Dirac Operators on Hypersurfaces as Large Mass Limits

Abstract

We show that the eigenvalues of the intrinsic Dirac operator on the boundary of a Euclidean domain can be obtained as the limits of eigenvalues of Euclidean Dirac operators, either in the domain with a MIT-bag type boundary condition or in the whole space, with a suitably chosen zero order mass term.

Appendices

A Basic Properties of Euclidean Dirac Operators

In the present section we would like to recall some key facts on the Euclidean Dirac operators $B_{m,M}$ and $A_m$. All the properties are very standard and well known for $n\in \{2,3\}$, but we were not able to find a suitable single reference covering arbitrary dimensions.

The analysis of $B_{m,M}$ will be provided in a slightly more general setting, with the hope that the constructions can be of use for other works. Denote

$$\begin{aligned} B_M:=B_{M,M}. \end{aligned}$$

Proposition A.1

Let $V\in L^\infty (\mathbb {R}^n)$ be real-valued with compact support. Consider the linear operator $C:=B_M+V\,\alpha _{n+1}$ in $L^2(\mathbb {R}^n,\mathbb {C}^N)$ with domain ${\mathcal {D}}(C)=H^1(\mathbb {R}^n,\mathbb {C}^N)$, then:

• (a) The operator C is self-adjoint, and it is essentially self-adjoint on $C^\infty _c(\mathbb {R}^n,\mathbb {C}^N)$.

• (b) The essential spectrum of C is $\big (-\infty ,-|M|\,\big ]\,\cup \,\big [\,|M|,+\infty \big )$, and there are at most finitely many discrete eigenvalues in $\big (-|M|,|M|\big )$.

• (c) Assume in addition that $n\notin 4\mathbb {Z}$. If E is an eigenvalue of C, then $(-E)$ is also an eigenvalue of C of the same multiplicity.

In particular, the three assertions hold for $C:=B_{m,M}\equiv B_M+(m-M)1_{\Omega }\alpha _{n+1}$.

Proof

• (a) As both C and $B_0$ are symmetric and only differ by the bounded symmetric operator $(M+V)\alpha _{n+1}$, the (essential) self-adjointness of C is equivalent to that of $B_0$. First, it is directly seen that $B_0$ is symmetric. Let S be the restriction of $B_0$ to $C^\infty _c(\mathbb {R}^n,\mathbb {C}^N)$, then S is a densely defined symmetric operator, and by applying the definitions one sees that its adjoint $S^*$ is given by

  $$\begin{aligned} S^*u=D_0 u, \quad {\mathcal {D}}(S^*)=\big \{u\in L^2(\mathbb {R}^n,\mathbb {C}^N):\, D_0 u\in L^2(\mathbb {R}^n,\mathbb {C}^N)\big \} \end{aligned}$$

with $D_0$ acting in the sense of distributions. In order to complete the proof we simply need to show that $S^*=B_0$. As both operators are given by the same differential expression $D_0$ and $S^*$ is clearly an extension of $B_0$, one only needs to check the inclusion ${\mathcal {D}}(S^*)\subset {\mathcal {D}}(B_0)$. Let a function u belong to ${\mathcal {D}}(S^*)$, i.e.

$$\begin{aligned} u\in L^2(\mathbb {R}^n,\mathbb {C}^N) \text { and } -\mathrm {i}\sum _{j=1}^n \alpha _j \partial _j u\in L^2(\mathbb {R}^n,\mathbb {C}^N). \end{aligned}$$

We need to show that this implies $u\in H^1(\mathbb {R}^n,\mathbb {C}^N)$. We remark first that the Fourier transform $\widehat{u}$ of u satisfies $\widehat{u}\in L^2(\mathbb {R}^n,\mathbb {C}^N)$ and $\Gamma (\xi ) \widehat{u} \in L^2(\mathbb {R}^n,\mathbb {C}^N)$. The matrices $\Gamma (\xi )$ are Hermitian, and, by construction, $\Gamma (\xi )^2=|\xi |^2 I_N$. In particular, one can represent $\Gamma (\xi )=|\xi | U(\xi )$ with the unitary matrices $U(\xi ):=\Gamma (\xi /|\xi |)$. As the pointwise multiplication by $U(\cdot )$ is an isomorphism of $L^2(\mathbb {R}^n,\mathbb {C}^N)$, it follows that the condition $\Gamma (\xi ) \widehat{u} \in L^2(\mathbb {R}^n,\mathbb {C}^N)$ yields $|\xi | \widehat{u} \in L^2(\mathbb {R}^n,\mathbb {C}^N)$. Due to $|\xi _j|\le |\xi |$ we obtain $\xi _j \widehat{u} \in L^2(\mathbb {R}^n,\mathbb {C}^N)$ for each $j=1,\dots n$. By applying the inverse Fourier transform this implies $\partial _j u \in L^2(\mathbb {R}^n,\mathbb {C}^N)$ for each $j=1,\dots n$, and then $u\in H^1(\mathbb {R}^n,\mathbb {C}^N)$.

• (b) Let us start by showing the inclusion

  $$\begin{aligned} \big (-\infty ,-|M|\big ]\cup \big [|M|,+\infty \big )\subset {\mathrm {spec}}_{\mathrm{ess}}C. \end{aligned}$$

  (59)

Denote $e_1:=(1,0,\dots ,0)\in \mathbb {R}^n$. For $\rho >0$, denote during the proof

$$\begin{aligned} b_\rho :=\{x\in \mathbb {R}^n: |x|<\rho \}. \end{aligned}$$

Let $\chi \in C^\infty _c(\mathbb {R}^n)$ with $\chi =1$ in $b_1$ and $\chi =0$ outside $b_2$, and $E\in \mathbb {R}$ with $|E|>|M|$. Let us choose $\eta \in \mathbb {C}^N$ with

$$\begin{aligned} \varphi :=(\sqrt{E^2-M^2}\,\alpha _1+M\alpha _{n+1}+E\,I_N)\eta \ne 0, \end{aligned}$$

which is possible as the matrix $\sqrt{E^2-M^2}\,\alpha _1+M\alpha _{n+1}+E\,I_N$ is non-zero (otherwise $\alpha _1$ could not anticommute with $\alpha _{n+1}$). For large $k\in \mathbb {N}$ consider the following functions $u_k\in H^1(\mathbb {R}^n,\mathbb {C}^N)$:

$$\begin{aligned} u_k(x)&=\chi \Big (\dfrac{x}{k}-ke_1\Big ) e^{\mathrm {i}\sqrt{E^2-M^2}\, x_1} \varphi \\&\equiv \chi \Big (\dfrac{x}{k}-ke_1\Big ) e^{\mathrm {i}\sqrt{E^2-M^2}\, x_1} (\sqrt{E^2-M^2}\,\alpha _1+M\alpha _{n+1}+E\,I_N)\eta , \end{aligned}$$

then one easily checks that $\Vert u_k\Vert _{L^2(\mathbb {R}^n,\mathbb {C}^N)}^2=c k^n>0$ with $c=\Vert \chi \Vert _{L^2(\mathbb {R}^n,\mathbb {C}^N)}^2>0$ independent of k. On the other hand, for large k one has $V=0$ on the support of $u_k$, hence,

$$\begin{aligned} (C u_k)(x)&=-\mathrm {i}\sum _{j=1}^n \alpha _j\partial _j u_k(x) + M\alpha _{n+1} u_k (x)\\&= -\dfrac{\mathrm {i}}{k}\sum _{j=1}^n \alpha _j \partial _j \chi \Big (\dfrac{x}{k}-ke_1\Big ) e^{\mathrm {i}\sqrt{E^2-M^2}\, x_1} \varphi \\&\qquad +\chi \Big (\dfrac{x}{k}-ke_1\Big ) \Big (\sqrt{E^2-M^2}\,\alpha _1+M\alpha _{n+1}\Big ) e^{\mathrm {i}\sqrt{E^2-M^2}\, x_1}\varphi , \end{aligned}$$

and then $(C-E)u_k=v_k+w_k$ with

$$\begin{aligned} v_k(x)&=-\dfrac{\mathrm {i}}{k}\sum _{j=1}^n \alpha _j \partial _j \chi \Big (\dfrac{x}{k}-ke_1\Big ) e^{\mathrm {i}\sqrt{E^2-M^2}\, x_1} \varphi ,\\ w_k(x)&=\chi \Big (\dfrac{x}{k}-ke_1\Big ) e^{\mathrm {i}\sqrt{E^2-M^2}\, x_1}\Big (\sqrt{E^2-M^2}\,\alpha _1+M\alpha _{n+1}-E\,I_N\Big ) \varphi \\&\equiv \chi \Big (\dfrac{x}{k}-ke_1\Big )e^{\mathrm {i}\sqrt{E^2-M^2}\, x_1}\Big (\sqrt{E^2-M^2}\,\alpha _1+M\alpha _{n+1}-E\,I_N\Big )\\&\qquad \times \Big (\sqrt{E^2-M^2}\,\alpha _1+M\alpha _{n+1}+E\Big )\eta =0. \end{aligned}$$

One estimates easily $\Vert v_k\Vert _{L^2(\mathbb {R}^n,\mathbb {C}^N)}^2=\mathcal {O}(k^{n-2})$ for $k\rightarrow +\infty$, and this shows that $\Vert (C-E)u_k\Vert _{L^2(\mathbb {R}^n,\mathbb {C}^N)}/\Vert u_k\Vert _{L^2(\mathbb {R}^n,\mathbb {C}^N)}=\mathcal {O}(1/k)\rightarrow 0$ and yields $E\in {\mathrm {spec}}C$. It follows that $\big (-\infty ,-|M|\big )\cup \big (|M|,+\infty \big )\subset {\mathrm {spec}}C$, and one can take the closure on the left-hand side as ${\mathrm {spec}}C$ is a closed set. Furthermore, as the set $\big (-\infty ,-|M|\big ]\cup \big [|M|,+\infty \big )$ has no isolated points, it is included into the essential spectrum of C. Hence, the claim (59) is proved.

Now it remains to check that C has no essential spectrum in $\big (-|M|,|M|\big )$ and that it has at most finitely many discrete eigenvalues, which will be done by an iterated application of the min–max principle. For $E\in \mathbb {R}$ and a self-adjoint semibounded from below operator T we will denote

$$\begin{aligned} \mathcal {N}(E,T):=\#\big \{j\in \mathbb {N}: E_j(T)<E\big \}. \end{aligned}$$

In other words, if ${\mathrm {spec}}_{\mathrm{ess}}T\,\cap \, (-\infty ,E)\ne \emptyset$, then $\mathcal {N}(E,T)=+\infty$, otherwise $\mathcal {N}(E,T)$ is the number of eigenvalues of T in $(-\infty ,E)$, where each eigenvalue counted according to its multiplicity. In these terms, we simply need to show that $\mathcal {N}(M^2,C^2)<+\infty$.

Choose $r>0$ large such that the support of V is contained in $b_r$, and then pick any $R>r$ and real-valued functions $\chi _1, \chi _2\in C^\infty (\mathbb {R}^2)$ with

$$\begin{aligned} \chi _1^2+\chi _2^2=1, \quad \chi _1=1 \text { in } b_r, \quad \chi _2=1 \text { in } \mathbb {R}^2{\setminus } b_R. \end{aligned}$$

Let $u\in {\mathcal {D}}(C)$, then for each $k\in \{1,2\}$ one also has $\chi _k u \in {\mathcal {D}}(C)$ and $C (\chi _k u)=\chi _k C u -\mathrm {i}\Gamma (\nabla \chi _k) u$, and

$$\begin{aligned} \big \Vert C(\chi _k u)\big \Vert ^2_{L^2(\mathbb {R}^n,\mathbb {C}^N)}&=\int _{\mathbb {R}^n} \Big (\chi _k^2|C u|^2+ \big |\mathrm {i}\Gamma (\nabla \chi _k) u\big |^2\Big )\,\mathrm {d}x + \mathcal {J}_k,\\ \mathcal {J}_k&=2 \mathfrak {R}\int _{\mathbb {R}^n} \, \big \langle \chi _k C u, -\mathrm {i}\Gamma (\nabla \chi _k) u\big \rangle _{\mathbb {C}^n}\,\mathrm {d}x\\&= \mathfrak {R}\int _{\mathbb {R}^n} \, \big \langle C u, -\mathrm {i}\Gamma (2\chi _k\nabla \chi _k) u\big \rangle _{\mathbb {C}^n}\,\mathrm {d}x = \mathfrak {R}\int _{\mathbb {R}^n} \, \big \langle C u,-\mathrm {i}\Gamma \big (\nabla (\chi _k^2)\big ) u\big \rangle _{\mathbb {C}^n}\,\mathrm {d}x. \end{aligned}$$

From $\chi _1^2+\chi _2^2=1$ we infer $\nabla (\chi _1^2+\chi _2^2)=0$ and then $\mathcal {J}_1+\mathcal {J}_2=0$. Therefore,

$$\begin{aligned}&\big \Vert C(\chi _1 u)\big \Vert ^2_{L^2(\mathbb {R}^n,\mathbb {C}^N)}+\big \Vert C(\chi _2 u)\big \Vert ^2_{L^2(\mathbb {R}^n,\mathbb {C}^N)}\nonumber \\&\quad =\int _{\mathbb {R}^n} ( \chi _1^2+\chi _2^2) |C u|^2\,\mathrm {d}x +\int _{\mathbb {R}^2} \big (|\nabla \chi _1|^2+|\nabla \chi _2|^2\big )|u|^2\big )\,\mathrm {d}x\nonumber \\&\quad = \int _{\mathbb {R}^n} |Cu|^2\,\mathrm {d}x +\int _{\mathbb {R}^n} W|u|^2\,\mathrm {d}x, \quad W:=|\nabla \chi _1|^2+|\nabla \chi _2|^2. \end{aligned}$$

(60)

Recall that W is supported in $\overline{b_R{\setminus } b_r}$, while the support of $\chi _2 u$ does not intersect the support of V, which gives

$$\begin{aligned} \big \Vert C(\chi _2 u)\Vert _{L^2(\mathbb {R}^n,\mathbb {C}^N)}^2=\big \Vert B_M(\chi _2 u)\Vert _{L^2(\mathbb {R}^n,\mathbb {C}^N)}^2=\int _{\mathbb {R}^n} \big (|\nabla (\chi _2 u)|^2+M^2|\chi _2 u|^2\big )\,\mathrm {d}x \end{aligned}$$

(for $u\in C^\infty _c(\mathbb {R}^n,\mathbb {C}^N)$ this is a simple integration by parts, and it is then extended by density to the whole of ${\mathcal {D}}(C)$ as $C^\infty _c(\mathbb {R}^n,\mathbb {C}^N)$ is a domain of essential self-adjointness as shown above). This allows one to rewrite (60) as

$$\begin{aligned} \big \Vert C u\Vert _{L^2(\mathbb {R}^n,\mathbb {C}^N)}^2= & {} \int _{b_R} \big ( \big |C(\chi _1 u)\big |^2- W|\chi _1 u|^2\big )\,\mathrm {d}x\nonumber \\&+\int _{\mathbb {R}^n} \big (|\nabla (\chi _2 u)|^2+M^2|\chi _2 u|^2 - W |\chi _2 u|^2 \big )\,\mathrm {d}x. \end{aligned}$$

(61)

For $v\in L^2(b_R,\mathbb {C}^N)$ let us denote by $v'$ its extension by zero to the whole of $\mathbb {R}^n$ and consider the following sesqulinear form $s_1$ in $L^2(b_R,\mathbb {C}^N)$:

$$\begin{aligned} s_1(v,v)= \int _{b_R} |C v'|^2\,\mathrm {d}x, \quad {\mathcal {D}}(s_1)=H^1_0(b_R,\mathbb {C}^N). \end{aligned}$$

The form $s_1$ is clearly non-negative and densely defined. Let us show that it is also closed. Let $(v_k)\subset {\mathcal {D}}(S_1)$ and $v\in L^2(b_R,\mathbb {C}^N)$ with $\Vert v_k-v\Vert _{L^2(b_R,\mathbb {C}^N)}\rightarrow 0$ for $k\rightarrow \infty$ and $s_1(v_k-v_l,v_k-v_l)\equiv \Vert C(v'_k-v'_l)\Vert ^2_{L^2(\mathbb {R}^n,\mathbb {C}^N)}\rightarrow 0$ as $k,l\rightarrow +\infty$. Then one has $\Vert v'_k-v'\Vert _{L^2(\mathbb {R}^n,\mathbb {C}^N)}\rightarrow 0$ for $k\rightarrow \infty$, and the closedness of the operator C implies that $v'\in {\mathcal {D}}(C)$ with $Cv'=\lim _{k\rightarrow \infty } Cv_k'$. Hence, the function v is such that its extension by zero belongs to $H^1(\mathbb {R}^n,\mathbb {C}^N)$, and this shows $v\in H^1_0(b_R,\mathbb {C}^N)\equiv {\mathcal {D}}(s_1)$ and then $s_1(v_k-v,v_k-v)\equiv \Vert C(v'_k-v')\Vert ^2_{L^2(\mathbb {R}^n,\mathbb {C}^N)}\rightarrow 0$. The semiboundedness and closedness of $s_1$ imply the existence of a self-adjoint operator $S_1$ in $L^2(b_R,\mathbb {C}^N)$ with $S_1[v,v]=s_1(v,v)$ for $v\in {\mathcal {Q}}(S_1)={\mathcal {D}}(s_1)$, and $S_1$ has compact resolvent as ${\mathcal {Q}}(S_1)$ is compactly embedded into $L^2(b_R,\mathbb {C}^N)$.

Denote by $S_0$ the (scalar) Schrödinger operator $-\Delta -W$ in $L^2(\mathbb {R}^n)$, i.e.

$$\begin{aligned} S_0[u,u]=\int _{\mathbb {R}^n}\big ( |\nabla u|^2-W|u|^2\big )\,\mathrm {d}x, \quad {\mathcal {Q}}(S_0)=H^1(\mathbb {R}^n), \end{aligned}$$

and consider the linear map $J:{\mathcal {D}}(C)\equiv {\mathcal {Q}}(C^2)\mapsto {\mathcal {Q}}(S_1)\times {\mathcal {Q}}(S_0\otimes I_N)$ given by $J u=\big ((\chi _1 u)|_{b_R}, \chi _2 u\big )$. For any $u\in {\mathcal {D}}(C)$ one has then $\Vert Ju\Vert _{L^2(b_R,\mathbb {C}^N)\times L^2(\mathbb {R}^n,\mathbb {C}^N)}=\Vert u\Vert _{L^2(\mathbb {R}^n,\mathbb {C}^N)}$, and the above representation (61) can be rewritten as

$$\begin{aligned} C^2[u,u]= & {} \big \Vert C u\Vert _{L^2(\mathbb {R}^n,\mathbb {C}^N)}^2\\= & {} (S_1-W)[(\chi _1 u)|_{b_R},(\chi _1 u)|_{b_R}] +\big ((S_0+M^2)\otimes I_N\big )[\chi _2u,\chi _2 u]\\\equiv & {} \Big ((S_1-W) \,\oplus \big ((S_0+M^2)\otimes I_N\big )\Big ) [Ju,Ju]. \end{aligned}$$

A standard application of the min–max principle (see e.g. Proposition 1.9) yields the inequality

$$\begin{aligned} \mathcal {N}(M^2,C^2)&\le \mathcal {N}\Big (M^2,(S_1-W) \,\oplus \big ((S_0+M^2)\otimes I_N\big )\Big )\\&\equiv \mathcal {N}(M^2,S_1-W) + N \mathcal {N}(0,S_0). \end{aligned}$$

As $S_1-W$ is semibounded from below with compact resolvent, one has $\mathcal {N}(M^2,S_1-W)<\infty$, and in order to show $\mathcal {N}(M^2,C^2)<\infty$ it is sufficient to prove that $\mathcal {N}(0,S_0)<\infty$.

Recall that W vanishes outside $b_R$. Let $T_R$ be the self-adjoint operator in $L^2(b_R)$ with

$$\begin{aligned} T_R[u,u]=\int _{b_R} \big ( |\nabla u|^2- W |u|^2\big )\,\mathrm {d}x, \quad {\mathcal {Q}}(T_R)=H^1(b_R), \end{aligned}$$

denote by $O_R$ the zero operator in $L^2(\mathbb {R}^2{\setminus } b_R)$, and consider the linear map

$$\begin{aligned} J':{\mathcal {Q}}(S_0)\rightarrow {\mathcal {Q}}(T_R)\times {\mathcal {Q}}(O_R), \quad J'u=(u|_{b_R}, u|_{\mathbb {R}^2{\setminus } b_R}). \end{aligned}$$

One has then

$$\begin{aligned} S_0[u,u]\ge \int _{b_R} \big ( |\nabla u|^2- W |u|^2\big )\,\mathrm {d}x\equiv (T_R \oplus O_R)[J'u,J'u] \end{aligned}$$

implying $\mathcal {N}(0,S_0)\le \mathcal {N}(0,T_R\oplus O_R)\equiv \mathcal {N}(0,T_R)+\mathcal {N}(0,O_R)\equiv \mathcal {N}(0,T_R) <\infty$, as $T_R$ is semibounded from below with compact resolvent. This shows that $\mathcal {N}(M^2,C^2)<\infty$ and finishes the proof of(b).

• (c) According to the general theory of Clifford algebras, see e.g. Theorem 15.19 in [9], for $n\notin 4\mathbb {Z}$ there exists an antilinear map $\theta :\mathbb {C}^N\rightarrow \mathbb {C}^N$ with $\theta ^2\in \{-1,+1\big \}$ and²

  $$\begin{aligned} \theta \alpha _j = \alpha _j \theta \text { for } j=1,\dots , n, \quad \theta (\mathrm {i}\alpha _{n+1})=(\mathrm {i}\alpha _{n+1})\theta . \end{aligned}$$

  (62)

The pointwise map $\Theta$ defined by $(\Theta u)(x)=\theta \big ( u(x)\big )$ is clearly an isomorphism of $H^1(\mathbb {R}^n,\mathbb {C}^N)$. Let $u\in \ker (C-E)$, $E\in \mathbb {R}$, then $\Theta u\in {\mathcal {D}}(C)$ and

$$\begin{aligned} E u=-\mathrm {i}\sum _{j=1}^n\alpha _j \dfrac{\partial u}{\partial x_j} + V\alpha _{n+1} u. \end{aligned}$$

The last equality in (62) rewrites as $\theta \alpha _{n+1}= - \alpha _{n+1}\theta$. We compute then

$$\begin{aligned} -E\Theta u&=\Theta (-Eu)=\Theta \Big (\mathrm {i}\sum _{j=1}^n\alpha _j \dfrac{\partial u}{\partial x_j} - V\alpha _{n+1} u\Big )\\&=-\mathrm {i}\sum _{j=1}^n\theta \alpha _j \dfrac{\partial u}{\partial x_j} - V\theta \alpha _{n+1} u =-\mathrm {i}\sum _{j=1}^n\alpha _j \theta \dfrac{\partial u}{\partial x_j} + V\alpha _{n+1} \theta u=C\Theta u, \end{aligned}$$

which shows that $\Theta u\in \ker (C+E)$. By construction one has $\Theta ^2\in \{1,-1\}$, hence, $\Theta$ is a bijection and the claim follows. $\square$

Let us now discuss the basic properties of the operator $A_m$.

Proposition A.2

The following assertions hold:

• (a) The operator $A_m$ is self-adjoint, and, in addition, it is essentially self-adjoint on ${\mathcal {D}}_\infty (A_m):= C^\infty (\overline{\Omega },\mathbb {C}^N)\cap {\mathcal {D}}(A_m)$,

• (b) The operator $A_m$ has compact resolvent, and its eigenfunctions belong to ${\mathcal {D}}_\infty (A_m)$,

• (c) Assume in addition that $n\notin 4\mathbb {Z}$, then the spectrum of $A_m$ is symmetric with respect to zero, i.e. $\dim \ker (A_{m}-E)=\dim \ker (A_{m}+E)$ for any $E\in \mathbb {R}$.

The above assertions also hold with $A_m$ replaced by $A_m'$.

Proof

The assertions (a) and (b) are standard properties of elliptic Dirac operators. One can use e.g. [3, Ex. 4.20] or [4, Ex. 7.26], by noting that $\mathcal {B}$ is a boundary chirality operator, which shows that $A_m$ belongs to the class of Dirac operators with local elliptic boundary conditions. Then [3, Thm. 4.11] implies that the restriction R of $A_m$ on ${\mathcal {D}}_\infty (A_m)$ is essentially self-adjoint, while the boundary regularity theorem [3, Thm. 4.9] implies that $\overline{R}\subset A_m$. As $A_m$ is clearly symmetric, this implies that $A_m$ is self-adjoint and proves (a). The same boundary regularity theorem [3, Thm. 4.9] implies that all eigenfunctions are smooth up to the boundary, and the compactness of the resolvent of $A_m$ follows then from the compactness of the embedding of $H^1(\Omega )$ into $L^2(\Omega )$. This proves (b).

The proof of (c) is very similar to the proof of Proposition A.1(c) with an additional attention given to the operator domain. Namely, for $n\notin 4\mathbb {Z}$ there exist an antilinear map $\theta :\mathbb {C}^N\rightarrow \mathbb {C}^N$ with $\theta ^2\in \{-1,+1\big \}$ satisfying (62), and the pointwise map $\Theta$ defined by $(\Theta u)(x)=\theta \big (u(x)\big )$ is again an isomorphism of $H^1(\Omega ,\mathbb {C}^N)$. Furthermore, if $v\in H^1(\Omega ,\mathbb {C}^N)$ satisfies the boundary condition $v=\mathcal {B}v$ on $\Sigma$, then due to

$$\begin{aligned} \Theta \mathcal {B}(s)=\Theta \big ( -\mathrm {i}\alpha _{n+1} \Gamma (s)\big )=-\mathrm {i}\alpha _{n+1} \Theta \Gamma (s)=-\mathrm {i}\alpha _{n+1}\Gamma (s)\Theta =\mathcal {B}\Theta (s) \end{aligned}$$

one also has $\Theta v=\mathcal {B}\Theta v$ on $\Sigma$. Therefore, if $v\in {\mathcal {D}}(A_m)$, then also $\Theta v\in {\mathcal {D}}(A_m)$. Finally, let $v\in \ker (A_m-E)$, $E\in \mathbb {R}$, then the same computation as in Proposition A.1(c) gives $\Theta v\in \ker (A_m+E)$.

The preceding arguments apply to $A_m'$ as well. First, $(-\mathcal {B})$ is still a boundary chirality operator in the sense of [3], which implies the assertions (a) and (b). The proof of (c) is the same by noting that $v=-\mathcal {B}v$ on $\Sigma$ yields $\Theta v=-\mathcal {B}\Theta v$ on $\Sigma$. $\square$

B Schrödinger–Lichnerowicz Formula for Euclidean Hypersurfaces

Let $\Sigma \subset \mathbb {R}^n$ be a smooth compact hypersurface with the outer unit normal field $\nu$ and endowed with the Riemannian metric induced by the embedding. Recall that the standard scalar product in $\mathbb {R}^n$ gives rise to the induced scalar product in $T\Sigma$, which we simply denote by $\langle \cdot ,\cdot \rangle$ in this section. Denote by W the Weingarten operator, $W X=\nabla _X \nu$ for $X\in T\Sigma$, with $\nabla$ being the gradient in $\mathbb {R}^n$. Recall that the Levi-Civita connection $\nabla '$ on $\Sigma$ is given by the Gauss formula

$$\begin{aligned} \nabla '_X Y=\nabla _X Y+\langle WX,Y\rangle \nu , \quad X,Y\in T\Sigma . \end{aligned}$$

We denote

$$\begin{aligned} H_1:={{\,\mathrm{tr}\,}}W, \quad |W|^2:={{\,\mathrm{tr}\,}}(W^2), \quad H_2:=\dfrac{H_1^2-|W|^2}{2}, \end{aligned}$$

i.e. $H_1$ is the mean curvature and $H_2$ is the half of the scalar curvature of $\Sigma$.

Let $N\in \mathbb {N}$ and $\gamma _1,\dots ,\gamma _n$ be $N\times N$ anticommuting Hermitian matrices satisfying $\gamma _j^2=I$, with I being the $N\times N$ identity matrix, then the matrices

$$\begin{aligned} \gamma (x):=\sum _{j=1}^n x_j \gamma _j, \quad x=(x_1,\dots ,x_n)\in \mathbb {R}^n, \end{aligned}$$

satisfy the Clifford commutation relation $\gamma (x)\gamma (y)+\gamma (y)\gamma (x)=2\langle x,y\rangle I$ for all $x,y\in \mathbb {R}^n$. Let us recall the definition of the associated extrinsically defined Dirac operator $D^\Sigma$ on $\Sigma$ following [14, Sec. 1–3].³ The induced spin connection $\nabla ^\Sigma$ on $\Sigma$ is defined by

$$\begin{aligned} \nabla ^\Sigma _X \psi =\nabla _X+\dfrac{1}{2}\,\gamma (\nu )\gamma (WX):\, C^\infty (\Sigma ,\mathbb {C}^N)\rightarrow C^\infty (\Sigma ,\mathbb {C}^N), \quad X\in T\Sigma , \end{aligned}$$

then $D^\Sigma$ acts on functions $\psi \in C^\infty (\Sigma ,\mathbb {C}^N)$ by

$$\begin{aligned} D^\Sigma \psi :=-\gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) \nabla ^\Sigma _{e_j} \psi \end{aligned}$$

with $(e_1,\dots ,e_{n-1})$ being an orthonormal frame of $T\Sigma$. Recall that $\gamma (e_j)$ anticommute with $\gamma (\nu )$ and, furthermore,

$$\begin{aligned} \sum _{j=1}^{n-1} \gamma (e_j) \gamma (We_j)=H_1\, I \end{aligned}$$

(63)

(which is seen by testing on an eigenbasis of W), and we may rewrite

$$\begin{aligned} D^\Sigma \psi =\dfrac{H_1}{2}\,\psi -\gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) \nabla _{e_j}\psi , \end{aligned}$$

Being viewed as an operator in $L^2(\Sigma ,\mathbb {C}^N)$, the operator $D^\Sigma$ is known to be essentially self-adjoint on $C^\infty (\Sigma ,\mathbb {R}^N)$. We would like to provide a elementary direct proof, adapted to the Euclidean setting, of the eminent Schrödinger–Lichnerowicz formula

$$\begin{aligned} (D^\Sigma )^2=(\nabla ^\Sigma )^*\nabla ^\Sigma +\dfrac{H_2}{2}\, I, \end{aligned}$$

(64)

where the first term on the right-hand side is the Bochner Laplacian associated with the above spin connection $\nabla ^\Sigma$, which is a self-adjoint operator in $L^2(\Sigma ,\mathbb {C}^N)$. (We refer to the original papers [17, 22] and other sources [7, 11,12,13] for a more general setting.)

In what follows we use the standard identification of $T\Sigma$ and $T^*\Sigma$ with the help of the musical isomorphism. For $\psi \in C^\infty (\Sigma ,\mathbb {C}^N)$ we have the decomposition

$$\begin{aligned} \nabla ^\Sigma \psi =\sum _{j=1}^{n-1} e_j\otimes \nabla ^\Sigma _{e_j} \psi = \sum _{j=1}^{n-1} e_j\otimes \Big (\nabla _{e_j}+\dfrac{1}{2}\,\gamma (\nu )\gamma (W e_j)\Big ) \psi . \end{aligned}$$

(65)

To compute the adjoint $(\nabla ^\Sigma )^*:T^*\Sigma \otimes C^\infty (\Sigma ,\mathbb {C}^N)\rightarrow C^\infty (\Sigma ,\mathbb {C}^N)$, let $X\in T\Sigma \simeq T^* \Sigma$ and $\varphi ,\psi \in C^\infty (\Sigma ,\mathbb {C}^N)$ then

$$\begin{aligned} \big \langle (\nabla ^\Sigma )^*(X\otimes \varphi ),\psi \big \rangle _{L^2(\Sigma ,\mathbb {C}^N)}&=\langle X\otimes \varphi ,\nabla ^\Sigma \psi \rangle _{T^*\Sigma \otimes L^2(\Sigma ,\mathbb {C}^N)}\\&=\big \langle \varphi , \nabla _X \psi +\dfrac{1}{2}\,\gamma (\nu )\gamma (WX)\psi \big \rangle _{L^2(\Sigma ,\mathbb {C}^N)}\\&=\big \langle \varphi , \nabla _X \psi \big \rangle _{L^2(\Sigma ,\mathbb {C}^N)} +\Big \langle \dfrac{1}{2}\,\gamma (WX) \gamma (\nu )\varphi ,\psi \Big \rangle _{L^2(\Sigma ,\mathbb {C}^N)}. \end{aligned}$$

Using Leibniz rule and the divergence theorem we have

$$\begin{aligned} \langle \varphi , \nabla _X \psi \rangle _{L^2(\Sigma ,\mathbb {C}^N)}&= \int _\Sigma X\langle \varphi ,\psi \rangle _{\mathbb {C}^N} \,\mathrm {d}s-\langle \nabla _X \varphi , \psi \rangle _{L^2(\Sigma ,\mathbb {C}^N)}\\&=-\big \langle ({{\,\mathrm{div}\,}}_\Sigma X) \varphi +\nabla _X \varphi ,\psi \big \rangle _{L^2(\Sigma ,\mathbb {C}^N)}, \end{aligned}$$

where ${{\,\mathrm{div}\,}}_\Sigma$ is the divergence on $\Sigma$,

$$\begin{aligned} {{\,\mathrm{div}\,}}_\Sigma X=\sum _{j=1}^{n-1} \langle e_j,\nabla '_{e_j} X\rangle . \end{aligned}$$

Therefore,

$$\begin{aligned} (\nabla ^\Sigma )^*(X\otimes \varphi )=-({{\,\mathrm{div}\,}}_\Sigma X)\,\varphi -\nabla _X \varphi + \frac{1}{2}\,\gamma (WX)\gamma (\nu )\varphi . \end{aligned}$$

By combining (65) with the last expression, for $\psi \in C^\infty (\Sigma ,\mathbb {C}^N)$ one obtains

$$\begin{aligned} (\nabla ^\Sigma )^* \nabla ^\Sigma \psi&=\sum _{j=1}^{n-1} (\nabla ^\Sigma )^* \Big [e_j\otimes \Big (\nabla _{e_j}+\dfrac{1}{2}\,\gamma (\nu )\gamma (W e_j)\Big ) \psi \Big ]\nonumber \\&=-\sum _{j=1}^{n-1} ({{\,\mathrm{div}\,}}_\Sigma e_j) \Big (\nabla _{e_j}+\dfrac{1}{2}\,\gamma (\nu )\gamma (W e_j)\Big ) \psi \nonumber \\&\quad +\sum _{j=1}^{n-1}\bigg \{ -\nabla _{e_j}\Big (\nabla _{e_j}+\dfrac{1}{2}\,\gamma (\nu )\gamma (W e_j)\Big )\psi \nonumber \\&\quad +\dfrac{1}{2}\,\gamma (We_j)\gamma (\nu )\Big (\nabla _{e_j}+\dfrac{1}{2}\,\gamma (\nu )\gamma (W e_j)\Big )\psi \bigg \}=:S_1+S_2. \end{aligned}$$

(66)

To simplify $S_1$ we first use the Leibniz rule and the orthogonality of $(e_j)$ to obtain

$$\begin{aligned} {{\,\mathrm{div}\,}}_\Sigma e_j&=\sum _{k=1}^{n-1} \langle e_k,\nabla '_{e_k} e_j\rangle =-\sum _{k=1}^{n-1} \langle \nabla '_{e_k} e_k, e_j\rangle ,\\ S_1&=\sum _{j,k=1}^{n-1} \langle \nabla '_{e_k} e_k, e_j\rangle \nabla _{e_j}\psi +\dfrac{1}{2}\sum _{j,k=1}^{n-1} \langle \nabla '_{e_k} e_k, e_j\rangle \gamma (\nu )\gamma (W e_j) \psi \\&=\sum _{k=1}^{n-1} \Big (\sum _{j=1}^{n-1}\langle \nabla '_{e_k} e_k, e_j\rangle \nabla _{e_j}\psi \Big ) +\dfrac{1}{2}\sum _{k=1}^{n-1} \gamma (\nu ) \gamma \bigg ( W\sum _{j=1}^{n-1}\langle \nabla '_{e_k} e_k, e_j\rangle e_j \bigg )\\&=\sum _{k=1}^{n-1} \nabla _{\nabla '_{e_k} e_k} \psi +\dfrac{1}{2}\sum _{k=1}^{n-1} \gamma (\nu )\gamma \Big ( W \nabla '_{e_k} e_k\Big ). \end{aligned}$$

Furthermore,

$$\begin{aligned} S_2&=\sum _{j=1}^{n-1} \bigg \{-\nabla _{e_j}\nabla _{e_j}\psi -\dfrac{1}{2}\,\gamma (We_j)\gamma (W e_j)\psi -\dfrac{1}{2}\,\gamma (\nu )\gamma \big (\nabla _{e_j}(W e_j)\big )\psi \\&\qquad -\dfrac{1}{2}\,\gamma (\nu )\gamma (We_j)\nabla _{e_j}\psi +\dfrac{1}{2}\,\gamma (We_j)\gamma (\nu )\nabla _{e_j}\psi + \dfrac{1}{4}\,\gamma (We_j)\gamma (\nu )\gamma (\nu )\gamma (W e_j)\psi \bigg \}\\&=\sum _{j=1}^{n-1} \bigg \{-\nabla _{e_j}\nabla _{e_j}\psi -\dfrac{1}{2}\,\gamma (\nu )\gamma \big (\nabla _{e_j}(W e_j)\big )\psi -\gamma (\nu )\gamma (We_j)\nabla _{e_j}\psi \bigg \}\psi - \dfrac{1}{4}\,|W|^2\psi , \end{aligned}$$

and then

$$\begin{aligned} (\nabla ^\Sigma )^* \nabla ^\Sigma \psi= & {} \sum _{j=1}^{n-1} \bigg [ \nabla _{\nabla '_{e_j} e_j} \psi -\nabla _{e_j}\nabla _{e_j}\psi \\&+\dfrac{1}{2}\gamma (\nu )\gamma \Big ( W \nabla '_{e_j} e_j-\nabla _{e_j}(W e_j)\Big )\psi -\gamma (\nu )\gamma (We_j)\nabla _{e_j}\psi \bigg ] - \dfrac{1}{4}\,|W|^2\psi . \end{aligned}$$

Using $\nabla '_{e_j} (W e_j)=\nabla _{e_j} (W e_j)+|W e_j|^2\nu$ and Leibniz rule we have

$$\begin{aligned} W \nabla '_{e_j} e_j-\nabla _{e_j}(W e_j)=W\nabla '_{e_j} e_j-\nabla '_{e_j} (W e_j)+|W e_j|^2\nu =-(\nabla '_{e_j} W)e_j +|W e_j|^2\nu \end{aligned}$$

implying $\gamma (\nu )\gamma \big ( W \nabla '_{e_j} e_j-\nabla _{e_j}(W e_j)\big )\psi =-\gamma (\nu )\gamma \big ((\nabla '_{e_j} W)e_j\big )\psi +|W e_j|^2\psi$, and then

$$\begin{aligned} (\nabla ^\Sigma )^* \nabla ^\Sigma \psi= & {} \sum _{j=1}^{n-1} \bigg [ \nabla _{\nabla '_{e_j} e_j} \psi -\nabla _{e_j}\nabla _{e_j}\psi \nonumber \\&-\dfrac{1}{2} \,\gamma (\nu )\,\gamma \big ((\nabla '_{e_j} W)e_j\big )\psi -\gamma (\nu )\,\gamma (We_j)\nabla _{e_j}\psi \bigg ] + \dfrac{1}{4}\,|W|^2\psi .\nonumber \\ \end{aligned}$$

(67)

On the other hand,

$$\begin{aligned} (D^\Sigma )^2 \psi&=\Big (\dfrac{H_1}{2}-\gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) \nabla _{e_j}\Big )\Big (\dfrac{H_1\,\psi }{2}-\gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) \nabla _{e_j}\psi \Big )\nonumber \\&=\dfrac{H_1^2}{4}\,\psi -\dfrac{1}{2} \Big ( \gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) \nabla _{e_j} H_1\Big )\, \psi - \dfrac{H_1}{2}\,\gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) \nabla _{e_j}\psi \nonumber \\&\quad -\dfrac{H_1}{2}\,\gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) \nabla _{e_j}\psi +\gamma (\nu )\sum _{j,k=1}^{n-1} \gamma (e_j)\gamma (We_j)\gamma (e_k)\nabla _{e_k}\psi \nonumber \\&\quad + \gamma (\nu )\sum _{j,k=1}^{n-1} \gamma (e_j)\gamma (\nu )\gamma (\nabla _{e_j}e_k)\nabla _{e_k}\psi +\gamma (\nu )\sum _{j,k=1}^{n-1}\gamma (e_j)\gamma (\nu )\gamma (e_k)\nabla _{e_j}\nabla _{e_k}\psi . \end{aligned}$$

(68)

The sum of the third, forth and fifth terms is zero, in fact,

$$\begin{aligned}&-\dfrac{H_1}{2}\,\gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) \nabla _{e_j}\psi -\dfrac{H_1}{2}\,\gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) \nabla _{e_j}\psi \\&\qquad +\gamma (\nu )\sum _{j,k=1}^{n-1} \gamma (e_j)\gamma (We_j)\gamma (e_k)\nabla _{e_k}\psi \\&\quad = -H_1\,\gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_k) \nabla _{e_k}\psi +\gamma (\nu )\sum _{j,k=1}^{n-1} \gamma (e_j)\gamma (We_j)\gamma (e_k)\nabla _{e_k}\psi \\&\quad =\gamma (\nu )\sum _{k=1}^{n-1} \Big ( -H_1 +\sum _{j=1}^{n-1}\gamma (e_j)\gamma (We_j)\Big )\gamma (e_k)\nabla _{e_k}\psi =0 \end{aligned}$$

as the term in the parentheses vanishes due to (63). Therefore, Eq. (68) reads as

$$\begin{aligned} (D^\Sigma )^2\psi&=\dfrac{H_1^2}{4}\,\psi -\dfrac{1}{2} \, \gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) (\nabla _{e_j} H_1)\, \psi - \sum _{j,k=1}^{n-1} \gamma (e_j)\gamma (\nabla _{e_j}e_k)\nabla _{e_k}\psi \nonumber \\&\quad -\sum _{j,k=1}^{n-1}\gamma (e_j)\gamma (e_k)\nabla _{e_j}\nabla _{e_k}\psi . \end{aligned}$$

(69)

We transform the last summand as follows:

$$\begin{aligned} \sum _{j,k=1}^{n-1}\gamma (e_j)\gamma (e_k)\nabla _{e_j}\nabla _{e_k}\psi&=\dfrac{1}{2}\,\sum _{j,k=1}^{n-1}\Big (\gamma (e_j)\gamma (e_k)\nabla _{e_j}\nabla _{e_k}\psi +\gamma (e_k)\gamma (e_j)\nabla _{e_k}\nabla _{e_j}\psi \Big )\\&=\dfrac{1}{2}\,\sum _{j,k=1}^{n-1}\Big (\gamma (e_j)\gamma (e_k)+\gamma (e_k)\gamma (e_j)\Big )\nabla _{e_j}\nabla _{e_k}\psi \\&\quad +\dfrac{1}{2}\sum _{j,k=1}^{n-1}\gamma (e_k)\gamma (e_j) \Big ( \nabla _{e_k}\nabla _{e_j}-\nabla _{e_j}\nabla _{e_k}\Big )\psi \\&=\sum _{j=1}^{n-1}\nabla _{e_j}\nabla _{e_j}\psi +\dfrac{1}{2}\,J, \end{aligned}$$

where

$$\begin{aligned} J:=\sum _{j,k=1}^{n-1}\gamma (e_j)\gamma (e_k) \big ( \nabla _{e_j}\nabla _{e_k}-\nabla _{e_k}\nabla _{e_j}\big )\psi \equiv \sum _{j,k=1}^{n-1}\gamma (e_j)\gamma (e_k) \nabla _{[e_j,e_k]}\psi . \end{aligned}$$

Representing $[e_j,e_k]=\sum _{l=1}^{n-1} \big \langle e_l,[e_j,e_k]\big \rangle e_k$ we have

$$\begin{aligned} J=\sum _{j,k,l=1}^{n-1} \gamma (e_j)\gamma (e_k)\Big [\langle e_l,\nabla '_{e_j}e_k\rangle -\langle e_l,\nabla '_{e_k}e_j\rangle \Big ]\nabla _{e_l}\psi , \end{aligned}$$

and using

$$\begin{aligned} \sum _{j=1}^{n-1} e_j\langle e_l,\nabla '_{e_k}e_j\rangle =-\sum _{j=1}^{n-1} e_j\langle \nabla '_{e_k} e_l,e_j\rangle =-\nabla '_{e_k} e_l \end{aligned}$$

we rewrite

$$\begin{aligned} J&=-\sum _{j,k=1}^{n-1}\gamma (e_j)\,\gamma (\nabla '_{e_j}e_k) \nabla _{e_k}\psi +\sum _{j,k=1}^{n-1}\gamma (\nabla '_{e_j}e_k)\,\gamma (e_j)\, \nabla _{e_k}\psi \\&=-2\sum _{j,k=1}^{n-1}\gamma (e_j)\,\gamma (\nabla '_{e_j}e_k) \nabla _{e_k}\psi \\&\quad +\sum _{j,k=1}^{n-1}\Big ( \gamma (e_j)\,\gamma (\nabla '_{e_j}e_k)+\gamma (\nabla '_{e_j}e_k)\,\gamma (e_j)\,\Big ) \nabla _{e_k}\psi \\&=-2\sum _{j,k=1}^{n-1}\gamma (e_j)\,\gamma (\nabla '_{e_j}e_k) \nabla _{e_k}\psi +2\sum _{j,k=1}^{n-1}\langle e_j,\nabla '_{e_j}e_k\rangle \nabla _{e_k}\psi \\&=-2\sum _{j,k=1}^{n-1}\gamma (e_j)\,\gamma (\nabla '_{e_j}e_k) \nabla _{e_k}\psi -2\sum _{j,k=1}^{n-1}\langle \nabla '_{e_j}e_j,e_k\rangle \nabla _{e_k}\psi \\&=-2\sum _{j,k=1}^{n-1}\gamma (e_j)\,\gamma (\nabla '_{e_j}e_k) \nabla _{e_k}\psi -2\sum _{j=1}^{n-1} \nabla _{\nabla '_{e_j}e_j}\psi . \end{aligned}$$

The substitution into (69) gives

$$\begin{aligned} (D^\Sigma )^2 \psi&=\dfrac{H_1^2}{4}\,\psi -\dfrac{1}{2} \, \gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) (\nabla _{e_j} H_1)\, \psi - \sum _{j,k=1}^{n-1} \gamma (e_j)\gamma (\nabla _{e_j}e_k)\nabla _{e_k}\psi \\&\quad -\sum _{j=1}^{n-1}\nabla _{e_j}\nabla _{e_j}\psi +\sum _{j,k=1}^{n-1}\gamma (e_j)\,\gamma (\nabla '_{e_j}e_k) \nabla _{e_k}\psi +\sum _{j=1}^{n-1} \nabla _{\nabla '_{e_j}e_j}\psi . \end{aligned}$$

The sum of the third and the fifth terms simplifies as

$$\begin{aligned}&- \sum _{j,k=1}^{n-1} \gamma (e_j)\gamma (\nabla _{e_j}e_k)\nabla _{e_k}\psi +\sum _{j,k=1}^{n-1}\gamma (e_j)\,\gamma (\nabla '_{e_j}e_k) \nabla _{e_k}\psi \\&\quad =\sum _{j,k=1}^{n-1} \gamma (e_j)\gamma (\nabla '_{e_j}e_k-\nabla _{e_j}e_k) \nabla _{e_k}\psi =\sum _{j,k=1}^{n-1} \gamma (e_j)\gamma \big (\langle W e_j,e_k\rangle \nu \big ) \nabla _{e_k}\psi \\&\quad =\sum _{j,k=1}^{n-1} \gamma \big (e_j \langle e_j, W e_k\rangle \big )\gamma (\nu ) \nabla _{e_k}\psi = \sum _{k=1}^{n-1}\gamma (W e_k) \gamma (\nu )\nabla _{e_k}\psi , \end{aligned}$$

hence,

$$\begin{aligned} (D^\Sigma )^2 \psi= & {} \dfrac{H_1^2}{4}\, \psi -\dfrac{1}{2} \, \gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) (\nabla _{e_j} H_1)\, \psi \\&+\sum _{j=1}^{n-1}\gamma (W e_j) \gamma (\nu )\nabla _{e_j}\psi -\sum _{j=1}^{n-1}\nabla _{e_j}\nabla _{e_j}\psi +\sum _{j=1}^{n-1} \nabla _{\nabla '_{e_j}e_j}\psi . \end{aligned}$$

By comparing the last expression with (67) we obtain

$$\begin{aligned}&D^2_\Sigma \psi - (\nabla ^\Sigma )^*\nabla ^\Sigma \psi \\&\quad = \dfrac{H_1^2}{4}\,\psi -\dfrac{1}{4} |W|^2\,\psi -\dfrac{1}{2} \, \gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) (\nabla _{e_j} H_1)\, \psi +\sum _{j=1}^{n-1}\gamma (W e_j) \gamma (\nu )\nabla _{e_j}\psi \\&\qquad +\dfrac{1}{2} \sum _{j=1}^{n-1}\,\gamma (\nu )\,\gamma \big ((\nabla '_{e_j} W)e_j\big )\psi +\sum _{j=1}^{n-1}\gamma (\nu )\,\gamma (We_j)\nabla _{e_j}\psi . \end{aligned}$$

The sum of the fourth and sixth term on the right hand is zero, hence,

$$\begin{aligned}&(D^\Sigma )^2 \psi - (\nabla ^\Sigma )^*\nabla ^\Sigma \psi \\&\quad = \dfrac{H_2}{2}\,\psi -\dfrac{1}{2} \, \gamma (\nu ) \sum _{j=1}^{n-1} \gamma (e_j) (\nabla _{e_j} H_1)\, \psi +\dfrac{1}{2} \sum _{j=1}^{n-1}\,\gamma (\nu )\,\gamma \big ((\nabla '_{e_j} W)e_j\big )\psi \\&\quad =\dfrac{H_2}{2}\,\psi + \dfrac{1}{2}\,\gamma (\nu ) \gamma \bigg (\sum _{j=1}^{n-1} (\nabla '_{e_j} W) e_j - \sum _{j=1}^{n-1} (\nabla _{e_j} H_1)e_j\bigg )\psi . \end{aligned}$$

Therefore, to show the sought identity (64) it is sufficient to prove the equality

$$\begin{aligned} \sum _{j=1}^{n-1} (\nabla '_{e_j} W)e_j=\sum _{j=1}^{n-1} (\nabla _{e_j} H_1)e_j. \end{aligned}$$

(70)

In order to check (70) let us remark that $\nabla _X \nabla _Y Z-\nabla _Y \nabla _X Z - \nabla _{[X,Y]} Z=0$ for any $X,Y,Z\in T\Sigma$. Using the definition of $\nabla '$ we have

$$\begin{aligned} 0&= \nabla _X \big ( \nabla '_Y Z - \langle WY,Z\rangle \nu \big )-\nabla _Y\big (\nabla '_X Z -\langle WX,Z\rangle \nu \big )- \nabla '_{[X,Y]} Z +\big \langle W[X,Y],Z\big \rangle \nu \\&= \nabla '_X\big ( \nabla '_Y Z - \langle WY,Z\rangle \nu \big )- \big \langle WX, \nabla '_Y Z - \langle WY,Z\rangle \nu \big \rangle - \nabla '_Y\big (\nabla '_X Z -\langle WX,Z\rangle \nu \big )\\&\quad +\big \langle WY, \nabla '_X Z -\langle WX,Z\rangle \nu \big \rangle - \nabla '_{[X,Y]} Z +\big \langle W[X,Y],Z\big \rangle \nu . \end{aligned}$$

Using $\nabla '_X \nu =\nabla _X \nu =WX$ we then arrive at

$$\begin{aligned} 0&= \nabla '_X\nabla '_Y Z - \big \langle (\nabla '_X W)Y,Z\big \rangle \nu -\big \langle W(\nabla '_X Y),Z\big \rangle \nu -\langle WY,\nabla '_X Z\rangle \nu -\langle WY,Z\rangle WX\\&\quad - \langle WX, \nabla '_Y Z\rangle \nu -\nabla '_Y \nabla '_X Z+ \big \langle (\nabla '_Y W)X,Z\big \rangle \nu +\langle W\nabla '_Y X,Z\rangle \nu \\&\quad +\langle WX,\nabla '_Y Z\rangle \nu +\langle WX,Z\rangle WY+\langle WY, \nabla '_X Z\rangle - \nabla '_{[X,Y]} Z +\big \langle W[X,Y],Z\big \rangle \nu \\&= \nabla '_X\nabla '_Y Z-\nabla '_Y \nabla '_X Z- \nabla '_{[X,Y]} Z -\langle WY,Z\rangle WX +\langle WX,Z\rangle WY\\&\quad +\big \langle (\nabla '_Y W)X,Z\big \rangle \nu - \big \langle (\nabla '_X W)Y,Z\big \rangle \nu \\&\quad +\langle W\nabla '_Y X,Z\rangle \nu -\big \langle W \nabla '_X Y,Z\big \rangle \nu +\big \langle W[X,Y],Z\big \rangle \nu . \end{aligned}$$

As $\nabla '_X Y-\nabla '_Y X=[X,Y]$, the sum of the three terms in the last line vanishes. Considering the normal components of the remaining equality we obtain $\big \langle (\nabla '_Y W)X,Z\big \rangle = \big \langle (\nabla '_X W)Y,Z\big \rangle$, and then $\big \langle (\nabla '_Y W)X,Z\big \rangle =\big \langle Y,(\nabla '_X W) Z\big \rangle$. Taking $Y=Z=e_k$ and summing over k we arrive at

$$\begin{aligned} \sum _{k=1}^{n-1}\big \langle (\nabla '_{e_k} W) X,e_k\big \rangle =\sum _{j=1}^{n-1} \big \langle e_k,(\nabla '_X W ) e_k\big \rangle , \text { i.e. } \sum _{k=1}^{n-1}\big \langle X, (\nabla '_{e_k} W) e_k\big \rangle =\nabla _X H_1. \end{aligned}$$

Using the last equality for $X=e_j$ we obtain

$$\begin{aligned} \sum _{j=1}^{n-1}\sum _{k=1}^{n-1}\big \langle e_j, (\nabla '_{e_k} W) e_k\big \rangle e_j=\sum _{j=1}^{n-1} (\nabla _{e_j} H_1) e_j. \end{aligned}$$

The left-hand side of the last equality simplifies to $\sum _{k=1}^{n-1} (\nabla '_{e_k} W) e_k$, which gives (70) and finishes the proof of (64).

C Dirac Operator on a Loop

Let $\Sigma \subset \mathbb {R}^2$ be a smooth loop of length $\ell >0$. We give an explicit computation for the eigenvalues of the intrinsic Dirac operator on $\Sigma$. Consider first the associated extrinsically defined Dirac operator $D^\Sigma$ using the notation introduced in Sect. 2.2. Denote by $\mathbb {T}:=\mathbb {R}/(\ell \mathbb {Z})$ and let $\gamma :\mathbb {T}\rightarrow \mathbb {R}^2$ be an arc-length parametrization of $\Sigma$, which provides a global coordinate of $\Sigma$. To be definite, assume that $\gamma (s)$ runs through $\Sigma$ in the anti-clockwise direction as s runs from 0 to $\ell$, which amounts to the choice of an orientation. We take (e) with $e=\tau :=\gamma '$ as an orthonormal frame tangent to $\Sigma$ and denote by $\nu$ the outer unit normal. In addition, let us make an explicit choice of $2\times 2$ matrices $\beta _1$ and $\beta _2$ satisfying the Clifford commutation relation: we choose them as the Pauli matrices,

$$\begin{aligned} \beta _1=\begin{pmatrix} 0 &{} 1\\ 1 &{} 0 \end{pmatrix}, \quad \beta _2=\begin{pmatrix} 0 &{} -\mathrm {i}\\ \mathrm {i}&{} 0 \end{pmatrix}, \end{aligned}$$

then, by setting $\mathcal {N}:=\nu _1+\mathrm {i}\nu _2$ and $\mathcal {T}:=\tau _1+\mathrm {i}\tau _2$,

$$\begin{aligned} \beta (\nu )=\begin{pmatrix} 0 &{} \overline{\mathcal {N}}\\ \mathcal {N}&{} 0 \end{pmatrix}, \quad \beta (e)= \begin{pmatrix} 0 &{} \overline{\mathcal {T}}\\ \mathcal {T}&{} 0 \end{pmatrix} \end{aligned}$$

Using (11) we realize $D^\Sigma$ as an operator in $L^2(\mathbb {T},\mathbb {C}^2)$: for $\psi \in C^\infty (\mathbb {T},\mathbb {C}^2)$ one has

$$\begin{aligned} D^\Sigma \psi =\Big (\dfrac{H_1}{2}-\beta (\nu ) \sum _{j=1}^{n-1} \beta (e_j) \nabla _{e_j}\Big ) \psi = \dfrac{\kappa }{2}\,\psi - \begin{pmatrix} 0 &{} \overline{\mathcal {N}}\\ \mathcal {N}&{} 0 \end{pmatrix} \begin{pmatrix} 0 &{} \overline{\mathcal {T}}\\ \mathcal {T}&{} 0 \end{pmatrix} \psi '. \end{aligned}$$

where $\kappa$ is the curvature of $\Sigma$. The above choice of orientation gives $\mathcal {T}=\mathrm {i}\, \mathcal {N}$ and

$$\begin{aligned} D^\Sigma \psi = \dfrac{\kappa }{2}\,\psi +\mathrm {i}\begin{pmatrix} 1 &{} 0 \\ 0 &{} -1\end{pmatrix} \psi '. \end{aligned}$$

Consider the function $K:\mathbb {R}\rightarrow \mathbb {R}$ given by

$$\begin{aligned} K(s):=\dfrac{1}{2}\int _0^s \kappa (t)\,\mathrm {d}t. \end{aligned}$$

Using the well-known identity

$$\begin{aligned} \int _0^\ell \kappa (t)\,\mathrm {d}t=2\pi \end{aligned}$$

we conclude that $K(\cdot +\ell )=\pi + K$. Hence, using the unitary transform

$$\begin{aligned} U:L^2( (0,\ell ), \mathbb {C}^2)\phi \mapsto \begin{pmatrix} e^{\mathrm {i}K} &{} 0 \\ 0 &{} e^{-\mathrm {i}K} \end{pmatrix}\,\phi \in L^2(\mathbb {T},\mathbb {C}^2) \end{aligned}$$

we rewrite

$$\begin{aligned} U^{-1}D^\Sigma U \phi = \mathrm {i}\begin{pmatrix} 1 &{} 0 \\ 0 &{} -1\end{pmatrix} \phi ', \end{aligned}$$

and remark that $U\phi \in C^\infty (\mathbb {T},\mathbb {C}^2)$ if and only if $\phi$ extends to a function from $C^\infty (\mathbb {R},\mathbb {C}^2)$ with $\phi (\cdot +\ell )=-\phi$.

It follows that $D^\Sigma$ is unitarily equivalent to $D\oplus (-D)$, where D is the operator $\phi \rightarrow -\mathrm {i}\phi '$ on $(0,\ell )$ with the antiperiodic boundary condition $\phi (\ell )=-\phi (0)$, and one easily shows that the eigenvalues of D are $(2r-1)\pi /\ell$, $r\in \mathbb {Z}$. In addition, as the dimension $n=2$ is even, the operator $D^\Sigma$ is also unitarily equivalent to , which means that has the same eigenvalues as D.