Proof of Lemma 2.5.

We may assume $X\subsetneq R$ . Now by the assumptions on the boundaries of $R$ and  $X$ , there is a closed collar $U$ about $\unicode[STIX]{x2202}R$ in $R$ such that $Y=X\setminus \mathring{U}$ is a deformation retract of $X$ in  $R$ . Observe that $Y$ does not intersect $\unicode[STIX]{x2202}R$ and that the boundaries of $R$ and $Y$ each are disjoint unions of circles.

If a component $D$ of $R\setminus \mathring{Y}$ were a closed disc, set $c=\unicode[STIX]{x2202}D$ , a circle in $\unicode[STIX]{x2202}Y$ . If $c$ were homotopic to zero in $Y$ , then there would be a closed disc $D^{\prime }$ in $Y$ with $\unicode[STIX]{x2202}D^{\prime }=c$ . Thus $D\cup D^{\prime }$ would be an embedded sphere in  $R$ . This is not possible since $R$ is connected and would have to be equal to that sphere. Thus $c$ is not homotopic to zero in  $Y$ , hence neither in $R$ since $Y$ is incompressible in  $R$ . This is a contradiction, and hence no component of $R\setminus \mathring{Y}$ is a disc. Note also that no component of $R\setminus \mathring{Y}$ is a closed surface since $R$ is connected and $Y$ is non-empty.

From the Mayer–Vietoris sequence, we obtain $\unicode[STIX]{x1D712}(R)=\unicode[STIX]{x1D712}(Y)+\unicode[STIX]{x1D712}(R\setminus \mathring{Y})$ . Since no component of $R\setminus \mathring{Y}$ is a disc or a closed surface, we have $\unicode[STIX]{x1D712}(R\setminus \mathring{Y})\leqslant 0$ and hence

$$\begin{eqnarray}\unicode[STIX]{x1D712}(R)\leqslant \unicode[STIX]{x1D712}(Y)=\unicode[STIX]{x1D712}(X).\end{eqnarray}$$

If $\unicode[STIX]{x1D712}(R)=\unicode[STIX]{x1D712}(Y)$ , we have $\unicode[STIX]{x1D712}(C)=0$ for each component $C$ of $R\setminus \mathring{Y}$ . Hence each such $C$ is an annulus or a cross cap. If $C$ does not intersect  $U$ , then $C$ is also a component of $R\setminus X$ .

If $C$ intersects $U$ , it contains the corresponding parts of the boundary of  $R$ . Since $R$ is connected and $Y$ is non-empty, $C$  also contains a part of the boundary of  $Y$ . Hence the boundary of $C$ has more than one component, and hence $C$ is an annulus. Therefore $C$ contains precisely one boundary circle of $R$ and intersects only the corresponding part of  $U$ .

Let $C^{\prime }$ be a component of $R\setminus \mathring{X}$ that is contained in  $C$ . If $C^{\prime }$ contains a component of $\unicode[STIX]{x2202}R$ , then $C=C^{\prime }$ and $C^{\prime }$ is an annulus. If $C^{\prime }$ intersects a component of $\unicode[STIX]{x2202}R$ but does not contain it, then $C\setminus C^{\prime }$ is a subdomain of $C$ whose boundary components intersects both boundary circles of  $C$ . This is possible only if $C\setminus \mathring{X}$ consists of attached lunes. ◻

Proof. Since $H_{0}^{1}(M)$ is dense in $L^{2}(M)$ , this follows immediately from the construction of the Friedrichs extension; compare with [Reference TaylorTay96, § A.8].◻

Proof. Since the boundary of $M$ is piecewise smooth, there is a sequence of functions $\unicode[STIX]{x1D712}_{n}$ in $C_{c}^{1}(M)$ such that $0\leqslant \unicode[STIX]{x1D712}_{n}\leqslant 1$ and $|\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D712}_{n}|\leqslant 1/n$ , such that $\{\unicode[STIX]{x1D712}_{n}=1\}$ contains the support of $\unicode[STIX]{x1D712}_{n-1}$ in its interior, and such that $\cup \{\unicode[STIX]{x1D712}_{n}=1\}=M$ . With such a sequence, we can reduce the assertion of Lemma 3.2 to the case of functions in $C_{c,0}^{0,1}(M)\cap H^{1}(M)$ .

Given a compact set $K\subseteq M$ , there is a sequence of functions $\unicode[STIX]{x1D712}_{n}$ in $C_{c}^{1}(M)$ such that $0\leqslant \unicode[STIX]{x1D712}_{n}\leqslant 1$ and $|\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D712}_{n}|\leqslant Cn$ for some constant $C=C(K)$ , such that $\unicode[STIX]{x1D712}_{n}=1$ on the set of $x\in K$ with $d(x,\unicode[STIX]{x2202}M)\geqslant 2/n$ , and such that $\unicode[STIX]{x1D712}_{n}=0$ on the set of $x\in K$ with $d(x,\unicode[STIX]{x2202}M)\leqslant 1/n$ .

Now let $\unicode[STIX]{x1D711}\in C_{c,0}^{0,1}(M)\cap H^{1}(M)$ and $K=\operatorname{supp}\unicode[STIX]{x1D711}$ . Choose a sequence of functions $\unicode[STIX]{x1D712}_{n}$ for $K$ as above. Then $\unicode[STIX]{x1D712}_{n}\unicode[STIX]{x1D711}\rightarrow \unicode[STIX]{x1D711}$ in $H^{1}(M)$ since the area of the set of $x\in K$ with $1/n\leqslant d(x,\unicode[STIX]{x2202}M)\leqslant 2/n$ , which contains $K\cap \operatorname{supp}\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D712}_{n}$ , is bounded by $A/n$ for some constant $A$ and since $\unicode[STIX]{x1D711}\leqslant 2B/n$ on this set, where $B$ is a Lipschitz constant for  $\unicode[STIX]{x1D711}$ . This reduces the assertion of Lemma 3.2 to the case where the support of $\unicode[STIX]{x1D711}$ is contained in  $\mathring{M}$ . In this case, the assertion follows from smoothing.◻

Proof. By the spectral theorem, we may represent $L^{2}(M)$ as the space $L^{2}(X)$ of square-integrable functions on a measured space $X$ such that $\unicode[STIX]{x1D6E5}$ corresponds to multiplication by a measurable function $f$ on  $X$ . By the definition of $\unicode[STIX]{x1D706}_{0}(M)$ , we have $f\geqslant \unicode[STIX]{x1D706}_{0}(M)\geqslant 0$ almost everywhere on  $X$ .◻

Proof. See the elementary argument in [Reference BärBär00, proof of Proposition 1]. ◻

Examples 3.7.

1. (i) Let $F=\{(x,y)\mid x\geqslant 0,y\in \mathbb{R}/L\mathbb{Z}\}$ be a funnel with the expanding hyperbolic metric $dx^{2}+\cosh (x)^{2}\,dy^{2}$ . Let $\unicode[STIX]{x1D705}:\mathbb{R}\rightarrow \mathbb{R}$ be a monotonic smooth function with $\unicode[STIX]{x1D705}(x)=-1$ for $x\leqslant 1$ and $\unicode[STIX]{x1D705}(x)\rightarrow -\infty$ as $x\rightarrow \infty$ . Suppose that $j:\mathbb{R}\rightarrow \mathbb{R}$ solves $j^{\prime \prime }+\unicode[STIX]{x1D705}j=0$ with initial condition $j(0)=1$ and $j^{\prime }(0)=0$ . Then $j(x)>\cosh x$ for all $x>1$ . Furthermore, the funnel $F$ with Riemannian metric $g=dx^{2}+j(x)^{2}\,dy^{2}$ has curvature $K(x,y)=\unicode[STIX]{x1D705}(x)\leqslant -1$ and infinite area. By comparison, the Rayleigh quotient with respect to $g$ of any smooth function $\unicode[STIX]{x1D711}$ with compact support in the part $\{x\geqslant x_{0}\}$ of the funnel is at least $-\unicode[STIX]{x1D705}(x_{0})/4$ .

   Let $S$ be a non-compact surface of finite type. Endow $S$ with a hyperbolic metric which is expanding along its funnels as above. Replace the hyperbolic metric on the funnels by the above Riemannian metric  $g$ . Then the new Riemannian metric on $S$ is complete with curvature $K\leqslant -1$ and infinite area. By Proposition 3.6 and by what we said above about the Rayleigh quotients, the essential spectrum of the new Riemannian metric is empty. Hence the eigenspaces of the new metric are finite dimensional and span $L^{2}(S)$ . Therefore $S$ has infinitely many eigenvalues, and all of them have finite multiplicity.

2. (ii) As a variation of example (i), suppose now that $j$ is the unique solution of $j^{\prime \prime }+\unicode[STIX]{x1D705}j=0$ which satisfies the boundary condition $j(0)=1$ and $j(\infty )=0$ . Then $j^{\prime }(0)<-1$ and $j(x)<\exp (-x)$ for all $x>0$ . The funnel $F$ with Riemannian metric $g=dx^{2}+j(x)^{2}\,dy^{2}$ has curvature $K(x,y)=\unicode[STIX]{x1D705}(x)$ and finite area. Again by comparison, the Rayleigh quotient with respect to $g$ of any smooth function $\unicode[STIX]{x1D711}$ with compact support in the part $\{x\geqslant x_{0}\}$ of the funnel is at least $-\unicode[STIX]{x1D705}(x_{0})/4$ .

   Let $S$ be a non-compact surface of finite type, and choose $r>0$ such that $\coth (r)=-j^{\prime }(0)$ . It is not hard to see that $S$ minus the parts $\{x\geqslant r\}$ of its funnels carries hyperbolic metrics which are equal to $dx^{2}+j_{0}(x)^{2}\,dy^{2}$ along the parts $\{x<r\}$ of its funnels, where $j_{0}(x)=\sinh (r-x)/\sinh (r)$ . Then $j_{0}(x)=j(x)$ for $x<\min \{1,r\}$ . Hence any such hyperbolic metric, restricted to $S$ minus the parts $\{x\geqslant \min \{1,r\}\}$ of its funnels, when combined with $g$ along the funnels, defines a smooth and complete Riemannian metric on $S$ which has curvature $K\leqslant -1$ and finite area. Again, its essential spectrum is empty, by Proposition 3.6 and by what we said above about the Rayleigh quotients. Again we obtain that $S$ has infinitely many eigenvalues, and all of them have finite multiplicity.

3. (iii) Let $S$ be a complete non-compact hyperbolic surface without boundary. Replace a simple closed geodesic $c$ in $S$ by a Euclidean cylinder $C=\{(x,y)\mid 0\leqslant x\leqslant h,y\in \mathbb{R}/L\mathbb{Z}\}$ of height $h$ and circumference $L=L(c)$ and smooth out the resulting Riemannian metric appropriately. Let $\unicode[STIX]{x1D711}=\unicode[STIX]{x1D711}(x)$ be a non-vanishing smooth function on $\mathbb{R}$ with support in $[-1,0]$ . Then the support of $\unicode[STIX]{x1D711}_{k,i}=\unicode[STIX]{x1D711}(x/k-i)$ is in $[(i-1)k,ik]$ and the Rayleigh quotient of $\unicode[STIX]{x1D711}_{k,i}$ is $R(\unicode[STIX]{x1D711})/k^{2}$ . Hence, given $\unicode[STIX]{x1D700}>0$ , we have $R(\unicode[STIX]{x1D711}_{k,i})<\unicode[STIX]{x1D700}$ if $k^{2}>R(\unicode[STIX]{x1D711})/\unicode[STIX]{x1D700}$ . We may also view $\unicode[STIX]{x1D711}_{k,i}$ as a smooth function on the cylinder $C$ and the surface $S$ if $h$ is sufficiently large. More specifically, given  $n$ , choose $h>nk$ . Then the functions $\unicode[STIX]{x1D711}_{k,1},\ldots ,\unicode[STIX]{x1D711}_{k,n}$ have disjoint supports in $C$ and Rayleigh quotients less than $\unicode[STIX]{x1D700}$ . Hence $S$ has at least $n$ eigenvalues which are less than $\unicode[STIX]{x1D700}$ . Since $C$ is a cylinder, we also have $\unicode[STIX]{x1D6EC}(S)<\unicode[STIX]{x1D700}$ . On the other hand, the essential spectrum of $S$ is still contained in $[1/4,\infty )$ , by Proposition 3.6 and the special geometry of the ends of  $S$ .

Proof. Since the ends of surfaces of finite type are funnels and funnels are diffeomorphic to annuli, we have $\unicode[STIX]{x1D6EC}(S)\leqslant \unicode[STIX]{x1D706}_{\text{ess}}(S)$ if $S$ is of finite type.

Under a Riemannian covering of complete and connected Riemannian manifolds, the bottom of the spectrum of the covered manifold is at most the bottom of the spectrum of the covering manifold; see e.g. [Reference BrooksBro85, p. 101]. Brooks showed that, under a normal Riemannian covering of complete and connected Riemannian manifolds with an amenable group of covering transformations, the bottom of the spectrum does not change, provided that the covered manifold is of finite topological type [Reference BrooksBro85, Theorem 1]. The relevant arguments of Brooks in [Reference BrooksBro85, proof of Theorem 1] and of Sullivan in [Reference SullivanSul87, proof of Theorem 2.1] remain valid in the more general case of complete Riemannian manifolds of finite topological type with boundary.

Now in our setting, embedded discs in $S$ lift to embedded discs in $\tilde{S}$ and embedded annuli and cross caps in $S$ lift to embedded annuli and cross caps in cyclic subcoverings $\bar{S}$ of  $\tilde{S}$ . Moreover, $\bar{S}$ is either an annulus or a cross cap, in particular of finite topological type. For any such  $\bar{S}$ , we have $\unicode[STIX]{x1D706}_{0}(\bar{S})=\unicode[STIX]{x1D706}_{0}(\tilde{S})$ since cyclic groups are amenable. Hence $\unicode[STIX]{x1D6EC}(S)\geqslant \unicode[STIX]{x1D706}_{0}(\tilde{S})$ by the definition of  $\unicode[STIX]{x1D6EC}(S)$ .◻

Proof of Theorem 1.7.

Recall that non-zero eigenfunctions of the Laplacian cannot vanish of infinite order at any point; see e.g. [Reference AronszajnAro57]. Hence by the main result of [Reference BersBer55], at any critical point $z\in Z_{\unicode[STIX]{x1D711}}\cap \mathring{S}$ of $\unicode[STIX]{x1D711}$ , there are Riemannian normal coordinates $(x,y)$ about  $z$ , a spherical harmonic $p=p(x,y)\neq 0$ of some order $n\geqslant 2$ , and a constant $\unicode[STIX]{x1D6FC}\in (0,1)$ such that

$$\begin{eqnarray}\unicode[STIX]{x1D711}(x,y)=p(x,y)+O(r^{n+\unicode[STIX]{x1D6FC}}),\end{eqnarray}$$

where we write $(x,y)=(r\cos \unicode[STIX]{x1D703},r\sin \unicode[STIX]{x1D703})$ . By [Reference ChengChe76, Lemma 2.4], there is a local $C^{1}$ -diffeomorphism $\unicode[STIX]{x1D6F7}$ about $0\in \mathbb{R}^{2}$ fixing 0 such that $\unicode[STIX]{x1D711}=p\circ \unicode[STIX]{x1D6F7}$ . Note that, up to a rotation of the $(x,y)$ -plane, we have

$$\begin{eqnarray}p=p(x,y)=cr^{n}\cos n\unicode[STIX]{x1D703}\end{eqnarray}$$

for some constant $c\neq 0$ . It follows that the interior nodal set $Z_{\unicode[STIX]{x1D711}}\cap \mathring{S}$ of $\unicode[STIX]{x1D711}$ is a locally finite graph with critical points of $\unicode[STIX]{x1D711}$ as vertices and that the valence of points on $Z_{\unicode[STIX]{x1D711}}$ is as asserted.

It remains to discuss points $z\in Z_{\unicode[STIX]{x1D711}}\cap \unicode[STIX]{x2202}S$ . Since $\dim S=2$ , there are isothermal coordinates around $z$ , that is, coordinates $(x,y)$ about $z$ in which the Riemannian metric $g$ of $S$ is conformal to the Euclidean metric $g_{0}$ : $g=fg_{0}$ with $f=f(x,y)>0$ . Then, again since $\dim S=2$ , the associated Laplacians satisfy $f\unicode[STIX]{x1D6E5}=\unicode[STIX]{x1D6E5}_{0}$ , and hence $\unicode[STIX]{x1D711}$ solves the Schrödinger equation $(\unicode[STIX]{x1D6E5}_{0}+fV)\unicode[STIX]{x1D711}=0$ in the domain of the coordinates.

After an appropriate further conformal change of the coordinates, we can assume that the domain of the coordinates is $B_{\unicode[STIX]{x1D700}}(0)\cap \{y\geqslant 0\}$ such that $\unicode[STIX]{x2202}S$ corresponds to $B_{\unicode[STIX]{x1D700}}(0)\cap \{y=0\}$ . We consider $\unicode[STIX]{x1D711}$ and $W=fV$ as functions on $B^{+}=B_{\unicode[STIX]{x1D700}}(0)\cap \{y\geqslant 0\}$ , where $\unicode[STIX]{x1D711}(x,0)=0$ , and extend them to functions on $B_{\unicode[STIX]{x1D700}}(0)$ by setting $\unicode[STIX]{x1D711}(x,y)=-\unicode[STIX]{x1D711}(x,-y)$ and $W(x,-y)=W(x,y)$ . Then $\unicode[STIX]{x1D711}$ and $W$ are $C^{1,1}$ and $C^{0,1}$ on $B_{\unicode[STIX]{x1D700}}(0)$ , respectively, and $\unicode[STIX]{x1D711}$ solves $(\unicode[STIX]{x1D6E5}_{0}+W)\unicode[STIX]{x1D711}=0$ in $B^{+}$ . Since the reflection about the $x$ -axis is an isometry of the Euclidean plane, we also have $(\unicode[STIX]{x1D6E5}_{0}\unicode[STIX]{x1D711})(x,y)=-(\unicode[STIX]{x1D6E5}_{0}\unicode[STIX]{x1D711})(x,-y)$ . Hence

$$\begin{eqnarray}(\unicode[STIX]{x1D6E5}_{0}+W)\unicode[STIX]{x1D711}(x,y)=-(\unicode[STIX]{x1D6E5}_{0}\unicode[STIX]{x1D711})(x,-y)-W(x,y)\unicode[STIX]{x1D711}(x,-y)=0\end{eqnarray}$$

in $B^{-}=B_{\unicode[STIX]{x1D700}}(0)\cap \{y\leqslant 0\}$ . Since $\unicode[STIX]{x1D711}=0$ along the $x$ -axis, all $x$ -derivatives of $\unicode[STIX]{x1D711}$ vanish along the $x$ -axis. Since $\unicode[STIX]{x1D711}$ solves $(\unicode[STIX]{x1D6E5}_{0}+W)\unicode[STIX]{x1D711}=0$ , the second derivative of $\unicode[STIX]{x1D711}$ in the $y$ -direction vanishes along the $x$ -axis as well, and hence $\unicode[STIX]{x1D711}$ is $C^{2,1}$ . We conclude that $\unicode[STIX]{x1D711}$ is a strong solution of $(\unicode[STIX]{x1D6E5}_{0}+W)\unicode[STIX]{x1D711}=0$ on $B_{\unicode[STIX]{x1D700}}(0)$ , and hence the main result of [Reference BersBer55] and [Reference ChengChe76, proof of Lemma 2.4] applies. The remaining assertions follow as in the case of $z\in Z_{\unicode[STIX]{x1D711}}\cap \mathring{S}$ above.◻

Proof. For all $x\in S$ , we have $|\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}(x)|\leqslant |\unicode[STIX]{x1D711}(x)|$ . Hence $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}$ is in $L^{2}(M)$ . Moreover, $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}(x)\rightarrow \unicode[STIX]{x1D711}(x)$ , and hence $\lim _{\unicode[STIX]{x1D700}\rightarrow 0}\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}=\unicode[STIX]{x1D711}$ in $L^{2}(M)$ . Furthermore, $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}$ has weak gradient

(4.4) $$\begin{eqnarray}\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}(x)=\left\{\begin{array}{@{}ll@{}}\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}(x)\quad & \text{if}~|\unicode[STIX]{x1D711}(x)|\geqslant \unicode[STIX]{x1D700},\\ 0\quad & \text{otherwise}.\end{array}\right.\end{eqnarray}$$

It follows that $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}$ is in $H^{1}(M)$ .

As for the claim about the $H^{1}$ -convergence $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}\rightarrow \unicode[STIX]{x1D711}$ , we note that $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}=\unicode[STIX]{x1D711}$ on $S\setminus Z_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ and that the family of sets $Z_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ is nested with $\bigcap _{\unicode[STIX]{x1D700}>0}Z_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})=Z_{\unicode[STIX]{x1D711}}$ . Moreover, $\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}$ vanishes on $Z_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ . On the other hand, with respect to the area element of  $S$ , the set of points of density of $Z_{\unicode[STIX]{x1D711}}$ has full measure in $Z_{\unicode[STIX]{x1D711}}$ and $\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}(x)=0$ in any such point. It follows that $\lim _{\unicode[STIX]{x1D700}\rightarrow 0}\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}=\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}$ in $L^{2}(M)$ and hence that $\lim _{\unicode[STIX]{x1D700}\rightarrow 0}\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}=\unicode[STIX]{x1D711}$ in $H^{1}(M)$ .◻

Proof. Part (i) follows immediately from what we said above. As for part (ii), suppose that there is a loop in a component $C$ of $Y_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ which is not contractible in  $C$ , but is contractible in $S$ . Then there is an embedded circle $c$ in the interior of $C$ with that property. By Theorem 2.3, there is a closed disc $D$ in $S$ with $\unicode[STIX]{x2202}D=c$ . Since $c$ is not contractible in  $C$ , the interior of $D$ contains a component of $S\setminus Y_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})\subseteq Z_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ . This contradicts part (i).◻

Proof. The first statement is clear since $\unicode[STIX]{x1D700}$ is $\unicode[STIX]{x1D711}$ -regular.

Fix an exhaustion of $S$ by compact subsurfaces $S_{n}$ such that $S\setminus S_{n}$ consists of cylindrical neighborhoods of the ends of $S$ and such that $\unicode[STIX]{x2202}S_{n}$ intersects the set $\{\unicode[STIX]{x1D711}=\pm \unicode[STIX]{x1D700}\}$ transversally. Then the boundary components of any $S_{n}$ are labeled by the ends of $S$ they belong to, any $S_{n}$ meets only finitely many components of the set $\{\unicode[STIX]{x1D711}=\pm \unicode[STIX]{x1D700}\}$ , and the sets $\{\unicode[STIX]{x1D711}=\pm \unicode[STIX]{x1D700}\}\cap \unicode[STIX]{x2202}S_{n}$ are finite.

Let $D_{1}\subseteq D_{2}\subseteq \cdots \,$ be an ascending chain of pairwise distinct $\unicode[STIX]{x1D700}$ -discs. For $n$ sufficiently large, we have $D_{1}\subseteq S_{n}$ . Then $D_{l}\cap S_{n}\neq \emptyset$ and $\unicode[STIX]{x2202}D_{l}\cap \unicode[STIX]{x2202}S_{n}\subseteq \{\unicode[STIX]{x1D711}=\pm \unicode[STIX]{x1D700}\}\cap \unicode[STIX]{x2202}S_{n}$ for all $l\geqslant 1$ . Moreover, if $\unicode[STIX]{x2202}D_{l}\cap \unicode[STIX]{x2202}S_{n}=\emptyset$ , then $D_{l}\subseteq S_{n}$ . Since $\unicode[STIX]{x1D700}$ is $\unicode[STIX]{x1D711}$ -regular, it follows that the chain of discs is finite.

Let $D$ and $D^{\prime }$ be maximal $\unicode[STIX]{x1D700}$ -discs and suppose that $D\cap D^{\prime }\neq \emptyset$ . Note that $D$ and $D^{\prime }$ each have only one boundary circle, $c$ and  $c^{\prime }$ . If $c=c^{\prime }$ , then $D=D^{\prime }$ by maximality. If $c$ is contained in the interior of  $D^{\prime }$ , then $c^{\prime }$ is contained in the interior of $D$ , since otherwise $D$ would be contained in the interior of  $D^{\prime }$ , contradicting maximality. But then $D\cup D^{\prime }$ is a subsurface of $S$ without boundary which is closed as a subset, and hence $D\cup D^{\prime }=S$ . This is impossible since then $S=D\cup D^{\prime }$ would be a sphere.◻

Proof. By Lemma 4.7, $Y_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ is the union of $S\setminus Z_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ with all maximal $\unicode[STIX]{x1D700}$ -discs. By the requirement on the normal derivative in (4.5), the boundary circle $c$ of each maximal $\unicode[STIX]{x1D700}$ -disc $D$ is attached to a component of $\{\unicode[STIX]{x1D711}\geqslant \unicode[STIX]{x1D700}\}$ if $\unicode[STIX]{x1D711}|_{c}=\unicode[STIX]{x1D700}$ and a component of $\{\unicode[STIX]{x1D711}\leqslant -\unicode[STIX]{x1D700}\}$ if $\unicode[STIX]{x1D711}|_{c}=-\unicode[STIX]{x1D700}$ , respectively.◻

Proof. We may assume that $Y_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})\neq S$ . We suppose first that the Rayleigh quotient $R(\unicode[STIX]{x1D711})<\unicode[STIX]{x1D6EC}(S)$ and choose $\unicode[STIX]{x1D6FF}>0$ such that

(4.11) $$\begin{eqnarray}R(\unicode[STIX]{x1D711})\leqslant \unicode[STIX]{x1D6EC}(S)-2\unicode[STIX]{x1D6FF}.\end{eqnarray}$$

By Lemma 4.3 and since $S\setminus Y_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})\subseteq Z_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ , we have, for any sufficiently small $\unicode[STIX]{x1D711}$ -regular $\unicode[STIX]{x1D700}>0$ ,

(4.12) $$\begin{eqnarray}\frac{\mathop{\sum }_{C}\int _{C}|\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}|^{2}}{\mathop{\sum }_{C}\int _{C}\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}^{2}}\leqslant \frac{\int _{S}|\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}|^{2}\,dv}{\int _{S}\unicode[STIX]{x1D711}^{2}\,dv}+\unicode[STIX]{x1D6FF}=R(\unicode[STIX]{x1D711})+\unicode[STIX]{x1D6FF}\leqslant \unicode[STIX]{x1D6EC}(S)-\unicode[STIX]{x1D6FF},\end{eqnarray}$$

where the sums run over the components $C$ of $Y_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ . We conclude that there is a component $C$ of $Y_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ such that

(4.13) $$\begin{eqnarray}R(\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}|_{C})=\frac{\int _{C}|\unicode[STIX]{x1D6FB}\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}|^{2}}{\int _{C}\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}^{2}}\leqslant \unicode[STIX]{x1D6EC}(S)-\unicode[STIX]{x1D6FF}.\end{eqnarray}$$

Now $\unicode[STIX]{x1D711}$ is smooth on $S$ , and hence $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}|_{C}$ is smooth on $C$ and vanishes along $\unicode[STIX]{x2202}C$ . Therefore $\unicode[STIX]{x1D711}_{\unicode[STIX]{x1D700}}|_{C}\in H_{0}^{1}(C)$ , by Lemma 3.2. Now it follows from the definition of $\unicode[STIX]{x1D6EC}(S)$ that the interior of $C$ cannot be diffeomorphic to an open disc, an open annulus, or an open cross cap. Thus the fundamental group of $C$ contains  $F_{2}$ .

Assume now that $R(\unicode[STIX]{x1D711})=\unicode[STIX]{x1D6EC}(S)$ . Recall that $\unicode[STIX]{x1D711}$ is a finite linear combination of eigenfunctions of  $S$ , $\unicode[STIX]{x1D711}=\sum c_{i}\unicode[STIX]{x1D711}_{i}$ , where $\unicode[STIX]{x1D711}_{i}\in E$ is a $\unicode[STIX]{x1D706}_{i}$ -eigenfunction with $\unicode[STIX]{x1D706}_{i}\leqslant \unicode[STIX]{x1D6EC}(S)$ . If there would be an $i$ with $c_{i}\neq 0$ and $\unicode[STIX]{x1D706}_{i}<\unicode[STIX]{x1D6EC}(S)$ , then we would have $R(\unicode[STIX]{x1D711})<\unicode[STIX]{x1D6EC}(S)$ , a contradiction. It follows that all $\unicode[STIX]{x1D706}_{i}$ with $c_{i}\neq 0$ are equal to $\unicode[STIX]{x1D6EC}(S)$ , and hence that $\unicode[STIX]{x1D711}$ is a $\unicode[STIX]{x1D6EC}(S)$ -eigenfunction.

Suppose now, more generally, that $\unicode[STIX]{x1D711}$ is an eigenfunction with corresponding eigenvalue $\unicode[STIX]{x1D706}\leqslant \unicode[STIX]{x1D6EC}(S)$ . Then $\unicode[STIX]{x1D711}$ is smooth on $S$ . By Theorem 1.7, the nodal domains of $\unicode[STIX]{x1D711}$ have piecewise smooth boundary. Hence Lemma 3.2 implies that, for any nodal domain $C$ of $\unicode[STIX]{x1D711}$ , we have $\unicode[STIX]{x1D711}|_{C}\in H_{0}^{1}(C)$ with $R(\unicode[STIX]{x1D711}|_{C})=\unicode[STIX]{x1D706}$ . In particular, $\unicode[STIX]{x1D706}_{0}(C)\leqslant \unicode[STIX]{x1D706}$ .

Let $C^{\prime }$ be a thickening of $C$ , that is, $C^{\prime }$ is a domain in $S$ with piecewise smooth boundary which contains $C$ in its interior and such that $C$ is a deformation retract of  $C^{\prime }$ . If the fundamental group of $C$ does not contain  $F_{2}$ , then neither does the fundamental group of  $C^{\prime }$ , and then

$$\begin{eqnarray}\unicode[STIX]{x1D6EC}(S)\leqslant \unicode[STIX]{x1D706}_{0}(C^{\prime })\leqslant \unicode[STIX]{x1D706}_{0}(C).\end{eqnarray}$$

The extension $\unicode[STIX]{x1D711}^{\prime }$ of $\unicode[STIX]{x1D711}|_{\unicode[STIX]{x2202}C}$ to $C^{\prime }$ , setting $\unicode[STIX]{x1D711}^{\prime }|_{C^{\prime }\setminus C}=0$ , is in $H_{0}^{1}(M)$ and in the domain of the Laplacian of $C^{\prime }$ . Moreover, it has Rayleigh quotient $R(\unicode[STIX]{x1D711}^{\prime })=\unicode[STIX]{x1D706}$ . Hence Lemma 3.5 applies and shows that $\unicode[STIX]{x1D706}_{0}(C^{\prime })=\unicode[STIX]{x1D6EC}(S)$ and $\unicode[STIX]{x1D6E5}\unicode[STIX]{x1D711}^{\prime }=\unicode[STIX]{x1D6EC}(S)\unicode[STIX]{x1D711}^{\prime }$ . Now $\unicode[STIX]{x1D711}^{\prime }$ does not vanish identically on  $C$ , but vanishes on $C^{\prime }\setminus C$ . This is in contradiction to the unique continuation property for Laplace operators. Hence the fundamental group of any nodal domain of $\unicode[STIX]{x1D711}$ contains  $F_{2}$ .

Let $C$ be a nodal domain of $\unicode[STIX]{x1D711}$ and suppose that $C$ is not incompressible in  $S$ . Then there is a loop $c$ in $C$ which is not homotopic to zero in  $C$ , but is homotopic to zero in  $S$ . Without loss of generality, we may assume that $c$ is a Jordan curve in  $\mathring{C}$ . Then $c$ bounds a disc in  $S$ , which is not contained in  $C$ , by Theorem 2.3. Again by Theorem 2.3, there would be a nodal domain $D$ of $\unicode[STIX]{x1D711}$ whose closure is a closed disc with piecewise smooth boundary and with $\unicode[STIX]{x1D706}_{0}(D)=\unicode[STIX]{x1D6EC}(S)$ . This is impossible, since $\unicode[STIX]{x1D6EC}(S)$ is not attained on (embedded) closed discs. Hence all nodal domains of $\unicode[STIX]{x1D711}$ are incompressible in  $S$ .

Let $C$ be a nodal domain and $c_{1},c_{2}:[0,1]\rightarrow C$ be two loops at a point $x\in C$ which generate a free subgroup $F_{2}\in \unicode[STIX]{x1D70B}_{1}(C,x)$ . By Theorem 1.7, we may assume that the images of $c_{1}$ and $c_{2}$ are contained in  $\mathring{C}$ . Without loss of generality, we may also assume that $\unicode[STIX]{x1D711}$ is positive on  $\mathring{C}$ . Then

$$\begin{eqnarray}\mathring{C}=\mathop{\bigcup }_{\unicode[STIX]{x1D700}>0}\{y\in C\mid \unicode[STIX]{x1D711}(y)>\unicode[STIX]{x1D700}\}.\end{eqnarray}$$

Therefore the image of $c_{0}$ and $c_{1}$ is contained in $\{y\in C\mid \unicode[STIX]{x1D711}(y)>\unicode[STIX]{x1D700}\}$ for all sufficiently small $\unicode[STIX]{x1D700}>0$ . Hence the fundamental group of the component $C_{0}\subseteq C$ of $Y_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ which contains $c_{0}$ and $c_{1}$ also contains  $F_{2}$ . From Lemma 4.8 we conclude that the component of $Y_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ containing $C_{0}$ contains  $F_{2}$ .◻

Proof of Lemma 4.14.

The components of $S\setminus K$ are diffeomorphic to open annuli, hence their fundamental group is infinite cyclic. Moreover, $C$  is incompressible in  $S$ , hence also in the component of $S\setminus K$ containing it. Therefore the fundamental group of $C$ is either trivial or infinite cyclic. Now, as a domain in a cylinder, $C$  is orientable. Hence the interior of $C$ is diffeomorphic to an open disc or an open annulus.◻

Proof. By definition, $Z_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700}^{\prime })\subseteq Z_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ . If a component of $Z_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})$ is contained in the interior of an embedded closed disc, then also all the components of $Z_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700}^{\prime })$ it contains. It follows that $Y_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})\subseteq Y_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700}^{\prime })$ . By Lemma 4.6(ii), if $C$ is an $F_{2}$ -component of $X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700})$ , then also the component $C^{\prime }$ of $X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700}^{\prime })$ containing it. Hence $X_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700})\subseteq X_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700}^{\prime })$ .

Now let $C$ be a component of $X_{\unicode[STIX]{x1D711}}^{+}(\unicode[STIX]{x1D700})$ and suppose that the component $C^{\prime }$ of $X_{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D700}^{\prime })$ containing $C$ belongs to $X_{\unicode[STIX]{x1D711}}^{-}(\unicode[STIX]{x1D700}^{\prime })$ . By Lemma 4.8, $C$ is the union of a component $C_{0}$ of $\{\unicode[STIX]{x1D711}\geqslant \unicode[STIX]{x1D700}\}$ with maximal $\unicode[STIX]{x1D700}$ -discs. Now let $x$ be a point in the interior of $C_{0}$ and choose loops $c_{0}$ and $c_{1}$ in the interior of $C_{0}$ generating an $F_{2}$ in $\unicode[STIX]{x1D70B}_{1}(C,x)$ ; compare with our discussion further up. In particular, $\unicode[STIX]{x1D711}>\unicode[STIX]{x1D700}$ along $c_{0}$ and  $c_{1}$ . Under the inclusion $C\rightarrow C^{\prime }$ , $c_{0}$ and $c_{1}$ cannot be contained in the maximal $\unicode[STIX]{x1D700}^{\prime }$ -discs belonging to $C^{\prime }$ because they would be homotopic to zero in $S$ otherwise. But then they must meet $\{\unicode[STIX]{x1D711}\leqslant -\unicode[STIX]{x1D700}^{\prime }\}$ , a contradiction. We conclude that $X_{\unicode[STIX]{x1D711}}^{+}(\unicode[STIX]{x1D700})\subseteq X_{\unicode[STIX]{x1D711}}^{+}(\unicode[STIX]{x1D700}^{\prime })$ ; similarly $X_{\unicode[STIX]{x1D711}}^{-}(\unicode[STIX]{x1D700})\subseteq X_{\unicode[STIX]{x1D711}}^{-}(\unicode[STIX]{x1D700}^{\prime })$ .◻

Proof. After the above discussion leading to the definition of $S_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700},K)$ , we have the following remaining issues.

If a boundary circle $c$ of $X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700},K)$ bounds a cross cap $C$ in  $S$ , then either already $C\subseteq X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700}^{\prime },K^{\prime })$ or else an annulus $A\subseteq C$ is attached to $c$ along one of its boundary circles and the other boundary circle $c^{\prime }$ belongs to the boundary of $X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700}^{\prime },K^{\prime })$ . Then $c^{\prime }$ bound a cross cap $C^{\prime }$ in the complement (of the interior) of $A$ in $C$ and $C=A\cup C^{\prime }$ .

Conversely, if a boundary circle $c$ of $X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700}^{\prime },K^{\prime })$ bounds a cross cap $C^{\prime }$ in  $S$ , then $c$ is contained in the interior of $K^{\prime }$ and thus $\unicode[STIX]{x1D711}=\pm \unicode[STIX]{x1D700}^{\prime }$ along  $c$ . We conclude that $c$ is a boundary circle of an annulus $A$ attached to $X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700},K)$ along the other boundary circle of  $A$ . Thus $C=A\cup C^{\prime }$ is a cross cap in $S$ with $\unicode[STIX]{x2202}C$ a boundary circle of $X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700},K)$ . By the discussion further up we obtain that $C$ is in the interior of  $K$ .

If boundary circles $c_{0}$ and $c_{1}$ of $X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700},K)$ bound an annulus $A$ in  $S$ , then either already $A\subseteq X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700}^{\prime },K^{\prime })$ or else disjoint annuli $A_{0},A_{1}\subseteq A$ are attached to $c_{0}$ and  $c_{1}$ , each along one of its boundary circles, and the other boundary circles $c_{0}^{\prime }$ and $c_{1}^{\prime }$ bound an annulus $A^{\prime }\subseteq A$ between $A_{0}$ and  $A_{1}$ . Then $A=A_{0}\cup A^{\prime }\cup A_{1}$ .

Conversely, if boundary circles $c_{0}$ and $c_{1}$ of $X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700}^{\prime },K^{\prime })$ bound an annulus $A^{\prime }$ in  $S$ , then $A^{\prime }$ is contained in the interior of $K^{\prime }$ and thus $\unicode[STIX]{x1D711}=\pm \unicode[STIX]{x1D700}^{\prime }$ along $\unicode[STIX]{x2202}A^{\prime }$ . Arguing as in the case of cross caps, we get annuli $A_{0}$ and $A_{1}$ with one boundary circle in $X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700},K)$ and the other equal to $c_{0}$ and  $c_{1}$ , respectively. Thus $A=A_{0}\cup A^{\prime }\cup A_{1}$ is an annulus in $S$ such that $\unicode[STIX]{x2202}A$ lies in $X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700},K)$ . By the discussion further up we obtain that $A$ is in the interior of  $K$ .◻

Proof. If $\unicode[STIX]{x1D713}$ is sufficiently $C^{2}$ -close to $\unicode[STIX]{x1D711}$ on $L$ , then $\pm \unicode[STIX]{x1D700}$ are regular values of $\unicode[STIX]{x1D713}|_{L}$ , the curves $\unicode[STIX]{x1D713}|_{L}=\pm \unicode[STIX]{x1D700}$ intersect $\unicode[STIX]{x2202}K$ transversally, and there is a small isotopy of $S$ which leaves $K$ and $\unicode[STIX]{x2202}K$ invariant which deforms the configuration of curves $\{\unicode[STIX]{x1D713}=\unicode[STIX]{x1D700}\}\cap L$ and $\{\unicode[STIX]{x1D713}=-\unicode[STIX]{x1D700}\}\cap L$ to the configuration of curves $\{\unicode[STIX]{x1D711}=\unicode[STIX]{x1D700}\}\cap L$ respectively $\{\unicode[STIX]{x1D711}=-\unicode[STIX]{x1D700}\}\cap L$ and, therefore, also the subsurfaces $\{\unicode[STIX]{x1D713}\geqslant \unicode[STIX]{x1D700}\}\cap K$ and $\{\unicode[STIX]{x1D713}\leqslant -\unicode[STIX]{x1D700}\}\cap K$ to the subsurfaces $\{\unicode[STIX]{x1D711}\geqslant \unicode[STIX]{x1D700}\}\cap K$ respectively $\{\unicode[STIX]{x1D711}\leqslant -\unicode[STIX]{x1D700}\}\cap K$ .

Clearly, if a boundary segment of the latter intersects an $\unicode[STIX]{x1D700}$ -disc $D$ of  $\unicode[STIX]{x1D711}$ , then $D$ is contained in $L$ and corresponds under the isotopy to an $\unicode[STIX]{x1D700}$ -disc $B$ of  $\unicode[STIX]{x1D713}$ . Attaching the parts $B\cap K$ of such discs, we get a surface $T^{\pm }$ such that the above isotopy deforms $T^{\pm }$ to $X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700},K)$ . In particular, the fundamental group of $T^{\pm }$ contains an  $F_{2}$ , $T^{\pm }$ is incompressible in $S$ and

$$\begin{eqnarray}\unicode[STIX]{x1D712}(T^{\pm })=\unicode[STIX]{x1D712}(X_{\unicode[STIX]{x1D711}}^{\pm }(\unicode[STIX]{x1D700},K)).\end{eqnarray}$$

By changing $\unicode[STIX]{x1D700}$ slightly, we can achieve that $\unicode[STIX]{x1D700}$ is also $\unicode[STIX]{x1D713}$ -regular. Then, by what we said, $T^{\pm }$ is a component of $X_{\unicode[STIX]{x1D713}}^{\pm }(\unicode[STIX]{x1D700},K)$ . Moreover, choosing a $\unicode[STIX]{x1D713}$ -regular $(\unicode[STIX]{x1D700},K^{\prime })$ with $K$ in the interior of $K^{\prime }$ , we have $T^{\pm }\subseteq X_{\unicode[STIX]{x1D713}}^{\pm }(\unicode[STIX]{x1D700},K^{\prime })$ . Hence $X_{\unicode[STIX]{x1D713}}^{\pm }(\unicode[STIX]{x1D700},K^{\prime })$ is obtained from $T^{\pm }$ by attaching annuli, cross caps, and lunes. Now annuli where both boundary curves are attached to $T^{\pm }$ and cross caps attached to $T^{\pm }$ are contained in the interior of $K$ and belong to $S_{\unicode[STIX]{x1D713}}^{\pm }(\unicode[STIX]{x1D700},K)$ . We (finally) conclude that $(S,S_{\unicode[STIX]{x1D713}}^{+}(\unicode[STIX]{x1D700},K^{\prime }),S_{\unicode[STIX]{x1D713}}^{-}(\unicode[STIX]{x1D700},K^{\prime }))$ is isotopic to the triple $(S,S_{\unicode[STIX]{x1D711}}^{+}(\unicode[STIX]{x1D700},K),S_{\unicode[STIX]{x1D711}}^{-}(\unicode[STIX]{x1D700},K))$ .◻

End of proof of Theorem 1.5.

Let $\mathbb{E}$ be a subspace of $L^{2}(M)$ which is generated by finitely many eigenfunctions with corresponding eigenvalues less than or equal to $\unicode[STIX]{x1D6EC}(S)$ and denote by $\mathbb{S}$ the unit sphere in $\mathbb{E}$ and by $\mathbb{P}$ the projective space of $\mathbb{E}$ . Theorem 1.5 follows if any such $\mathbb{E}$ has dimension at most  $-\unicode[STIX]{x1D712}(S)$ .