Multiple brake orbits in $m$-dimensional disks

Abstract

Let $(M,g)$ be a (complete) Riemannian surface, and let $\Omega \subset M$ be an open subset whose closure is homeomorphic to a disk. We prove that if $\partial \Omega$ is smooth and it satisfies a strong concavity assumption, then there are at least two distinct orthogonal geodesics in $\overline{\Omega }=\Omega \bigcup \partial \Omega$. Using the results given in Giambò et al. (Adv Differ Eq 10:931–960, 2005), we then obtain a proof of the existence of two distinct brake orbits for a class of Hamiltonian systems. In our proof we shall use recent deformation results proved in Giambò et al. (Nonlinear Anal Ser A Theory Methods Appl 73:290–337, 2010).

Appendix A. An estimate on the relative category

Let $n\ge 1$ be an integer; $\mathbb S^n$ is the $n$-dimensional sphere, and $\Delta ^n\subset \mathbb S^n\times \mathbb S^n$ is the diagonal. We want to estimate the relative Lusternik–Schnirelman category of the pair $(\mathbb S^n\times \mathbb S^n,\Delta ^n)$, and to this aim we will prove an estimate on the relative cuplength of the pair.

For a topological space $X$ and an integer $k\ge 0$, we will denote by $H^k(X)$ and $\widetilde{H}^k(X)$ respectively the $k$th singular cohomology and the $k$th reduced singular cohomology group of $X$. For a topological pair $(X,Y)$, $H^k(X,Y)$ is the $k$th relative singular cohomology group of the pair; in particular, $H^k(X,\emptyset )=H^k(X)$. Given $\alpha \in H^p(X,Y)$ and $\beta \in H^q(X,Z)$, $\alpha \cup \beta \in H^{p+q}(X,Y\bigcup Z)$ will denote the cup product of $\alpha$ and $\beta$; recall that $\alpha \cup \beta =(-1)^{pq}\beta \cup \alpha$.

The notion of relative cuplength, here recalled, will be also used.

Definition 10.1

The number ${\mathrm {cuplength}}(X,Y)$ is the largest positive integer $k$ for which there exists $\alpha _0\in H^{q_0}(X,Y)$ ($q_0\ge 0$) and $\alpha _i\in H^{q_i}(X)$, $i=1,\ldots ,k$ such that

$$\begin{aligned} q_i\ge 1,\quad \forall \; i=1,\ldots ,k, \end{aligned}$$

and

$$\begin{aligned} \alpha _0\cup \alpha _1\cup \cdots \cup \alpha _k\ne 0\quad \text {in}\quad H^{q_0+q_1+\cdots +q_k}(X,Y), \end{aligned}$$

where $\cup$ denotes the cup product.

As for the absolute Lusternik–Schirelmann category, we have the following estimate of relative category by means of relative cuplenght, cf. e.g. [3, 4]

Proposition 10.2

${\mathrm {cat}}_{\mathbb S^n \times \mathbb S^n,\Delta ^n}(\mathbb S^n \times \mathbb S^n) \ge {\mathrm {cuplength}}(\mathbb S^n\times \mathbb S^n,\Delta ^n) + 1$.

Therefore, to prove that $cat_{\mathbb S^n \times \mathbb S^n,\Delta ^n}(\mathbb S^n \times \mathbb S^n) \ge 2$ it will be sufficient to prove the following

Proposition 10.3

For all $n\ge 1$, ${\mathrm {cuplength}}(\mathbb S^n\times \mathbb S^n,\Delta ^n)\ge 1$.

Proof

The statement is equivalent to proving the existence of $p\ge 0$, $q\ge 1$, $\alpha \in H^p(\mathbb S^n\times \mathbb S^n,\Delta ^n)$ and $\beta \in H^q(\mathbb S^n\times \mathbb S^n)$ such that $\alpha \cup \beta \ne 0$. This will follow immediately from the Lemma below. $\square$

Lemma 10.4

For $n\ge 1$, the group $H^{2n}(\mathbb S^n\times \mathbb S^n, \Delta ^n)$ is isomorphic to $\mathbb {Z}$, and the map $H^n(\mathbb S^n\times \mathbb S^n,\Delta ^n) \times H^n(\mathbb S^n\times \mathbb S^n)\ni (\alpha ,\beta )\mapsto \alpha \cup \beta \in H^{2n}(\mathbb S^n\times \mathbb S^n, \Delta ^n)$ is surjective.

Proof

It is well known that $H^k(\mathbb S^n)\cong \mathbb {Z}$ for $k=0,n$, and $H^k(\mathbb S^n)=0$ if $k\ne 0,n$. It follows $H^n(\mathbb {S}^n\times \mathbb {S}^n)\cong \bigoplus _{k=0}^nH^k(\mathbb S^n)\otimes H^{n-k}(\mathbb S^n)\cong \mathbb {Z}\oplus \mathbb {Z}$. If $\omega$ is a generator of $H^n(\mathbb S^n)$, then the two generators of $H^n(\mathbb S^n\times \mathbb S^n)\cong \mathbb {Z}\oplus \mathbb {Z}$ are $\pi _1^*(\omega )$ and $\pi _2^*(\omega )$, where $\pi _1,\pi _2:\mathbb S^n\times \mathbb S^n\rightarrow \mathbb S^n$ are the projections.

For the computation of $H^n(\mathbb S^n\times \mathbb S^n,\Delta ^n)$, we use the long exact sequence of the pair $(\mathbb S^n\times \mathbb S^n,\Delta ^n)$ in reduced cohomology:

$$\begin{aligned} \cdots \longrightarrow \widetilde{H}^{n-1}(\Delta ^n)\longrightarrow H^n(\mathbb S^n\times \mathbb S^n,\Delta ^n)\longrightarrow \widetilde{H}^n(\mathbb S^n\times \mathbb S^n)\mathop {\longrightarrow }\limits ^{\mathfrak i^*}\widetilde{H}^n(\Delta ^n)\longrightarrow \cdots \end{aligned}$$

Since $\Delta ^n$ is homeomorphic to $\mathbb S^n$, then $\widetilde{H}^{n-1}(\Delta ^n)=0$. Thus, the group $H^n(\mathbb S^n\times \mathbb S^n,\Delta ^n)$ can be identified with the subgroup of $\widetilde{H}^n(\mathbb S^n\times \mathbb S^n)$ given by the kernel of the map $\mathfrak i^*:\widetilde{H}^n(\mathbb S^n\times \mathbb S^n)\rightarrow \widetilde{H}^n(\Delta ^n)$. This map takes each of the two generators $\pi _i^*(\omega )$, $i=1,2$, to $\omega$ (here we identify $\Delta ^n$ with $\mathbb S^n$), so that $H^n(\mathbb S^n\times \mathbb S^n,\Delta ^n)$ is the subgroup of $\widetilde{H}^n(\mathbb S^n\times \mathbb S^n)$ generated by $\pi _1^*(\omega )-\pi _2^*(\omega )$, which is isomorphic to $\mathbb {Z}$.

Finally, let us compute $H^{2n}(\mathbb S^n\times \mathbb S^n,\Delta )$ using again the long exact sequence of the pair $(\mathbb S^n\times \mathbb S^n,\Delta ^n)$ in reduced cohomology:

$$\begin{aligned} \cdots \longrightarrow \widetilde{H}^{2n-1}(\Delta ^n)\longrightarrow H^{2n}(\mathbb S^n\times \mathbb S^n,\Delta ^n)\longrightarrow \widetilde{H}^{2n}(\mathbb S^n\times \mathbb S^n)\mathop {\longrightarrow }\limits ^{\mathfrak i^*}\widetilde{H}^{2n}(\Delta ^n)\longrightarrow \cdots \end{aligned}$$

Clearly, $\widetilde{H}^{2n}(\Delta ^n)=0$, and if $n>1$, also $\widetilde{H}^{2n-1}(\Delta ^n)=0$. When $n=1$, then $\widetilde{H}^{2n-1}(\Delta ^n)=\widetilde{H}^1(\Delta ^1)\cong \mathbb {Z}$, however the map $\widetilde{H}^1(\Delta ^1)\rightarrow \widetilde{H}^2(\mathbb S^1\times \mathbb S^1)$ is identically zero, because the previous map of the exact sequence $\widetilde{H}^1(\mathbb S^1\times \mathbb S^1)\rightarrow \widetilde{H}^1(\Delta ^1)$ is clearly surjective.⁴ In both cases, $n=1$ or $n>1$, we obtain $H^{2n}(\mathbb S^n\times \mathbb S^n,\Delta ^n)\cong \widetilde{H}^{2n}(\mathbb S^n\times \mathbb S^n)\cong \mathbb {Z}$. A generator of $\widetilde{H}^{2n}(\mathbb S^n\times \mathbb S^n)$ is $\pi _1^*(\omega )\cup \pi _2^*(\omega )$.

In conclusion, using the above identifications, the map $H^n(\mathbb S^n\times \mathbb S^n,\Delta ^n) \times H^n(\mathbb S^n\times \mathbb S^n)\ni (\alpha ,\beta )\mapsto \alpha \cup \beta \in H^{2n}(\mathbb S^n\times \mathbb S^n, \Delta ^n)$ reads as the bilinear map $\mathbb {Z}\times (\mathbb {Z}\oplus \mathbb {Z})\rightarrow \mathbb {Z}$ that takes $\big (1,(1,0)\big )$ to $(-1)^{n+1}$ and $\big (1,(0,1)\big )$ to $1$. This is clearly surjective. $\square$

From Propositions 10.2 and 10.3 we get:

Corollary 10.5

For all $n\ge 1$, ${\mathrm {cat}}_{\mathbb S^n \times \mathbb S^n}(\mathbb S^n \times \mathbb S^n,\Delta ^n)\ge 2$.