Multiple orthogonal geodesic chords in nonconvex Riemannian disks using obstacles

Abstract

We use nonsmooth critical point theory and the theory of geodesics with obstacle to show a multiplicity result about orthogonal geodesic chords in a Riemannian manifold (with boundary) which is homeomorphic to an N-disk. This applies to brake orbits in a potential well of a natural Hamiltonian system, providing a further step towards the proof of a celebrated conjecture by Seifert (Math Z 51:197–216, 1948).

Appendix A. On the transversality of the normal bundle and the tangent bundle of submanifolds

Let (M, g) be a Riemannian manifold, and let $S\subset M$ be an embedded submanifold. For the application we aim at, S will be the boundary of M.

The normal bundle of S will be denoted by ${\mathcal {N}}(S)$, and for $p\in S$, we set ${\mathcal {N}}_p(S)={\mathcal {N}}(S)\cap T_pM$. The normal bundle ${\mathcal {N}}(S)$ and the tangent bundle TS will be considered submanifolds of the tangent bundle TM. Given a point $p\in S$ and a normal vector $v\in {\mathcal {N}}_p(S)$, we will denote by $\alpha ^S_p:T_pS\times T_pS\rightarrow {\mathcal {N}}_p(S)$ the second fundamental form of S at p, and by $A_v^S:T_pS\rightarrow T_pS$ the shape operator of S at p in the direction v, defined by:

$$\begin{aligned}\phantom {,\quad \forall \,u,u'\in T_pS.}g\big (A_v^S(u),u'\big )=-g\big (\alpha ^S(u,u'),v\big ),\quad \forall \,u,u'\in T_pS.\end{aligned}$$

For $v\in T_pM$, let us identify $T_v(TM)$ with $T_pM\oplus T_pM$, with the first summand being the horizontal space of the Levi–Civita connection of g and the second summand being the vertical subspace.

1.1 A.1. A transversality condition

Proposition A.1

Let $S_1,S_2\subset M$ hypersurfaces that meet transversally at $p\in S_1\cap S_2$. Assume that $v\in T_pM\setminus \{0\}$ belongs to the intersection ${\mathcal {N}}(S_1)\cap TS_2$ and denote by ${\mathcal {A}}_v\subset T_pM$ the subspace:

$$\begin{aligned} {\mathcal {A}}_v=\big \{A_v^{S_1}(w)-\alpha ^{S_2}(w,v):w\in T_pS_1\cap T_pS_2\big \}. \end{aligned}$$

(A.1)

An element $(z,z')\in T_pM\oplus T_pM\cong T_v(TM)$ belongs to the sum $T_v({\mathcal {N}}_p(S_1)\big )+T_v(TS_2)$ iff $z'$ belongs to the affine subspace of $T_pM$:

$$\begin{aligned} \big [A^1_v(u_1)+\alpha ^2_p(u_2,v)\big ]+{\mathcal {A}}_v+T_pS_2, \end{aligned}$$

where $u_1\in T_pS_1$, $u_2\in T_pS_2$ are chosen in such a way that $z=u_1+u_2$. In particular, ${\mathcal {N}}(S_1)$ and $TS_2$ meet transversally at v if and only if:

$$\begin{aligned} {\mathcal {A}}_v\not \subset T_pS_2. \end{aligned}$$

(A.2)

Proof

Using the identification $T_v(TM)\cong T_pM\oplus T_pM$, the tangent spaces $T_v\big ({\mathcal {N}}(S_1)\big )$ and $T_v(TS_2)$ are given by:

$$\begin{aligned}\begin{aligned}&T_v\big ({\mathcal {N}}(S_1)\big )=\big \{\big (u_1,A_v^{S_1}(u_1)+\lambda \cdot v\big ):u_1\in T_pS_1,\ \lambda \in {\mathbb {R}}\big \},\\&T_v(TS_2)=\big \{\big (u_2,u_2'+\alpha ^{S_2}(u_2,v)\big ):u_2,u_2'\in T_pS_2\big \}. \end{aligned} \end{aligned}$$

From these equalities, one obtains readily a proof of the first statement.

As to the transversality of ${\mathcal {N}}(S_1)$ and $TS_2$ at v, i.e., $T_v\big ({\mathcal {N}}(S_1)\big )+T_v(TS_2)=T_v(TM)$, the proof follows from a linear algebra argument (apply next Lemma to $V=T_pM$, $V_1=T_pS_1$, $V_2=T_pS_2$, ${\widetilde{V}}_2=\mathbb {R}\cdot v$, $A=A^{S_1}_v$ and $\alpha =\alpha ^{S_2}(\cdot ,v)$). $\square$

Lemma A.2

Let V be a finite dimensional vector space, and let $V_1,V_2\subset V$ be subspaces with $V_1+V_2=V$, and $V_2$ of codimension 1. Let ${\widetilde{V}}_2\subset V_2$ be an arbitrary subspace, let $A:V_1\rightarrow V_1$ and $\alpha :V_2\rightarrow V$ be linear maps, and define subspaces $W_1,W_2\subset V\oplus V$ and ${\mathcal {A}}\subset V$ by:

$$\begin{aligned}\begin{aligned}&W_1=\big \{\big (u_1,A(u_1)+{\widetilde{v}}_2\big ):u_1\in V_1,\ {\widetilde{v}}_2\in {\widetilde{V}}_2\big \},\\&W_2=\big \{\big (u_2,v_2+\alpha (u_2)\big ):u_2,v_2\in V_2\big \},\\&{\mathcal {A}}=\big \{Aw-\alpha w:w\in V_1\cap V_2\big \}. \end{aligned} \end{aligned}$$

Then, $W_1+W_2=V\oplus V$ if and only if ${\mathcal {A}}\not \subset V_2$.

Proof

The equality $W_1+W_2=V\oplus V$ holds iff for all $s,t\in V$, there exist $u_1\in V_1$, ${\widetilde{v}}_2\in {\widetilde{V}}_2$ and $u_2,v_2\in V_2$ such that:

$$\begin{aligned} u_1+u_2=s,\quad \text {and}\quad A(u_1)+{\widetilde{v}}_2+v_2+\alpha (u_2)=t. \end{aligned}$$

Since $V_1+V_2=V$, there exist ${\overline{u}}_1\in V_1$ and ${\overline{u}}_2\in V_2$ such that ${\overline{u}}_1+{\overline{u}}_2=s$; every other pair $(u_1,u_2)\in V_1\times V_2$ with $u_1+u_2=s$ is of the form $u_1={\overline{u}}_1+h$ and $u_2={\overline{u}}_2-h$ for some $h\in V_1\cap V_2$. Thus, the desired transversality holds iff there exists $h\in V_1\cap V_2$, ${\widetilde{v}}_2\in {\widetilde{V}}_2$ and $v_2\in V_2$ such that:

$$\begin{aligned} A({\overline{u}}_1)+A(h)+{\widetilde{v}}_2+v_2+\alpha ({\overline{u}}_2)-\alpha (h)=t, \end{aligned}$$

i.e.,

$$\begin{aligned} A(h)-\alpha (h)+v_2=t-{\widetilde{v}}_2-A({\overline{u}}_1)-\alpha ({\overline{u}}_2). \end{aligned}$$

Clearly, when t is arbitrary in V, the right hand side of the above equality is an arbitrary vector in V. Thus, $W_1+W_2=V\oplus V$ holds iff the linear map

$$\begin{aligned} (V_1\cap V_2)\times V_2\ni (h,v_2)\longmapsto A(h)-\alpha (h)+v_2\in V \end{aligned}$$

is surjective. Since $V_2$ has codimension 1 in V, this holds iff the image of the linear map $V_1\cap V_2\ni h\mapsto A(h)-\alpha (h)\in V$ is not contained in $V_2$, i.e., iff ${\mathcal {A}}\not \subset V_2$. $\square$

1.2 A.2. The case of a 1-parameter family of submanifolds

We will now study a transversality problem between the tangent bundle of a hypersurface $S_2$, and the union of the normal bundles of a smooth 1-parameter family of hypersurfaces $S_1(t)$, $t\in ]-\varepsilon ,\varepsilon [$.

More precisely, let us consider the following setup.

• $S_1,S_2\subset M$ are hypersurfaces, and $p\in S_1\cap S_2$;

• there exists a nonzero $v\in {\mathcal {N}}_p(S_1)\cap T_pS_2$. In particular, $S_1$ and $S_2$ meet transversally at p.

• ${\mathbf {n}}$ is a smooth unit vector field along $S_1$;

• $F:]-\varepsilon ,\varepsilon [\times S_1\rightarrow M$ is defined by:

  $$\begin{aligned}F(t,x)=\exp _x\big (t\cdot {\mathbf {n}}(x)\big );\end{aligned}$$

• $S_1(t)$ is defined as the image of the map $F(t,\cdot ):S_1\rightarrow M$.

Under suitable non-focality assumption for p, we can assume that F is an embedding of $]-\varepsilon ,\varepsilon [\times S_1$ onto an open subset of M. Its image is foliated by the smooth hypersurfaces $S_1(t)$.

We consider the union:

$$\begin{aligned} \bigcup _{t\in ]-\varepsilon ,\varepsilon [}{\mathcal {N}}\big (S_1(t)\big )\subset TM; \end{aligned}$$

this is a smooth submanifold of TM. Under the above assumptions, $\frac{\partial F}{\partial t}(t,x)$ is a unit vector normal to $S_1(t)$ at F(t, x), and we have a parameterization of the union $\bigcup \nolimits _{t\in ]-\varepsilon ,\varepsilon [}{\mathcal {N}}\big (S_1(t)\big )$ given by the smooth map:

$$\begin{aligned} G:]-\varepsilon ,\varepsilon [\times S_1\times {\mathbb {R}}\longrightarrow TM \end{aligned}$$

defined by:

$$\begin{aligned} \qquad G(t,y,\lambda )=\lambda \cdot \frac{\partial F}{\partial t}(t,y),\quad (t,y,\lambda )\in ]-\varepsilon ,\varepsilon [\times S_1\times {\mathbb {R}}. \end{aligned}$$

Our aim is to determine when the map G is transversal in TM to $TS_2$ at the point $(0,p,\lambda )$, where $G(0,p,\lambda )=\lambda \cdot {\mathbf {n}}(p)=v$.

Clearly, the image of ${\text {dG}}(0,p,\lambda )$ is given by:

$$\begin{aligned} \text {Im}\big ({\text {dG}}(0,p,\lambda )\big )={\mathbb {R}}\cdot \frac{\partial G}{\partial t}(0,p,\lambda )+T_v\big ({\mathcal {N}}(S_1)\big ). \end{aligned}$$

As we have observed above, if $T_v\big ({\mathcal {N}}(S_1)\big )+T_v(TS_2)=T_v(TM)$, i.e., if ${\mathcal {A}}_v\not \subset T_pS_2$, then G is transversal to $TS_2$ at $(0,p,\lambda )$.

On the other hand, if $T_v\big ({\mathcal {N}}(S_1)\big )+T_v(TS_2)\ne T_v(TM)$, then it is easy to check that $T_v\big ({\mathcal {N}}(S_1)\big )+T_v(TS_2)$ has codimension 1 in $T_v(TM)$. This follows immediately from the observation the the projection onto the first factor $T_v(TM)\cong T_pM\times T_pM\rightarrow T_pM$ induces an isomorphism between the quotient:

$$\begin{aligned} \left( T_v\big ({\mathcal {N}}(S_1)\big )+T_v(TS_1)\right) \big /\left( \{0\}\times T_pS_2\right) \end{aligned}$$

and $T_pM$.

Thus, G is transversal to $TS_2$ at $(0,p,\lambda )$ iff

$$\begin{aligned} \frac{\partial G}{\partial t}(0,p,\lambda )\not \in T_v\big ({\mathcal {N}}(S_2)\big )+T_v(TS_1). \end{aligned}$$

Now, $\frac{\partial G}{\partial t}(0,p,\lambda )=\big ({\mathbf {n}}(p),0\big )$, because the $t\mapsto \frac{\partial F}{\partial t}(t,p)$ is parallel along the geodesic $t\mapsto F(t,p)$. Since $v=\lambda \cdot {\mathbf {n}}(p)$, we conclude that G is transversal to $TS_2$ at $(0,p,\lambda )$ iff $(v,0)\not \in T_v\big ({\mathcal {N}}(S_1)\big )+T_v(TS_1)$. This happens iff $\alpha _p^2(v,v)\not \in T_pS_2$, i.e., iff $\alpha _p^2(v,v)\ne 0$.

In conclusion, we have proved the following:

Proposition A.3

In the above notations, the submanifold

$$\begin{aligned} \bigcup \limits _{t\in ]-\varepsilon ,\varepsilon [}{\mathcal {N}}\big (S_1(t)\big ) \end{aligned}$$

is transversal to $TS_2$ at v iff either one of the two conditions holds:

• (a) ${\mathcal {A}}_v\not \subset T_pS_2$;

• (b) $\alpha _p^2(v,v)\ne 0$.

$\square$

Note that conditions (a) and (b) above are stable by $C^2$-small perturbations of the metric g.

Corollary A.4

Let $N\ge 2$ and let ${\mathbb {D}}^N$ be the N-dimensional unit disk in ${\mathbb {R}}^n$. The set of Riemannian metrics on ${\mathbb {D}}^N$ that admit O–T chords has nonempty interior relatively to the $C^2$-topology.

Proof

Given a metric g on ${\mathbb {D}}^N$, an O–T chord in ${\mathbb {D}}^N$ relative to g corresponds to the intersection of the tangent bundle of some open portion $S_2$ of $\partial {\mathbb {D}}^N$ and the normal bundle of some other open portion $S_1$ of $\partial \mathbb {D}^N$. Evidently, one can easily construct a (smooth) metric g for which such intersection exists, and for which either one of conditions (a) and (b) of Proposition A.3 holds. For instance, condition (b) is rather easy to obtain. In this situation, the transversality implies that sufficiently $C^2$-small perturbations of the metric will preserve the existence of some intersection between the normal and the tangent bundle of $\partial {\mathbb {D}}^N$. This says that O–T chords will persist by sufficiently $C^2$-small perturbations of the metric, i.e., the set of metrics in ${\mathbb {D}}^N$ admitting O–T chords has nonempty interior. $\square$

It is not hard to see that, in fact, the conclusion of Corollary A.4 holds more generally for arbitrary smooth (compact) manifolds with boundary, whose dimension is greater than or equal to 2.