\cap [0,1]^3\) has at most one point (see Lemma 6), 15 shows that \(M^{\boldsymbol{C}}_{sgn}(Q)\cap \mathbb{T}^3\) is finite. This proves (1).
(2) The relation \(M^{\boldsymbol{C}}_{reg}(Q)=M_{\boldsymbol{C}}^+ \cup M_{\boldsymbol{C}}^-\) is then obvious. In addition, we have \(M_{\boldsymbol{C}}^\pm=(\sin_\mathbb{T}^2)^{-1}(M^{\boldsymbol{C}}(P^\pm)) \setminus M^{\boldsymbol{C}}_{sgn}(Q)\). Since \(M^{\boldsymbol{C}}(P^\pm)\) is a two-dimensional submanifold of \({\boldsymbol{C}}^3\) and \(M^{\boldsymbol{C}}_{sgn}(Q)\) has dimension one, \(M_{\boldsymbol{C}}^\pm\) is a two-dimensional submanifold of \(\mathbb{T}_{\boldsymbol{C}}^3\).
(3) and (4) for \(M_{\boldsymbol{C}}^+\). It is similar to the corresponding assertion in Lemma 9.
(4) for \(M_{\boldsymbol{C}}^-\). This is a direct consequence of (3-3), Lemma 6. ◻
Lemma 11. Assume \(\Pi_{j=1}^3\beta_j\neq0\). Let \(|\lambda| \in (0,\lambda_+)\) and set \(Q(x)=\lambda^{-2}p(z;\lambda)\).
(1) The set \(M^{\boldsymbol{C}}_{sgn}(Q)\) is finite.
(2) Assume \(|\lambda| \in (0,\lambda_+)\). Then, each connected component of \(M^{\boldsymbol{C}}_{reg}(Q)\) intersects \(\mathbb{T}^3\), and the
intersection is a two-dimensional real manifold.
Proof. Put \(P(z):=Q(x)\) so \(\partial_{x_j}Q(x)=0\) iff \(\partial_j P(z)=0\) or \(z_j\in \{0,1\}\). Hence
\(M^{\boldsymbol{C}}_{sgn}(Q) = (\sin_\mathbb{T}^2)^{-1}(F)\) with \[\begin{align} F &:=& M^{\boldsymbol{C}}_{sgn}(P) \cup_{j=1}^3 F_j \cup_{j\neq k} F_{j,k} \cup (\{0,1\}^3\cap
M^{\boldsymbol{C}}(P)),\\ F_j &:=& \{z\in M^{\boldsymbol{C}}(P)\; \mid \; z_j\in\{0,1\}^3, \; \partial_{z_k}P(z)=0 \quad k\neq j\},\\ F_{j,k} &:=& \{z\in M^{\boldsymbol{C}}(P)\; \mid \; z_j,z_k\in\{0,1\}^3, \; \partial_{z_l}P(z)=0 \quad
l\neq j,k\}.
\end{align}\] (1) Thanks to Lemma 7 we have \(M^{\boldsymbol{C}}_{sgn}(P)=\emptyset\). We denote by \(A_j\) the
\(j^{\mathrm{th}}\) column of the matrix \(A\) defined by 11 , by \(E_j\) the column of the coefficients of \(e_j\) in the canonical basis \((e_1,e_2,e_3)\), by \(\tilde{A}_j\) the \(3\times 3\) matrix obtained from \(A\) by replacing the column \(A_j\) by \(E_j\) and by \(\tilde{A}_{j,k}\) (with \(j\neq k\))
the \(3\times 3\) matrix obtained from \(A\) by replacing the columns \(A_j\) and \(A_k\) respectively by the column \(E_j\) and \(E_k\). Since \(P(z) = Az\cdot z + \boldsymbol{b}\cdot z + c\), then \(\partial_j P(z)= 2A_j\cdot z +
\boldsymbol{b}_j\).
Let us consider \(F_1\). We have \(F_1=F_1(0)\cup F_1(1)\), where \(F_1(\xi)\) is the intersection of the three hyperplanes \(z_1=\xi\), \(2A_k\cdot z + \boldsymbol{b}_k=0\), \(k=2,3\). The matrix of the above system is \(B_1:=(E_1|A_2|A_3)\) where \(E_1:=(1\: 0\: 0)^T\). Its determinant is \[\det(B_1)=\gamma_2\gamma_3-(\varepsilon_1\mu_1\alpha_1)^2 = (\gamma_1-\alpha_1^2)(\varepsilon_1\mu_1)^2 = -\frac{1}{4} \beta_1^2 (\varepsilon_1\mu_1)^2
<0.\] Hence \(F_1(\xi)\) is reduced to one point, and then \(F_1\) is a couple of points. Similarly, each \(F_j\) is reduced to two points. Let us
consider \(F_{1,2}\). We have \(F_{1,2}=\cup_{\xi,\xi'\in \{0,1\}} F_{1,2}(\xi,\xi')\) where \(F_{1,2}(\xi,\xi')\) is the intersection of the
three hyperplanes \(z_1=\xi\), \(z_2=\xi'\), \(2A_3\cdot z + \boldsymbol{b}_3=0\). The matrix of the above system is \(B_{1,2}:=(E_1|E_2|A_3)\) where \(E_2:=(0\: 1\: 0)^T\). Its determinant is \(\det(B_{1,2})=\gamma_3\neq 0\). Hence \(F_{1,2}\) is
finite. Then, \(F\) is finite. Now we observe that, for any finite set \(E\subset{\boldsymbol{C}}^3\), \((\sin_\mathbb{T})^{-1}(E)\) is a finite subset of
\(\mathbb{T}_{\boldsymbol{C}}^3\). Consequently, \(M^{\boldsymbol{C}}_{sgn}(Q)\) is finite.
(2) We have \(M^{\boldsymbol{C}}_{reg}(Q) = (\sin_\mathbb{T}^2)^{-1}(E)\) with \(E := M^{\boldsymbol{C}}_{reg}(P) \setminus F\). Thanks to Lemma 7, \(M^{\boldsymbol{C}}_{reg}(P)\) is a connected two-dimensional submanifold of \({\boldsymbol{C}}^3\). So, since \(\dim F \le
1\), then \(E\) is also a connected two-dimensional submanifold of \({\boldsymbol{C}}^3\). Thus \(M^{\boldsymbol{C}}_{reg}(Q)\) is a two-dimensional
submanifold of \(\mathbb{T}_{\boldsymbol{C}}^3\). Let \(C_x\) be a connected component of \(M^{\boldsymbol{C}}_{reg}(Q)\). Lemma 8 says that \(\sin_\mathbb{T}^2(C_x)=E\). We have \(E\cap (0,1)^3= M^{\boldsymbol{C}}_{reg}(P)\cap(0,1)^3\), since \(F \cap(0,1)^3\) is empty. Thanks to Lemma 7, \(M^{\boldsymbol{C}}_{reg}(P)\cap(0,1)^3\) has dimension two. Thus \(E\) intersects \((0,1)^3\). The end of the proof is similar to the end of the proof of Lemma 9. ◻
Theorem 1 is a straight consequence of Theorem 8, Lemmas 9, 10, 11.
We need
Lemma 12. Assume \(\beta \neq 0\) and \(\beta_1\beta_2\beta_3 =0\). Let \(\lambda \neq 0\).
(1) The set \(M^{\boldsymbol{C}}_{reg}(q(\cdot;\lambda))\) is a disjoint union of the following two complex manifolds of dimension 2: \[M_{\boldsymbol{C}}^{\pm} := \{x \in
{\mathbb{T}}^3_{\boldsymbol{C}} \; \mid \; z \not\in \{0,1\}^3, \tau^{\pm}(z) = \lambda^2 \neq \tau^{\mp}(z)\}.\] (2) Assume \(|\lambda| > \lambda_-\). Then, \(M_{\boldsymbol{C}}^-\cap
\mathbb{T}^3\) is finite,
The assertions (1) and (2) of Lemma 12 are a straight consequence of Lemma 10.
Set \[B(x)=- \boldsymbol{\mu}\tilde{M}(y) \boldsymbol{\varepsilon}\tilde{M}(y),\] and \(u=(u_E,u_H)\) with \[u_E(x) := \frac{1}{\lambda}
\boldsymbol{\varepsilon}\tilde{M}(y) u_H(x) , \quad u_H(x) := (\tau^-(z)-\lambda^2)^{-1} {\rm comat}(B(x)-\lambda^2) v_H,\] where \(v_H\) is a non-null constant column-vector of length 3. Since \(u_H\in \mathcal{C}^\infty(\mathbb{T}^3;{\boldsymbol{C}}^3)\) so does \(u_E\). Thus \(\hat{u}\in L^2({\boldsymbol{Z}}^3)\subset
\mathcal{B}_0^*({\boldsymbol{Z}}^3)\). Then, \((B(x)-\lambda^2)u_H(x)= \lambda^2 (\tau^+(z)-\lambda^2)v_H\) is a trigonometric polynomial, and so is \((H^D-\lambda)u=(0,\lambda^{-1}(B-\lambda^2)u_H)\).
Let us prove that we can choose \(v_H\) such that \(u\) is not a trigonometric polynomial. We remember that the family of eigenvalues of the symmetric matrix \(-\boldsymbol{\varepsilon}\tilde{M}(y) \boldsymbol{\mu}\tilde{M}(y)\) is \((0,\tau^+(z),\tau^-(z))\) and we denote by \(\Pi^0(y)\), \(\Pi^+(y)\), \(\Pi^-(y)\), respectively, the associated spectral eigenprojectors. Thus, we have \[\begin{align} {\rm comat}(B(x)-t) &=& -t(\tau^+(z)-t) \Pi^-(y)
-t(\tau^-(z)-t) \Pi^+(y)\\ && +(\tau^+(z)-t)(\tau^-(z)-t)\Pi^0(y),
\end{align}\] therefore, \[\label{rel46comB} {\rm comat}(B(x)-\tau^-(z)) = -2\tau^-(z) \sqrt{K_0(z)} \Pi^-(y).\tag{16}\] Hence, \({\rm
comat}(B(x)-\tau^-(z))\) has rank one at any \(z\in {\boldsymbol{C}}^3\) such that \(K_0(z)\neq 0\). If each coefficient \(c_{j,k}(y)\) of \({\rm comat}(B(x)-\lambda^2)\) were reducible by \((\tau^-(z)-\lambda^2)\), i.e, if \((\tau^-(z)-\lambda^2)^{-1}c_{j,k}(y)\) were a trigonometrical polynomial,
then \({\rm comat}(B(x)-\lambda^2)\) would vanish at any \(z\in {\boldsymbol{C}}^3\) such that \(\tau^-(z)-\lambda^2=0\). This is in contradiction with 16 . Hence there exists \(v_H\) such that \(u\) is not a trigonometrical polynomial. This ends the proof of Theorem 2.
We complete this section by proving the assertion of Remark 1. Set \(Q^\pm(x)=(\tau^\pm(z)-\lambda^2)v(x)\), \(v:=Q^-u\). Since \(\tau^-\) is linear, \(Q^-\) is then smooth in the \(x\) variable, so \(v\in \mathcal{B}_0^*(\mathbb{T}^3)\). In addition we have \(Q^+ = \lambda^{-2} q(\cdot;\lambda)u\) which is a trigonometric polynomial. Since \(Q^+\) satisfies Conditions (A-1)-(A-2) of Theorem 8, the result follows.
I. We put \[\mathcal{P}_{disc}^-(\xi,l):= \{n\in{\boldsymbol{Z}}^3\; \mid \; \xi \cdot n \le l\} , \quad (\xi,l)\in {\boldsymbol{R}}^3\times \mathbf{R},\] and \[G^* :=\{\xi \in \mathbb{S}^2\; \mid \; |\xi_i|> |\xi_j|\ge |\xi_l|for some permutation (i,j,l) of (1,2,3)\},\] where \(\mathbb{S}^2\) denotes the euclidian unit sphere of \({\boldsymbol{R}}^3\). By definition we have \({\boldsymbol{K}}_{\rm int}=\Omega\cap{\boldsymbol{Z}}^3\) where \(\Omega\) is a bounded convex set of \({\boldsymbol{R}}^3\).
Lemma 13. There exists a finite family \(F\subset G^* \times \mathbf{R}\) such that \[\label{val46Kint} {\boldsymbol{K}}_{\rm int}= \cap_{(\xi,l)\in F} \mathcal{P}_{disc}^-(\xi,l) .\qquad{(6)}\]
Proof. Notice that the characterization ?? of \({\boldsymbol{K}}_{\rm int}\) is well-known with \(G^*\) replaced by \(\mathbb{S}^2\). The bounded convex set \(\Omega\subset {\boldsymbol{R}}^3\) can be assumed closed and written \(\Omega= \cap_{(\xi,r)\in F_0} \mathcal{P}^-(\xi,r)\), where \(\mathcal{P}^-(\xi,r):=\{\nu\in{\boldsymbol{R}}^3\) \(\mid\) \(\nu \cdot \xi \le r\}\) and \(F_0\subset \mathbb{S}^2 \times \mathbf{R}\). Since \({\boldsymbol{K}}_{\rm int}\) is bounded then \({\boldsymbol{K}}_{\rm int}\subset K_1\), \(K_1:= [-l_0,l_0]^3\cap {\boldsymbol{Z}}^3\) for some \(l_0>0\). Since \({\boldsymbol{K}}_{\rm int}\) is finite we then have \[{\boldsymbol{K}}_{\rm int}= \cap_{(\xi,r)\in F_1} \mathcal{P}^-(\xi,r) \cap K_1,\] where \(F_1\subset F_0\) is a finite family. Obviously, we can assume that \(F_1\) is non void and that \(K_1 \not\subset \mathcal{P}^-(\nu,r)\), \((\nu,r)\in F_1\). (If \(K_1\subset \mathcal{P}^-(\nu,r)\) we suppress \((\nu,r)\) in \(F_1\), that is, we replace \(F_1\) by \(F_1\setminus\{(\nu,r)\}\).) Fix \((\nu,r)\in F_1\). Since \(K_1\setminus\mathcal{P}^-(\nu,r)\) is finite and non void, we put \(\delta:= \sup\{ \nu \cdot n -r\) \(\mid\) \(n\in K_1\setminus\mathcal{P}^-(\nu,r)\}>0\). We then have \[\mathcal{P}^-(\nu,r) \cap K_1 =\mathcal{P}^-(\nu,r+\delta/2) \cap K_1 ,\] \[K_1\setminus\mathcal{P}^-(\nu,r) = K_1\setminus\mathcal{P}^-(\nu,r-\delta/2).\] Since \(G^*\) is dense in \(\mathbb{S}^2\) and since \(K_1\) is finite, there then exists \(\xi\in G^*\) sufficiently closed to \(\nu\) such that \[\begin{align} \mathcal{P}^-(\xi,r) \cap K_1 \subset \mathcal{P}^-(\nu,r+\delta/2) \cap K_1,\\ K_1\setminus\mathcal{P}^-(\xi,r) \subset K_1\setminus\mathcal{P}^-(\nu,r-\delta/2). \end{align}\] Then we obtain \[\mathcal{P}^-(\nu,r) \cap K_1 = \mathcal{P}^-(\xi,r)\cap K_1 = \mathcal{P}_{disc}^-(\xi,r)\cap K_1.\] Hence, ?? holds with \(F=F_1 \cup_{j=1}^3 (e_j,l_0) \cup_{j=1}^3 (-e_j,l_0)\). 0◻
II. Letting \(\hat{v}:\:{\boldsymbol{Z}}^3 \ni n \mapsto \hat{v}(n) \in {\boldsymbol{C}}^d\) be a sequence with compact support, we set \[\hat{v}_S(n)=\overline{\hat{v}(n)},\] so \(v_S(x) = \overline{v(-x)}\). In addition, letting \(\xi\in{\boldsymbol{R}}^3\), \(\xi\neq 0\), we define \[N_\xi^{max}(v) = N_\xi^{max}(\hat{v}) := \max\{n\cdot\xi \; \mid \; \hat{v}(n)\neq 0\} ,\] with the convention \(\max(\emptyset)=-\infty\).
We put \(|\xi|_\infty:=\max(|\xi_j|;\; j\in\{1,2,3\})\), where the \(\xi_l\)’s are the usual coordinates of \(\xi\).
Lemma 14. Let \(\lambda\neq 0\) and \(\xi\in G^*\). Let \(v\) and \(w\) are two trigonometric polynomials.
1) We have \[N_\xi^{max}(v+w) = N_\xi^{max}(v) \quad {\rm if} \quad N_\xi^{max}(v)> N_\xi^{max}(w),\] \[N_\xi^{max}(vw) \le N_\xi^{max}(v) + N_\xi^{max}(w)\] holds true.
2) We have \[\begin{align} N_\xi^{max}(e^{ix_j})&=N_\xi^{max}(\delta_{e_j}) = |\xi_j|, \\ N_\xi^{max}(z_j) &= 2|\xi_j| ,\\ N_\xi^{max}(z_j^2) &= 4|\xi_j|. \end{align}\] 3) For \(t\neq 0\) and all sequence \(\hat{v}\) we have \[\label{val46Nqv} N_\xi^{max}(q_t v) = 4|\xi|_\infty + N_\xi^{max}(v).\qquad{(7)}\] 4) We have \[\begin{align} N_\xi^{max}(v_S) &=& N_\xi^{max}(v),\\ N_\xi^{max}(<v,\overline{v_S(x)}>_{{\boldsymbol{C}}^m}) &=& 2N_\xi^{max}(v) . \end{align}\]
Proof. 1) Obvious.
2) Since \(z_j= (-4)^{-1}(e^{2ix_j}+e^{-2ix_j}-2)\), the computation is straightforward.
3) Thanks to 10 and 11 we have \[t^{-2} q_t(x) = \sum_{j=1}^3 \gamma_j z_j^2 + \sum_{j\neq l} A_{j,l}z_j z_l + \boldsymbol{b}\cdot z+ t^2 ,\] with \(t\gamma_j\neq 0\). Since \(|\xi|_\infty = \xi_{i^*}>\max(|\xi_j|,|\xi_k|)\) for some \(i^*\) where \(\{i^*,j,k\}=\{1,2,3\}\),
then, \[N_\xi^{max}(q_t) = N_\xi^{max}(z_i^2)= 4\xi_{i^*} = 4|\xi|_\infty.\] In addition, we then have \[N_\xi^{max}(q_t v) \le N_\xi^{max}(q_t) + N_\xi^{max}(v) = 4|\xi|_\infty +
N_\xi^{max}(v).\] We have \[\begin{align} N_\xi^{max}(e^{\pm 2ix_j} v) &=& \max\{n\cdot\xi \; \mid \; U(e^{\pm 2ix_j} v)(n)\neq 0\} = \max\{n\cdot\xi \; \mid \; v(n\pm 2e_j)\neq 0\} \\ &=& \max\{(n\mp
2e_j)\cdot\xi \; \mid \; v(n)\neq 0\} = N_\xi^{max}(v) \mp 2 \xi_j.
\end{align}\] Hence, \(N_\xi^{max}(z_jv) = N_\xi^{max}(v) + 2|\xi_j|\), \(N_\xi^{max}(z_jz_k v) = N_\xi^{max}(v) + 2|\xi_j| + 2|\xi_k|\). Thus, since \(\xi\in G^*\), \[N_\xi^{max}(z_jz_k v) < N_\xi^{max}(v) + 4|\xi| \quad {\rm if}\quad (j,k)\neq (i^*,i^*).\] Hence, \[N_\xi^{max}(q_tv) = N_\xi^{max}(z_{i^*}^2 v) =
N_\xi^{max}(v) + 4|\xi| .\] 4) The function \(<v(x),\overline{v_S(x)}>_{{\boldsymbol{C}}^d}=(2\pi)^{-3}\sum_{n,m}e^{-i(n+m)x}\hat{v}(n)\overline{\hat{v}(m)}\) is a trigonometric polynomial, and, by a simple
computation, the conclusion follows. 0◻
III. We put \(L_t(x)=L(x;t):={\rm comat}(H^D(x)-t)\), and \(q_t(x):=\det(H^D(x)-t)\). The symmetric real matrix \(L(x;t)\) has size \(5\times 5\) and \(L\) is a homogeneous polynomial of degree 5 in the variables \(t,y_1,y_2,y_3\). Since \(H^D(x)\) has rank
\(\le 4\) then, \(L_0=0\) and, consequently, \(L_t\) is a polynomial of the variables \(y_1,y_2,y_3\) with degree \(\le 4\). Hence \(L_t\) has the form \[\label{rel46B1vS} L_t(x) = tB_1(x) + t^2 B_2(x;t),\tag{17}\] where \(B_1(x)=-\Phi_0(z)\Pi_0(y)\), \(\Pi_0(y)\) is the orthogonal projection on \(\ker H^D(x)\), \(\Phi_0:=\Psi_0^2-K_0\) is a
homogeneous polynomial of degree 2 of the variables \(z_1,z_2,z_3\), and \(B_2(\cdot;t)\) is a trigonometric polynomial of degree 3. Thus, if \(t\neq0\),
then \(L_t\) is a trigonometric polynomial of degree \(4\). Since \(B_1(x)=\lim_{t\to 0} t^{-1}L_t(x)\) we then see that \(B_1\) is also a trigonometric polynomial of degree 4.
We have
Lemma 15. Let \(t\neq 0\) and \(\xi\in G^*\). We then have \[\label{est46NLt} N_\xi^{max}(L_t) = N_\xi^{max}(B_1) = N_\xi^{max}(q_t) = 4|\xi|_\infty > 3|\xi|_\infty \ge N_\xi^{max}(B_2(t)).\qquad{(8)}\]
Proof. Without loss of generality we assume \(\xi_1>|\xi_j|\), \(j\in\{2,3\}\). Since \(q_t I = (H^D-t)L_t\), we then have, thanks to Lemma 14, \[\label{rel46qtHDLt} 4|\xi|_\infty = 4\xi_1 = N_\xi^{max}(q_t) \le N_\xi^{max}(H^D) +N_\xi^{max}(L_t).\tag{18}\] Since \(H^D-t\) (respect., \(L_t\)) is a polynomial of degree \(1\) (respect;, of degree 4) of the variables \(y_j=(2i)^{-1}(e^{ix_j}-e^{-ix_j})\), then \(N_\xi^{max}(H^D-t)\le \xi_1\) and \(N_\xi^{max}(L_t)\le 4\xi_1\). Hence the inequalities in 18 are equalities. Finally, since \(B_2(t)\) is a polynomial of degree \(\le 3\) of the variables \(y_j=(2i)^{-1}(e^{ix_j}-e^{-ix_j})\), then \(N_\xi^{max}(B_2(t))\le 3\xi_1\). ◻
IV. We have the following
Lemma 16. Let \(\lambda\neq 0\) and \(\xi\in G^*\). Let \(l\in\mathbf{R}\) and \(\hat{u}\) be compactly supported in \({\boldsymbol{Z}}^3\) such that \((\hat{H}^{D_p}-\lambda)\hat{u}(n)=0\) in \(\{n\in{\boldsymbol{Z}}^3\) \(\mid\) \(\xi\cdot n>l\}\). Then, \(\hat{u}(n)=0\) in \(\{n\in{\boldsymbol{Z}}^3\) \(\mid\) \(\xi\cdot n>l\}\).
Proof. It is sufficient to consider the case \(l=0\). The sequence \(\hat{u}\) satisfies \[\label{eq46HDpvu61f} (H^{D_p}-\lambda) u= f,\tag{19}\] where \(\hat{f}\) is compactly supported in \(\{n\in{\boldsymbol{Z}}^3\) \(\mid\) \(\xi\cdot n\le 0\}\). Thus, \(N_\xi^{max}(f)\le 0\). The operator \(\hat{D}(\hat{D}_p)^{-1}\) is bounded and invertible
in \(l^2({\boldsymbol{Z}}^3)\) and is an operator of multiplication by a diagonal matrix of the form \(I+\hat{K}\), so \(\hat{K}=\hat{D}(\hat{D}_p)^{-1} -
I\) is an operator of multiplication by a diagonal matrix and has compact support. We then have \[\label{eq46HDu} (H^D -\lambda) u = \lambda K u + g,\tag{20}\] where \(\hat{g}:=\hat{D}(\hat{D}_p)^{-1}\hat{f}\) vanishes in \(\{n\in{\boldsymbol{Z}}^3\) \(\mid\) \(n\cdot\xi>0\}\), so \(N_\xi^{max}(g)\le 0\).
Since \(\hat{D}_p\) is real-valued we have \[D_p v_S(x) = (2\pi)^{-3/2}\sum_{n\in{\boldsymbol{Z}}^3} \hat{D}_p(n)\overline{\hat{v}(n)}e^{-inx} = \overline{D_p v(-x)} .\] In addition, since
\(\overline{H^D(-x)} = H^D(-x) =-H^D(x)\), then 19 implies \[-D_p H_0 u_S = \lambda u_S + f_S.\] Since \(B_1(y)=-\Phi_0(z)\Pi_0(y)\) we have \(B_1 H_0=0\). Then, multiplying 17 from the left by \(B_1 D_p^{-1}\) and observing that \(\Pi_0 H_0=0\), we get \[\label{rel246B1v} B_1D_p^{-1} u_S = -\lambda^{-1} B_1 D_p^{-1} f_S .\tag{21}\] Multiplying 20 from
the left by \(L_\lambda\) and using 17 , we get \[q_\lambda u = \lambda L_\lambda K u + L_\lambda g = \lambda^2 B_1K u + R_0,\] where
\[\label{def46R0} R_0:=\lambda^3 B_2(\cdot;\lambda)K u + L_\lambda g.\tag{22}\] Put \(T(x):=\sum_{n\in{\boldsymbol{Z}}^3} \hat{D}_p^{-1} (n)
|\hat{u}(n)|^2 e^{2inx}\). On one hand, since \(\hat{D}_p^{-1} (n) |\hat{u}(n)|^2\neq 0\) if \(\hat{u}(n)\neq 0\), then \[\label{val46NT} N_\xi^{max}(T) = 2N_\xi^{max}(u).\tag{23}\] On the other hand, we have \[T(x) = < D_p^{-1}u_S(x),u(x)>_{{\boldsymbol{R}}^6}.\] Thus, \[\begin{align} q_\lambda(x) T(x) &=& < D_p^{-1}u_S(x),q_\lambda u(x)>_{{\boldsymbol{R}}^6} \\ &=& \lambda^2 < D_p^{-1}u_S(x),B_1(x) K u(x)>_{{\boldsymbol{R}}^6} + R(x),
\end{align}\] where \(R(x):= < D_p^{-1}u_S(x),R_0(x)>_{{\boldsymbol{R}}^6}\). Then, since \(B_1(x)\) and \(D\) are real-symmetric matrices, by
virtue of 21 , \[\begin{align}
\nonumber q_\lambda(x) T(x) &=& \lambda^2 < B_1D_p^{-1}u_S(x),K u(x)>_{{\boldsymbol{R}}^6} + R(x)\\
\label{val46qT} &=& -\lambda < B_1 D_p^{-1} f_S(x),K u(x)>_{{\boldsymbol{R}}^6} +R(x).
\end{align}\tag{24}\] We have, in view of 22 , ?? , \[\begin{align}
\nonumber N_\xi^{max}(R) \le N_\xi^{max}(D_p^{-1}u_S) + \max(N_\xi^{max}(B_2(\cdot;\lambda)K u),N_\xi^{max}(L_\lambda g)) \\
\label{maj46NR} \le N_\xi^{max}(u)+ \max(3|\xi|_\infty+N_\xi^{max}(u),4|\xi|_\infty),
\end{align}\tag{25}\] and, in addition, \[\begin{align}
\label{maj46NB1DpKu} N_\xi^{max}(< B_1 D_p^{-1} f_S,K u>_{{\boldsymbol{R}}^6} ) \le 4|\xi|_\infty + N_\xi^{max}(u).
\end{align}\tag{26}\] Thus, in view of 24 , 25 , 26 , and ?? , \[N_\xi^{max}(q_\lambda T) \le \max(4|\xi|_\infty + N_\xi^{max}(u),
3|\xi|_\infty+2N_\xi^{max}(u)).\] Using ?? , 23 and ?? , we obtain \[\begin{align}
\nonumber N_\xi^{max}(T) = N_\xi^{max}(q_\lambda T) - N_\xi^{max}(q_\lambda) \\
\label{maj46T} \le \max(N_\xi^{max}(u), -|\xi|_\infty+2N_\xi^{max}(u)).
\end{align}\tag{27}\] Thus, 27 and 23 show that \[N_\xi^{max}(u) = N_\xi^{max}(T)-N_\xi^{max}(u) \le 0.\] ◻
V. (Last step.) Theorem 4 is a direct consequence of Lemma 16. In fact, putting \(H(\xi,l):=\{n\in{\boldsymbol{Z}}^3\) \(\mid\) \(\xi\cdot n> l\}\), \(l\in\mathbf{R}\), we have \(H(\xi,l)= {\boldsymbol{Z}}^3\setminus\mathcal{P}_{disc}^-(\xi,l)\). Hence, if \(u\) satisfies the conditions of Theorem 4, then \(\hat{u}(n)=0\) in \(\cup_{(\xi,l)\in F} H(\xi,l)= {\boldsymbol{Z}}^3\setminus{\boldsymbol{K}}_{\rm int}={\boldsymbol{K}}_{\rm ext}\). 0◻
As above we put \(\hat{K}=\hat{D}(\hat{D}_p)^{-1} - I\), which has compact support. The distribution \(u\) satisfies the above relation 20 with \(g=0\). Hence, the distribution \(v\in \mathcal{B}_0^*(\mathbb{T}^3)\) defined by \(v(x)=(\tau^-(z)-\lambda^2)u(x)\) is a trigonometric polynomial. By observing that \(\tau^-(z)-\lambda^2\le \lambda_-^2-\lambda^2<0\), we thus obtain \(u(x)= (\tau^-(z)-\lambda^2)^{-1}v(x)\), so \(u\in L^2(\mathbb{T}^3,{\boldsymbol{C}}^6)\).
Put \(f:=(H^{D_p}-\lambda)u\). Assume that a finite convex set \({\boldsymbol{K}}_{\rm int}\) contains \({\rm supp}\,(\hat{D}^p-\hat{D})\). Then, \(\hat{f}\) has support in \({\boldsymbol{K}}_{\rm int}\). In addition, we have \((\hat{H}^D-\lambda)\hat{u}=\lambda \hat{K} \hat{u} + \hat{D}^{-1} \hat{D}_p \hat{f}\), where \(\hat{K}:= \hat{D} (\hat{D}_p)^{-1}-I\) has compact support. Since (RT) for \(\hat{H}^D-\lambda\) holds, then \(\hat{u}\) has compact support. Consequently, thanks to Theorem 4, \(\hat{u}\) vanishes in \({\boldsymbol{K}}_{\rm ext}\). 0◻