On the index of a free-boundary minimal surface in Riemannian Schwarzschild-AdS

Abstract

We consider the index of a certain non-compact free-boundary minimal surface with boundary on the rotationally symmetric minimal sphere in the Schwarzschild-AdS geometry with $m>0$. As in the Schwarzschild case, we show that in dimensions $n\ge 4$, the surface is stable, whereas in dimension three, the stability depends on the value of the mass $m>0$ and the cosmological constant $\Lambda <0$ via the parameter $\mu :=m\sqrt{-\Lambda /3}$. We show that while for $\mu \ge \tfrac{5}{27}$ the surface is stable, there exist positive numbers $\mu _0$ and $\mu _1$, with $\mu _1<\tfrac{5}{27}$, such that for $0<\mu <\mu _0$, the surface is unstable, while for all $\mu \ge \mu _1$, the index is at most one.

Appendices

Appendix A: Curvature calculations

We record the Ricci curvature for warped products $(B\times F, g_B + \psi ^2 g_F)$, where $\psi$ is a positive function on B. In verifying curvature calculations in the paper, we have applied this to metrics which have the forms: (i) $h(r) \, \text {d}r^2 + r^2\mathring{g}_{{\mathbb {S}}^{n-1}}$, with base $(I, h(r)\, \text {d}r^2)$ on an interval $I\subset \mathbb (0, +\infty )$ and fibre ${\mathbb {S}}^{n-1}$; (ii) $\text {d}\ell ^2 + (r(\ell ))^2 \mathring{g}_{{\mathbb {S}}^{n-1}}$ with base $(I, \text {d}s^2)$; and (iii) Lorentzian warped products $-f(r)\, \text {d}t^2 + g$, with base $(M^n, g)$ and fibre ${\mathbb {R}}$. We work out (i) as an example below.

We suppose X and Y are (lifts) of tangent vectors to B, and V and W are lifts of tangent vectors to F, $d= \textrm{dim}(F)$, and where we note that for a function like $\psi$ on the base, then with $g=g_B + \psi ^2 g_F$, $\textrm{Hess}_g\psi (X, Y)= \textrm{Hess}_{g_B}\psi (X, Y)$, and $g_B( \mathrm {grad_{g_B}}\psi , \mathrm {grad_{g_B}}\psi )= g( \mathrm {grad_{g}}\psi , \mathrm {grad_{g}}\psi )=:\langle \text {d}\psi , d\psi \rangle$. The Ricci curvature is determined from the following [7, Corollary 43]:

$$\begin{aligned} \textrm{Ric}_g(X, Y)&= \textrm{Ric}_{g_B}(X, Y) - \frac{ d \;\textrm{Hess}_{g_B}\psi (X, Y)}{\psi } \nonumber \\ \textrm{Ric}_g(X, V)&= 0 \nonumber \\ \textrm{Ric}_g(V, W)&= \textrm{Ric}_{g_F}(V, W) - g( V, W) \left( \frac{ \Delta _{g_B} \psi }{\psi } + (d-1) \frac{ \langle \text {d}\psi , \text {d}\psi \rangle }{\psi ^2}\right) . \end{aligned}$$

(A.1)

We begin with the Levi-Civita connection $\nabla$ for

$$\begin{aligned} g&= h(r) \, \text {d}r^2 + r^2\mathring{g}_{{\mathbb {S}}^{n-1}} \qquad \qquad \qquad \qquad (n\ge 2)\\&= h(r) \, \text {d}r^2 + r^2\, ( \text {d}\phi ^2 +\sin ^2\phi \;\mathring{g}_{{\mathbb {S}}^{n-2}})\qquad \qquad (n\ge 3)\\&= h(r) \, \text {d}r^2 + r^2\, (d\phi ^2 + \sin ^2\phi (d\theta ^2 + \sin ^2\theta \mathring{g}_{{\mathbb {S}}^{n-3}})) \qquad (n\ge 4). \end{aligned}$$

For $n\ge 4$, we recursively employ corresponding angular coordinates on ${\mathbb {S}}^{n-3}$, and in these coordinates, the metric components and Ricci components comprise diagonal matrices. We let $\vartheta$ denote the general angular coordinate, and we abbreviate $\mathring{g}_{{\mathbb {S}}^{n-1}}=:\mathring{g}$. The nonzero Christoffel symbols $\Gamma ^k_{rj}$ are

$$\begin{aligned} \Gamma ^r_{rr} = \frac{ h'(r) }{2 h(r)}, \qquad \Gamma ^\vartheta _{r\vartheta } = \frac{1}{r} \end{aligned}$$

i.e.

$$\begin{aligned} \nabla _{\frac{\partial }{\partial r}} \tfrac{\partial }{\partial r} = \tfrac{ h'(r) }{2 h(r)} \tfrac{\partial }{\partial r}, \qquad \nabla _{\frac{\partial }{\partial r}} \tfrac{\partial }{\partial \vartheta } = \tfrac{ 1 }{r} \tfrac{\partial }{\partial \vartheta }. \end{aligned}$$

The nonzero $\Gamma ^r_{jk}$ for j and k angular indices are

$$\begin{aligned} \Gamma ^r_{\vartheta \vartheta }= -\frac{r\mathring{g}_{\vartheta \vartheta }}{h(r)} = -\frac{g_{\vartheta \vartheta }}{rh(r)}, \end{aligned}$$

i.e. the vector-valued second fundamental form ${I\hspace{-0.1cm}I}$ of the spherical fibre is

$$\begin{aligned} {I\hspace{-0.1cm}I}(V, W)= - \frac{g(V, W)}{r h(r)} \frac{\partial }{\partial r} = - \frac{g(V, W)}{r } \textrm{grad}(r)=- \frac{g(V, W)}{r \sqrt{h(r)}} {\textbf{n}} \end{aligned}$$

where ${\textbf{n}}= \frac{ 1}{\sqrt{h(r)}} \frac{\partial }{\partial r}$ is a unit normal to a spherical fibre.

The Christoffel symbols $\Gamma ^k_{ij}$ with all angular indices are just the Christoffel symbols of the round sphere (which are invariant under a constant scaling, i.e. independent of radius). If the angular coordinates are numbered $\vartheta _1=\phi$, $\vartheta _2=\theta , \ldots , \vartheta _{n-1}$, such that for $j\ge 2$, $g_{\vartheta _j \vartheta _j} = r^2 \sin ^2 \vartheta _1 \cdots \sin ^2\vartheta _{j-1}$, then $\Gamma _{\vartheta _j\vartheta _j}^{\vartheta _i}=0$ for $i\ge j$, whereas for $1\le i <j$, $\Gamma _{\vartheta _j\vartheta _j}^{\vartheta _i}=-\sin \vartheta _i\cos \vartheta _i \sin ^2\vartheta _{i+1} \cdots \sin ^2\vartheta _{j-1}$, while for $i<j$, $\Gamma _{\vartheta _i \vartheta _j}^{\vartheta _j}=\cot \vartheta _i$.

Observe that $\textrm{Hess}_g r(\tfrac{\partial }{\partial r}, \frac{\partial }{\partial r}) = - \Gamma _{rr}^r= -\frac{ h'(r) }{2\,h(r)}=\textrm{Hess}_{g_B} r(\tfrac{\partial }{\partial r}, \frac{\partial }{\partial r})$, the Hessian of r with respect to $g_B:=h(r) \, \text {d}r^2$.

As for the Ricci curvature components, we apply the above-warped product formulas, with the warping function $\psi =r$. Since the base is one-dimensional, its curvature vanishes, and

$$\begin{aligned} \textrm{Ric}_{rr} = - \frac{n-1}{r} \textrm{Hess}_g r( \tfrac{\partial }{\partial r}, \tfrac{\partial }{\partial r})= \frac{(n-1) h'(r)}{2rh(r)}. \end{aligned}$$

Now, on ${\mathbb {S}}^{n-1}$ we have $\textrm{Ric}(\mathring{g}) = (n-2)\mathring{g}$. Thus the remaining nonzero components of Ricci are

$$\begin{aligned} \textrm{Ric}_{\vartheta \vartheta }&= (n-2) \mathring{g}_{\vartheta \vartheta } - r^2 \mathring{g}_{\vartheta \vartheta } \left( -\frac{ h'(r) }{2r (h(r))^2} + (n-2) \frac{1}{r^2h(r)} \right) \\ {}&= \mathring{g}_{\vartheta \vartheta }\left( (n-2)\Big ( 1-\frac{1}{h(r)} \Big ) + \frac{ rh'(r) }{2(h(r))^2} \right) . \end{aligned}$$

The scalar curvature of g is given by

$$\begin{aligned} R(g)&= \frac{(n-1) h'(r)}{2r(h(r))^2}+\frac{ (n-1) (n-2)}{r^2} \left( 1- \frac{1}{h(r)}\right) + \frac{ (n-1) h'(r) }{2r (h(r))^2}\\&= \frac{(n-1) h'(r)}{r(h(r))^2}+\frac{ (n-1) (n-2)}{r^2} \left( 1- \frac{1}{h(r)}\right) . \end{aligned}$$

Example A.1

We apply these formulas to $g_{m, \Lambda }= \tfrac{\text {d}r^2}{1-\frac{2\Lambda }{n(n-1)} r^2-\frac{2\,m}{r^{n-2}}} + r^2 \mathring{g}_{{\mathbb {S}}^{n-1}}$. In this case we have $(h(r))^{-1} = 1-\tfrac{2\Lambda }{n(n-1)} r^2-\tfrac{2\,m}{r^{n-2}},$ and so

$$\begin{aligned} 1-\frac{1}{h(r)}&= \frac{2\Lambda }{n(n-1)} r^2+\frac{2m}{r^{n-2}}\\ \frac{2}{rh(r)} \Gamma ^r_{rr} = \frac{h'(r)}{r(h(r))^2}&= \frac{4\Lambda }{n(n-1)} -\frac{2(n-2)m}{r^n}. \end{aligned}$$

Thus

$$\begin{aligned} \textrm{Ric}_{rr}&= \frac{n-1}{2} h(r) \left( \frac{4\Lambda }{n(n-1)} -\frac{2(n-2)m}{r^n} \right) \\ \textrm{Ric}_{\vartheta \vartheta }&= \mathring{g}_{\vartheta \vartheta } \left( (n-2) \left( \frac{2\Lambda }{n(n-1)} r^2+\frac{2m}{r^{n-2}}\right) + \frac{2\Lambda }{n(n-1)}r^2 -\frac{(n-2)m}{r^{n-2}}\right) \\&= r^2\mathring{g}_{\vartheta \vartheta } \left( \frac{2\Lambda }{n} +\frac{(n-2)m}{r^{n}}\right) \\ R(g)&= (n-1) \Big (\frac{2\Lambda }{n(n-1)} -\frac{(n-2)m}{r^n}\Big ) + (n-1) \left( \frac{2\Lambda }{n} +\frac{(n-2)m}{r^{n}}\right) \\&=2\Lambda . \end{aligned}$$

A unit normal $\nu$ along the surface $x^n=0$ is given by $\nu =\frac{1}{r} \frac{\partial }{\partial \phi }$, and so we have $\textrm{Ric}_g(\nu , \nu ) = \frac{2\Lambda }{n} + \frac{(n-2)m}{r^n}$. Furthermore, the second fundamental form of the spherical fibre at fixed r satisfies, with $\eta = -{\textbf{n}}= - \frac{ 1}{\sqrt{h(r)}} \frac{\partial }{\partial r}$,

$$\begin{aligned} \langle {I\hspace{-0.1cm}I}(\nu , \nu ), \eta \rangle = \frac{\sqrt{1-\tfrac{2\Lambda }{n(n-1)} r^2-\tfrac{2m}{r^{n-2}}}}{r}. \end{aligned}$$

This remains bounded as r increases when $\Lambda <0$, and tends to 0 as $r\nearrow \infty$ for $\Lambda =0$.

Appendix B: Schwarzschild test function

Consider the Schwarzschild metric $g_m^S = (1+\tfrac{m}{2\rho })^4 (d\rho ^2 + \rho ^2 \mathring{g}_{{\mathbb {S}}^2})$ in isotropic coordinates $x\in {\mathbb {R}}^3{\setminus } \{ 0\}$, $\rho = |x|$ (Euclidean length). We let $\Sigma$ be the free-boundary minimal surface $x^3=0$ ($\phi =\frac{\pi }{2}$) and $\rho \ge \tfrac{m}{2}$. For any $\alpha >\max ( 1, \tfrac{m}{2})$, let $\Sigma ^\alpha \subset \Sigma$ be the subset with $\alpha \le \rho \le \alpha ^2$, and let $\Sigma '_R\subset \Sigma$ be given by $\tfrac{m}{2} \le \rho \le R$.

If u is a radial eigenfunction of ${\mathcal {L}}_{\Sigma }$ on $\Sigma '_R$ with eigenvalue $\lambda$, we let $w= \sqrt{\rho } \; u$, to obtain

$$\begin{aligned} w''(\rho ) + w(\rho ) \left( \tfrac{1}{4\rho ^2} + (1+ \tfrac{m}{2\rho })^{-2} \tfrac{m}{\rho ^3} + \lambda (1+ \tfrac{m}{2\rho })^{4}\right) =0. \end{aligned}$$

The Neumann condition on u at $\rho = \tfrac{m}{2}$ translates to $w'(\tfrac{m}{2})= m^{-1} w(\tfrac{m}{2})$. We let $w_\lambda$ be the solution to the above equation with initial conditions $w(\tfrac{m}{2})=1$ and $w'(\tfrac{m}{2})=m^{-1}$. For $\lambda =0$ one can solve to get $w_0(\rho )= \sqrt{\frac{2\rho }{m}} \left( 1- \frac{ 2\rho -m}{2\rho +m} \log \sqrt{\frac{2\rho }{m}} \right)$, as observed in [6]. This is eventually negative, so that 0 is not an eigenvalue on $(\Sigma '_R, g_m^S)$ for R large (recall Lemma 3.4). That said, as in [6], as $\lambda \nearrow 0$, $w_\lambda$ must have a zero, just as $w_0$ does, by perturbation, and thus $\Sigma$ is unstable. We recall a different argument here, using a log cut-off, cf. [8, p. 54].

The idea is to modify the test function $\psi =1$, which is not in $L^2(\Sigma )$, and likewise does not have the appropriate boundary conditions on $\Sigma '_R$. It is easy to show that a simple linear cut-off over $\Sigma ^\alpha$ will not suffice, so we use the following test function:

$$\begin{aligned} \psi _\alpha (x) = {\left\{ \begin{array}{ll} 1, \qquad \qquad \quad |x|< \alpha \\ \frac{\ln (\alpha ^2 |x|^{-1})}{\ln \alpha }, \quad x\in \Sigma ^{\alpha } \\ 0, \qquad \qquad \quad |x|>\alpha ^2 \end{array}\right. }. \end{aligned}$$

This test function is in $W^{1,2}(\Sigma )$, and on $\Sigma ^{\alpha }$,

$$\begin{aligned} |\nabla \psi _\alpha |^2_{g_m^S} =( 1+ \tfrac{m}{2|x|})^{-4}|d\psi _\alpha |^2_{ g_{{\mathbb {E}}^2}}=( 1+ \tfrac{m}{2|x|})^{-4} \tfrac{1}{|x|^2 (\ln \alpha )^2}. \end{aligned}$$

Thus we estimate

$$\begin{aligned} \int \limits _{\Sigma ^\alpha } \Big (&|\nabla \psi _\alpha |^2_{g_m^S} -\textrm{Ric}_{g_m^S} (\nu , \nu )\psi _\alpha ^2\Big ) \; \text {d}A_{g_m^S}\\&= \int \limits _0^{2\pi } \int \limits _{\alpha }^{\alpha ^2} ( 1+ \tfrac{m}{2 \rho })^{-4} \tfrac{1}{\rho ^2 (\ln \alpha )^2}\rho \; d\rho \; \text {d}\theta -\int \limits _{\Sigma ^\alpha } \textrm{Ric}_{g_m^S} (\nu , \nu )\psi _\alpha ^2 \; \text {d}A_{g_m^S}\\&\le 2\pi \frac{\ln \rho }{(\ln \alpha )^2} \Big |_{\rho =\alpha }^{\rho =\alpha ^2} -\int \limits _{\Sigma ^\alpha } \textrm{Ric}_{g_m^S} (\nu , \nu )\psi _\alpha ^2 \; \text {d}A_{g_m^S}\\&= \frac{2\pi }{\ln \alpha } -\int \limits _{\Sigma ^\alpha } \textrm{Ric}_{g_m^S} (\nu , \nu )\psi _\alpha ^2 \; \text {d}A_{g_m^S}. \end{aligned}$$

We conclude

$$\begin{aligned} {\mathcal {Q}}^{\Sigma '_{\alpha ^2}}(\psi _\alpha )&=\frac{2\pi }{\ln \alpha } -\int \limits _{\Sigma ^\alpha } \textrm{Ric}_{g_m^S} (\nu , \nu )\psi _\alpha ^2 \; \text {d}A_{g_m^S}- \int \limits _{\Sigma '_\alpha } \textrm{Ric}_{g_m^S} (\nu , \nu ) \; \text {d}A_{g_m^S} \\ {}&{\mathop {\longrightarrow }\limits ^{\alpha \nearrow \infty }}\int _{\Sigma }-\textrm{Ric}_{g_m^S} (\nu , \nu ) \; \text {d}A_{g_m^S}. \end{aligned}$$

As $\textrm{Ric}_{g_m^S}(\nu , \nu )>0$ along $\Sigma$, for $\alpha$ large enough, ${\mathcal {Q}}^{\Sigma '_{\alpha ^2}}(\psi _\alpha )<0$, as desired.

Appendix C: On eigenfunctions on $\Sigma$

In this section we discuss an application of the arguments in [5] to the setting we consider. We take $\Sigma \subset (M,g)$ to be a complete non-compact free-boundary minimal surface in (M, g), with smooth unit normal field $\nu$, and with free boundary $\partial \Sigma$ lying on a totally geodesic hypersurface S. We also consider an exhaustion $\Sigma _R$ of $\Sigma$ by smooth regions, with $\partial \Sigma _R= \partial \Sigma \sqcup \partial \Sigma _R^+$, and let $\eta$ be the outward-pointing unit conormal along the boundary of $\Sigma _R$. The Robin boundary condition along $\partial \Sigma$ reduces to the Neumann condition. Recall the eigenvalue equation ${\mathcal {L}}_\Sigma u + \lambda u=0$. We start with a simple lemma.

Lemma C.1

Suppose $W\subset L^2(\Sigma )$ is a finite-dimensional subspace such that for all $\varphi \in C^{\infty }_c(\Sigma ) \cap W^{\perp }$, ${\mathcal {Q}}^\Sigma (\varphi )\ge 0$. Then $\textrm{ind}(\Sigma )\le \textrm{dim}(W)$.

Proof

If $\textrm{ind}(\Sigma _R)>\textrm{dim}(W)$, there is a non-trivial test function $\varphi$ which is a linear combination of eigenfunctions with negative eigenvalues for $\Sigma _R$ (Neumann on $\partial \Sigma$, Dirichlet on $\partial \Sigma _R^+$ and extended to vanish outside $\Sigma _R$), and is in the kernel of the $L^2(\Sigma )$-orthogonal projection map onto W, i.e. $\varphi$ is $L^2(\Sigma )$-orthogonal to W. But since $\mathcal Q^\Sigma (\varphi )<0$, this is a contradiction. Thus $\textrm{ind}(\Sigma ) \le \textrm{dim}(W)$. $\square$

The next proposition addresses the converse question, for which we make two additional assumptions, both of which hold in case $g= g_{m, \Lambda }$ with $m>0$ (recall the proof of Proposition 3.9 for (ii)):

• (i) For some constant $c>0$, $q:=\textrm{Ric}_g(\nu ,\nu )+| I\hspace{-0.1cm}I^\Sigma (\nu , \nu )|^2_g$ satisfies $|q|\le c$.

• (ii) Given any $\ell \in {\mathbb {Z}}_+$ and $u_1, \ldots , u_\ell \in W^{1,2}(\Sigma )$, there is a sequence $r_j\nearrow \infty$ such that for each $1\le i\le \ell$, $\int \limits _{\partial \Sigma ^+_{r_j}} (u_i^2 + |du_i|^2_g)\; d\sigma _g\rightarrow 0$.

Under these conditions, the next proposition is a corollary of the proof of Proposition 2 in [5], utilizing the uniform bound on |q|, which was not assumed in [5].

Proposition C.2

Suppose $\Sigma \subset (M,g)$ is a FBMS with free boundary $\partial \Sigma$ lying on a totally geodesic hypersurface S, and that conditions (i) and (ii) above hold. If $\textrm{ind}(\Sigma )=k$ is finite, then there is a k-dimensional subspace $W\subset W^{1,2}(\Sigma )$ spanned by $L^2(\Sigma )$-orthonormal eigenfunctions $f_1, \ldots , f_k$ with negative eigenvalues $\lambda _1, \ldots , \lambda _k$ and Neumann boundary condition along $\partial \Sigma$, such that for any $\varphi \in W^{1,2}(\Sigma )$ which is $L^2(\Sigma )$-orthogonal to W, we have $\mathcal Q^\Sigma (\varphi )\ge 0$. Moreover, ${\mathcal {Q}}^{\Sigma }$ is negative definite on W, and ${\mathcal {Q}}^{\Sigma }(f_i)=\lambda _i$. As such, if $\widetilde{W} \subset W^{1,2}(\Sigma )$ is a subspace such that ${\mathcal {Q}}^{\Sigma }$ is negative definite on $\widetilde{W}$, then $\textrm{dim}(\widetilde{W})\le \textrm{ind}(\Sigma )$.

Proof

We will follow closely the proof of [5, Proposition 2]. As S is totally geodesic, ${\mathcal {Q}}^\Sigma (\varphi )= \int \limits _\Sigma ( |\nabla ^\Sigma \varphi |^2_g - q \varphi ^2) \; dA_{g}$, which converges for all $\varphi \in W^{1,2}(\Sigma )$ by the bound on q. In case $\Sigma$ is stable, $\mathcal Q^{\Sigma }(\varphi )\ge 0$ for all $\varphi \in C^\infty _c(\Sigma )$, so that by approximation we have ${\mathcal {Q}}^\Sigma (\varphi )\ge 0$ for all $\varphi \in W^{1,2}(\Sigma )$ (where we use that |q| is bounded). Thus $W=\{0\}$ in case $k=0$.

We now suppose $\textrm{ind}(\Sigma )=k\in {\mathbb {Z}}_+$. The first part of the proof of Proposition 1 in [5] shows that for large R, say $R\ge R_0$, $\Sigma {\setminus } \Sigma _R$ is stable (of course in the setting we consider above in $g_{m, \Lambda }$ with $\Lambda <0$, this is obvious). Just as in [5], for each R there is a cut-off function $\eta =\eta _R$ which is zero on $\Sigma _R$ and identically 1 outside $\Sigma _{2R}$, with $0\le \eta \le 1$ and $|\nabla ^\Sigma \eta |_g^2 \le \frac{5(1-\eta ^2)}{R^2}$. We apply the stability inequality on $\Sigma \setminus \Sigma _R$ to $\eta \varphi$ for $\varphi \in C^\infty _c(\Sigma )$ to find

$$\begin{aligned} 0&\le \int \limits _\Sigma ( |\nabla ^\Sigma (\eta \varphi )|^2_g-q\eta ^2 \varphi ^2)\; \text {d}A_{g} \\&= \int \limits _\Sigma (\eta ^2 |\nabla ^\Sigma \varphi |^2_g + 2 \eta \varphi \langle \nabla ^\Sigma \eta , \nabla ^\Sigma \varphi \rangle + \varphi ^2|\nabla ^\Sigma \eta |^2_g -q\eta ^2 \varphi ^2)\; \text {d}A_{g}\\&\le \int \limits _\Sigma (\eta ^2 |\nabla ^\Sigma \varphi |^2_g + 2 | \nabla ^\Sigma \eta |^2_g | \nabla ^\Sigma \varphi |_g^2 + \varphi ^2|\nabla ^\Sigma \eta |^2_g +(\tfrac{1}{2} -q) \eta ^2\varphi ^2)\; dA_{g}.\\ \end{aligned}$$

We add ${\mathcal {Q}}^\Sigma (\varphi ) = \int _\Sigma ( |\nabla ^\Sigma \varphi |^2_g -q \varphi ^2)\; \text {d}A_{g}$ to each side and rearrange, as in [5],

$$\begin{aligned} \int \limits _\Sigma (1&-\eta ^2) |\nabla ^\Sigma \varphi |^2_g \; dA_{g} \\&\le {\mathcal {Q}}^{\Sigma }(\varphi )+ \int \limits _\Sigma ( 2 | \nabla ^\Sigma \eta |^2_g | \nabla ^\Sigma \varphi |_g^2 + \varphi ^2|\nabla ^\Sigma \eta |^2_g +(\tfrac{1}{2} \eta ^2 + (1-\eta ^2) q) \varphi ^2)\; \text {d}A_{g}. \end{aligned}$$

Hence we can use the above bound on $|\nabla ^{\Sigma } \eta |_g$ to see that for $R\ge \max (R_0, \sqrt{20})$,

$$\begin{aligned} \int \limits _{\Sigma _R} |\nabla ^\Sigma \varphi |^2_g \; \text {d}A_{g}&\le \int \limits _\Sigma (1 -\eta ^2) |\nabla ^\Sigma \varphi |^2_g \; dA_{g} \le 2 {\mathcal {Q}}^{\Sigma }(\varphi )+ (1 + 2c) \int \limits _\Sigma \varphi ^2\; \text {d}A_{g}. \end{aligned}$$

(C.1)

By approximation, ${\mathcal {Q}}^\Sigma (\varphi ) \ge - ( \tfrac{1}{2} + c) \int \limits _\Sigma \varphi ^2\; \text {d}A_{g}$ holds for all $\varphi \in W^{1,2}(\Sigma )$.

Since $k=\textrm{ind}(\Sigma )$, then for R large, say $R\ge R_1$, there are precisely k negative eigenvalues (with multiplicity) for $\Sigma _R$, with Neumann boundary condition on $\partial \Sigma$, and Dirichlet on $\partial \Sigma _R^+$. We take $f_{i, R}$, $1\le i\le k$, to be $L^2(\Sigma _R)$-orthonormal eigenfunctions with respective eigenvalues $\lambda _{i, R}<0$, and extend $f_{i,R}$ by zero to $\Sigma$. By the mini-max characterization of eigenvalues, we have that the maximum of these negative eigenvalues decreases with R, so that there is a $\varepsilon _0>0$ so that $\lambda _{i,R} \le -\varepsilon _0<0$ for all $R\ge R_1$. On the other hand, from the last paragraph above we can infer that the eigenvalues are uniformly bounded below: $\lambda _{i,R}\ge -c_0:=-(\tfrac{1}{2} +c)$.

From the stability inequality, with $\eta =\eta _R$ for $R\ge R_0$ and $\rho \ge R_1$,

$$\begin{aligned} 0&\le \int \limits _\Sigma \Big (\eta ^2 |\nabla ^\Sigma f_{i, \rho }|^2_g + 2 \eta f_{i, \rho } \langle \nabla ^\Sigma \eta , \nabla ^\Sigma f_{i, \rho } \rangle + f_{i, \rho }^2|\nabla ^\Sigma \eta |^2_g -q\eta ^2 f_{i, \rho }^2\Big )\; \text {d}A_{g}\\&= \int \limits _\Sigma \Big (\eta ^2 |\nabla ^\Sigma f_{i,\rho }|^2_g + \tfrac{1}{2} \langle \nabla ^\Sigma ( \eta ^2), \nabla ^\Sigma (f_{i, \rho }^2) \rangle + f_{i,\rho }^2|\nabla ^\Sigma \eta |^2_g -q\eta ^2 f_{i, \rho }^2\Big )\; \text {d}A_{g}\\&= \int \limits _\Sigma \Big (\eta ^2 |\nabla ^\Sigma f_{i,\rho }|^2_g - \eta ^2( f_{i,\rho } \Delta _\Sigma f_{i,\rho } + | \nabla ^\Sigma f_{i, \rho }|_g^2) + f_{i,\rho }^2|\nabla ^\Sigma \eta |^2_g -q\eta ^2 f_{i, \rho }^2\Big )\; \text {d}A_{g}, \end{aligned}$$

which we can rearrange to obtain

$$\begin{aligned} -\lambda _{i,\rho } \int \limits _\Sigma \eta ^2 f_{i,\rho }^2 \; \text {d}A_{g} \le \int \limits _\Sigma f_{i,\rho }^2|\nabla ^\Sigma \eta |^2_g\; dA_{g}\le 5R^{-2} . \end{aligned}$$

Using the upper bound on the eigenvalues we find

$$\begin{aligned} \int \limits _{\Sigma \setminus \Sigma _{2R}} f_{i,\rho }^2\; \text {d}A_{g}\le 5\varepsilon _0^{-1}R^{-2}. \end{aligned}$$

(C.2)

Using $\varphi =f_{i, \rho }$ in (C.1), we find

$$\begin{aligned} \int \limits _{\Sigma _R} |\nabla ^\Sigma f_{i,\rho }|^2_g \; \text {d}A_{g} \le 2 {\mathcal {Q}}^{\Sigma }(f_{i,\rho })+ (1 + 2c) \le -2\varepsilon _0+(1 + 2c), \end{aligned}$$

and thus $\{f_{i,\rho }: 1\le i\le k, \; \rho \ge R_1\}$ is bounded in $W^{1,2}(\Sigma _R)$. By Rellich’s lemma (along with Banach-Alaoglu and Riesz Representation), we may arrange a sequence $\rho _j\nearrow \infty$ such that for each $1\le i\le k$, $f_{i, \rho _j}$ converges in $L^2(\Sigma _{R})$ and weakly in $W^{1,2}(\Sigma _{R})$ for each $R\ge R_0$, and the limit is given by a function $f_i\in W^{1,2}_{\textrm{loc}}(\Sigma )$. Using (C.2), we see that $f_i\in L^2(\Sigma )$ and that $f_{i, \rho _j}$ converges to $f_i$ in $L^2(\Sigma )$. As such, $f_1, \ldots , f_k$ is an $L^2(\Sigma )$-orthonormal collection. We can also arrange by boundedness of the eigenvalues that for each i, $\lambda _{i, \rho _j}$ converges, say to $\lambda _i\in [-c_0, -\varepsilon _0]$.

Moreover, by the uniform bound on $\int _{\Sigma _R} |\nabla ^\Sigma f_{i, \rho _j}|^2_g \; \text {d}A_{g}$ and the lower semicontinuity of the norm under weak convergence, we see that $f_i\in W^{1,2}(\Sigma )$. By the weak convergence, we have for any test function $\psi \in C^\infty _c(\Sigma )$, and for j large enough so that $\psi$ is supported in $\Sigma _{\rho _j}$,

$$\begin{aligned} 0&=\int \limits _{\Sigma } ( \langle \nabla ^\Sigma \psi , \nabla ^\Sigma f_{i, \rho _j} \rangle -q\psi f_{i, \rho _j}-\lambda _{i, \rho _j} \psi f_{i, \rho _j})\; \text {d}A_{g}\\ {}&\qquad \longrightarrow \int \limits _{\Sigma } ( \langle \nabla ^\Sigma \psi , \nabla ^\Sigma f_{i} \rangle -q\psi f_{i}-\lambda _i \psi f_i)\; \text {d}A_{g}. \end{aligned}$$

This is the weak formulation of the boundary-value problem $\mathcal L_\Sigma f_i + \lambda _i f_i=0$, $\frac{\partial f_i}{\partial \eta } =0$ on $\partial \Sigma$. By regularity for the elliptic boundary-value problem, we have that $f_i$ is a classical solution.

That ${\mathcal {Q}}^{\Sigma }(f_i)=\lambda _i$ now follows by (ii) and Cauchy-Schwarz:

$$\begin{aligned} \int \limits _{\Sigma _{r_j}}( |\nabla ^\Sigma f_i|^2_g - qf_i^2)\; dA_{g}= \lambda _i \int \limits _{\Sigma _{r_j}} f_i^2 \; \text {d}A_{g} + \int \limits _{\partial \Sigma _{r_j}^+} f_i \frac{\partial f_i}{\partial \eta }\; \text {d}\sigma _{g}\rightarrow \lambda _i. \end{aligned}$$

That ${\mathcal {Q}}^\Sigma \Big (\sum \limits _{i=1}^k c_i f_i \Big ) = \sum \limits _{i=1}^k \lambda _i c_i^2$ then follows by a similar argument.

Finally, let W be the span of $f_1, \ldots , f_k$, and suppose $\varphi \in W^{1,2}(\Sigma )\cap W^\perp$. We follow [5] to show ${\mathcal {Q}}^\Sigma (\varphi )\ge 0$. Let $\varepsilon >0$, and let $\varphi _{\ell }\in C^\infty _c(\Sigma )$ approach $\varphi$ in $W^{1,2}(\Sigma )$. For sufficiently large $\ell$, $|{\mathcal {Q}}^\Sigma (\varphi )-\mathcal Q^{\Sigma }(\varphi _{\ell })|< \varepsilon$, and for each $1\le i\le k$, $|\int \limits _\Sigma \varphi _{\ell } f_i dA_{g_\Sigma }|<\sqrt{\varepsilon }$. Fix such an $\ell$, and consider j so large such that $\varphi _{\ell }$ is supported in $\Sigma _{\rho _j}$. Decompose $\varphi _{\ell } = \sum \limits _{i=1}^k {a_{ij\ell }} f_{i, \rho _j} + \psi _{\ell }$, where $\psi _{\ell }\perp f_{i, \rho _j}$ in $L^2(\Sigma )$. Then since $\varphi _{\ell }$ and $f_{i, \rho _j}$ vanish outside $\Sigma _{\rho _j}$, $\psi _{\ell }$ does as well, and so ${\mathcal {Q}}^{\Sigma }(\psi _{\ell })\ge 0$ and

$$\begin{aligned} {\mathcal {Q}}^{\Sigma }(\varphi _{\ell })&= {\mathcal {Q}}^{\Sigma } \Big ( \sum \limits _{i=1}^k a_{ij\ell } f_{i, \rho _j}\Big ) + \mathcal Q^{\Sigma } (\psi _{\ell }) + 2 \sum \limits _{i=1}^k a_{ij\ell } \mathcal B^\Sigma (f_{i, \rho _j}, \psi _{\ell }) \ge \sum \limits _{i=1}^k \lambda _{i, \rho _j} a^2_{ij\ell } . \end{aligned}$$

Furthermore, $a_{ij\ell }= \int \limits _{\Sigma } \varphi _{\ell } f_{i, \rho _j} \text {d}A_{g_\Sigma } {\mathop {\longrightarrow }\limits ^{j\nearrow \infty }} \int \limits _{\Sigma } \varphi _{\ell } f_{i} \text {d}A_{g_\Sigma },$ so that for large enough j, $|a_{ij\ell }|< \sqrt{\varepsilon }$. By the uniform bound on the eigenvalues $\lambda _{i, \rho _j}$, we see for some constant $C>0$, ${\mathcal {Q}}^{\Sigma }(\varphi )\ge - C \varepsilon$. Since $\varepsilon >0$ was arbitrary, we conclude ${\mathcal {Q}}^{\Sigma } (\varphi )\ge 0$. $\square$