# Multiplicity and stability of closed geodesics on positively curved Finsler
$4$-spheres

Huagui Duan, Dong Xie

School of Mathematical Sciences and LPMC, Nankai University,
Tianjin 300071, P. R. China
Partially supported by National Key R&D Program of China (Grant No.
2020YFA0713300), NNSFC (Nos. 12271268 and 11790271), and the Fundamental
Research Funds for the Central Universities. E-mail:
<EMAIL_ADDRESS>supported by NNSFC (No. 12271268). E-mail:
<EMAIL_ADDRESS>
###### Abstract

In this paper, we prove that for every Finsler $4$-dimensional sphere
$(S^{4},F)$ with reversibility $\lambda$ and flag curvature $K$ satisfying
$\frac{25}{9}\left(\frac{\lambda}{1+\lambda}\right)^{2}<K\leq 1$ with
$\lambda<\frac{3}{2}$, either there exist at least four prime closed
geodesics, or there exist exactly three prime non-hyperbolic closed geodesics
and at least two of them are irrationally elliptic.

Key words: Positively curved, closed geodesic, irrationally elliptic, Finsler
metric, spheres.

2020 Mathematics Subject Classification: 53C22, 58E05, 58E10.

##  1 Introduction and main result

Let $(M,F)$ be a Finsler manifold. A closed curve on $(M,F)$ is a closed
geodesic if it is locally the shortest path connecting any two nearby points
on this curve. As usual, a closed geodesic $c:S^{1}={\bf R}/{\bf Z}\to M$ is
prime if it is not a multiple covering (i.e., iteration) of any other closed
geodesics. Here the $m$-th iteration $c^{m}$ of $c$ is defined by
$c^{m}(t)=c(mt)$. The inverse curve $c^{-1}$ of $c$ is defined by
$c^{-1}(t)=c(1-t)$ for $t\in{\bf R}$. Note that unlike Riemannian manifold,
the inverse curve $c^{-1}$ of a closed geodesic $c$ on an irreversible Finsler
manifold need not be a geodesic. We call two prime closed geodesics $c$ and
$d$ distinct if there is no ${\theta}\in(0,1)$ such that $c(t)=d(t+{\theta})$
for all $t\in{\bf R}$. On a reversible Finsler (or Riemannian) manifold, two
closed geodesics $c$ and $d$ are called geometrically distinct if
$c(S^{1})\neq d(S^{1})$, i.e., they have different image sets in $M$. We shall
omit the word distinct when we talk about more than one prime closed geodesic.

For a closed geodesic $c$ on an $(n+1)$-dimensional manifold $M$, denote by
$P_{c}$ the linearized Poincaré map of $c$, which is a symplectic matrix,
i.e., $P_{c}\in{\rm Sp}(2n)$. we define the elliptic height  $e(P_{c})$ of
$P_{c}$ to be the total algebraic multiplicity of all eigenvalues of $P_{c}$
on the unit circle ${\bf U}=\\{z\in{\bf C}|\;|z|=1\\}$ in the complex plane
${\bf C}$. Since $P_{c}$ is symplectic, $e(P_{c})$ is even and $0\leq
e(P_{c})\leq 2n$. A closed geodesic $c$ is called elliptic if $e(P_{c})=2n$,
i.e., all the eigenvalues of $P_{c}$ locate on ${\bf U}$; irrationally
elliptic if, in the homotopy component ${\Omega}^{0}(P_{c})$ of $P_{c}$ (cf.
Section 2 below for the definition), $P_{c}$ can be connected to the
${\diamond}$-product of $n$ rotation matrices $R({\theta}_{i})$ with
${\theta}_{i}$ being irrational multiple of $\pi$ for $1\leq i\leq n$;
hyperbolic if $e(P_{c})=0$, i.e., all the eigenvalues of $P_{c}$ locate away
from ${\bf U}$; non-degenerate if $1$ is not an eigenvalue of $P_{c}$. A
Finsler metric $F$ is called bumpy if all the closed geodesics on $(M,F)$ are
non-degenerate.

There is a famous conjecture in Riemannian geometry which claims the existence
of infinitely many closed geodesics on any compact Riemannian manifold. This
conjecture has been proved for many cases, but not yet for compact rank one
symmetric spaces except for $S^{2}$. The results of Franks in [Fra92] and
Bangert in [Ban93] imply that this conjecture is true for any Riemannian
2-sphere (cf. [Hin93] and [Hin97]). However for a Finsler manifold, the above
conjecture does not hold due to the Katok’s examples. It was quite surprising
when Katok in [Kat73] found some irreversible Finsler metrics on spheres with
only finitely many closed geodesics and all of them are non-degenerate and
irrationally elliptic (cf. [Zil82]).

Based on Katok’s examples, Anosov in [Ano74] proposed the following conjecture
(cf. [Lon06])

$\displaystyle{\cal N}(S^{n},F)\geq 2\left[\frac{n+1}{2}\right]\quad\mbox{for
any Finsler metric $F$ on}\ S^{n},$ (1.1)

where, denote by ${\cal N}(M,F)$ the number of the distinct closed geodesics
on $(M,F)$, and $[a]=\max\\{k\in{\bf Z}\,|\,k\leq a\\}$. In 2005, Bangert and
Long in [BL10] proved this conjecture for any Finsler $2$-dimensional sphere
$(S^{2},F)$. Since then, the index iteration theory of closed geodesics (cf.
[Bot56] and [Lon02]) has been applied to study the closed geodesic problem on
Finsler manifolds.

When $n\geq 3$, the above conjectures in the Riemannian or Finsler case is
still widely open in full of generality. About the multiplicity and stability
problem of closed geodesics, two classes of typical conditions, including the
positively curved condition and the non-degenerate (or bumpy) condition, have
been used widely.

In [Rad04], Rademacher has introduced the reversibility $\lambda=\lambda(M,F)$
of a compact Finsler manifold defined by

$\displaystyle\lambda=\max\\{F(-X)\ |\ X\in TM,\ F(X)=1\\}\geq 1.$

Then Rademacher in [Rad07] has obtained some results about the multiplicity
and stability of closed geodesics. For example, let $F$ be a Finsler metric on
$S^{n}$ with reversibility ${\lambda}$ and flag curvature $K$ satisfying
$\left(\frac{{\lambda}}{1+{\lambda}}\right)^{2}<K\leq 1$, then there exist at
least $n/2-1$ closed geodesics with length $<2n\pi$. If
$\frac{9{\lambda}^{2}}{4(1+{\lambda})^{2}}<K\leq 1$ with ${\lambda}<2$, then
there exists a closed geodesic of elliptic-parabolic, i.e., its linearized
Poincaré map split into $2$-dimensional rotations and a part whose eigenvalues
are $\pm 1$. These results are some generalization of those in [BTZ82] and
[BTZ83] in the Riemannian case.

Recently, Wang in [Wan12] proved the conjecture (1.1) for $(S^{n},F)$ provided
that $F$ is bumpy and its flag curvature $K$ satisfies
$\left(\frac{\lambda}{1+\lambda}\right)^{2}<K\leq 1$. Also in [Wan12], Wang
showed that for every bumpy Finsler metric $F$ on $S^{n}$ satisfying
$\frac{9{\lambda}^{2}}{4(1+{\lambda})^{2}}<K\leq 1$, there exist two prime
elliptic closed geodesics provided the number of closed geodesics on
$(S^{n},F)$ is finite. As some further generalization, Duan, Long and Wang in
[DLW16] obtained the optimal lower bound of the number of distinct closed
geodesics on a compact simply-connected Finsler manifold $(M,F)$ if $F$ is
bumpy and some much weak index conditions or positively curved conditions are
satisfied.

The first author in [Dua15] and [Dua16] proved that for every Finsler
$(S^{n},F)$ for $n\geq 3$ with reversibility ${\lambda}$ and flag curvature
$K$ satisfying $\left(\frac{{\lambda}}{1+{\lambda}}\right)^{2}<K\leq 1$,
either there exist infinitely many prime closed geodesics, or there exist
exactly three prime closed geodesics and at least two of them are elliptic. In
fact, the multiplicity and stability problem on high dimensional manifolds
without the assumption of bumpy metrics is much difficult.

In this paper, we further consider the positively curved Finsler
$4$-dimensional sphere $(S^{4},F)$ without the bumpy assumption, and obtain
the following new progress about the multiplicity and stability of closed
geodesics on $(S^{4},F)$.

Theorem 1.1. For every Finsler metric $F$ on a $4$-dimensional sphere $S^{4}$
with reversibility ${\lambda}$ and flag curvature $K$ satisfying
$\frac{25}{9}\left(\frac{\lambda}{1+\lambda}\right)^{2}<K\leq 1$ with
$\lambda<\frac{3}{2}$, either there exist at least four prime closed
geodesics, or there exist exactly three prime non-hyperbolic closed geodesics
and at least two of them are irrationally elliptic.

First, under the assumption of positively curved condition
$\frac{25}{9}\left(\frac{\lambda}{1+\lambda}\right)^{2}<K\leq 1$ with
$\lambda<\frac{3}{2}$, Theorem 1 and Theorem 4 in [Rad04] established the
lower bound of the length of closed geodesics, which in turn gives the lower
bound of $i(c^{m})$ and mean index $\hat{i}(c)$ for any prime closed geodesic
$c$ on such $S^{4}$ (cf. Lemma 3.1 below). Second, we shall make full use of
the enhanced common index jump theorem established in [DLW16], which
generalized the common index jump theorem in [LZ02], to obtain some crucial
precise estimates of $i(c^{m})$ and $\nu(c^{m})$ (cf. Section 3.1 below). Note
that Theorem 1.1 in [Dua15] showed the existence of three prime closed
geodesics on $(S^{4},F)$ with reversibility ${\lambda}$ and flag curvature $K$
satisfying $\left(\frac{{\lambda}}{1+{\lambda}}\right)^{2}<K\leq 1$. So, in
order to prove our above Theorem 1.1, we assume the existence of exactly three
prime closed geodesics on such $(S^{4},F)$ (cf. the assumption (TCG) below).
Finally, together with the Morse theory, under (TCG) we will carefully analyze
the local and global information of these three prime closed geodesics and
their iterates to complete the proof of Theorem 1.1 in Section 3.2.

In addition, under the assumption (TCG), in Theorem 4.2 in Section 4, we
obtain more precise information about the third prime closed geodesic except
that it is non-hyperbolic. These information will be greatly helpful to
completely solve the conjecture (1.1) on the positively cured Finsler
$(S^{4},F)$ in the future.

Remark 1.2. (i) The irrationally ellipticity of closed geodesics is a kind of
very important stability, in fact, it specially implies the non-degeneracy of
closed geodesics. It is conjectured that all closed geodesics are irrationally
elliptic if the number of prime closed geodesics on Finsler sphere $(S^{n},F)$
with $n\geq 2$ is finite (cf. Conjecture of [DLW16]). In this direction, there
are several well-known results. For example, Long and Wang in [LW08] proved
that on every Finsler $S^{2}$ with only finitely many closed geodesics, there
exist at least two irrationally elliptic one. Duan and Liu in [DL16] showed
that if there exist exactly three prime closed geodesics on every Finsler
$S^{3}$ satisfying
$\frac{9}{4}\left(\frac{\lambda}{1+\lambda}\right)^{2}<K\leq 1$ with
$\lambda<2$, then two of them are irrationally elliptic.

(ii) This paper continues to consider the case of $n=4$ of $S^{n}$. Compared
to previous works, and in particular to [Dua15] and [Dua16], the first
substantial progress in this paper is to prove the existence of two prime
irrationally elliptic closed geodesics on the above positively curved Finsler
$S^{4}$ if there exist exactly three prime closed geodesics on it. Except for
this, the second progress in this paper is to obtain five possible
characterizations about the Morse index and the linearized Poincaré map of the
third closed geodesic if there exist exactly three prime closed geodesics on
such $S^{4}$ (cf. Theorem 4.2 below). These results is very difficult due to
the high dimension of $n\geq 4$. In fact, in the case of $n=4$, there possibly
exist three $2\times 2$ rotation matrices in the decomposition of $P_{c}$ (cf.
(2.14) of Theorem 2.6 below), which brings high degeneracy of closed geodesics
so that some methods and technical tools in [Dua15], [Dua16] and [DL16] are
not enough to deal with these difficulties. In this paper, we make full use of
the enhanced common index jump theorem established in [DLW16], together with
some new ideas and estimates about indices of closed geodesics.

(iii) On one hand, note that the curvature pinching condition in Theorem 1.1
and Theorem 4.2 in this paper is a little restrictive than the corresponding
assumption in [Dua15] and [Dua16], this is because that we need good index
estimates in Lemma 3.1 and Lemma 3.2 to count the contributions of iterated
closed geodesics to the non-zero Morse-type numbers, and the current pinching
condition is a sufficient one for these estimates. On the other hand, we do
not consider the case $n\geq 5$ because of more higher degeneracy. This maybe
need some new ideas and different tools to deal with more difficulties.

In this paper, let ${\bf N}$, ${\bf N}_{0}$, ${\bf Z}$, ${\bf Q}$, ${\bf R}$,
and ${\bf C}$ denote the sets of natural integers, non-negative integers,
integers, rational numbers, real numbers, and complex numbers respectively. We
use only singular homology modules with ${\bf Q}$-coefficients. For an
$S^{1}$-space $X$, we denote by $\overline{X}$ the quotient space $X/S^{1}$.
We define the functions

$\displaystyle E(a)=\min\\{k\in{\bf Z}\,|\,k\geq
a\\},\quad\varphi(a)=E(a)-[a],\quad\\{a\\}=a-[a].$ (1.2)

Especially, $\varphi(a)=0$ if $a\in{\bf Z}\,$, and $\varphi(a)=1$ if
$a\notin{\bf Z}\,$.

##  2 Morse theory and Morse indices of closed geodesics

###  2.1 Morse theory for closed geodesics

Let $M=(M,F)$ be a compact Finsler manifold, the space $\Lambda=\Lambda M$ of
$H^{1}$-maps $\gamma:S^{1}\rightarrow M$ has a natural structure of Riemannian
Hilbert manifolds on which the group $S^{1}={\bf R}/{\bf Z}$ acts continuously
by isometries. This action is defined by $(s\cdot\gamma)(t)=\gamma(t+s)$ for
all $\gamma\in{\Lambda}$ and $s,t\in S^{1}$. For any $\gamma\in\Lambda$, the
energy functional is defined by

$E(\gamma)=\frac{1}{2}\int_{S^{1}}F(\gamma(t),\dot{\gamma}(t))^{2}dt.$ (2.1)

It is $C^{1,1}$ and invariant under the $S^{1}$-action. The critical points of
$E$ of positive energies are precisely the closed geodesics $\gamma:S^{1}\to
M$. The index form of the functional $E$ is well defined along any closed
geodesic $c$ on $M$, which we denote by $E^{\prime\prime}(c)$. As usual, we
denote by $i(c)$ and $\nu(c)$ the Morse index and nullity of $E$ at $c$. In
the following, we denote by

${\Lambda}^{\kappa}=\\{d\in{\Lambda}\;|\;E(d)\leq\kappa\\},\quad{\Lambda}^{\kappa-}=\\{d\in{\Lambda}\;|\;E(d)<\kappa\\},\quad\forall\kappa\geq
0.$

For a closed geodesic $c$ we set ${\Lambda}(c)=\\{{\gamma}\in{\Lambda}\mid
E({\gamma})<E(c)\\}$.

Recall that respectively the mean index $\hat{i}(c)$ and the $S^{1}$-critical
modules of $c^{m}$ are defined by

$\hat{i}(c)=\lim_{m\rightarrow\infty}\frac{i(c^{m})}{m},\quad\overline{C}_{*}(E,c^{m})=H_{*}\left(({\Lambda}(c^{m})\cup
S^{1}\cdot c^{m})/S^{1},{\Lambda}(c^{m})/S^{1};{\bf Q}\right).$ (2.2)

We call a closed geodesic satisfying the isolation condition, if the following
holds:

(Iso) For all $m\in{\bf N}$ the orbit $S^{1}\cdot c^{m}$ is an isolated
critical orbit of $E$.

Note that if the number of prime closed geodesics on a Finsler manifold is
finite, then all the closed geodesics satisfy (Iso).

If $c$ has multiplicity $m$, then the subgroup ${\bf Z}_{m}=\\{\frac{n}{m}\mid
0\leq n<m\\}$ of $S^{1}$ acts on $\overline{C}_{*}(E,c)$. As studied in p.59
of [Rad92], for all $m\in{\bf N}$, let $H_{\ast}(X,A)^{\pm{\bf
Z}_{m}}=\\{[\xi]\in H_{\ast}(X,A)\,|\,T_{\ast}[\xi]=\pm[\xi]\\}$, where $T$ is
a generator of the ${\bf Z}_{m}$-action. On $S^{1}$-critical modules of
$c^{m}$, the following lemma holds:

Lemma 2.1. (cf. Satz 6.11 of [Rad92] and [BL10]) Suppose $c$ is a prime closed
geodesic on a Finsler manifold $M$ satisfying (Iso). Then there exist
$U_{c^{m}}^{-}$ and $N_{c^{m}}$, the so-called local negative disk and the
local characteristic manifold at $c^{m}$ respectively, such that
$\nu(c^{m})=\dim N_{c^{m}}$ and

$\displaystyle\overline{C}_{q}(E,c^{m})$ $\displaystyle\equiv$ $\displaystyle
H_{q}\left(({\Lambda}(c^{m})\cup S^{1}\cdot
c^{m})/S^{1},{\Lambda}(c^{m})/S^{1}\right)$ $\displaystyle=$
$\displaystyle\left(H_{i(c^{m})}(U_{c^{m}}^{-}\cup\\{c^{m}\\},U_{c^{m}}^{-})\otimes
H_{q-i(c^{m})}(N_{c^{m}}\cup\\{c^{m}\\},N_{c^{m}})\right)^{+{\bf Z}_{m}},$

(i) When $\nu(c^{m})=0$, there holds

$\overline{C}_{q}(E,c^{m})=\left\\{\matrix{{\bf Q},&\quad{\it
if}\;\;i(c^{m})-i(c)\in 2{\bf Z}\;\;{\it and}\;\;q=i(c^{m}),\;\cr 0,&\quad{\it
otherwise},\cr}\right.$

(ii) When $\nu(c^{m})>0$, there holds

$\overline{C}_{q}(E,c^{m})=H_{q-i(c^{m})}(N_{c^{m}}\cup\\{c^{m}\\},N_{c^{m}})^{{\epsilon}(c^{m}){\bf
Z}_{m}},$

where ${\epsilon}(c^{m})=(-1)^{i(c^{m})-i(c)}$.

Define

$k_{j}(c^{m})\equiv\dim\,H_{j}(N_{c^{m}}\cup\\{c^{m}\\},N_{c^{m}}),\quad
k_{j}^{\pm
1}(c^{m})\equiv\dim\,H_{j}(N_{c^{m}}\cup\\{c^{m}\\},N_{c^{m}})^{\pm{\bf
Z}_{m}}.$ (2.3)

Then we have

Lemma 2.2. (cf. [Rad92], [LD09], [Wan12]) Let $c$ be a prime closed geodesic
on a Finsler manifold $(M,F)$. Then

(i) For any $m\in{\bf N}$, there holds $k_{j}(c^{m})=0$ for
$j\not\in[0,\nu(c^{m})]$.

(ii) For any $m\in{\bf N}$, $k_{0}(c^{m})+k_{\nu(c^{m})}(c^{m})\leq 1$ and if
$k_{0}(c^{m})+k_{\nu(c^{m})}(c^{m})=1$ then there holds $k_{j}(c^{m})=0$ for
$j\in(0,\nu(c^{m}))$.

(iii) For any $m\in{\bf N}$, there holds $k_{0}^{+1}(c^{m})=k_{0}(c^{m})$ and
$k_{0}^{-1}(c^{m})=0$. In particular, if $c^{m}$ is non-degenerate, there
holds $k_{0}^{+1}(c^{m})=k_{0}(c^{m})=1$, and $k_{0}^{-1}(c^{m})=k_{j}^{\pm
1}(c^{m})=0$ for all $j\neq 0$.

(iv) Suppose for some integer $m=np\geq 2$ with $n$ and $p\in{\bf N}$ the
nullities satisfy $\nu(c^{m})=\nu(c^{n})$. Then there holds
$k_{j}(c^{m})=k_{j}(c^{n})$ and ${k}_{j}^{\pm 1}(c^{m})={k}^{\pm
1}_{j}(c^{n})$ for any integer $j$.

Let $(M,F)$ be a compact simply connected Finsler manifold with finitely many
closed geodesics. It is well known that for every prime closed geodesic $c$ on
$(M,F)$, there holds either $\hat{i}(c)>0$ and then $i(c^{m})\to+\infty$ as
$m\to+\infty$, or $\hat{i}(c)=0$ and then $i(c^{m})=0$ for all $m\in{\bf N}$.
Denote those prime closed geodesics on $(M,F)$ with positive mean indices by
$\\{c_{j}\\}_{1\leq j\leq k}$. Rademacher in [Rad89] and [Rad92] established a
celebrated mean index identity relating all the $c_{j}$s with the global
homology of $M$ for compact simply connected Finsler manifolds (especially for
$S^{4}$) as follows.

Theorem 2.3. (Satz 7.9 of [Rad92], cf. also [DuL10], [LD09] and [Wan12])
Assume that there exist finitely many prime closed geodesics on $(S^{4},F)$
and denote prime closed geodesics with positive mean indices by
$\\{c_{j}\\}_{1\leq j\leq k}$ for some $k\in{\bf N}$. Then the following
identity holds

$\sum_{j=1}^{k}\frac{\hat{\chi}(c_{j})}{\hat{i}(c_{j})}=-\frac{2}{3},$ (2.4)

where

$\hat{\chi}(c_{j})=\frac{1}{n(c_{j})}\sum_{1\leq m\leq
n(c_{j})}\chi(c_{j}^{m})=\frac{1}{n(c_{j})}\sum_{1\leq m\leq n(c_{j})\atop
0\leq l\leq
2(n-1)}(-1)^{i(c_{j}^{m})+l}k_{l}^{{\epsilon}(c_{j}^{m})}(c_{j}^{m})\in{\bf
Q},$ (2.5)

and the analytical period $n(c_{j})$ of $c_{j}$ is defined by (cf. [LD09])

$n(c_{j})=\min\\{l\in{\bf N}\,|\,\nu(c_{j}^{l})=\max_{m\geq
1}\nu(c_{j}^{m}),\;\;i(c_{j}^{m+l})-i(c_{j}^{m})\in 2{\bf
Z},\;\;\forall\,m\in{\bf N}\\}.$ (2.6)

Set $\overline{{\Lambda}}^{0}=\overline{\Lambda}^{0}S^{4}=\\{{\rm
constant\;point\;curves\;in\;}S^{4}\\}\cong S^{4}$. Let $(X,Y)$ be a space
pair such that the Betti numbers $b_{i}=b_{i}(X,Y)=\dim H_{i}(X,Y;{\bf Q})$
are finite for all $i\in{\bf Z}$. As usual the Poincaré series of $(X,Y)$ is
defined by the formal power series $P(X,Y)=\sum_{i=0}^{\infty}b_{i}t^{i}$. We
need the following well known version of results on Betti numbers and the
Morse inequality.

Lemma 2.4. (cf. Theorem 2.4 and Remark 2.5 of [Rad89] and [Hin84], Lemma 2.5
of [DuL10]) Let $(S^{4},F)$ be a $4$-dimensional Finsler sphere. Then, the
Betti numbers are given by

$\displaystyle b_{j}$ $\displaystyle=$ $\displaystyle{\rm
rank}H_{j}({\Lambda}S^{4}/S^{1},{\Lambda}^{0}S^{4}/S^{1};{\bf Q})$ (2.7)
$\displaystyle=$ $\displaystyle\left\\{\matrix{2,&\quad{\it if}\quad j\in{\cal
K}\equiv\\{3k\,|\,3\leq k\in 2{\bf N}+1\\},\cr 1,&\quad{\it if}\quad
j\in\\{2k+3\,|\,k\in{\bf N}_{0}\\}\setminus{\cal K},\cr 0,&\quad{\it
otherwise}.\cr}\right.$

Theorem 2.5. (cf. Theorem I.4.3 of [Cha93]) Let $(M,F)$ be a Finsler manifold
with finitely many prime closed geodesics, denoted by $\\{c_{j}\\}_{1\leq
j\leq k}$. Set

$\displaystyle M_{q}=\sum_{1\leq j\leq k,\;m\geq
1}\dim{\overline{C}}_{q}(E,c^{m}_{j}),\quad q\in{\bf Z}.$

Then for every integer $q\geq 0$ there holds

$\displaystyle M_{q}-M_{q-1}+\cdots+(-1)^{q}M_{0}$ $\displaystyle\geq$
$\displaystyle{b}_{q}-{b}_{q-1}+\cdots+(-1)^{q}{b}_{0},$ (2.8) $\displaystyle
M_{q}$ $\displaystyle\geq$ $\displaystyle{b}_{q}.$ (2.9)

###  2.2 Index iteration theory of closed geodesics

In [Lon99] of 1999, Y. Long established the basic normal form decomposition of
symplectic matrices. Based on this result he further established the precise
iteration formulae of indices of symplectic paths in [Lon00] of 2000. Note
that this index iteration formulae works for Morse indices of iterated closed
geodesics (cf. [Liu05] and Chap. 12 of [Lon02]). Since every closed geodesic
on a sphere must be orientable. Then by Theorem 1.1 of [Liu05], the initial
Morse index of a closed geodesic on a Finsler $S^{4}$ coincides with the index
of a corresponding symplectic path.

As in [Lon00], denote by

$\displaystyle N_{1}({\lambda},b)$ $\displaystyle=$
$\displaystyle\left(\matrix{{\lambda}&b\cr 0&{\lambda}\cr}\right),\qquad{\rm
for\;}{\lambda}=\pm 1,\;b\in{\bf R},$ (2.10) $\displaystyle D({\lambda})$
$\displaystyle=$ $\displaystyle\left(\matrix{{\lambda}&0\cr
0&{\lambda}^{-1}\cr}\right),\qquad{\rm for\;}{\lambda}\in{\bf
R}\setminus\\{0,\pm 1\\},$ (2.11) $\displaystyle R({\theta})$ $\displaystyle=$
$\displaystyle\left(\matrix{\cos{\theta}&-\sin{\theta}\cr\sin{\theta}&\cos{\theta}\cr}\right),\qquad{\rm
for\;}{\theta}\in(0,\pi)\cup(\pi,2\pi),$ (2.12) $\displaystyle
N_{2}(e^{{\theta}\sqrt{-1}},B)$ $\displaystyle=$
$\displaystyle\left(\matrix{R({\theta})&B\cr
0&R({\theta})\cr}\right),\qquad{\rm
for\;}{\theta}\in(0,\pi)\cup(\pi,2\pi)\;\;{\rm and}\;$ (2.13)
$\displaystyle\qquad B=\left(\matrix{b_{1}&b_{2}\cr
b_{3}&b_{4}\cr}\right)\;{\rm with}\;b_{j}\in{\bf R},\;\;{\rm
and}\;\;b_{2}\not=b_{3}.$

Here $N_{2}(e^{{\theta}\sqrt{-1}},B)$ is non-trivial if
$(b_{2}-b_{3})\sin\theta<0$, and trivial if $(b_{2}-b_{3})\sin\theta>0$.

As in [Lon00], the $\diamond$-sum (direct sum) of any two real matrices is
defined by

$\left(\matrix{A_{1}&B_{1}\cr C_{1}&D_{1}\cr}\right)_{2i\times
2i}\diamond\left(\matrix{A_{2}&B_{2}\cr C_{2}&D_{2}\cr}\right)_{2j\times
2j}=\left(\matrix{A_{1}&0&B_{1}&0\cr 0&A_{2}&0&B_{2}\cr C_{1}&0&D_{1}&0\cr
0&C_{2}&0&D_{2}}\right).$

For every $M\in{\rm Sp}(2n)$, the homotopy set $\Omega(M)$ of $M$ in ${\rm
Sp}(2n)$ is defined by

${\Omega}(M)=\\{N\in{\rm Sp}(2n)\,|\,{\sigma}(N)\cap{\bf
U}={\sigma}(M)\cap{\bf
U}\equiv\Gamma\;\mbox{and}\;\nu_{{\omega}}(N)=\nu_{{\omega}}(M),\
\forall{\omega}\in\Gamma\\},$

where ${\sigma}(M)$ denotes the spectrum of $M$,
$\nu_{{\omega}}(M)\equiv\dim_{{\bf C}}\ker_{{\bf C}}(M-{\omega}I)$ for
${\omega}\in{\bf U}$. The component ${\Omega}^{0}(M)$ of $P$ in ${\rm Sp}(2n)$
is defined by the path connected component of ${\Omega}(M)$ containing $M$.

Theorem 2.6. (cf. Theorem 7.8 of [Lon99], Theorems 1.2 and 1.3 of [Lon00], cf.
also Theorem 1.8.10, Lemma 2.3.5 and Theorem 8.3.1 of [Lon02]) For every
$P\in{\rm Sp}(2n-2)$, there exists a continuous path $f\in{\Omega}^{0}(P)$
such that $f(0)=P$ and

$\displaystyle f(1)$ $\displaystyle=$ $\displaystyle
N_{1}(1,1)^{{\diamond}p_{-}}\,{\diamond}\,I_{2p_{0}}\,{\diamond}\,N_{1}(1,-1)^{{\diamond}p_{+}}{\diamond}\,N_{1}(-1,1)^{{\diamond}q_{-}}\,{\diamond}\,(-I_{2q_{0}})\,{\diamond}\,N_{1}(-1,-1)^{{\diamond}q_{+}}$
(2.14)
$\displaystyle{\diamond}\,N_{2}(e^{{\alpha}_{1}\sqrt{-1}},A_{1})\,{\diamond}\,\cdots\,{\diamond}\,N_{2}(e^{{\alpha}_{r_{\ast}}\sqrt{-1}},A_{r_{\ast}}){\diamond}\,N_{2}(e^{{\beta}_{1}\sqrt{-1}},B_{1})\,{\diamond}\,\cdots\,{\diamond}\,N_{2}(e^{{\beta}_{r_{0}}\sqrt{-1}},B_{r_{0}})$
$\displaystyle{\diamond}\,R({\theta}_{1})\,{\diamond}\,\cdots\,{\diamond}\,R({\theta}_{r^{\prime}})\,{\diamond}\,R({\theta}_{r^{\prime}+1})\,{\diamond}\,\cdots\,{\diamond}\,R({\theta}_{r}){\diamond}\,H(2)^{{\diamond}h},$

where $\frac{{\theta}_{j}}{2\pi}\in{\bf Q}\cap(0,1)\setminus\\{\frac{1}{2}\\}$
for $1\leq j\leq r^{\prime}$ and $\frac{{\theta}_{j}}{2\pi}\notin{\bf
Q}\cap(0,1)$ for $r^{\prime}+1\leq j\leq r$;
$N_{2}(e^{{\alpha}_{j}\sqrt{-1}},A_{j})$’s are nontrivial and
$N_{2}(e^{{\beta}_{j}\sqrt{-1}},B_{j})$’s are trivial, and non-negative
integers $p_{-},p_{0},p_{+},q_{-},q_{0},q_{+},r,r_{\ast},r_{0},h$ satisfy the
equality

$p_{-}+p_{0}+p_{+}+q_{-}+q_{0}+q_{+}+r+2r_{\ast}+2r_{0}+h=n-1.$ (2.15)

Let ${\gamma}\in{\cal P}_{\tau}(2n-2)=\\{{\gamma}\in C([0,\tau],{\rm
Sp}(2n-2))\,|\,{\gamma}(0)=I\\}$. we extend ${\gamma}(t)$ to $t\in[0,m\tau]$
for every $m\in{\bf N}$ by

$\displaystyle{\gamma}^{m}(t)={\gamma}(t-j\tau){\gamma}(\tau)^{j}\qquad\forall
j\tau\leq t\leq(j+1)\tau\ \mbox{and}\ j=0,1,\cdots,m-1.$ (2.16)

Denote the basic normal form decomposition of $P\equiv{\gamma}(\tau)$ by
(2.14). Then we have

$\displaystyle i({\gamma}^{m})$ $\displaystyle=$ $\displaystyle
m(i({\gamma})+p_{-}+p_{0}-r)+2\sum_{j=1}^{r}{E}\left(\frac{m{\theta}_{j}}{2\pi}\right)-r$
(2.17) $\displaystyle-p_{-}-p_{0}-{{1+(-1)^{m}}\over
2}(q_{0}+q_{+})+2\sum_{j=1}^{r_{\ast}}{\varphi}\left(\frac{m{\alpha}_{j}}{2\pi}\right)-2r_{\ast},$
$\displaystyle\nu({\gamma}^{m})$ $\displaystyle=$
$\displaystyle\nu({\gamma})+{{1+(-1)^{m}}\over
2}(q_{-}+2q_{0}+q_{+})+2{\varsigma}(m,{\gamma}(\tau)),$ (2.18)

where we denote by

${\varsigma}(m,{\gamma}(\tau))=r-\sum_{j=1}^{r}{\varphi}(\frac{m{\theta}_{j}}{2\pi})+r_{\ast}-\sum_{j=1}^{r_{\ast}}{\varphi}(\frac{m{\alpha}_{j}}{2\pi})+r_{0}-\sum_{j=1}^{r_{0}}{\varphi}(\frac{m{\beta}_{j}}{2\pi}).$
(2.19)

Let

${\cal M}\equiv\\{N_{1}(1,1);\;\;N_{1}(-1,a_{2}),\,a_{2}=\pm
1;\;\;R({\theta}),\,{\theta}\in[0,2\pi);\,H(-2)\\}.$ (2.20)

By Theorems 8.1.4-8.1.7 and 8.2.1-8.2.4 of [Lon02], we have

Proposition 2.7. Every path ${\gamma}\in{\cal P}_{\tau}(2)$ with end matrix
homotopic to some matrix in ${\cal M}$ has odd index $i({\gamma})$. Every path
$\xi\in{\cal P}_{\tau}(2)$ with end matrix homotopic to $N_{1}(1,-1)$ or
$H(2)$, and every path $\eta\in{\cal P}_{\tau}(4)$ with end matrix homotopic
to $N_{2}({\omega},B)$ has even indices $i(\xi)$ and $i(\eta)$.

The common index jump theorem (cf. Theorem 4.3 of [LZ02]) for symplectic paths
has become one of the main tools in studying the multiplicity and stability of
periodic orbits in Hamiltonian and symplectic dynamics. Recently, the
following enhanced common index jump theorem has been obtained by Duan, Long
and Wang in [DLW16].

Theorem 2.8. (cf. Theorem 3.5 of [DLW16]) Let
$\gamma_{k}\in\mathcal{P}_{\tau_{k}}(2n)$ for $k=1,\cdots,q$ be a finite
collection of symplectic paths. Let $M_{k}={\gamma}_{k}(\tau_{k})$. We extend
${\gamma}_{k}$ to $[0,+\infty)$ by (2.16) inductively. Suppose

$\hat{i}({\gamma}_{k},1)>0,\qquad\forall\ k=1,\cdots,q.$ (2.21)

Then for any fixed integer $\bar{m}\in{\bf N}$, there exist infinitely many
$(q+1)$-tuples $(N,m_{1},\cdots,m_{q})\in{\bf N}^{q+1}$ such that for all
$1\leq k\leq q$ and $1\leq m\leq\bar{m}$, there holds

$\displaystyle\nu({\gamma}_{k},2m_{k}-m)$ $\displaystyle=$
$\displaystyle\nu({\gamma}_{k},2m_{k}+m)=\nu({\gamma}_{k},m),$ (2.22)
$\displaystyle i({\gamma}_{k},2m_{k}+m)$ $\displaystyle=$ $\displaystyle
2N+i({\gamma}_{k},m),$ (2.23) $\displaystyle i({\gamma}_{k},2m_{k}-m)$
$\displaystyle=$ $\displaystyle
2N-i({\gamma}_{k},m)-2(S^{+}_{M_{k}}(1)+Q_{k}(m)),$ (2.24) $\displaystyle
i({\gamma}_{k},2m_{k})$ $\displaystyle=$ $\displaystyle
2N-(S^{+}_{M_{k}}(1)+C(M_{k})-2\Delta_{k}),$ (2.25)

where $S_{M_{k}}^{\pm}({\omega})$ is the splitting number of $M_{k}$ at
${\omega}$ (cf. Definition 9.1.4 of [Lon02]) and

$\displaystyle
C(M_{k})=\sum\limits_{0<\theta<2\pi}S^{-}_{M_{k}}(e^{\sqrt{-1}\theta}),\
\Delta_{k}=\sum_{0<\\{m_{k}{\theta}/\pi\\}<\delta}S^{-}_{M_{k}}(e^{\sqrt{-1}{\theta}}),$
$\displaystyle
Q_{k}(m)=\sum_{\theta\in(0,2\pi),e^{\sqrt{-1}{\theta}}\in{\sigma}(M_{k}),\atop\\{\frac{m_{k}{\theta}}{\pi}\\}=\\{\frac{m{\theta}}{2\pi}\\}=0}S^{-}_{M_{k}}(e^{\sqrt{-1}{\theta}}).$
(2.26)

More precisely, by (4.10), (4.40) and (4.41) in [LZ02], we have

$\displaystyle
m_{k}=\left(\left[\frac{N}{\bar{M}\hat{i}(\gamma_{k},1)}\right]+\chi_{k}\right)\bar{M},\quad
1\leq k\leq q,$ (2.27)

where $\chi_{k}=0$ or $1$ for $1\leq k\leq q$ and
$\frac{\bar{M}\theta}{\pi}\in{\bf Z}$ whenever
$e^{\sqrt{-1}\theta}\in\sigma(M_{k})$ and $\frac{\theta}{\pi}\in{\bf Q}$ for
some $1\leq k\leq q$. Furthermore, for any fixed $M_{0}\in{\bf N}$, we may
further require $M_{0}|N$, and for any $\epsilon>0$, we can choose $N$ and
$\\{\chi_{k}\\}_{1\leq k\leq q}$ such that

$\displaystyle\left|\left\\{\frac{N}{\bar{M}\hat{i}(\gamma_{k},1)}\right\\}-\chi_{k}\right|<\epsilon,\quad
1\leq k\leq q.$ (2.28)

We also have the following properties in the index iteration theory.

Theorem 2.9. (cf. Theorem 2.2 of [LZ02] or Theorem 10.2.3 of [Lon02]) Let
${\gamma}\in P_{\tau}(2n)$. Then, for any $m\in N$, there holds

$\nu({\gamma},m)-\frac{e(M)}{2}\leq
i({\gamma},m+1)-i({\gamma},m)-i({\gamma},1)\leq\nu({\gamma},1)-\nu({\gamma},m+1)+\frac{e(M)}{2},$

where $e(M)$ is the elliptic height defined in Section 1.

##  3 Some index estimates and proof of Theorem 1.1

###  3.1 Some index estimates for closed geodesics

Firstly we make the following assumption

(FCG) Suppose that there exist only finitely many prime closed geodesics
$\\{c_{k}\\}_{k=1}^{q}$ on $(S^{4},F)$ with reversibility $\lambda$ and flag
curvature $K$ satisfying
$\frac{25}{9}\left(\frac{{\lambda}}{1+{\lambda}}\right)^{2}<K\leq 1$ with
$\lambda<\frac{3}{2}$.

For any $1\leq k\leq q$, we rewrite (2.14) as follows

$\displaystyle f_{k}(1)$ $\displaystyle=$ $\displaystyle
N_{1}(1,1)^{{\diamond}p_{k,-}}\,{\diamond}\,I_{2p_{k,0}}\,{\diamond}\,N_{1}(1,-1)^{{\diamond}p_{k,+}}$
(3.1)
$\displaystyle{\diamond}\,N_{1}(-1,1)^{{\diamond}q_{k,-}}\,{\diamond}\,(-I_{2q_{k,0}})\,{\diamond}\,N_{1}(-1,-1)^{{\diamond}q_{k,+}}$
$\displaystyle{\diamond}\,R({\theta}_{k,1})\,{\diamond}\,\cdots\,{\diamond}\,R({\theta}_{k,r_{k,1}})\,{\diamond}\,R(\tilde{{\theta}}_{k,1})\,{\diamond}\,\cdots\,{\diamond}\,R(\tilde{{\theta}}_{k,r_{k,2}})$
$\displaystyle{\diamond}\,N_{2}(e^{\sqrt{-1}{\alpha}_{k,1}},A_{k,1})\,{\diamond}\,\cdots\,{\diamond}\,N_{2}(e^{\sqrt{-1}{\alpha}_{k,r_{k,3}}},A_{k,r_{k,3}})$
$\displaystyle{\diamond}\,N_{2}(e^{\sqrt{-1}\tilde{{\alpha}}_{k,1}},\tilde{A}_{k,1})\,{\diamond}\,\cdots\,{\diamond}\,N_{2}(e^{\sqrt{-1}\tilde{{\alpha}}_{k,r_{k,4}}},\tilde{A}_{k,r_{k,4}})$
$\displaystyle{\diamond}\,N_{2}(e^{\sqrt{-1}{\beta}_{k,1}},B_{k,1})\,{\diamond}\,\cdots\,{\diamond}\,N_{2}(e^{\sqrt{-1}{\beta}_{k,r_{k,5}}},B_{k,r_{k,5}})$
$\displaystyle{\diamond}\,N_{2}(e^{\sqrt{-1}\tilde{{\beta}}_{k,1}},\tilde{B}_{k,1})\,{\diamond}\,\cdots\,{\diamond}\,N_{2}(e^{\sqrt{-1}\tilde{{\beta}}_{k,r_{k,6}}},\tilde{B}_{k,r_{k,6}}){\diamond}\,H(2)^{{\diamond}h_{k,+}}{\diamond}\,H(-2)^{{\diamond}h_{k,-}},$

where $\frac{{\theta}_{k,j}}{2\pi}\in{\bf
Q}\cap(0,1)\setminus\\{\frac{1}{2}\\}$ for $1\leq j\leq r_{k,1}$,
$\frac{\tilde{{\theta}}_{k,j}}{2\pi}\in(0,1)\setminus{\bf Q}$ for $1\leq j\leq
r_{k,2}$, $\frac{{\alpha}_{k,j}}{2\pi}\in{\bf
Q}\cap(0,1)\setminus\\{\frac{1}{2}\\}$ for $1\leq j\leq r_{k,3}$,
$\frac{\tilde{{\alpha}}_{k,j}}{2\pi}\in(0,1)\setminus{\bf Q}$ for $1\leq j\leq
r_{k,4}$, $\frac{{\beta}_{k,j}}{2\pi}\in{\bf
Q}\cap(0,1)\setminus\\{\frac{1}{2}\\}$ for $1\leq j\leq r_{k,5}$,
$\frac{\tilde{{\beta}}_{k,j}}{2\pi}\in(0,1)\setminus{\bf Q}$ for $1\leq j\leq
r_{k,6}$; $N_{2}(e^{\sqrt{-1}{\alpha}_{k,j}},A_{k,j})$’s and
$N_{2}(e^{\sqrt{-1}\tilde{{\alpha}}_{k,j}},\tilde{A}_{k,j})$’s are nontrivial
and $N_{2}(e^{\sqrt{-1}{\beta}_{k,j}},B_{k,j})$’s and
$N_{2}(e^{\sqrt{-1}\tilde{{\beta}}_{k,j}},\tilde{B}_{k,j})$’s are trivial, and
non-negative integers $p_{k,-}$, $p_{k,0}$, $p_{k,+}$, $q_{k,-}$, $q_{k,0}$,
$q_{k,+}$, $r_{k,1}$, $r_{k,2}$, $r_{k,3}$, $r_{k,4}$, $r_{k,5}$, $r_{k,6}$,
$h_{k}=h_{k,+}+h_{k,-}$ satisfy the equality

$\displaystyle
p_{k,-}+p_{k,0}+p_{k,+}+q_{k,-}+q_{k,0}+q_{k,+}+r_{k,1}+r_{k,2}+2\sum_{j=3}^{6}r_{k,j}+h_{k}=3.$
(3.2)

Lemma 3.1. Under the assumption (FCG), for any prime closed geodesic $c_{k}$,
$1\leq k\leq q$, there holds

$i(c_{k}^{m})\geq 3\left[\frac{5m}{3}\right],\qquad\forall\ m\in{\bf N}$ (3.3)

and

$\hat{i}(c_{k})>5.$ (3.4)

Proof. By the assumption (FCG), since the flag curvature $K$ satisfies
$\frac{25}{9}\left(\frac{{\lambda}}{1+{\lambda}}\right)^{2}<K\leq 1$, we can
choose $\frac{25}{9}\left(\frac{\lambda}{\lambda+1}\right)^{2}<\delta\leq
K\leq 1$. Then by Lemma 2 in [Rad07], it yields

$\hat{i}(c_{k})\geq 3\sqrt{\delta}\frac{1+\lambda}{\lambda}>5.$

Note that it follows from Theorem 3 of [Rad04] that
$L(c_{k}^{m})=mL(c_{k})\geq
m\pi\frac{1+\lambda}{\lambda}>\frac{5m}{3}\pi/\sqrt{\delta}$ for $m\geq 1$ and
$1\leq k\leq q$. Then it follows from Lemma 3 of [Rad04] that
$i(c_{k}^{m})\geq 3[\frac{5m}{3}]$.

Combining Lemma 3.1 with Theorem 2.9, it follows that

$\displaystyle i(c_{k}^{m+1})-i(c_{k}^{m})-\nu(c_{k}^{m})\geq
i(c_{k})-\frac{e(P_{c_{k}})}{2}\geq 0,\quad\forall\ m\in{\bf N},\ 1\leq k\leq
q.$ (3.5)

Here the last inequality holds by the fact that $e(P_{c_{k}})\leq 6$ and
$i(c_{k})\geq 3$.

It follows from (3.4), Theorem 4.3 in [LZ02] and Theorem 2.8 that for any
fixed integer $\bar{m}\in{\bf N}$, there exist infinitely many $(q+1)$-tuples
$(N,m_{1},\cdots,m_{q})\in{\bf N}^{q+1}$ such that for any $1\leq k\leq q$ and
$1\leq m\leq\bar{m}$, there holds

$\displaystyle i(c_{k}^{2m_{k}-m})+\nu(c_{k}^{2m_{k}-m})$ $\displaystyle=$
$\displaystyle
2N-i(c_{k}^{m})-\left(2S^{+}_{P_{c_{k}}}(1)+2Q_{k}(m)-\nu(c_{k}^{m})\right),$
(3.6) $\displaystyle i(c_{k}^{2m_{k}})$ $\displaystyle\geq$ $\displaystyle
2N-\frac{e(P_{c_{k}})}{2},$ (3.7) $\displaystyle
i(c_{k}^{2m_{k}})+\nu(c_{k}^{2m_{k}})$ $\displaystyle\leq$ $\displaystyle
2N+\frac{e(P_{c_{k}})}{2},$ (3.8) $\displaystyle i(c_{k}^{2m_{k}+m})$
$\displaystyle=$ $\displaystyle 2N+i(c_{k}^{m}),$ (3.9)

where (3.7) and (3.8) follow from (4.32) and (4.33) in Theorem 4.3 in [LZ02]
respectively.

Note that by List 9.1.12 of [Lon02], (2.26), (2.18) and
$\nu(c_{k})=p_{k,-}+2p_{k,0}+p_{k,+}$, we have

$\displaystyle S^{+}_{P_{c_{k}}}(1)=p_{k,-}+p_{k,0},$ (3.10) $\displaystyle
C(P_{c_{k}})=q_{k,0}+q_{k,+}+r_{k,1}+r_{k,2}+2r_{k,3}+2r_{k,4},$ (3.11)
$\displaystyle
Q_{k}(m)=\frac{1+(-1)^{m}}{2}(q_{k,0}+q_{k,+})+(r_{k,1}+r_{k,3})-\sum_{j=1}^{r_{k,1}}{\varphi}\left(\frac{m\theta_{k,j}}{2\pi}\right)-\sum_{j=1}^{r_{k,3}}{\varphi}\left(\frac{m{\alpha}_{k,j}}{2\pi}\right),$
(3.12)
$\displaystyle\nu(c_{k}^{m})=(p_{k,-}+2p_{k,0}+p_{k,+})+\frac{1+(-1)^{m}}{2}(q_{k,-}+2q_{k,0}+q_{k,+})+2(r_{k,1}+r_{k,3}+r_{k,5})$
$\displaystyle\qquad\qquad-2\left(\sum_{j=1}^{r_{k,1}}{\varphi}\left(\frac{m{\theta}_{k,j}}{2\pi}\right)+\sum_{j=1}^{r_{k,3}}{\varphi}\left(\frac{m{\alpha}_{k,j}}{2\pi}\right)+\sum_{j=1}^{r_{k,5}}{\varphi}\left(\frac{m{\beta}_{k,j}}{2\pi}\right)\right).$
(3.13)

By (3.10), (3.12) and (3.13), we obtain

$\displaystyle
2S^{+}_{P_{c_{k}}}(1)+2Q_{k}(m)-\nu(c_{k}^{m})=p_{k,-}-p_{k,+}-\frac{1+(-1)^{m}}{2}(q_{k,-}-q_{k,+})-2r_{k,5}+2\sum_{j=1}^{r_{k,5}}{\varphi}\left(\frac{m{\beta}_{k,j}}{2\pi}\right),$

which, together with (3.6), gives

$\displaystyle i(c_{k}^{2m_{k}-m})+\nu(c_{k}^{2m_{k}-m})$ $\displaystyle=$
$\displaystyle
2N-i(c_{k}^{m})-p_{k,-}+p_{k,+}+\frac{1+(-1)^{m}}{2}(q_{k,-}-q_{k,+})$ (3.14)
$\displaystyle\quad+2r_{k,5}-2\sum_{j=1}^{r_{k,5}}{\varphi}\left(\frac{m{\beta}_{k,j}}{2\pi}\right),\quad\forall\
1\leq m\leq\bar{m}.$

By (3.7)-(3.9), (3.14), (3.2), (3.3) and the fact $e(P_{c_{k}})\leq 6$, there
holds

$\displaystyle i(c_{k}^{2m_{k}-m})+\nu(c_{k}^{2m_{k}-m})$ $\displaystyle\leq$
$\displaystyle 2N+3-3\left[\frac{5m}{3}\right],\quad\forall\ 1\leq
m\leq\bar{m},$ (3.15) $\displaystyle 2N-3$ $\displaystyle\leq$ $\displaystyle
2N-\frac{e(P_{c_{k}})}{2}\leq i(c_{k}^{2m_{k}}),$ (3.16) $\displaystyle
i(c_{k}^{2m_{k}})+\nu(c_{k}^{2m_{k}})$ $\displaystyle\leq$ $\displaystyle
2N+\frac{e(P_{c_{k}})}{2}\leq 2N+3,$ (3.17) $\displaystyle
2N+3\left[\frac{5m}{3}\right]$ $\displaystyle\leq$ $\displaystyle
i(c_{k}^{2m_{k}+m}),\qquad\forall\ 1\leq m\leq\bar{m}.$ (3.18)

Note that by (3.5), we have

$\displaystyle i(c_{k}^{m})\leq i(c_{k}^{m+1}),\qquad
i(c_{k}^{m})+\nu(c_{k}^{m})\leq i(c_{k}^{m+1})+\nu(c_{k}^{m+1}),\quad\forall
m\in{\bf N},$

which, together with (3.15) and (3.18), implies

$\displaystyle i(c_{k}^{m})+\nu(c_{k}^{m})\leq
i(c_{k}^{2m_{k}-\bar{m}})+\nu(c_{k}^{2m_{k}-\bar{m}})\leq
2N+3-3\left[\frac{5\bar{m}}{3}\right],\,\forall\ 1\leq m\leq 2m_{k}-\bar{m},$
(3.19) $\displaystyle 2N+3\left[\frac{5\bar{m}}{3}\right]\leq
i(c_{k}^{2m_{k}+\bar{m}})\leq i(c_{k}^{m}),\,\forall\ m\geq 2m_{k}+\bar{m}.$
(3.20)

In addition, by (2.25), (3.10), (3.11) and (3.13), the precise formulae of
$i(c_{k}^{2m_{k}})+\nu(c_{k}^{2m_{k}})$ can be computed as follows

$\displaystyle i(c_{k}^{2m_{k}})+\nu(c_{k}^{2m_{k}})$ $\displaystyle=$
$\displaystyle
2N+2\Delta_{k}-(p_{k,-}+p_{k,0}+q_{k,0}+q_{k,+}+r_{k,1}+r_{k,2}+2r_{k,3}+2r_{k,4})$
(3.21)
$\displaystyle+p_{k,+}+2p_{k,0}+p_{k,-}+q_{k,+}+2q_{k,0}+q_{k,-}+2r_{k,1}+2r_{k,3}+2r_{k,5}$
$\displaystyle=$ $\displaystyle 2N+p_{k,0}+p_{k,+}+q_{k,-}+q_{k,0}$
$\displaystyle\qquad+r_{k,1}+2r_{k,5}-r_{k,2}-2r_{k,4}+2\Delta_{k},\quad
k=1,\cdots,q,$

where

$\displaystyle\Delta_{k}\equiv\sum_{0<\\{m_{k}{\theta}/\pi\\}<\delta}S^{-}_{M_{k}}(e^{\sqrt{-1}{\theta}})\leq
r_{k,2}+r_{k,4}.$ (3.22)

by (2.26) and List 9.1.12 of [Lon02].

Lemma 3.2. Under the assumption (FCG), for $k=1,\cdots,q$ and $1\leq
m\leq\bar{m}$, we have

$\displaystyle i(c_{k}^{2m_{k}-1})+\nu(c_{k}^{2m_{k}-1})$ $\displaystyle\leq$
$\displaystyle 2N-3,$ (3.23) $\displaystyle
i(c_{k}^{2m_{k}-2})+\nu(c_{k}^{2m_{k}-2})$ $\displaystyle\leq$ $\displaystyle
2N-9.$ (3.24)

Proof. By (2.17), we have

$\displaystyle\hat{i}(c_{k})$ $\displaystyle=$ $\displaystyle
i(c_{k})+p_{k,-}+p_{k,0}-r_{k,1}-r_{k,2}+\sum_{j=1}^{r_{k,1}}\frac{\theta_{k,j}}{\pi}+\sum_{j=1}^{r_{k,2}}\frac{\tilde{\theta}_{k,j}}{\pi}$
(3.25) $\displaystyle<$ $\displaystyle
i(c_{k})+p_{k,-}+p_{k,0}+r_{k,1}+r_{k,2}.$

Combining (3.25) with (3.4), there holds

$\displaystyle i(c_{k})+p_{k,-}+p_{k,0}+r_{k,1}+r_{k,2}\geq 6.$ (3.26)

Then by (3.14), (3.26) and (3.2), we obtain

$\displaystyle i(c_{k}^{2m_{k}-1})+\nu(c_{k}^{2m_{k}-1})$ $\displaystyle=$
$\displaystyle 2N-i(c_{k})-p_{k,-}+p_{k,+}$ (3.27) $\displaystyle\leq$
$\displaystyle 2N-6+p_{k,0}+r_{k,1}+r_{k,2}+p_{k,+}\leq 2N-3.$

By (3.14), (3.3) and (3.2), we get

$\displaystyle
i(c_{k}^{2m_{k}-2})+\nu(c_{k}^{2m_{k}-2})=2N-i(c_{k}^{2})-p_{k,-}+p_{k,+}+q_{k,-}-q_{k,+}\leq
2N-9+3=2N-6.$ (3.28)

Now we assume $i(c_{k}^{2m_{k}-2})+\nu(c_{k}^{2m_{k}-2})\geq 2N-8$, $\forall\
k=1,\cdots,q$, which, together with (3.28), gives
$i(c_{k}^{2m_{k}-2})+\nu(c_{k}^{2m_{k}-2})\in\\{2N-6,2N-7,2N-8\\}$.

We continue the proof by distinguishing three cases.

Case 1: $i(c_{k}^{2m_{k}-2})+\nu(c_{k}^{2m_{k}-2})=2N-6$.

In this case, by (3.3) and (3.28), we know that $p_{k,+}+q_{k,-}=3$ and
$i(c_{k}^{2})=9$. It follows from (2.17) and (3.2) that $i(c_{k}^{2})=2i(c)\in
2{\bf N}$ since $p_{k,+}+q_{k,-}=3$, thus Case 1 cannot happen.

Case 2: $i(c_{k}^{2m_{k}-2})+\nu(c_{k}^{2m_{k}-2})=2N-7$.

In this case, by (3.3) and (3.28), one of the following cases may happen.

(i) $i(c_{k}^{2})=10$ and $p_{k,+}+q_{k,-}=3$.

(ii) $i(c_{k}^{2})=9$, $p_{k,+}+q_{k,-}=2$ and $p_{k,-}+q_{k,+}=0$.

For (i), by (2.17), we have $i(c_{k}^{2})=2i(c_{k})$ and
$\hat{i}(c_{k})=i(c_{k})$ since $p_{k,+}+q_{k,-}=3$. However, there holds
$\hat{i}(c_{k})>5$ by (3.4). So we have
$i(c_{k}^{2})=2i(c_{k})=2\hat{i}(c_{k})>10$, thus (i) of Case 2 cannot happen.

For (ii), by (3.2), there holds

$\displaystyle p_{k,0}+q_{k,0}+r_{k,1}+r_{k,2}+h_{k}=1.$ (3.29)

It follows from (2.17) and (1.2) that

$\displaystyle i(c_{k}^{2})$ $\displaystyle=$ $\displaystyle
2i(c_{k})+p_{k,0}-q_{k,0}-3(r_{k,1}+r_{k,2})+2\sum_{j=1}^{r_{k,1}}E\left(\frac{\theta_{k,j}}{\pi}\right)+2\sum_{j=1}^{r_{k,2}}E\left(\frac{\tilde{\theta}_{k,j}}{\pi}\right)$
(3.30) $\displaystyle>$ $\displaystyle
2i(c_{k})+p_{k,0}-q_{k,0}-3(r_{k,1}+r_{k,2})+2\sum_{j=1}^{r_{k,1}}\frac{\theta_{k,j}}{\pi}+2\sum_{j=1}^{r_{k,2}}\frac{\tilde{\theta}_{k,j}}{\pi}.$

Combining (3.30) and (3.29), it yields

$\displaystyle\hat{i}(c_{k})$ $\displaystyle=$ $\displaystyle
i(c_{k})+p_{k,0}-r_{k,1}-r_{k,2}+\sum_{j=1}^{r_{k,1}}\frac{\theta_{k,j}}{\pi}+\sum_{j=1}^{r_{k,2}}\frac{\tilde{\theta}_{k,j}}{\pi}$
(3.31) $\displaystyle<$
$\displaystyle\frac{1}{2}\left(i(c_{k}^{2})+p_{k,0}+q_{k,0}+r_{k,1}+r_{k,2}\right)\leq\frac{1}{2}\left(i(c_{k}^{2})+1\right),$

which, together with (3.4), yields $i(c_{k}^{2})>9$, thus Case (ii) cannot
happen.

Case 3: $i(c_{k}^{2m_{k}-2})+\nu(c_{k}^{2m_{k}-2})=2N-8$.

In this case, by (3.3) and (3.28), one of the following cases may happen.

(i) $i(c_{k}^{2})=11$ and $p_{k,+}+q_{k,-}=3$.

(ii) $i(c_{k}^{2})=10$, $p_{k,+}+q_{k,-}=2$ and $p_{k,-}+q_{k,+}=0$.

(iii) $i(c_{k}^{2})=9$, $p_{k,+}+q_{k,-}=2$ and $p_{k,-}+q_{k,+}=1$.

(iv) $i(c_{k}^{2})=9$, $p_{k,-}+q_{k,+}=0$ and $p_{k,+}+q_{k,-}=1$.

For (i), similar to the arguments in Case 1, it can be shown that this case
cannot happen.

For (ii), similar to (3.29) and the first equality in (3.30), we have

$\displaystyle p_{k,0}+q_{k,0}+r_{k,1}+r_{k,2}+h_{k}=1,$ (3.32) $\displaystyle
i(c_{k}^{2})=p_{k,0}+q_{k,0}+r_{k,1}+r_{k,2}\quad({\rm mod}\ 2),$ (3.33)

which yields

$\displaystyle i(c_{k}^{2})=1+h_{k}\quad({\rm mod}\ 2).$ (3.34)

Then we get $h_{k}=1$ by $i(c_{k}^{2})=10$ and (3.32). By (2.17) and (3.2), we
have $i(c_{k}^{2})=2i(c_{k})$ and $\hat{i}(c_{k})=i(c_{k})$ since
$p_{k,+}+q_{k,-}=2$ and $h_{k}=1$. However $\hat{i}(c_{k})>5$ by (3.4), thus
Case (ii) cannot happen.

For (iii), by (2.17) and (3.2), we have
$i(c_{k}^{2})=2i(c_{k})+p_{k,-}-q_{k,+}$ and $\hat{i}(c_{k})=i(c_{k})+p_{k,-}$
since $p_{k,-}+q_{k,+}=1$ and $p_{k,+}+q_{k,-}=2$. Then we obtain
$2\hat{i}(c_{k})=i(c_{k}^{2})+1=10$. However $\hat{i}(c_{k})>5$ by (3.4), thus
Case (iii) cannot happen.

For (iv), similar to the proof of (3.33) in (ii) of Case 3, we have

$\displaystyle i(c_{k}^{2})=p_{k,0}+q_{k,0}+r_{k,1}+r_{k,2}\quad({\rm mod}\
2).$ (3.35)

Therefore by (3.2), (3.35), $i(c_{k}^{2})=9$ and $p_{k,+}+q_{k,-}=1$, we get

$\displaystyle p_{k,0}+q_{k,0}+r_{k,1}+r_{k,2}=1.$ (3.36)

Then similar to the proof of (3.31) in (ii) of Case 2, we have

$\displaystyle\hat{i}(c_{k})<\frac{1}{2}\left(i(c_{k}^{2})+p_{k,0}+q_{k,0}+r_{k,1}+r_{k,2}\right)=\frac{1}{2}\left(i(c_{k}^{2})+1\right),$
(3.37)

which, together with (3.4), yields $i(c_{k}^{2})>9$, thus Case (iv) cannot
happen.

This completes the proof of Lemma 3.2.

Under the assumption (FCG), Theorem 1.1 of [Dua16] shows that there exist at
least two elliptic closed geodesics $c_{1}$ and $c_{2}$ on $(S^{4},F)$ whose
flag curvature satisfies $\left(\frac{{\lambda}}{1+{\lambda}}\right)^{2}<K\leq
1$. The following Lemma gives some properties of these two closed geodesics
which will be useful in the proof of Theorem 1.1.

Lemma 3.3. (cf. Lemma 3.1 and Lemma 3.3 of [Dua15] and Section 3 of [Dua16])
Under the assumption (FCG), there exist at least two prime elliptic closed
geodesics $c_{1}$ and $c_{2}$ on $(S^{4},F)$ whose flag curvature satisfies
$\left(\frac{{\lambda}}{1+{\lambda}}\right)^{2}<K\leq 1$. Moreover, there
exist infinitely many pairs of $(q+1)$-tuples
$(N,m_{1},m_{2},\cdots,m_{q})\in{\bf N}^{q+1}$ and
$(N^{\prime},m_{1}^{\prime},m_{2}^{\prime},\cdots,m_{q}^{\prime})\in{\bf
N}^{q+1}$ such that

$\displaystyle
i(c_{1}^{2m_{1}})+\nu(c_{1}^{2m_{1}})=2N+3,\quad\overline{C}_{2N+3}(E,c_{1}^{2m_{1}})={\bf
Q},$ (3.38) $\displaystyle
i(c_{2}^{2m_{2}^{\prime}})+\nu(c_{2}^{2m_{2}^{\prime}})=2N^{\prime}+3,\quad\overline{C}_{2N^{\prime}+3}(E,c_{2}^{2m_{2}^{\prime}})={\bf
Q},$ (3.39) $\displaystyle
p_{k,-}=q_{k,+}=r_{k,3}=r_{k,4}=r_{k,6}=h_{k}=0,\quad k=1,2,$ (3.40)
$\displaystyle r_{1,2}=\Delta_{1}\geq 1,\qquad r_{2,2}=\Delta_{2}^{\prime}\geq
1,$ (3.41) $\displaystyle\Delta_{k}+\Delta_{k}^{\prime}=r_{k,2},\quad k=1,2,$
(3.42)

where we can require $3|N$ or $3|N^{\prime}$ as remarked in Theorem 2.8 and

$\displaystyle\Delta_{k}^{\prime}\equiv\sum_{0<\\{m_{k}^{\prime}{\theta}/\pi\\}<\delta}S^{-}_{M_{k}}(e^{\sqrt{-1}{\theta}}),\quad
k=1,2.$ (3.43)

In addition, for these two closed geodesics $c_{1}$ and $c_{2}$, there holds

$\displaystyle
k_{\nu(c_{k}^{n(c_{k})})}^{{\epsilon}(c_{k}^{n(c_{k})})}(c_{k}^{n(c_{k})})=1,\quad
k_{j}^{{\epsilon}(c_{k}^{n(c_{k})})}(c_{k}^{n(c_{k})})=0,\quad\forall\ 0\leq
j<\nu(c_{k}^{n(c_{k})}),\ k=1,2.$ (3.44)

###  3.2 Proof of Theorem 1.1

In this section, let $(S^{4},F)$ be a Finsler sphere of dimension $4$ with its
reversibility $\lambda$ and flag curvature $K$ satisfying
$\frac{25}{9}\left(\frac{{\lambda}}{1+{\lambda}}\right)^{2}<K\leq 1$ with
$\lambda<\frac{3}{2}$. In order to prove Theorem 1.1, according to Lemma 3.3
and Theorem 1.1 of [Dua15], we make the following assumption.

(TCG) Suppose that there exist exactly two prime elliptic closed geodesics
$c_{1}$ and $c_{2}$ possessing all properties listed in Lemma 3.3, and the
third prime closed geodesic $c_{3}$ on such $(S^{4},F)$.

In order to count the contribution of $c_{k}^{m}$ to the Morse-type number
$M_{q}$, for the sake of convenience, we set

$\displaystyle M_{q}(k,m)=\dim{\overline{C}}_{q}(E,c^{m}_{k}),\quad\forall\
1\leq k\leq 3,\ m\geq 1,\ q\in{\bf N}_{0}.$ (3.45)

Next we fix $\bar{m}=4$. Before proving Theorem 1.1, firstly we establish
several crucial lemmas.

Lemma 3.4. For an integer $q$ satisfying $2N-9\leq q\leq 2N+17$, there holds

$\displaystyle M_{q}=\left\\{\matrix{\sum_{1\leq k\leq 3,\atop 1\leq m\leq
2}M_{q}(k,2m_{k}-m),\quad&&{\it if}\quad q=2N-9,\cr\sum_{1\leq k\leq
3}M_{q}(k,2m_{k}-1),\quad&&{\it if}\quad 2N-8\leq q\leq 2N-4,\cr\sum_{1\leq
k\leq 3,\atop 0\leq m\leq 1}M_{q}(k,2m_{k}-m),\quad&&{\it if}\quad
q=2N-3,\cr\sum_{1\leq k\leq 3}M_{q}(k,2m_{k}),\quad&&{\it if}\quad 2N-2\leq
q\leq 2N+2,\cr\sum_{1\leq k\leq 3,\atop 0\leq m\leq
1}M_{q}(k,2m_{k}+m),\quad&&{\it if}\quad q=2N+3,\cr\sum_{1\leq k\leq
3}M_{q}(k,2m_{k}+1),\quad&&{\it if}\quad 2N+4\leq q\leq 2N+8,\cr\sum_{1\leq
k\leq 3,\atop 1\leq m\leq 2}M_{q}(k,2m_{k}+m),\quad&&{\it if}\quad 2N+9\leq
q\leq 2N+14,\cr\sum_{1\leq k\leq 3,\atop 1\leq m\leq
3}M_{q}(k,2m_{k}+m),\quad&&{\it if}\quad 2N+15\leq q\leq 2N+17.\cr}\right.$
(3.46)

Proof. According to Lemma 2.1, (2.3) and (i) of Lemma 2.2, we have

$\displaystyle M_{q}$ $\displaystyle=$ $\displaystyle\sum_{1\leq k\leq 3,\atop
m\geq 1}M_{q}(k,m)=\sum_{1\leq k\leq 3,\atop m\geq
1}k_{q-i(c_{k}^{m})}^{{\epsilon}(c_{k}^{m})}(c_{k}^{m})$ (3.47)
$\displaystyle=$ $\displaystyle\sum_{1\leq k\leq 3,\atop m\in\left\\{m\in{\bf
N}|i(c_{k}^{m})\leq q\leq
i(c_{k}^{m})+\nu(c_{k}^{m})\right\\}}k_{q-i(c_{k}^{m})}^{{\epsilon}(c_{k}^{m})}(c_{k}^{m})=\sum_{1\leq
k\leq 3,\atop m\in\left\\{m\in{\bf N}|i(c_{k}^{m})\leq q\leq
i(c_{k}^{m})+\nu(c_{k}^{m})\right\\}}M_{q}(k,m).$

On one hand, by (3.19), (3.15), (3.24), (3.23) and (3.17), it yields

$\displaystyle
i(c_{k}^{m})+\nu(c_{k}^{m})\leq\left\\{\matrix{2N-15,&&\quad{\it if}\quad
1\leq m\leq 2m_{k}-4,\cr 2N-12,&&\quad{\it if}\quad m=2m_{k}-3,\cr
2N-9,&&\quad{\it if}\quad m=2m_{k}-2,\cr 2N-3,&&\quad{\it if}\quad
m=2m_{k}-1,\cr 2N+3,&&\quad{\it if}\quad m=2m_{k}.\cr}\right.$ (3.48)

On the other hand, by (3.16), (3.18) and (3.20), it yields

$\displaystyle i(c_{k}^{m})\geq\left\\{\matrix{2N-3,&&\quad{\it if}\quad
m=2m_{k},\cr 2N+3,&&\quad{\it if}\quad m=2m_{k}+1,\cr 2N+9,&&\quad{\it
if}\quad m=2m_{k}+2,\cr 2N+15,&&\quad{\it if}\quad m=2m_{k}+3,\cr
2N+18,&&\quad{\it if}\quad m\geq 2m_{k}+4.\cr}\right.$ (3.49)

Combining (3.47)-(3.49), we get Lemma 3.4.

Lemma 3.5. For some tuple $(k,m)$ with $k=1,2,3$ and $m\in{\bf N}$, if there
exist some integers $q_{1},q_{2}\in{\bf N}$ satisfying

$\displaystyle q_{1}\leq i(c_{k}^{m})\quad\mbox{and}\quad
i(c_{k}^{m})+\nu(c_{k}^{m})\leq q_{2},$

then there holds

$\displaystyle M_{q_{1}}(k,m)+M_{q_{2}}(k,m)\leq 1.$ (3.50)

Furthermore, if $M_{q_{1}}(k,m)+M_{q_{2}}(k,m)=1$, then

$\displaystyle M_{q}(k,m)=0,\quad\forall\ q\neq q_{1},q_{2}.$ (3.51)

Proof. This follows directly from Lemma 2.1, (2.3), (i) and (ii) of Lemma 2.2.

Lemma 3.6. For some $k\in\\{1,2,3\\}$, assume that either $M_{2N-q\pm
1}(k,2m_{k}-1)\geq 1$ or $M_{2N+q\pm 1}(k,2m_{k}+1)\geq 1$ for some even
$q\in{\bf N}$, then there exists a continuous path $f_{k}\in
C([0,1],{\Omega}^{0}(P_{c_{k}}))$ such that $f_{k}(0)=P_{c_{k}}$ and
$f_{k}(1)$ belongs to one of the following five cases:

(i) $I_{4}\,{\diamond}\,H(2)$,

(ii) $N_{1}(1,1)\,{\diamond}\,I_{2}\,{\diamond}\,N_{1}(1,-1)$,

(iii) $I_{4}\,{\diamond}\,N_{1}(1,-1)$,

(iv) $N_{1}(1,1)\,{\diamond}\,I_{4}$,

(v) $I_{6}$.

And in either case, the index iteration formula of $c_{k}^{m}$ can be written
as follows:

$\displaystyle i(c_{k}^{m})=qm-p_{k,-}-p_{k,0},\quad\forall\ m\geq 1.$ (3.52)

Proof. We only give the proof under the assumption $M_{2N-q\pm
1}(k,2m_{k}-1)\geq 1$. The proof under the assumption $M_{2N+q\pm
1}(k,2m_{k}+1)\geq 1$ is similar.

First, by Lemma 3.5 and the assumption $M_{2N-q\pm 1}(k,2m_{k}-1)\geq 1$ in
Lemma 3.6, we have

$\displaystyle i(c_{k}^{2m_{k}-1})\leq 2N-q-2,\qquad
i(c_{k}^{2m_{k}-1})+\nu(c_{k}^{2m_{k}-1})\geq 2N-q+2,$ (3.53)

which, together with $\nu(c_{k}^{2m_{k}-1})=\nu(c_{k})$ by (2.22), implies
$\nu(c_{k})=p_{k,-}+2p_{k,0}+p_{k,+}\in\\{4,5,6\\}$.

If $\nu(c_{k})=4$, by (3.53), we have $i(c_{k}^{2m_{k}-1})=2N-q-2$ and
$i(c_{k}^{2m_{k}-1})+\nu(c_{k}^{2m_{k}-1})=2N-q+2$. And we also have
$p_{k,-}+p_{k,0}+p_{k,+}=2$ or $3$ by (3.2). Since
$i(c_{k}^{2m_{k}-1})=2N-q-2\in 2{\bf N}$, we get $i(c_{k})\in 2{\bf N}$ by
(2.24), then by Proposition 2.7 and the symplectic additivity of symplectic
paths (cf. Theorem 6.2.6 of [Lon02]), we must have $p_{k,+}+h_{k,+}=1$.
Therefore, if $p_{k,-}+p_{k,0}+p_{k,+}=2$, we must have $p_{k,0}=2$ since
$\nu(c_{k})=4$ and $h_{k,+}=1$, i.e., $f_{k}(1)=I_{4}\,{\diamond}\,H(2)$. If
$p_{k,-}+p_{k,0}+p_{k,+}=3$, we must have $p_{k,0}=1$, $p_{k,+}=1$ and
$p_{k,-}=1$, i.e.,
$f_{k}(1)=N_{1}(1,1)\,{\diamond}\,I_{2}\,{\diamond}\,N_{1}(1,-1)$.

If $\nu(c_{k})=5$, we have $p_{k,-}+p_{k,0}+p_{k,+}=3$. Then we must have
$p_{k,0}=2$ and $p_{k,-}+p_{k,+}=1$. So either $i(c_{k}^{2m_{k}-1})=2N-q-2$
when $p_{k,+}=1$, or $i(c_{k}^{2m_{k}-1})=2N-q-3$ when $p_{k,-}=1$ by
Proposition 2.7, the symplectic additivity and (3.53), i.e.,
$f_{k}(1)=I_{4}\,{\diamond}\,N_{1}(1,-1)$ or $N_{1}(1,1)\,{\diamond}\,I_{4}$.

If $\nu(c_{k})=6$, we must have $p_{k,0}=3$, and then we have
$i(c_{k}^{2m_{k}-1})=2N-q-3$ by Proposition 2.7, the symplectic additivity and
(3.53), i.e., $f_{k}(1)=I_{6}$.

Note that by (2.24), (3.10) and (3.12), we get

$i(c_{k}^{2m_{k}-1})=2N-i(c_{k})-2p_{k,-}-2p_{k,0},$

and by the above arguments in either case, we have
$i(c_{k})=q-p_{k,-}-p_{k,0}$. Then by (2.17), we have

$i(c_{k}^{m})=qm-p_{k,-}-p_{k,0}.$

This completes the proof of Lemma 3.6.

Proof of Theorem 1.1.

At first, we consider the contribution of $c_{k}^{m}$ with $k=1,2,3$ and
$m\in{\bf N}$ to the Morse-type numbers in Claim 1 and Claim 2. And then we
use Claim 3 and Claim 4 to complete the proof of Theorem 1.1.

Claim 1: For $2N-2\leq q\leq 2N+2$, there holds: (i) $M_{q}(1,m)=0$ for any
$m\in{\bf N}$, and (ii) $M_{q}(k,m)=0$ for $k=2,3$ and $m\neq 2m_{k}$. In
addition, $M_{q}(2,2m_{2})=0$ for $q=2N-2,2N,2N+2$ and
$M_{2N-1}(2,2m_{2})+M_{2N+1}(2,2m_{2})\leq 1$.

Proof. By Lemma 3.4, we know that $M_{q}(k,m)=0$ for $2N-2\leq q\leq 2N+2$,
$m\neq 2m_{k}$ and $k=1,2,3$. Also note that by (3.38), (3.44), Lemma 2.1 and
(2.3), we have

$M_{q}(1,2m_{1})=0\quad\mbox{for}\quad 2N-2\leq q\leq 2N+2.$

On one hand, there holds

$\nu(c_{2}^{2m_{2}})=\nu(c_{2}^{2m^{\prime}_{2}})$

by the choices of $m_{2}$ and $m_{2}^{\prime}$ in (2.27) of Theorem 2.8. On
the other hand, it yields

$i(c_{2}^{2m_{2}})=i(c_{2}^{2m_{2}^{\prime}})\quad({\rm mod}\ 2)$

by (2.17) of Theorem 2.6. So, $i(c_{2}^{2m_{2}})+\nu(c_{2}^{2m_{2}})$ is odd
since $i(c_{2}^{2m^{\prime}_{2}})+\nu(c_{2}^{2m^{\prime}_{2}})$ is odd by
(3.39) of Lemma 3.3, which implies that $M_{q}(2,2m_{2})=0$ for
$q=2N-2,2N,2N+2$ and $M_{2N-1}(2,2m_{2})+M_{2N+1}(2,2m_{2})\leq 1$ by (3.44),
Lemma 2.1 and (2.3). Hence, Claim 1 holds.

Claim 2: $M_{2N-1}(2,2m_{2})=M_{2N+1}(2,2m_{2})=0$.

Proof. Otherwise, by Claim 1 we have

$\displaystyle M_{2N-1}(2,2m_{2})+M_{2N+1}(2,2m_{2})=1.$ (3.54)

Then by (3.16) and Lemma 3.5, it yields

$\displaystyle M_{2N-3}(2,2m_{2})=0.$ (3.55)

By (2.7) and Theorem 2.5, we have $M_{2N-1}\geq b_{2N-1}=1$ and $M_{2N+1}\geq
b_{2N+1}=1$, then we get $M_{2N-1}(3,2m_{3})+M_{2N+1}(3,2m_{3})\geq 1$ by
Lemma 3.4, (i) of Claim 1 and (3.54). Thus by (3.16), (3.17) and Lemma 3.5, it
yields

$\displaystyle M_{2N-3}(3,2m_{3})=M_{2N+3}(3,2m_{3})=0.$ (3.56)

So, by Lemma 3.4, Claim 1, (3.54) and (3.56), we obtain

$\displaystyle\sum_{q=2N-2}^{2N+2}(-1)^{q}M_{q}=\sum_{q=2N-2}^{2N+2}(-1)^{q}\sum_{1\leq
k\leq 3}M_{q}(k,2m_{k})=\sum_{q=2N-3}^{2N+3}(-1)^{q}M_{q}(3,2m_{3})-1.$ (3.57)

And by (3.16), (3.17), Lemma 2.1 and (2.3), we have

$\displaystyle\sum_{q=2N-3}^{2N+3}(-1)^{q}M_{q}(3,2m_{3})=\sum_{0\leq l\leq
6}(-1)^{i(c_{3}^{2m_{3}})+l}k_{l}^{{\epsilon}(c_{3}^{2m_{3}})}(c_{3}^{2m_{3}}).$
(3.58)

On the other hand, by (2.7) and Theorem 2.5, we have

$\displaystyle\sum_{q=2N-2}^{2N+2}(-1)^{q}M_{q}\geq\sum_{q=2N-2}^{2N+2}(-1)^{q}b_{q}=-2.$
(3.59)

Combining (3.57)-(3.59), we get

$\displaystyle\chi(c_{3}^{2m_{3}})=\sum_{0\leq l\leq
6}(-1)^{i(c_{3}^{2m_{3}})+l}k_{l}^{{\epsilon}(c_{3}^{2m_{3}})}(c_{3}^{2m_{3}})\geq-1.$
(3.60)

Note that since $n(c_{3})|2m_{3}$ and
$\nu(c_{3}^{2m_{3}})=\nu(c_{3}^{n(c_{3})})$ by (2.6) and (2.27), there holds

$\displaystyle\chi(c_{3}^{n(c_{3})})=\chi(c_{3}^{2m_{3}})\geq-1.$ (3.61)

By (3.38), (3.44), Lemma 2.1 and (2.3), it yields $M_{2N-3}(1,2m_{1})=0$.
Together with (3.55) and (3.56), we have

$\displaystyle\sum_{1\leq k\leq 3}M_{2N-3}(k,2m_{k})=0.$ (3.62)

Then combining Lemma 3.4 and (3.62), we obtain that

$\displaystyle M_{2N-3}=\sum_{1\leq k\leq 3}M_{2N-3}(k,2m_{k}-1).$ (3.63)

One one hand, by (3.23) and Lemma 3.5 it yields

$\displaystyle M_{2N-3}(k,2m_{k}-1)\leq 1,\quad\forall\ k=1,2,3,$ (3.64)

then it follows from (3.63) and (3.64) that $M_{2N-3}\leq 3$. On the other
hand, we have $M_{2N-3}\geq b_{2N-3}=2$ by (2.7) and Theorem 2.5. So it yields
$M_{2N-3}\in\\{2,3\\}$.

We continue the proof by distinguishing two cases.

Case 1: $M_{2N-3}=3$.

In this case, it follows from (3.63) and (3.64) that $M_{2N-3}(k,2m_{k}-1)=1$
for $k=1,2,3$. Then according to (3.23) and Lemma 3.5, there holds

$\displaystyle M_{2N-5}(k,2m_{k}-1)=0,\quad\forall\ k=1,2,3.$ (3.65)

Combining (3.65) and Lemma 3.4, we get $M_{2N-5}=0$. But by (2.7) and Theorem
2.5, we have $M_{2N-5}\geq b_{2N-5}=1$. This is a contradiction.

Case 2: $M_{2N-3}=2$.

In this case, it follows from (3.63) and (3.64) that one of the following
cases may happen:

(i) $M_{2N-3}(3,2m_{3}-1)=0$ and
$M_{2N-3}(1,2m_{1}-1)=M_{2N-3}(2,2m_{1}-1)=1$.

(ii) $M_{2N-3}(3,2m_{3}-1)=1$ and
$M_{2N-3}(1,2m_{1}-1)+M_{2N-3}(2,2m_{2}-1)=1$.

For (i), by (3.23) and Lemma 3.5, there holds

$\displaystyle M_{q}(k,2m_{k}-1)=0,\quad\forall\ q\neq 2N-3,\ k=1,2.$ (3.66)

So, according to Lemma 3.4 and (3.66), we have

$\displaystyle M_{2N-5}$ $\displaystyle=$ $\displaystyle\sum_{1\leq k\leq
3}M_{2N-5}(k,2m_{k}-1)=M_{2N-5}(3,2m_{3}-1),$ (3.67) $\displaystyle M_{2N-7}$
$\displaystyle=$ $\displaystyle\sum_{1\leq k\leq
3}M_{2N-7}(k,2m_{k}-1)=M_{2N-7}(3,2m_{3}-1).$ (3.68)

By (2.7) and Theorem 2.5, we have $M_{2N-5}\geq b_{2N-5}=1$ and $M_{2N-7}\geq
b_{2N-7}=1$, then it follows from (3.67) and (3.68) that

$\displaystyle M_{2N-5}(3,2m_{3}-1)\geq 1,\quad M_{2N-7}(3,2m_{3}-1)\geq 1.$
(3.69)

So the assumption with $q=6$ in Lemma 3.6 is satisfied, and then by (3.52) and
(2.6), we have

$\displaystyle i(c_{3}^{m})$ $\displaystyle=$ $\displaystyle
6m-p_{3,-}-p_{3,0},$ (3.70) $\displaystyle n(c_{3})$ $\displaystyle=$
$\displaystyle 1.$ (3.71)

Notice that $\nu(c_{3}^{2m_{3}-1})=\nu(c_{3}^{2m_{3}-2})=\nu(c_{3})$ by (2.18)
for each of five cases in Lemma 3.6. Together with Lemma 2.1, (2.3), (iv) of
Lemma 2.2, (3.70), (3.71) and (i) of Case 2, it yields

$\displaystyle M_{2N-9}(3,2m_{3}-2)$ $\displaystyle=$ $\displaystyle
k_{2N-9-i(c_{3}^{2m_{3}-2})}^{{\epsilon}(c_{3}^{2m_{3}-2})}(c_{3}^{2m_{3}-2})$
(3.72) $\displaystyle=$ $\displaystyle
k_{2N-9-i(c_{3}^{2m_{3}-2})}^{{\epsilon}(c_{3}^{2m_{3}-1})}(c_{3}^{2m_{3}-1})$
$\displaystyle=$ $\displaystyle
M_{2N-9-i(c_{3}^{2m_{3}-2})+i(c_{3}^{2m_{3}-1})}(3,2m_{3}-1)$ $\displaystyle=$
$\displaystyle M_{2N-3}(3,2m_{3}-1)=0.$

Through comparing (3.9) and (3.70), we get $2N=12m_{3}$. Then by (3.70) and
(3.2) it yields

$\displaystyle i(c_{3}^{2m_{3}-1})=12m_{3}-6-p_{3,-}-p_{3,0}\geq 2N-9,$ (3.73)

Then by (3.69) and Lemma 3.5, there holds

$\displaystyle M_{2N-9}(3,2m_{3}-1)=0.$ (3.74)

Finally, combining (3.66), (3.72), (3.74) and Lemma 3.4, we obtain

$\displaystyle M_{2N-9}=\sum_{1\leq k\leq 3,\atop 1\leq m\leq
2}M_{2N-9}(k,2m_{k}-m)=M_{2N-9}(1,2m_{1}-2)+M_{2N-9}(2,2m_{2}-2).$ (3.75)

By (3.24) and (ii) of Lemma 2.2, we get $M_{2N-9}(k,2m_{k}-2)\leq 1$ for
$k=1,2$. Then it follows from (3.75) that $M_{2N-9}\leq 2$. However by (2.7)
and Theorem 2.5, we have $M_{2N-9}\geq b_{2N-9}=2$. Thus we get $M_{2N-9}=2$.

Now, in this case, we have

$\displaystyle M_{2N-3}=b_{2N-3},\quad M_{2N-9}=b_{2N-9}.$ (3.76)

Combining (2.7), Theorem 2.5 and (3.76), we obtain

$\displaystyle\sum_{q=2N-8}^{2N-4}(-1)^{q}M_{q}=\sum_{q=2N-8}^{2N-4}(-1)^{q}b_{q}=-2.$
(3.77)

Note that for $2N-8\leq q\leq 2N-4$, it follows from (3.23) and (i) of this
case that $M_{q}(k,2m_{k}-1)=0$ for $k=1,2$. So we get
$M_{q}=M_{q}(3,2m_{3}-1)$ by Lemma 3.4, and then it yields

$\displaystyle\sum_{q=2N-8}^{2N-4}(-1)^{q}M_{q}=\sum_{q=2N-8}^{2N-4}(-1)^{q}M_{q}(3,2m_{3}-1).$
(3.78)

Note that $2N-9\leq i(c_{3}^{2m_{3}-1})\leq
i(c_{3}^{2m_{3}-1})+\nu(c_{3}^{2m_{3}-1})\leq 2N-3$ by (3.73) and (3.23),
according to (3.74), (i) of this case, Lemma 2.1 and (2.3), we obtain

$\displaystyle\sum_{q=2N-8}^{2N-4}(-1)^{q}M_{q}(3,2m_{3}-1)$ $\displaystyle=$
$\displaystyle\sum_{q=2N-9}^{2N-3}(-1)^{q}M_{q}(3,2m_{3}-1)$ (3.79)
$\displaystyle=$ $\displaystyle\sum_{0\leq l\leq
6}(-1)^{i(c_{3}^{2m_{3}-1})+l}k_{l}^{{\epsilon}(c_{3}^{2m_{3}-1})}(c_{3}^{2m_{3}-1}).$

Combining (3.77)-(3.79), we get

$\displaystyle\chi(c_{3}^{2m_{3}-1})=\sum_{0\leq l\leq
6}(-1)^{i(c_{3}^{2m_{3}-1})+l}k_{l}^{{\epsilon}(c_{3}^{2m_{3}-1})}(c_{3}^{2m_{3}-1})=-2.$
(3.80)

However, since $n(c_{3})=1$ by (3.71), it follows from (iv) of Lemma 2.2 and
(3.80) that $\chi(c_{3})=\chi(c_{3}^{2m_{3}-1})=-2$, which contradicts to
(3.61), thus Case (i) cannot happen.

For (ii), without loss of generality, we assume that $M_{2N-3}(1,2m_{1}-1)=0$.
Then by (3.23) and Lemma 3.5, $M_{q}(k,2m_{k}-1)=0$ for $q\neq 2N-3$ and
$k=2,3$. So, according to Lemma 3.4, we have

$\displaystyle M_{2N-5}=M_{2N-5}(1,2m_{1}-1),\quad
M_{2N-7}=M_{2N-7}(1,2m_{1}-1).$ (3.81)

By (2.7) and Theorem 2.5, we have $M_{2N-5}\geq b_{2N-5}=1$ and $M_{2N-7}\geq
b_{2N-7}=1$, then it follows from (3.81) that $M_{2N-5}(1,2m_{1}-1)\geq 1$ and
$M_{2N-7}(1,2m_{1}-1)\geq 1$. Thus by Lemma 3.6, there holds
$f_{1}(0)=P_{c_{1}}$ and $f_{1}(1)$ belongs to one of five cases in Lemma 3.6,
which contradicts (3.40) in Lemma 3.3.

This completes the proof of Claim 2.

Claim 3: $c_{1}$ and $c_{2}$ are irrationally elliptic.

Proof. By (3.41) and (3.42), there holds $\Delta_{2}=0$. Then, together with
the fact that $r_{2,3}=r_{2,4}=0$ from (3.40), it follows from (3.41) and
(3.21) that

$\displaystyle 2N+1$ $\displaystyle\geq$ $\displaystyle
i(c_{2}^{2m_{2}})+\nu(c_{2}^{2m_{2}})$ $\displaystyle=$ $\displaystyle
2N+(p_{2,0}+p_{2,+}+q_{2,-}+q_{2,0}+r_{2,1}+2r_{2,5}-r_{2,2})$
$\displaystyle\geq$ $\displaystyle 2N-3,$ (3.83)

where (3.2) holds by the fact that
$p_{2,0}+p_{2,+}+q_{2,-}+q_{2,0}+r_{2,1}+2r_{2,5}\leq 2$ from (3.2) and
(3.41), and $r_{2,2}\geq 1$ from (3.41), and the equality in (3.83) holds if
and only if $r_{2,2}=3$.

On the other hand, by Claim 2, we have
$i(c_{2}^{2m_{2}})+\nu(c_{2}^{2m_{2}})\notin\\{2N-1,2N+1\\}$. Note that
$i(c_{2}^{2m_{2}})+\nu(c_{2}^{2m_{2}})=i(c_{2}^{2m^{\prime}_{2}})+\nu(c_{2}^{2m^{\prime}_{2}})=1\
({\rm mod}\ 2)$ by (3.39). Thus, by (3.2), we obtain
$i(c_{2}^{2m_{2}})+\nu(c_{2}^{2m_{2}})\leq 2N-3$, which together with (3.83)
implies $r_{2,2}=3$, i.e., $c_{2}$ is irrationally elliptic. By the symmetric
properties of $c_{1}$ and $c_{2}$ in Lemma 3.3 (or, more precisely, replacing
$N$ with $N^{\prime}$ in the above arguments), we conclude that $c_{1}$ is
also irrationally elliptic.

This completes the proof of Claim 3.

Claim 4: $c_{3}$ is non-hyperbolic.

Proof. Assume that $c_{3}$ is hyperbolic, which, together with the assumption
(TCG) and Claim 3, implies the Finsler metric $F$ on $S^{4}$ is bumpy. Then by
Theorem 1.1 in [DLW16], we know that there exist at least four distinct non-
hyperbolic prime closed geodesics, which contradicts the assumption (TCG).
Thus by $c_{3}$ is non-hyperbolic.

Therefore, by the assumption (TCG), Claims 3 and 4 complete the proof of
Theorem 1.1.

##  4 Some further information about the third closed geodesic

In this section, under the assumption (TCG), we further study the third closed
geodesic $c_{3}$ and obtain some much precise information about it (cf.
Theorem 4.2 below). At first, we establish a similar result as Lemma 3.6.

Lemma 4.1. For some $k\in\\{1,2,3\\}$, assume that either $M_{2N-q\pm
1}(k,2m_{k}-2)\geq 1$ or $M_{2N+q\pm 1}(k,2m_{k}+2)\geq 1$ for some even
$q\in{\bf N}$, then there exists a continuous path $f_{k}\in
C([0,1],{\Omega}^{0}(P_{c_{k}}))$ such that $f_{k}(0)=P_{c_{k}}$ and
$f_{k}(1)$ belongs to one of the following cases:

(i) $I_{2p_{k,0}}\,{\diamond}\,(-I_{2q_{k,0}})\,{\diamond}\,H(2)$ with
$p_{k,0}+q_{k,0}=2$,

(ii)
$N_{1}(1,1)^{{\diamond}p_{k,-}}\,{\diamond}\,N_{1}(-1,-1)^{{\diamond}q_{k,+}}\,{\diamond}\,I_{2p_{k,0}}\,{\diamond}\,(-I_{2q_{k,0}})\,{\diamond}\,N_{1}(1,-1)^{{\diamond}p_{k,+}}\,{\diamond}\,N_{1}(-1,1)^{{\diamond}q_{k,-}}$
with $p_{k,-}+q_{k,+}=1$, $p_{k,0}+q_{k,0}=1$ and $p_{k,+}+q_{k,-}=1$,

(iii)
$I_{2p_{k,0}}\,{\diamond}\,(-I_{2q_{k,0}})\,{\diamond}\,N_{1}(1,-1)^{{\diamond}p_{k,+}}\,{\diamond}\,N_{1}(-1,1)^{{\diamond}q_{k,-}}$
with $p_{k,0}+q_{k,0}=2$ and $p_{k,+}+q_{k,-}=1$,

(iv)
$N_{1}(1,1)^{{\diamond}p_{k,-}}\,{\diamond}\,N_{1}(-1,-1)^{{\diamond}q_{k,+}}\,{\diamond}\,I_{2p_{k,0}}\,{\diamond}\,(-I_{2q_{k,0}})$
with $p_{k,0}+q_{k,0}=2$ and $p_{k,-}+q_{k,+}=1$,

(v) $I_{2p_{k,0}}\,{\diamond}\,(-I_{2q_{k,0}})$ with $p_{k,0}+q_{k,0}=3$.

And in either case, the index iteration formula of $c_{k}^{m}$ can be written
as follows:

$\displaystyle
i(c_{k}^{m})=\frac{q}{2}m-(p_{k,-}+p_{k,0})-\frac{1+(-1)^{m}}{2}(q_{k,0}+q_{k,+}).$
(4.1)

Proof. We only give the proof under the assumption $M_{2N+q\pm
1}(k,2m_{k}+2)\geq 1$. The proof under the assumption $M_{2N-q\pm
1}(k,2m_{k}-2)\geq 1$ is similar.

First, by Lemma 3.5 and the assumption $M_{2N+q\pm 1}(k,2m_{k}+2)\geq 1$ in
Lemma 4.1, we have

$\displaystyle i(c_{k}^{2m_{k}+2})\leq 2N+q-2,\quad
i(c_{k}^{2m_{k}+2})+\nu(c_{k}^{2m_{k}+2})\geq 2N+q+2,$ (4.2)

which, together with $\nu(c_{k}^{2m_{k}+2})=\nu(c_{k}^{2})$ by (2.22), implies
$\nu(c_{k}^{2})=p_{k,-}+2p_{k,0}+p_{k,+}+q_{k,-}+2q_{k,0}+q_{k,+}\in\\{4,5,6\\}$.

If $\nu(c_{k}^{2})=4$, by (4.2), we have $i(c_{k}^{2m_{k}+2})=2N+q-2$ and
$i(c_{k}^{2m_{k}+2})+\nu(c_{k}^{2m_{k}+2})=2N+q+2$. And by (3.2) we also have

$\displaystyle p_{k,-}+p_{k,0}+p_{k,+}+q_{k,-}+q_{k,0}+q_{k,+}\in\\{2,3\\}.$
(4.3)

Since $i(c_{k}^{2m_{k}+2})=2N+q-2\in 2{\bf N}$, we get $i(c_{k}^{2})\in 2{\bf
N}$ by (3.9). Note that by (2.17) and (3.2), we have

$\displaystyle i(c_{k}^{2})$ $\displaystyle=$ $\displaystyle
p_{k,-}+p_{k,0}+q_{k,0}+q_{k,+}+r_{k,1}+r_{k,2}\quad({\rm mod}\ 2)$ (4.4)
$\displaystyle=$ $\displaystyle 1+p_{k,+}+q_{k,-}+h_{k}\quad({\rm mod}\ 2).$

So, by (4.3) and (3.2), we get $p_{k,+}+q_{k,-}+h_{k}=1$ since
$\nu(c_{k}^{2})=4$. Therefore, if
$p_{k,-}+p_{k,0}+p_{k,+}+q_{k,-}+q_{k,0}+q_{k,+}=2$, we must have
$p_{k,0}+q_{k,0}=2$ and $h_{k}=1$, if
$p_{k,-}+p_{k,0}+p_{k,+}+q_{k,-}+q_{k,0}+q_{k,+}=3$, we must have
$p_{k,0}+q_{k,0}=1$, $p_{k,+}+q_{k,-}=1$ and $p_{k,-}+q_{k,+}=1$.

If $\nu(c_{k}^{2})=5$, we have
$p_{k,-}+p_{k,0}+p_{k,+}+q_{k,-}+q_{k,0}+q_{k,+}=3$. Then we must have
$p_{k,0}+q_{k,0}=2$ and $p_{k,-}+q_{k,+}+p_{k,+}+q_{k,-}=1$. And by (3.9) and
(4.4) we have $i(c_{k}^{2m_{k}+2})=2N+q-2$ when $p_{k,+}+q_{k,-}=1$,
$i(c_{k}^{2m_{k}+2})=2N+q-3$ when $p_{k,-}+q_{k,+}=1$.

If $\nu(c_{k})=6$, we must have $p_{k,0}+q_{k,0}=3$, and then we have
$i(c_{k}^{2m_{k}+2})=2N+q-3$ by (4.4) and (3.9).

Note that by (3.9), we get $i(c_{k}^{2m_{k}+2})=2N+i(c_{k}^{2})$, and in
either case, we have

$\displaystyle i(c_{k}^{2})=q-p_{k,-}-p_{k,0}-q_{k,+}-q_{k,0}.$ (4.5)

Then by (2.17) and according to the precise cases in Lemma 4.1, we have

$\displaystyle i(c_{k}^{2})=2i(c_{k})+p_{k,-}+p_{k,0}-q_{k,0}-q_{k,+}.$ (4.6)

Combining (2.17), (4.5) and (4.6), it yields

$i(c_{k}^{m})=\frac{q}{2}m-(p_{k,-}+p_{k,0})-\frac{1+(-1)^{m}}{2}(q_{k,0}+q_{k,+}).$

This completes the proof of Lemma 4.1.

Theorem 4.2. For every Finsler metric F on $S^{4}$ with reversibility
${\lambda}$ and flag curvature $K$ satisfying
$\frac{25}{9}(\frac{{\lambda}}{{\lambda}+1})^{2}<K\leq 1$ with
$\lambda<\frac{3}{2}$, suppose that there exist precisely three prime closed
geodesics $c_{1}$, $c_{2}$ and $c_{3}$, then both $c_{1}$ and $c_{2}$ are
irrationally elliptic with $i(c_{1})=3$ and $i(c_{2})=9$, and $c_{3}$ is non-
hyperbolic and must belong to one of the following precise classes:

(i) $i(c_{3})=3$ and $P_{c_{3}}\approx N_{1}(1,1)\,{\diamond}\,I_{4}$,

(ii) $i(c_{3})=3$ and $P_{c_{3}}\approx I_{6}$,

(iii) $i(c_{3})=4$ and $P_{c_{3}}\approx
N_{1}(1,1)\,{\diamond}\,I_{2}\,{\diamond}\,N_{1}(1,-1)$,

(iv) $i(c_{3})=4$ and $P_{c_{3}}\approx I_{4}\,{\diamond}\,N_{1}(1,-1)$,

(v) $i(c_{3})=4$ and $P_{c_{3}}\approx I_{4}\,{\diamond}\,H(2)$,

where and below, “$P_{c_{k}}\approx A$” means that there exists a continuous
path $f_{k}\in C([0,1],{\Omega}^{0}(P_{c_{k}}))$ such that
$f_{k}(0)=P_{c_{k}}$ and $f_{k}(1)=A$ in Theorem 2.6.

Proof. Under the assumption (TCG), it follows from Theorem 1.1 that both
$c_{1}$ and $c_{2}$ are irrationally elliptic. Then there holds
$P_{c_{k}}\approx
R(\tilde{\theta}_{k,1})\,{\diamond}\,R(\tilde{\theta}_{k,2})\,{\diamond}\,R(\tilde{\theta}_{k,3})$
for some
$\frac{\tilde{\theta}_{k,1}}{2\pi},\frac{\tilde{\theta}_{k,2}}{2\pi},\frac{\tilde{\theta}_{k,3}}{2\pi}\in(0,1)\setminus{\bf
Q}$ for $k=1,2$, respectively.

Then by (2.17), we have

$\displaystyle i(c_{k}^{m})$ $\displaystyle=$ $\displaystyle
m\left(i(c_{k})-3\right)+2\sum_{j=1}^{3}E\left(\frac{m\tilde{\theta}_{k,j}}{2\pi}\right)-3,\quad\nu(c_{k}^{m})=0,\quad
k=1,2,$ (4.7) $\displaystyle\hat{i}(c_{k})$ $\displaystyle=$ $\displaystyle
i(c_{k})-3+\sum_{j=1}^{3}\frac{\tilde{\theta}_{k,j}}{\pi},\quad k=1,2,$ (4.8)

and then by (2.6), we get

$\displaystyle n(c_{k})=1,\quad k=1,2.$ (4.9)

By (2.5), (iii) of Lemma 2.2, (4.7) and (4.9), for the average Euler numbers
of $c_{1}$ and $c_{2}$ we have

$\displaystyle\hat{\chi}(c_{1})=-1,\quad\hat{\chi}(c_{2})=-1.$ (4.10)

Noticing that $\nu(c_{k}^{m})=0,\,\forall\ m\in{\bf N}$ and $k=1,2$, so by
Lemma 2.1, (2.3), and (iii) of Lemma 2.2, we have

$\displaystyle M_{q}(k,m)=\left\\{\matrix{1,&&\quad{\it if}\quad
q=i(c_{k}^{m}),\cr 0,&&\quad{\it if}\quad q\neq
i(c_{k}^{m}),\cr}\right.\quad\mbox{for}\ k=1,2,\ m\in{\bf N}.$ (4.11)

By Claim 3 of the proof of Theorem 1.1, (4.7) and (4.11), we have

$\displaystyle
i(c_{2}^{2m_{2}})+\nu(c_{2}^{2m_{2}})=i(c_{2}^{2m_{2}})=2N-3,\quad
M_{2N-3}(2,2m_{2})=1.$ (4.12)

By (3.38) and (4.7), we have

$\displaystyle
i(c_{1}^{2m_{1}})+\nu(c_{1}^{2m_{1}})=i(c_{1}^{2m_{1}})=2N+3,\quad
M_{2N+3}(1,2m_{1})=1.$ (4.13)

It follows from (4.11)-(4.13) and Lemma 3.4 that

$\displaystyle M_{q}=\sum_{1\leq k\leq
3}M_{q}(k,2m_{k})=M_{q}(3,2m_{3})\quad\mbox{for}\ 2N-2\leq q\leq 2N+2$ (4.14)

and

$\displaystyle M_{2N+3}(2,2m_{2})=0.$ (4.15)

By (2.7), Theorem 2.5 and (4.14), we have $M_{2N-1}=M_{2N-1}(3,2m_{3})\geq
b_{2N-1}=1$ and $M_{2N+1}=M_{2N+1}(3,2m_{3})\geq b_{2N+1}=1$, then by (3.16),
(3.17) and Lemma 3.5, there holds

$\displaystyle M_{2N-3}(3,2m_{3})=M_{2N+3}(3,2m_{3})=0.$ (4.16)

By (4.13), (4.15), (4.16) and Lemma 3.4, we have

$\displaystyle M_{2N+3}=\sum_{1\leq k\leq 3}M_{2N+3}(k,2m_{k}+1)+1.$ (4.17)

By (3.18) and Lemma 3.5, $M_{2N+3}(k,2m_{k}+1)\leq 1$ for $k=1,2,3$. Then it
follows from (4.17) that $M_{2N+3}\leq 4$. On the other hand, by (2.7) and
Theorem 2.5, we have $M_{2N+3}\geq b_{2N+3}=2$. Thus $M_{2N+3}\in\\{2,3,4\\}$.

We continue the proof by distinguishing three cases.

Case 1: $M_{2N+3}=4$.

In this case, by (4.17), it yields

$\displaystyle M_{2N+3}(k,2m_{k}+1)=1,\quad\forall\ k=1,2,3.$ (4.18)

Then by (3.18) and Lemma 3.5, we have

$\displaystyle M_{2N+5}(k,2m_{k}+1)=0,\quad\forall\ k=1,2,3.$ (4.19)

It follows from (4.19) and Lemma 3.4 that $M_{2N+5}=0$. However by (2.7) and
Theorem 2.5, we have $M_{2N+5}\geq b_{2N+5}=1$, which is a contradiction.

Case 2: $M_{2N+3}=3$.

In this case, by (4.17), one of the following cases may happen.

(i) $M_{2N+3}(3,2m_{3}+1)=0$ and
$M_{2N+3}(1,2m_{1}+1)=M_{2N+3}(2,2m_{2}+1)=1$.

(ii) $M_{2N+3}(3,2m_{3}+1)=1$ and
$M_{2N+3}(1,2m_{1}+1)+M_{2N+3}(2,2m_{2}+1)=1$.

For (i), by (4.11), we have $i(c_{1}^{2m_{1}+1})=2N+3$ and
$i(c_{2}^{2m_{2}+1})=2N+3$, then by (3.9), it yields $i(c_{1})=i(c_{2})=3$.
So, by (3.4) and (4.8), we get

$\displaystyle 5<\hat{i}(c_{1})<6,\quad 5<\hat{i}(c_{2})<6.$ (4.20)

By (3.18), Lemma 3.5 and (i) of Case 2, there holds

$\displaystyle M_{q}(k,2m_{k}+1)=0\quad\mbox{for}\ q\neq 2N+3\ \mbox{and}\
k=1,2.$ (4.21)

So, according to Lemma 3.4 and (4.21), we have

$\displaystyle M_{2N+5}$ $\displaystyle=$ $\displaystyle\sum_{1\leq k\leq
3}M_{2N+5}(k,2m_{k}+1)=M_{2N+5}(3,2m_{3}+1),$ (4.22) $\displaystyle M_{2N+7}$
$\displaystyle=$ $\displaystyle\sum_{1\leq k\leq
3}M_{2N+7}(k,2m_{k}+1)=M_{2N+7}(3,2m_{3}+1).$ (4.23)

By (2.7) and Theorem 2.5, we have $M_{2N+5}\geq b_{2N+5}=1$ and $M_{2N+7}\geq
b_{2N+7}=1$, then it follows from (4.22) and (4.23) that

$\displaystyle M_{2N+5}(3,2m_{3}+1)\geq 1,\quad M_{2N+7}(3,2m_{3}+1)\geq 1.$
(4.24)

Thus the assupmtion with $q=6$ in Lemma 3.6 is satisfied, and then by (3.52)
we have

$\displaystyle i(c_{3}^{m})=6m-p_{3,-}-p_{3,0}.$ (4.25)

Then we have $n(c_{3})=1$ in either case of $P_{c_{3}}$ by (2.6). By (4.25),
we have

$\displaystyle\hat{i}(c_{3})=6.$ (4.26)

By (2.4), we have the following identity

$\displaystyle\sum_{k=1}^{3}\frac{\hat{\chi}(c_{k})}{\hat{i}(c_{k})}=B(4,1)=-\frac{2}{3}.$
(4.27)

Combining (4.10), (4.20), (4.26) and (4.27), we obtain
$-2<\hat{\chi}(c_{3})=-4+\frac{6}{\hat{i}(c_{1})}+\frac{6}{\hat{i}(c_{2})}<-\frac{8}{5}$,
which contradicts to $\hat{\chi}(c_{3})\in{\bf Z}$, where the latter is due to
$n(c_{3})=1$ and the definition of $\hat{\chi}(c_{3})$.

For (ii), without loss of generality, we assume that $M_{2N+3}(1,2m_{1}+1)=0$
and $M_{2N+3}(2,2m_{1}+1)=1$. Thus, similarly, by (3.18) and Lemma 3.5 and
Lemma 3.4, we have

$\displaystyle M_{2N+5}=M_{2N+5}(1,2m_{1}+1),\quad
M_{2N+7}=M_{2N+7}(1,2m_{1}+1).$ (4.28)

By (2.7) and Theorem 2.5, we have $M_{2N+5}\geq b_{2N+5}=1$ and $M_{2N+7}\geq
b_{2N+7}=1$. Then it follows from (4.28) that $M_{2N+5}(1,2m_{1}+1)\geq 1$ and
$M_{2N+7}(1,2m_{1}+1)\geq 1$. Thus by Lemma 3.6, there are five cases for
$P_{c_{1}}$, which contradicts the fact that $c_{1}$ is irrationally elliptic.

Case 3: $M_{2N+3}=2$.

In this case, by (4.17), one of the following cases may happen:

(i) $M_{2N+3}(3,2m_{3}+1)=1$ and
$M_{2N+3}(1,2m_{1}+1)=M_{2N+3}(2,2m_{2}+1)=0$.

(ii) $M_{2N+3}(3,2m_{3}+1)=0$ and
$M_{2N+3}(1,2m_{1}+1)+M_{2N+3}(2,2m_{2}+1)=1$.

For (i), by (3.18) and Lemma 3.5, there holds

$\displaystyle M_{q}(3,2m_{3}+1)=0\quad\mbox{for}\ q\neq 2N+3,$ (4.29)

and then by Lemma 3.4, for $q=2N+5,2N+7$, we have

$\displaystyle M_{q}=\sum_{1\leq k\leq
3}M_{q}(k,2m_{k}+1)=M_{q}(1,2m_{1}+1)+M_{q}(2,2m_{2}+1).$ (4.30)

On the other hand, by (2.7) and Theorem 2.5, we have $M_{2N+5}\geq b_{2N+5}=1$
and $M_{2N+7}\geq b_{2N+7}=1$, which, together with (4.11), implies

$\displaystyle M_{q}(1,2m_{1}+1)=\left\\{\matrix{1,\quad{\it if}\quad
q=2N+5,\cr 0,\quad{\it if}\quad q\neq 2N+5,\cr}\right.\quad
M_{q}(2,2m_{2}+1)=\left\\{\matrix{1,\quad{\it if}\quad q=2N+7,\cr 0,\quad{\it
if}\quad q\neq 2N+7,\cr}\right.$ (4.31)

or

$\displaystyle M_{q}(1,2m_{1}+1)=\left\\{\matrix{1,\quad{\it if}\quad
q=2N+7,\cr 0,\quad{\it if}\quad q\neq 2N+7,\cr}\right.\quad
M_{q}(2,2m_{2}+1)=\left\\{\matrix{1,\quad{\it if}\quad q=2N+5,\cr 0,\quad{\it
if}\quad q\neq 2N+5.\cr}\right.$ (4.32)

So it follows from Lemma 3.4, (4.29) and (4.31)-(4.32) that

$\displaystyle M_{2N+9}=\sum_{1\leq k\leq 3,\atop 1\leq m\leq
2}M_{2N+9}(k,2m_{k}+m)=\sum_{1\leq k\leq 3}M_{2N+9}(k,2m_{k}+2).$ (4.33)

Without loss of generality, we assume that (4.31) holds. So we get
$i(c_{1}^{2m_{1}+1})=2N+5$ and $i(c_{2}^{2m_{2}+1})=2N+7$ by (4.11). Then by
(3.9), it yields $i(c_{1})=5$ and $i(c_{2})=7$.

Since $i(c_{2})=7$, by (4.7), we have the index iteration formula of $c_{2}$
as follows

$\displaystyle
i(c_{2}^{m})=4m-3+2\sum_{j=1}^{3}E\left(\frac{m\tilde{\theta}_{2,j}}{2\pi}\right).$
(4.34)

By (4.34), it yields $i(c_{2}^{2})\geq 11$, then by (3.9), we get
$i(c_{2}^{2m_{2}+2})\geq 2N+11$. Thus by (4.11), we have

$\displaystyle M_{2N+9}(2,2m_{2}+2)=0.$ (4.35)

By (3.18) and Lemma 3.5, we have

$\displaystyle M_{2N+9}(k,2m_{k}+2)\leq 1,\quad\forall\ k=1,2,3,$ (4.36)

which, together with (4.33) and (4.35), implies $M_{2N+9}\leq 2$. However by
(2.7) and Theorem 2.5, we have $M_{2N+9}\geq b_{2N+9}=2$, which implies
$M_{2N+9}=2$. So we have

$\displaystyle M_{2N+9}(1,2m_{1}+2)=M_{2N+9}(3,2m_{3}+2)=1.$ (4.37)

It follows from (4.11) and (4.37) that

$\displaystyle M_{q}(1,2m_{1}+2)=M_{q}(3,2m_{3}+2)=0\quad\mbox{for}\
q=2N+11,2N+13.$ (4.38)

Combining (4.29), (4.31), (4.38) and Lemma 3.4, we obtain

$\displaystyle M_{q}=\sum_{1\leq k\leq 3,\atop 1\leq m\leq
2}M_{q}(k,2m_{k}+m)=M_{q}(2,2m_{2}+2)\quad\mbox{for}\ q=2N+11,2N+13.$ (4.39)

Then by (4.11), we obtain that $M_{2N+11}=0$ or $M_{2N+13}=0$, which
contradicts to $M_{2N+11}\geq b_{2N+11}=1$ and $M_{2N+13}\geq b_{2N+13}=1$ by
(2.7) and Theorem 2.5.

For (ii), without loss of generality, we assume that

$\displaystyle M_{2N+3}(2,2m_{2}+1)=0,\quad M_{2N+3}(1,2m_{1}+1)=1.$ (4.40)

By (4.11), we have $i(c_{1}^{2m_{1}+1})=2N+3$ and

$\displaystyle M_{q}(1,2m_{1}+1)=0,\quad\forall\ q\neq 2N+3.$ (4.41)

Then by (3.9), it yields $i(c_{1})=3$. Then by (4.7), we obtain

$\displaystyle
i(c_{1}^{m})=2\sum_{j=1}^{3}E\left(\frac{m\tilde{\theta}_{1,j}}{2\pi}\right)-3.$
(4.42)

By (4.42), it yields $i(c_{1}^{2})\leq 9$. Then by (3.3), we get
$i(c_{1}^{2})=9$. And then $i(c_{1}^{2m_{1}+2})=2N+9$ by (3.9). Thus by
(4.11), we have

$\displaystyle M_{q}(1,2m_{1}+2)=\left\\{\matrix{1,\quad{\it if}\quad
q=2N+9,\cr 0,\quad{\it if}\quad q\neq 2N+9.\cr}\right.$ (4.43)

Claim 1: $i(c_{2}^{2m_{2}+1})\leq 2N+9$, or equivalently, $i(c_{2})\leq 9$ by
(3.9).

If $i(c_{2}^{2m_{2}+1})>2N+9$, by (3.5) we get $i(c_{2}^{2m_{2}+2})\geq
i(c_{2}^{2m_{2}+1})>2N+9$. Then by (4.11) we know

$\displaystyle M_{q}(2,2m_{2}+1)=M_{q}(2,2m_{2}+2)=0\quad\mbox{for}\
q=2N+5,2N+7,2N+9.$ (4.44)

Combining Lemma 3.4, (4.41) and (4.44), we have

$\displaystyle M_{2N+5}$ $\displaystyle=$ $\displaystyle\sum_{1\leq k\leq
3}M_{2N+5}(k,2m_{k}+1)=M_{2N+5}(3,2m_{3}+1),$ (4.45) $\displaystyle M_{2N+7}$
$\displaystyle=$ $\displaystyle\sum_{1\leq k\leq
3}M_{2N+7}(k,2m_{k}+1)=M_{2N+7}(3,2m_{3}+1).$ (4.46)

On the other hand, by (2.7) and Theorem 2.5, we have $M_{2N+5}\geq b_{2N+5}=1$
and $M_{2N+7}\geq b_{2N+7}=1$. Then by (4.45) and (4.46), we know that

$\displaystyle M_{2N+5}(3,2m_{3}+1)\geq 1,\quad M_{2N+7}(3,2m_{3}+1)\geq 1.$
(4.47)

Thus the assumption with $q=6$ in Lemma 3.6 is satisfied, and then we have the
index iteration formula of $c_{3}$ as follows

$\displaystyle i(c_{3}^{m})=6m-p_{3,-}-p_{3,0}.$ (4.48)

Then by the fact $\nu(c_{3})=p_{3,-}+2p_{3,0}+p_{3,+}$ and (3.2), we get

$\displaystyle i(c_{3})+\nu(c_{3})=6+p_{3.0}+p_{3,+}\leq 9.$ (4.49)

Then by (3.9) and (2.22), it yields
$i(c_{3}^{2m_{3}+1})+\nu(c_{3}^{2m_{3}+1})\leq 2N+9$. So, according to Lemma
3.5 and (4.47), there holds

$\displaystyle M_{2N+9}(3,2m_{3}+1)=0.$ (4.50)

Similar to (3.72), we obtain

$\displaystyle M_{2N+9}(3,2m_{3}+2)=M_{2N+3}(3,2m_{3}+1)=0.$ (4.51)

Combining (4.41), (4.43), (4.44), (4.50), (4.51) and Lemma 3.4, we obtain that

$\displaystyle M_{2N+9}=\sum_{1\leq k\leq 3,\atop 1\leq m\leq
2}M_{2N+9}(k,2m_{k}+m)=M_{2N+9}(1,2m_{1}+2)=1,$ (4.52)

which gives a contradiction $1=M_{2N+9}\geq b_{2N+9}=2$ by (2.7) and Theorem
2.5. This finished the proof of Claim 1.

Note that $i(c_{2})\neq 3$ by (4.40), (4.11) and (3.9), it yields
$i(c_{2})\in\\{5,7,9\\}$ by Claim 1 since $i(c_{2})$ is odd. Next we have
three subcases according to the value of $i(c_{2})$.

Subcase 3.1: $i(c_{2})=5$.

In this subcase, by (4.11) and (3.9), we have

$\displaystyle M_{q}(2,2m_{2}+1)=\left\\{\matrix{1,\quad{\it if}\quad
q=2N+5,\cr 0,\quad{\it if}\quad q\neq 2N+5.\cr}\right.$ (4.53)

By Lemma 3.4, (4.41) and (4.53), we obtain that

$\displaystyle M_{2N+7}=\sum_{1\leq k\leq
3}M_{2N+7}(k,2m_{k}+1)=M_{2N+7}(3,2m_{3}+1).$ (4.54)

Together with $M_{2N+7}\geq b_{2N+7}=1$ by (2.7) and Theorem 2.5, we get

$\displaystyle M_{2N+7}(3,2m_{3}+1)\geq 1.$ (4.55)

Claim 2: $M_{q}(3,2m_{3}+1)=0$ for $q\geq 2N+11$.

If $M_{q_{0}}(3,2m_{3}+1)\geq 1$ for some $q_{0}\geq 2N+11$, then by (4.55)
and Lemma 3.5, we know that $i(c_{3}^{2m_{3}+1})\leq 2N+6$ and
$i(c_{3}^{2m_{3}+1})+\nu(c_{3}^{2m_{3}+1})\geq q_{0}+1\geq 2N+12$, which,
together with the fact $\nu(c_{3}^{2m_{3}+1})\leq 6$, implies
$i(c_{3}^{2m_{3}+1})=2N+6$ and $\nu(c_{3}^{2m_{3}+1})=6$ and $q_{0}$ only can
be $2N+11$. Now by (3.9) and (2.22), we have $i(c_{3})=6$ and $\nu(c_{3})=6$,
which implies that $P_{c_{3}}\approx I_{6}$ by (3.1), (3.2) and the fact
$\nu(c_{3})=p_{3,-}+2p_{3,0}+p_{3,+}$. Then $i(c_{3})$ must be odd by
Proposition 2.7 and the symplectic additivity of symplectic paths. This
contradicts to $i(c_{3})=6$ and completes the proof of Claim 2.

In summary, by Lemma 3.4, (4.41), (4.53), Claim 2 and (4.43), there holds

$\displaystyle M_{q}=\sum_{1\leq k\leq
3}M_{q}(k,2m_{k}+1)=M_{q}(3,2m_{3}+1)\quad\mbox{for}\ 2N+4\leq q\leq 2N+8,\
q\neq 2N+5,$ (4.56) $\displaystyle M_{2N+5}=\sum_{1\leq k\leq
3}M_{2N+5}(k,2m_{k}+1)=1+M_{2N+5}(3,2m_{3}+1),$ (4.57) $\displaystyle
M_{2N+9}=\sum_{1\leq k\leq 3,\atop 1\leq m\leq
2}M_{2N+9}(k,2m_{k}+m)=1+M_{2N+9}(3,2m_{3}+1)+\sum_{2\leq k\leq
3}M_{2N+9}(k,2m_{k}+2),$ (4.58) $\displaystyle M_{2N+10}=\sum_{1\leq k\leq
3,\atop 1\leq m\leq 2}M_{2N+10}(k,2m_{k}+m)=M_{2N+10}(3,2m_{3}+1)+\sum_{2\leq
k\leq 3}M_{2N+10}(k,2m_{k}+2),$ (4.59) $\displaystyle M_{q}=\sum_{1\leq k\leq
3,\atop 1\leq m\leq 2}M_{q}(k,2m_{k}+m)=M_{q}(2,2m_{2}+2)+M_{q}(3,2m_{3}+2),\
2N+11\leq q\leq 2N+14.$ (4.60)

Note that by (4.7) and $i(c_{2})=5$, we have the index iteration formula of
$c_{2}$ as follows

$\displaystyle
i(c_{2}^{m})=2m-3+2\sum_{j=1}^{3}E\left(\frac{m\tilde{\theta}_{2,j}}{2\pi}\right).$
(4.61)

It follows from (4.61) that $i(c_{2}^{2})\leq 13$. Then by (3.3) and the fact
that $i(c_{2}^{2})$ is odd, we get $i(c_{2}^{2})\in\\{9,11,13\\}$.

We continue the proof by distinguishing three values of $i(c_{2}^{2})$.

Subcase 3.1.1: $i(c_{2}^{2})=9$.

In this subcase, by (3.9), we have $i(c_{2}^{2m_{2}+2})=2N+9$, then by (4.11),
we have

$\displaystyle M_{q}(2,2m_{2}+2)=\left\\{\matrix{1,\quad{\it if}\quad
q=2N+9,\cr 0,\quad{\it if}\quad q\neq 2N+9.\cr}\right.$ (4.62)

It follows from (4.60) and (4.62) that

$\displaystyle M_{2N+11}=M_{2N+11}(3,2m_{3}+2),\quad
M_{2N+13}=M_{2N+13}(3,2m_{3}+2),$ (4.63)

which, together with $M_{2N+11}\geq b_{2N+11}=1$ and $M_{2N+13}\geq
b_{2N+13}=1$ by (2.7) and Theorem 2.5, implies

$\displaystyle M_{2N+11}(3,2m_{3}+2)\geq 1,\quad M_{2N+13}(3,2m_{3}+2)\geq 1.$
(4.64)

Thus the assumption with $q=12$ in Lemma 4.1 is satisfied, and then we have

$\displaystyle i(c_{3}^{m})$ $\displaystyle=$ $\displaystyle
6m-(p_{3,-}+p_{3,0})-\frac{1+(-1)^{m}}{2}(q_{3,0}+q_{3,+}),$ (4.65)
$\displaystyle\nu(c_{3}^{m})$ $\displaystyle=$ $\displaystyle
p_{3,-}+2p_{3,0}+p_{3,+}+\frac{1+(-1)^{m}}{2}(q_{3,-}+2q_{3,0}+q_{3,+}).$
(4.66)

Then we can know that

$\displaystyle\hat{i}(c_{3})=6$ (4.67)

and

$\displaystyle n(c_{3})\in\\{1,2\\}.$ (4.68)

By (3.2), we obtain
$i(c_{3}^{2})+\nu(c_{3}^{2})=12+p_{3,0}+p_{3,+}+q_{3,-}+q_{3,0}\leq 15$, which
implies

$\displaystyle i(c_{3}^{2m_{3}+2})+\nu(c_{3}^{2m_{3}+2})\leq 2N+15$ (4.69)

by (3.9) and (2.22). Then by Lemma 3.5 and (4.64), there holds

$\displaystyle M_{2N+15}(3,2m_{3}+2)=0,\quad M_{2N+17}(3,2m_{3}+2)=0.$ (4.70)

It follows from Lemma 3.4, (4.41), (4.43), (4.53), Claim 2, (4.62) and (4.70)
that

$\displaystyle M_{2N+15}$ $\displaystyle=$ $\displaystyle\sum_{1\leq k\leq
3,\atop 1\leq m\leq 3}M_{2N+15}(k,2m_{k}+m)=\sum_{1\leq k\leq
3}M_{2N+15}(k,2m_{k}+3),$ (4.71) $\displaystyle M_{2N+17}$ $\displaystyle=$
$\displaystyle\sum_{1\leq k\leq 3,\atop 1\leq m\leq
3}M_{2N+17}(k,2m_{k}+m)=\sum_{1\leq k\leq 3}M_{2N+17}(k,2m_{k}+3).$ (4.72)

By (3.18) and Lemma 3.5, we have

$\displaystyle M_{2N+15}(k,2m_{k}+3)\leq 1,\quad\forall\ k=1,2,3.$ (4.73)

Then by (4.71) we get $M_{2N+15}\leq 3$. We claim that $M_{2N+15}\neq 3$. In
fact, if $M_{2N+15}=3$, we have $M_{2N+15}(k,2m_{k}+3)=1,\,k=1,2,3$. Then by
(3.18) and Lemma 3.5, there holds $M_{2N+17}(k,2m_{k}+3)=0,\,k=1,2,3$, which,
together with (4.72), implies $M_{2N+17}=0$. This gives a contradiction
$0=M_{2N+17}\geq b_{2N+17}=1$ by (2.7) and Theroem 2.5. Hence $M_{2N+15}\leq
2$. However, again by (2.7) and Theroem 2.5, we have $M_{2N+15}\geq
b_{2N+15}=2$. So we get $M_{2N+15}=2$.

In summary, in Case 3, we have

$\displaystyle M_{2N+3}=b_{2N+3},\quad M_{2N+15}=b_{2N+15}.$ (4.74)

Combining (4.74), (2.7) and Theorem 2.5, we obtain that

$\displaystyle\sum_{q=2N+4}^{2N+14}(-1)^{q}M_{q}=\sum_{q=2N+4}^{2N+14}(-1)^{q}b_{q}=-6.$
(4.75)

By (4.56)-(4.60), (ii) of Case 3, Claim 2, (4.62) and (4.70), we have

$\displaystyle\sum_{q=2N+4}^{2N+14}(-1)^{q}M_{q}$ $\displaystyle=$
$\displaystyle-2+\sum_{q=2N+4}^{2N+10}(-1)^{q}M_{q}(3,2m_{3}+1)+\sum_{2N+9}^{2N+14}(-1)^{q}\sum_{2\leq
k\leq 3}M_{q}(k,2m_{k}+2)$ (4.76) $\displaystyle=$
$\displaystyle-3+\sum_{q=2N+3}^{2N+15}(-1)^{q}M_{q}(3,2m_{3}+1)+\sum_{2N+9}^{2N+15}(-1)^{q}M_{q}(3,2m_{3}+2).$

Note that $2N+3\leq i(c_{3}^{2m_{3}+1})\leq
i(c_{3}^{2m_{3}+1})+\nu(c_{3}^{2m_{3}+1})\leq 2N+15$ and $2N+9\leq
i(c_{3}^{2m_{3}+2})\leq i(c_{3}^{2m_{3}+2})+\nu(c_{3}^{2m_{3}+2})\leq 2N+15$
by (3.18), (3.5) and (4.69), then according to Lemma 2.1 and (2.3), we obtain

$\displaystyle\sum_{q=2N+3}^{2N+15}(-1)^{q}M_{q}(3,2m_{3}+1)+\sum_{2N+9}^{2N+15}(-1)^{q}M_{q}(3,2m_{3}+2)$
(4.77) $\displaystyle=$ $\displaystyle\sum_{0\leq l\leq
6}(-1)^{i(c_{3}^{2m_{3}+1})+l}k_{l}^{{\epsilon}(c_{3}^{2m_{3}+1})}(c_{3}^{2m_{3}+1})+\sum_{0\leq
l\leq
6}(-1)^{i(c_{3}^{2m_{3}+2})+l}k_{l}^{{\epsilon}(c_{3}^{2m_{3}+2})}(c_{3}^{2m_{3}+2}).$

Combining (4.75), (4.76) and (4.77), by (2.5), we get

$\displaystyle\chi(c_{3}^{2m_{3}+1})+\chi(c_{3}^{2m_{3}+2})$ $\displaystyle=$
$\displaystyle\sum_{0\leq l\leq
6}(-1)^{i(c_{3}^{2m_{3}+1})+l}k_{l}^{{\epsilon}(c_{3}^{2m_{3}+1})}(c_{3}^{2m_{3}+1})$
(4.78) $\displaystyle+\sum_{0\leq l\leq
6}(-1)^{i(c_{3}^{2m_{3}+2})+l}k_{l}^{{\epsilon}(c_{3}^{2m_{3}+2})}(c_{3}^{2m_{3}+2})=-3.$

Note that by (4.68) and (iv) of Lemma 2.2, there holds

$\displaystyle
k_{j}^{{\epsilon}(c_{3}^{2m_{3}+1})}(c_{3}^{2m_{3}+1})=k_{j}^{{\epsilon}(c_{3})}(c_{3}),\quad
k_{j}^{{\epsilon}(c_{3}^{2m_{3}+2})}(c_{3}^{2m_{3}+2})=k_{j}^{{\epsilon}(c_{3}^{2})}(c_{3}^{2}),\quad\forall\
0\leq j\leq 6.$ (4.79)

Then by (2.5) it yields

$\displaystyle\chi(c_{3})=\chi(c_{3}^{2m_{3}+1}),\quad\chi(c_{3}^{2})=\chi(c_{3}^{2m_{3}+2}),$
(4.80)

which, together with (4.78) and (4.68), implies

$\displaystyle\chi(c_{3})\neq\chi(c_{3}^{2})$ (4.81)

since $\chi(c_{3}^{m})\in{\bf Z}$, and

$\displaystyle n(c_{3})=2.$ (4.82)

It follows from (2.5), (4.78), (4.80) and (4.82) that

$\displaystyle\hat{\chi}(c_{3})=\frac{1}{2}\left(\chi(c_{3})+\chi(c_{3}^{2})\right)=\frac{1}{2}\left(\chi(c_{3}^{2m_{3}+1})+\chi(c_{3}^{2m_{3}+2})\right)=-\frac{3}{2}.$
(4.83)

Note that $\hat{i}(c_{k})>5$, $k=1,2$ by (3.4), so it follows from (4.10),
(4.67) and (4.83) that

$\displaystyle\frac{\hat{\chi}(c_{1})}{\hat{i}(c_{1})}=-\frac{1}{\hat{i}(c_{1})}>-\frac{1}{5},\quad\frac{\hat{\chi}(c_{2})}{\hat{i}(c_{2})}=-\frac{1}{\hat{i}(c_{2})}>-\frac{1}{5},\quad\frac{\hat{\chi}(c_{3})}{\hat{i}(c_{3})}=-\frac{1}{4},$
(4.84)

which, together with (2.4), yields

$\displaystyle-\frac{2}{3}=\frac{\hat{\chi}(c_{1})}{\hat{i}(c_{1})}+\frac{\hat{\chi}(c_{2})}{\hat{i}(c_{2})}+\frac{\hat{\chi}(c_{3})}{\hat{i}(c_{3})}>-\frac{13}{20}.$
(4.85)

This is a contradiction.

Subcase 3.1.2: $i(c_{2}^{2})=11$.

In this subcase, by (3.9), it yields $i(c_{2}^{2m_{2}+2})=2N+11$ and then by
(4.11), we have

$\displaystyle M_{q}(2,2m_{2}+2)=\left\\{\matrix{1,\quad{\it if}\quad
q=2N+11,\cr 0,\quad{\it if}\quad q\neq 2N+11,\cr}\right.$ (4.86)

which, together with (4.60), implies

$\displaystyle M_{2N+13}=M_{2N+13}(3,2m_{3}+2).$ (4.87)

However by (2.7) and Theorem 2.5, we have $M_{2N+13}\geq b_{2N+13}=1$, so we
have

$\displaystyle M_{2N+13}(3,2m_{3}+2)\geq 1.$ (4.88)

Thus by (3.18) and Lemma 3.5, there holds

$\displaystyle M_{2N+9}(3,2m_{3}+2)=0.$ (4.89)

Comnining (4.58), (4.86) and (4.89), we obtain that

$\displaystyle M_{2N+9}=1+M_{2N+9}(3,2m_{3}+1).$ (4.90)

Since $M_{2N+9}\geq b_{2N+9}=2$ by (2.7) and Theorem 2.5, by (4.90), we know
that

$\displaystyle M_{2N+9}(3,2m_{3}+1)\geq 1.$ (4.91)

Noticing that we also have $M_{2N+7}(3,2m_{3}+1)\geq 1$ by (4.55), then by
Lemma 3.6, we obtain that

$\displaystyle i(c_{3}^{m})=8m-p_{3,-}-p_{3,0}.$ (4.92)

and

$\displaystyle n(c_{3})=1.$ (4.93)

Similar to (3.72), we have

$\displaystyle M_{2N+5}(3,2m_{3}+1)$ $\displaystyle=$ $\displaystyle
k_{2N+5-i(c_{3}^{2m_{3}+1})}^{{\epsilon}(c_{3}^{2m_{3}+1})}(c_{3}^{2m_{3}+1})$
(4.94) $\displaystyle=$ $\displaystyle
k_{2N+5-i(c_{3}^{2m_{3}+1})}^{{\epsilon}(c_{3}^{2m_{3}+2})}(c_{3}^{2m_{3}+2})$
$\displaystyle=$ $\displaystyle
M_{2N+5-i(c_{3}^{2m_{3}+1})+i(c_{3}^{2m_{3}+2})}(3,2m_{3}+2)$ $\displaystyle=$
$\displaystyle M_{2N+13}(3,2m_{3}+2).$

On the other hand, by (4.92) and (3.2), it yields $i(c_{3})\geq 5$, then by
(3.9), we have $i(c_{3}^{2m_{3}+1})\geq 2N+5$. So, by (4.91) and Lemma 3.5,
there holds

$\displaystyle M_{2N+5}(3,2m_{3}+1)=0,$ (4.95)

which, together with (4.94), contradicts to (4.88).

Subcase 3.1.3: $i(c_{2}^{2})=13$.

In this subcase, by (3.9), it yields $i(c_{2}^{2m_{2}+2})=2N+13$ and then by
(4.11), we have

$\displaystyle M_{q}(2,2m_{2}+2)=\left\\{\matrix{1,\quad{\it if}\quad
q=2N+13,\cr 0,\quad{\it if}\quad q\neq 2N+13,\cr}\right.$ (4.96)

which, together with (4.60), implies

$\displaystyle M_{2N+11}=M_{2N+11}(3,2m_{3}+2).$ (4.97)

However by (2.7) and Theorem 2.5, we have $M_{2N+11}\geq b_{2N+11}=1$, so we
have

$\displaystyle M_{2N+11}(3,2m_{3}+2)\geq 1.$ (4.98)

Thus by (3.18) and Lemma 3.5, there holds

$\displaystyle M_{2N+9}(3,2m_{3}+2)=0.$ (4.99)

Similar to Subcase 3.1.2, we have

$\displaystyle i(c_{3}^{m})=8m-p_{3,-}-p_{3,0},$ (4.100)

and we obtain a contradiction

$\displaystyle 0=M_{2N+3}(3,2m_{3}+1)=M_{2N+11}(3,2m_{3}+2)\geq 1.$ (4.101)

Subcase 3.2: $i(c_{2})=7$.

In this subcase, by (3.9), it yields $i(c_{2}^{2m_{2}+1})=2N+7$ and then by
(4.11), we have

$\displaystyle M_{q}(2,2m_{2}+1)=\left\\{\matrix{1,\quad{\it if}\quad
q=2N+7,\cr 0,\quad{\it if}\quad q\neq 2N+7,\cr}\right.$ (4.102)

which, together with (4.41) and Lemma 3.4, implies

$\displaystyle M_{2N+5}=\sum_{1\leq k\leq
3}M_{2N+5}(k,2m_{k}+1)=M_{2N+5}(3,2m_{3}+1).$ (4.103)

However by (2.7) and Theorem 2.5, we have $M_{2N+5}\geq b_{2N+5}=1$, so we
have

$\displaystyle M_{2N+5}(3,2m_{3}+1)\geq 1.$ (4.104)

Claim 3: $M_{q}(3,2m_{3}+1)=0$, $\forall\ q\geq 2N+9$.

If $M_{q_{0}}(3,2m_{3}+1)\geq 1$ for some $q_{0}\geq 2N+9$, then by (4.104)
and Lemma 3.5, we know that $i(c_{3}^{2m_{3}+1})\leq 2N+4$ and
$i(c_{3}^{2m_{3}+1})+\nu(c_{3}^{2m_{3}+1})\geq q_{0}+1\geq 2N+10$, which,
together with the fact $\nu(c_{3}^{2m_{3}+1})\leq 6$, implies
$i(c_{3}^{2m_{3}+1})=2N+4$ and $\nu(c_{3}^{2m_{3}+1})=6$ and $q_{0}$ only can
be $2N+9$. Now by (3.9) and (2.22), we have $i(c_{3})=4$ and $\nu(c_{3})=6$,
which implies that $P_{c_{3}}\approx I_{6}$ by (3.1), (3.2) and the fact
$\nu(c_{3})=p_{3,-}+2p_{3,0}+p_{3,+}$. So $i(c_{3})$ must be odd by
Proposition 2.7 and the symplectic additivity of symplectic paths. This
contradicts to $i(c_{3})=4$ and completes the proof of Claim 3.

Since $i(c_{2})=7$ in this subcase, similar to the proof of (4.35), we have

$\displaystyle M_{2N+9}(2,2m_{2}+2)=0.$ (4.105)

Then by Lemma 3.4, (4.41), (4.102), Claim 3, (4.43) and (4.105), we obtain

$\displaystyle M_{2N+9}=\sum_{1\leq k\leq 3,\atop 1\leq m\leq
2}M_{2N+9}(k,2m_{k}+m)=1+M_{2N+9}(3,2m_{3}+2),$ (4.106)

which, together with $M_{2N+9}\geq b_{2N+9}=2$ by (2.7) and Theorem 2.5,
implies that

$\displaystyle M_{2N+9}(3,2m_{3}+2)\geq 1.$ (4.107)

Then by (3.18) and Lemma 3.5 we get

$\displaystyle M_{q}(3,2m_{3}+2)=0,\quad\forall\ q\neq 2N+9.$ (4.108)

By Lemma 3.4, (4.41), (4.102), Claim 3, (4.43) and (4.108), we obtain

$\displaystyle M_{q}=\sum_{1\leq k\leq 3,\atop 1\leq m\leq
2}M_{q}(k,2m_{k}+m)=M_{q}(2,2m_{2}+2),\quad\mbox{for}\ q=2N+11,2N+13.$ (4.109)

So there holds $M_{2N+11}=0$ or $M_{2N+13}=0$ by (4.11), which contradicts to
$M_{2N+11}\geq b_{2N+11}=1$ and $M_{2N+13}\geq b_{2N+13}=1$ by (2.7) and
Theorem 2.5.

Subcase 3.3: $i(c_{2})=9$.

In this subcase, by (3.9), it yields $i(c_{2}^{2m_{2}+1})=2N+9$ and then by
(4.11), we have

$\displaystyle M_{q}(2,2m_{2}+1)=\left\\{\matrix{1,\quad{\it if}\quad
q=2N+9,\cr 0,\quad{\it if}\quad q\neq 2N+9,\cr}\right.$ (4.110)

which, together with (4.41) and Lemma 3.4, implies

$\displaystyle M_{q}=\sum_{1\leq k\leq
3}M_{q}(k,2m_{k}+1)=M_{q}(3,2m_{3}+1),\quad\mbox{for}\ q=2N+5,2N+7.$ (4.111)

However by (2.7) and Theorem 2.5, we have $M_{2N+5}\geq b_{2N+5}=1$ and
$M_{2N+7}\geq b_{2N+7}=1$, so we have $M_{2N+5}(3,2m_{3}+1)\geq 1$ and
$M_{2N+7}(3,2m_{3}+1)\geq 1$. Thus the assumption with $q=6$ of Lemma 3.6 is
satisfied, and then by Lemma 3.6 we conclude that $c_{3}$ must belong to one
of the classes in Theorem 4.2 and the index of $c_{3}$ is the following

$\displaystyle i(c_{3})=6-p_{3,-}-p_{3,0}.$ (4.112)

This completes the proof of Theorem 4.2.

Acknowledgments

The authors would like to thank sincerely Professor Yiming Long and Professor
Wei Wang for their careful reading of the manuscript and valuable comments.

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