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\(L^p\)-spectrum of the Dirac operator on products with hyperbolic spaces

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Abstract

We study the \(L^p\)-spectrum of the Dirac operator on complete manifolds. One of the main questions in this context is whether this spectrum depends on p. As a first example where p-independence fails we compute explicitly the \(L^p\)-spectrum for the hyperbolic space and its product with compact spaces.

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Acknowledgments

We thank an anonymous referee for many helpful comments.

Author information

Correspondence to Nadine Große.

Additional information

Communicated by J. Jost.

Appendices

Appendix A: Function spaces

We want to recall some analytical facts which are helpful to define spinorial function spaces on manifolds.

Let \((M^n,g)\) be an n-dimensional Riemannian spin manifold with the classical Dirac operator \(D:H_1^2(M, \Sigma _M)=\mathrm{dom}\, D\subset L^2(M, \Sigma _M) \rightarrow L^2(M, \Sigma _M)\). The set of compactly supported spinors \(C_c^\infty (M, \Sigma _M)\) is a core of D, i.e., D is the closure of \(D|_{C_c^\infty (M, \Sigma _M)}\) w.r.t. the graph norm \(H_1^2\).

A distributional spinor (or distribution with spinor values) is a linear map \(C_c^\infty (M, \Sigma _M)\rightarrow \mathbb {C}\) with the usual continuity properties of distributions. Any spinor with regularity \(L^1_{loc}\) defines a distributional spinor by using the standard \(L^2\)-scalar product on spinors.

Then \(D\varphi \) can be defined in the sense of distributions. Let \(H_1^s(M, \Sigma _M)\) be the set of distributional spinors \(\varphi \), such that \(\varphi \) und \(D\varphi \) are in \(L^s\), \(s\in [1,\infty ]\). Equipped with the norm \(\Vert \varphi \Vert _{H_1^s}:=\Vert \varphi \Vert _s+ \Vert D\varphi \Vert _s\) this is a Banach space. This norm is the graph norm of D viewed as an operator in \(L^s\) to \(L^s\). If we consider the operator \(D:H_1^s\subset L^s\rightarrow L^s\), we will shortly write \(D_s\) in the following.

Lemma A.1

Let \(1\le s<\infty \). \(C^\infty _c(M, \Sigma _M)\) is dense in \(H_1^s(M, \Sigma _M)\).

Proof

Assume that \(\varphi \in H_1^s(M, \Sigma _M)\), \(s<\infty \), is given. For a given point \(p\in M\) and for any \(R>0\) one can find a compactly supported smooth function \(\eta _R:M\rightarrow [0,1]\) such that \(\eta _R\equiv 1\) on \(B_R(p)\) and such that \(|\nabla \eta _R|\le R^{-1}\). Then one easily sees \(\lim _{R\rightarrow \infty } \Vert \varphi -\eta _R \varphi \Vert _s=0\). Further we calculate

$$\begin{aligned} \Vert D(\varphi -\eta _R \varphi )\Vert _s\le \Vert \nabla \eta _R\cdot \varphi \Vert _s + \Vert (1-\eta _R)D\varphi \Vert _s \rightarrow 0 \quad \mathrm{as\ }\; R\rightarrow \infty . \end{aligned}$$

Thus the elements with compact support are dense in \(H_1^s(M, \Sigma _M)\). Now if \(\psi \in H_1^s(M, \Sigma _M)\) has compact support, it follows from standard results that it can be approximated by smooth compactly supported spinors. \(\square \)

Thus, for \(s<\infty \), \(H_1^s(M, \Sigma _M)\) is equal to the completion of \(C^\infty _c(M, \Sigma _M)\) with respect to the graph norm of \(D:L^s\rightarrow L^s\). In particular this implies that for \(s<\infty \) the operator \(D_s\) is a closed extension of \(D|_{C_c^\infty (M, \Sigma _M)}\) with core \(C_c^\infty (M, \Sigma _M)\). Note that \(D_\infty \) is a closed extension of \(D|_{C_c^\infty (M, \Sigma _M)}\) as well but \(C_c^\infty (M, \Sigma _M)\) is in general no longer a core for this operator. Moreover, in the standard literature for \(L^p\)-theory of the Laplacian, e.g. [18], the operator for \(s=\infty \) is directly defined to be the adjoint operator for \(s=1\).

Lemma A.2

Let \(1<s<\infty \). On manifolds with bounded geometry, the \(H_1^s\)-norm is equivalent to the norm \(\Vert \varphi \Vert _s+\Vert \nabla \varphi \Vert _s\).

The proof of the lemma relies on local elliptic estimates which follow from the Calderon-Zygmund inequality, e.g. [20, Theorem 9.9], see also [2, Lemma 3.2.2] for the geometric adaptation.

Appendix B: General notes on the \(L^p\)-spectrum

In this section we collect general facts on the \(L^p\)-spectrum of the Dirac operator. Unless stated otherwise, we only assume that (Mg) is complete.

We examine the adjoint of the operator \(D_s:L^s\rightarrow L^s\) with respect to the duality pairing \(\langle .,.\rangle :L^s\times (L^s)^*\rightarrow \mathbb {C}\) whose restriction to compactly supported spinors coincides with the hermitian \(L^2\)-product. We use the convention that this pairing is antilinear in the second component. The adjoint \(D_s^*\) is an operator in \((L^s)^*\). For \(1\le s<\infty \) and \(s^{-1}+(s^*)^{-1}=1\), \((L^s)^*=L^{s^*}\) whereas \((L^\infty )^*\) is larger than \(L^1\). From the formal self-adjointness of D we see, that \(D_{s^*}|_{C_c^\infty (M, \Sigma _M)} = D_s^*|_{C_c^\infty (M, \Sigma _M)}\). Moreover, we have

Lemma B.1

For all \(\varphi \in H_1^s\) and \(\psi \in H_1^{s^*}\), \(1\le s \le \infty \), we have

$$\begin{aligned} (D_s\varphi ,\psi )=(\varphi , D_{s^*}\psi ). \end{aligned}$$

Proof

For \(1<s< \infty \), let \(\varphi _i, \psi _j\in C_c^\infty (M, \Sigma _M)\) with \(\varphi _i\rightarrow \varphi \) in \(H_1^s\) and \(\psi _j\rightarrow \psi \) in \(H_1^{s^*}\). Then,

$$\begin{aligned} \int _M \langle D_s\varphi ,\psi \rangle \mathrm {dvol}_g\leftarrow \int _M \langle D_s\varphi _i,\psi _j\rangle \mathrm {dvol}_g = \int _M \langle \varphi _i,D_{s^*}\psi _j\rangle \mathrm {dvol}_g\rightarrow \int _M \langle \varphi ,D_{s^*}\psi \rangle \mathrm {dvol}_g \end{aligned}$$

as \(i,j\rightarrow \infty \). Let now \(s=1\). For \(\varphi \in C_c^\infty (M, \Sigma _M)\) the equality follows from the distributional definition of \(D_\infty \). The rest follows since \(C_c^\infty (M, \Sigma _M)\) is dense in \(H_1^1\). The remaining case \(s=\infty \) just follows from the last one by interchanging s and \(s^*\). \(\square \)

Lemma B.2

For all \(1\le s<\infty \) the operators \(D_{s^*}\) and \(D_s^*\) coincide.

Proof

For \(\psi \in H_1^{s^*}\) Lemma B.1 yields \((D_s\varphi , \psi )=(\varphi , D_{s^*}\psi )\) for all \(\varphi \in H_1^s=\mathrm{dom}\, D_s\). This implies \(\psi \in \mathrm{dom}\, D_s^*\) and \(D_s^*\psi =D_{s^*}\psi .\) Hence, \(H_1^{s^*}\subset \mathrm{dom}\, D_s^*\) and \(D_s^*|_{H_1^{s^*}}=D_{s^*}:H_1^{s^*}\subset L^{s^*}\rightarrow L^{s^*}\). It remains to show that \(\mathrm{dom}\, D_s^*\subset H_1^{s^*}\): Let \(\psi \in \mathrm{dom}\, D_s^*\subset (L^s)^*=L^{s^*}\). Then there is a \(\rho \in L^{s^*}\) such that for all \(\varphi \in \mathrm{dom}\, D_s\) it holds \((D_s\varphi , \psi )= (\varphi , \rho )\). In particular, this is true for all \(\varphi \in C_c^\infty (M, \Sigma _M)\). In other words \(D_s^*\psi =\rho \) in the sense of distributions. Thus, \(\psi \in H_1^{s^*}\). \(\square \)

Since \(\varphi \in H_1^s\cap H_1^r\) implies \(D_s\varphi =D_r\varphi \) we often denote all those Dirac operators in the following just by D.

Moreover, a closed operator \(P:\mathrm{dom}\, P\subset V_1\rightarrow V_2\) between Banach spaces \(V_i\), and with dense domain \(\mathrm{dom}\, P\), will be called invertible if there exists a bounded inverse \(P^{-1}:V_2\rightarrow V_1\). We will use the phrase “P has a bounded inverse” synonymously.

Lemma B.3

Let \(1 \le s<\infty \).

  1. (i)

    If \(\overline{\mu }\) is in the \(L^{s^*}-\)spectrum of the Dirac operator where \((s^*)^{-1}+s^{-1}=1\), then \(\mu \) is in its \(L^s-\)spectrum.

  2. (ii)

    Let \(D_s-\mu \) be invertible. Then, \((D_{s^*}-\bar{\mu })^{-1}=((D_s-\mu )^{-1})^*\) and \(\Vert (D_{s^*}-\bar{\mu })^{-1}\Vert =\Vert (D_s-\mu )^{-1}\Vert \).

Proof

We prove this for \(\mu =0\). For arbitrary \(\mu \) this is done analogously. Assume that 0 is not in the \(L^s\)-spectrum of D, i.e., it has a bounded inverse \(E=D^{-1}:L^s\rightarrow L^s\) with range \(\mathrm{ran}\, E=H_1^s\). Let \(\varphi \in L^{s^*}\). Since E is bounded, \(f:L^s \rightarrow \mathbb {C}, \rho \mapsto (E\rho , \varphi )\) is a bounded functional and, thus, f is in the dual space of \(L^s\), i.e., there is \(\psi \in L^{s^*}\) with \((\rho ,\psi )=f(\rho )=(E\rho ,\varphi )\) for all \(\rho \in L^s\). Hence, \(\varphi \in \mathrm{dom}\, E^*\), i.e., \(\mathrm{dom}\, E^*=L^{s^*}\).

Now we can estimate for all \(\varphi \in H_1^s\) and all \(\psi \in L^{s^*}\) that \((D\varphi , E^*\psi )=(ED\varphi ,\psi )=(\varphi , \psi )\) which implies \(E^*\psi \in \mathrm{dom}\, D^*\) and \(D^*E^*\psi =\psi \). Thus, \(\mathrm{ran}\, D^*=L^{s^*}\) and \(D^*E^*={\text {Id}}:L^{s^*}\rightarrow L^{s^*}\).

If \(\rho \in L^s\) and \(\varphi \in \mathrm{dom}\, D^*\), we get \((\rho , E^*D^*\varphi )= (E\rho , D^*\varphi )=(DE\rho ,\varphi )=(\rho ,\varphi )\). Hence, \(E^*D^*={\text {Id}}:\mathrm{dom}\, D^*\rightarrow \mathrm{dom}\, D^*\). Together with the corresponding statement from above this gives that \((D^{-1})^*=(D^*)^{-1}\). Thus, 0 is not in the \(L^{s^*}\)-spectrum of D. This proves (i) and the first claim of (ii). The operator norm of an operator and its adjoint coincide, see [31, Thm VI.2]. Thus, the equality of the operator norms follows. \(\square \)

Corollary B.4

If \(D:H_1^q\rightarrow L^q\) has a bounded inverse for some \(q\in (1,\infty )\). Then as an operator from \(H_1^s \rightarrow L^s\) it has a bounded inverse for all \(s\in [q_1, q_2]\) where \(q_1=\min \{q,q^*\}\), \(q_2=\max \{q,q^*\}\), and \((q^*)^{-1}+q^{-1}=1\). In particular, the \(L^2\)-spectrum of D is a subset of the \(L^q\)-spectrum.

Proof

This Lemma follows directly from the Riesz–Thorin Interpolation Theorem 2.2 (using \(\mathcal {D}=C_c^\infty (M, \Sigma _M)\)) and Lemma B.3. \(\square \)

Lemma B.5

Let \(1\le s\le \infty \). Let \(R_s=\mathbb {C}{\setminus } \mathrm Spec_{L^s}(D)\) be the resolvent set of \(D:L^s\rightarrow L^s\). Then, the resolvent

$$\begin{aligned} \mu \in R_s \mapsto (D-\mu )^{-1}\in \mathcal {B}(L^s) \end{aligned}$$

is analytic, i.e., the map is locally given by a convergent power series with coefficients in \(\mathcal {B}(L^s)\). Here, \(\mathcal {B}(L^s)\) denotes the set of bounded operators from \(L^s\) to itself.

See [42, VIII.2 Theorem 1] for a proof.

For rounding up our presentation we will next add a lemma not needed in our context but maybe helpful to other applications.

Lemma B.6

  1. (1)

    The operator \(D:H_1^s\subset L^s\rightarrow L^s\), \(s\in [1,\infty ]\), is an invertible map onto its image if and only if there is a constant \(C>0\) with \(\Vert D\varphi \Vert _{s}\ge C\Vert \varphi \Vert _s\) for all \(\varphi \in H_1^s\).

  2. (2)

    Under the above conditions the image \(D(H_1^s)\) is closed.

  3. (3)

    Let \(s^{-1}+(s^*)^{-1}=1\), \(s<\infty \), and assume the conditions from above. Then D is surjective if and only if there is a \(C>0\) with \(\Vert D\varphi \Vert _{s^*}\ge C\Vert \varphi \Vert _{s^*}\) for all \(\varphi \in H_1^{s^*}\).

Proof

  1. (1)

    The proof is straightforward.

  2. (2)

    The operator \(D:H_{1}^s\rightarrow D(H_{1}^s)\), where the latter space is equipped with the \(L^s\)-norm, is a bijective bounded linear map. Hence, \(D(H_{1}^s)\) is a complete subspace of \(L^s\) and thus closed.

  3. (3)

    Suppose that \(D(H_1^s)\) is a proper subspace of \(L^s\). Due to Hahn-Banach there is a non-zero continuous functional \(\psi :L^s\rightarrow \mathbb {C}\) vanishing on \(D(H_1^s)\). We interpret \(\psi \) as an element in \(L^{s^*}\) using the Riesz representation theorem, i.e. \(\psi \in L^{s^*}\) is orthogonal on \(D(H_1^s)\). Then, \(\psi \in \mathrm{dom}\, (D_{s})^*\), and we even have \(D_{s}^*\psi =0\). Hence, by Lemma B.2 \(\psi \in H_1^{s*}\). This contradicts the estimate. Now assume that D is surjective. Then there is a bounded operator \(D^{-1}:L^s\rightarrow L^s\), inverse to D. Thus \((D^{-1})^*:L^{s^*}\rightarrow L^{s^*}\) is bounded as well, and \((D^{-1})^*\) is the inverse of \(D^*:H_1^{s^*}\rightarrow L^{s^*}\). The fact that the latter map has a bounded inverse is equivalent to the existence of a constant \(C>0\) with \(\Vert D\varphi \Vert _{s^*}\ge C\Vert \varphi \Vert _{s^*}\).

\(\square \)

Remark B.7

The \(L^s\)-spectrum of the Dirac operator D on a closed manifold \((M^m,g)\) is independent of s. We sketch the proof: Let \(\varphi \) be an \(L^2\)-eigenspinor of D. Then regularity theory implies that \(\varphi \in C^\infty (M, \Sigma _M)\) and, hence, \(\varphi \in L^s\) for all \(1\le s\le \infty \). In particular, \(\mathrm Spec_{L^2}^M(D)\subset \mathrm Spec_{L^s}^M(D)\). Let now \(\mu \not \in \mathrm Spec_{L^2}^M(D)\), i.e., \((D-\mu )^{-1}:L^2\rightarrow L^2\) is bounded. Let G(xy) be the unique Green function of \(D-\mu \), see Proposition 3.2. Then, \(\int _M |G(.,y)|^2 dy\) is bounded uniformly in y. Hölder’s inequality implies that also \(\int _M |G(.,y)| dy\) is bounded uniformly in y. Hence, \((D-\mu )^{-1}:L^1\rightarrow L^1\) is a bounded operator. Then interpolation gives that \((D-\mu )^{-1}:L^s\rightarrow L^s\) is bounded for all \(1\le s<2\). Because of \(\mathrm Spec_{L^2}(D)\subset \mathbb {R}\) the same is true for \((D-\bar{\mu })^{-1}:L^s\rightarrow L^s\), and by using Lemma B.3 we get that \((D-\mu )^{-1}:L^s\rightarrow L^s\) is bounded for all \(2< s<\infty \). It remains \(s=\infty \): Let \(r>m\). Then by the Sobolev Embedding Theorem \(H_1^r\hookrightarrow L^\infty \) is bounded. Moreover, by the discussion above and using the fact that \(H_1^r\) carries the graph norm of D we know that \((D-\mu )^{-1}:L^r\rightarrow H_1^r\) is bounded for \(\mu \not \in \mathrm Spec_{L^2}^M (D)\) the Hölder inequality gives that

$$\begin{aligned} (D-\mu )^{-1}:\, L^\infty \rightarrow L^r \rightarrow H_1^r\rightarrow L^\infty \end{aligned}$$

is bounded.

Lemma B.8

Let \(1\le s\le \infty \), and let \(\mathrm Spec_{L^s}^M(D)\ne \mathbb {C}\). Then the complex number \(\mu ^2\) is in the \(L^s\)-spectrum of \(D^2\) if and only if \(\mu \) or \(-\mu \) is in the \(L^s\)-spectrum of D.

Proof

We start with the “only if” part. So assume that both \(\mu \) and \(-\mu \) are not in the \(L^s\)-spectrum of D. Then we have bounded operators \((D-\mu )^{-1}:L^s\rightarrow L^s\) and \((D+\mu )^{-1}:L^s\rightarrow L^s\). It is then easy to verify that \((D-\mu )^{-1}\circ (D+\mu )^{-1}:L^s\rightarrow L^s\) is a bounded inverse of \(D^2-\mu ^2=(D+\mu )\circ (D-\mu )\). Thus \(\mu ^2\) is not in the \(L^s\)-spectrum of \(D^2\).

In order to prove the “if” statement, we assume that \(\mu ^2\) is not in the spectrum of \(D^2\). Then \(D^2-\mu ^2\) has a bounded inverse \(P:=(D^2-\mu ^2)^{-1}:L^s\rightarrow L^s\). Let \(\psi \in P(L^s)\). Then \(\psi \in L^s\) and \(D^2\psi \in L^s\). Next we will show that this implies \(D\psi \in L^s\). For that we choose \(\lambda \not \in \mathrm Spec_{L^s}^M(D)\). Then \(D\psi =(D-\lambda )^{-1}(D^2-\lambda ^2)\psi -\lambda \psi \), and hence \(D\psi \in L^s\). Thus, \(P(L^s)\subset H_{1}^s\). Hence \(Q_1:=(D\pm \mu )\circ P\) is a bounded operator with \(\mathrm{dom}\, Q_1=L^s\), and one easily checks that this a right inverse to \((D\mp \mu )\). Similarly, one shows that \(Q_2:=P\circ (D\pm \mu )\) is a left inverse of \((D\mp \mu )\). A priori \(Q_2\) is only defined on \(H_{1}^s\), but using \(Q_1=Q_1\circ (D\mp \mu )\circ Q_2=Q_2\) it is clear that \(Q_2\) and \(Q_1\) coincide on \(H_1^s\). So the integral kernels of \(Q_1\) and \(Q_2\) have to coincide, so \(Q_1\) is a left and right inverse of \((D\mp \mu )\) and thus \(\pm \mu \) is not in the spectrum of D. \(\square \)

Remark B.9

In the case \(1<s<\infty \) and M of bounded geometry, one can also prove that \(\mathrm Spec_{L^s}^M(D)=\mathbb {C}\) implies \(\mathrm Spec_{L^s}^M(D^2)=\mathbb {C}\): As in the proof of the “if” statement from above one has to show that \(D\psi \in L^s\). This can be proven using regularity theory on manifolds of bounded geometry.

Lemma B.10

(Pointwise symmetries) Let \(1\le s\le \infty \). Let (Mg) be an m-dimensional Riemannian spin manifold.

  1. (i)

    \(m\equiv 0 \mathrm{mod} 2\): The number \(\mu \) is in the \(L^s\)-spectrum of D if and only if \(-\mu \) is in the \(L^s\)-spectrum of D if and only if \(\bar{\mu }\) is in the \(L^s\)-spectrum of D.

  2. (ii)

    \(m\equiv 1 \mathrm{mod} 4\): The number \(\mu \) is in the \(L^s\)-spectrum of D if and only if \(-\bar{\mu }\) is in the \(L^s\)-spectrum of D.

  3. (iii)

    \(m\equiv 3 \mathrm{mod} 4\): The number \(\mu \) is in the \(L^s\)-spectrum of D if and only if \(\bar{\mu }\) is in the \(L^s\)-spectrum of D.

Proof

By [19, Prop. p. 31] we have a map \(\alpha :\Sigma _m\rightarrow \Sigma _m\) that is

  • a \(\mathrm{Spin}(m)\)-equivariant real structure that anticommutes with Clifford multiplication if \(m\equiv 0,1 \mathrm{mod} 8\).

  • a \(\mathrm{Spin}(m)\)-equivariant quaternionic structure that commutes with Clifford multiplication if \(m\equiv 2,3 \mathrm{mod} 8\).

  • a \(\mathrm{Spin}(m)\)-equivariant quaternionic structure that anticommutes with Clifford multiplication if \(m\equiv 4,5 \mathrm{mod} 8\).

  • a \(\mathrm{Spin}(m)\)-equivariant real structure that commutes with Clifford multiplication if \(m\equiv 6,7 \mathrm{mod} 8\).

Note that by definition real structure means that \(\alpha ^2={\text {Id}}\) and \(\alpha (\mathrm{i} v)=-\mathrm{i}\alpha (v)\). Moreover, quaternionic structure means that \(\alpha ^2=-{\text {Id}}\) and \(\alpha (\mathrm{i} v)=-\mathrm{i}\alpha (v)\).

Due to the \(\mathrm{Spin}(m)\)-equivariance \(\alpha \) induces a fiber preserving map \(\tilde{\alpha }\) on the spinor bundle with the same properties as above. Thus,

$$\begin{aligned} (D-\mu )\circ \tilde{\alpha } (\varphi ) =\left\{ \begin{array}{ll} \tilde{\alpha }\circ (-D-\bar{\mu })(\varphi ) &{}\quad m\equiv 0,1\;\; \mathrm{mod}\;\; 4\\ \tilde{\alpha }\circ (D-\bar{\mu }) (\varphi ) &{}\quad m\equiv 2,3\;\; \mathrm{mod}\;\; 4. \end{array} \right. \end{aligned}$$

Thus, if \(\mu \) is in the \(L^s\)-spectrum of D then \(-\bar{\mu }\) (resp. \(\bar{\mu }\)) in the \(L^s\)-spectrum of D for \(m\equiv 0,1\) (resp. 2,3) \(\mathrm{mod}\) 4. This gives (ii) and (iii).

If m is even, then \(D (\omega _M\cdot \varphi )=-\omega _M\cdot D\varphi \). Thus, the spectrum is symmetric when reflected on the imaginary axis. Together with the symmetries from above, (i) follows. \(\square \)

Lemma B.11

(Orientation reversing isometry) Let \(1\le s\le \infty \). Assume there is an orientation reversing isometry \(f:M^m\rightarrow M^m\) that “lifts” to the spin structure as described in the proof. Then \(\mu \) is in the \(L^s\)-spectrum of D if and only if \(-\mu \) is in the \(L^s\)-spectrum of D.

Proof

The proof follows the lines of [4, Appendix A]. In this reference, f is required to be a reflection at a hyperplane of M. But this doesn’t change the part we need: We lift f to the bundle \(P_{\mathrm{SO}(m)}M\) of oriented orthonormal frames by mapping the frame \(\mathcal {E}=(e_1,\ldots , e_m)\) to \(f_*\mathcal {E}=(-\mathrm {d}f(e_1),\mathrm {d}f(e_2),\ldots , \mathrm {d}f(e_m))\), so \(f_*:P_{\mathrm{SO}(m)}M\rightarrow P_{\mathrm{SO}(m)}M\). Since f is an orientation reserving isometry,

$$\begin{aligned} f_*(\mathcal {E}A)=f_*(\mathcal {E})JAJ \quad \text {for}\,\,\mathrm{all}\; A\in \mathrm{SO}(m) \end{aligned}$$

where \(J=\mathrm{diag}(-1, 1,1,\ldots , 1)\). The map f is assumed to lift to the spin structure, i.e., there is a lift \(\tilde{f}_* :P_{\mathrm{Spin}(m)}(M)\rightarrow P_{\mathrm{Spin}(m)}(M)\) with \(\vartheta \circ \tilde{f}_*=f_*\circ \vartheta \) where \(\vartheta \) denotes the double covering \(\vartheta :P_{\mathrm{Spin}(m)}(M)\rightarrow P_{\mathrm{SO}(m)}(M)\). By [4, Lemma A.1 and Lemma A.4], f then lifts to a map \(f_\sharp :\Sigma _M\rightarrow \Sigma _M\) on the spinor bundle which fulfils \(f_\sharp (D\varphi )=-D (f_\sharp \varphi )\). \(\square \)

Example B.12

  1. (i)

    Let \(M^{n+1}\) be a Riemannian spin manifold with a spin structure \(\vartheta \) as above. Assume that up to isomorphism this is the unique spin structure on M. Let \(f:M\rightarrow M\) be an orientation reversing isometry. By pulling back the double covering \(P_{\mathrm{Spin}}M\rightarrow P_{\mathrm{SO}}M\) by \(f_*\) we obtain the double covering \(f^*\vartheta : f^*P_{\mathrm{Spin}}M\rightarrow P_{\mathrm{SO}}M\). We then turn \(f^*P_{\mathrm{Spin}}M\) into a \(\mathrm{Spin}(n+1)\)-principal bundle by conjugating the action of \(\mathrm{Spin}(n+1)\) on \(P_{\mathrm{Spin}}M\) with Clifford multiplication with \(e_0\). Then \(f^*\vartheta \) is a spin structure on M. Thus an isomorphism from \(\vartheta \) to \(f^*\vartheta \) yields a map \(f_\sharp \) as above.

  2. (ii)

    Consider the map \(f=f_1\times \mathrm{id}:\mathbb {M}_c^m=\mathbb {H}_c^{k+1}\times N^n\rightarrow \mathbb {M}_c^{m,k}\) where \(f_1\) is an orientation reversing isometry as in (i). Then, f is again an orientation reversing isometry. Using \(P_\mathrm{SO}(\mathbb {H}_c\times N)= (P_\mathrm{SO}(\mathbb {H}_c^{k+1})\times P_\mathrm{SO}(N))\times _{\bar{\xi }} \mathrm{SO}(m)\) where \(\bar{\xi }:\mathrm{SO}(k+1)\times \mathrm{SO}(n)\rightarrow \mathrm{SO}(m)\) is the standard embedding and using the analogous describtion for \(P_{\mathrm{Spin}}(\mathbb {H}_c\times N)\), see Sect. 2.5, one see that also f lifts to the spin structure.

Appendix C: Dirac eigenvalues of generic metrics

Proposition C.1

Let (Mg) be a closed, connected Riemannian spin manifold, let \(\mu \in \mathbb {R}\). Let \(U\subset M\) be a nonempty open subset. In case that \(\mu =0\), assume additionally that the \(\alpha \)-genus of M is zero. Then, there is a metric \(\tilde{g}\) on M with \(\tilde{g}=g\) on \(M{\setminus } U\) and \(\ker \, (D^{\tilde{g}}-\mu )=\{0\}\).

Proof

For \(\mu =0\), the proposition follows from [4, Theorem 1.1]. For \(\mu \ne 0\), the proof is a direct consequence of the following lemma. \(\square \)

Lemma C.2

Let (Mg) be a closed, connected Riemannian spin manifold, let \(\mu \in \mathbb {R}{\setminus } \{0\}\), and let \(U\subset M\) be a nonempty open subset. Then there is a function \(f\in C^\infty (M, \mathbb {R}^+)\) with \(f|_{M{\setminus } U}\equiv 1\) such that \(\ker \, (D^{fg}-\mu )=\{0\}\).

Proof

Choose \(f\in C^\infty (M, \mathbb {R}^+)\) with \(f|_{M{\setminus } U}\equiv 1\) such that \(d=\dim (E_{f,\mu }:=\ker \, (D^{fg}-\mu ))\) is minimal. Assume \(d>0\), and set \(g_0=fg\). For \(\alpha \in C^\infty (M)\) with \(\mathrm supp\, \alpha \subset U\) and t close to 0 we define \(g_t:=(1+t\alpha )fg\). Then by [10] there are real analytic functions \(\mu _1, \ldots , \mu _d:(-\epsilon , \epsilon )\rightarrow \mathbb {R}\) with \(\mu _i(0)=\mu \) such that \(\mathrm Spec_{L^2}^M(D^{g_t})\cap (\mu -\delta , \mu +\delta )=\{\mu _1(t), \ldots , \mu _d(t)\}\) including multiplicities. It is shown in [10] that there is an orthonormal basis \((\psi ^{(1)}, \ldots , \psi ^{(d)})\) of \(E_{f,\mu }\) depending on the choice of \(\alpha \) such that

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}|_{t=0} \mu _i(t)=-\frac{1}{2} \int _M \langle \alpha g_0, Q_{\psi ^{(i)}}\rangle \mathrm {dvol}_{g_0} \end{aligned}$$

where \(Q_{\psi }(X,Y)=\frac{1}{2}\mathrm {Re\,}\langle X\cdot \nabla _Y \psi + Y\cdot \nabla _X \psi , \psi \rangle .\) Thus,

$$\begin{aligned} \langle g_0, Q_{\psi ^{(i)}}\rangle = \sum _r \langle e_r\cdot \nabla _{e_r} \psi ^{(i)}, \psi ^{(i)}\rangle = \mu |\psi ^{(i)}|^2. \end{aligned}$$

As d is minimal, we see that \( \frac{\mathrm {d}}{\mathrm {d}t}|_{t=0} \mu _i(t)=0\), and thus for all \(\alpha \) as above

$$\begin{aligned} -\frac{1}{2}\int _M \alpha \mu \sum _{i=1}^d |\psi ^{(i)}|^2 \mathrm {dvol}_{g_0}=0. \end{aligned}$$

Note that \(\varphi :=\sum _{i=1}^d |\psi ^{(i)}|^2\in C^\infty (M)\) does not depend on the choice of \(\alpha \). This can be seen by direct calculation with base change matrices or alternatively by observing that \(\varphi \) is the pointwise trace of the integral kernel of the projection to \(E_{f,\mu }\). With \(\mu \ne 0\) this implies that \(\varphi \) and thus all \(\psi ^{(i)}\) vanish on U. The unique continuation principle implies then \(\psi ^{(i)}\equiv 0\) which gives a contradiction. \(\square \)

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Ammann, B., Große, N. \(L^p\)-spectrum of the Dirac operator on products with hyperbolic spaces. Calc. Var. 55, 127 (2016). https://doi.org/10.1007/s00526-016-1046-z

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Mathematics Subject Classification

  • 58J50
  • 34B27