# Isoparametric functions on exotic spheres

Chao Qian  School of Mathematical Sciences, Laboratory of Mathematics and
Complex Systems, Beijing Normal University, Beijing 100875, P.R.China
<EMAIL_ADDRESS>and  Zizhou Tang  School of Mathematical Sciences,
Laboratory of Mathematics and Complex Systems, Beijing Normal University,
Beijing 100875, P.R.China<EMAIL_ADDRESS>
###### Abstract.

This paper extends widely the work in [GT12]. Existence or non-existence
results of isoparametric functions on exotic spheres and Eells-Kuiper
projective planes are established. In particular, every homotopy $n$-sphere
($n>4$) is proved to carry an isoparametric function (with certain metric)
with 2 points as the focal set. As an application, we improve a beautiful
result of Bérard-Bergery [BB77] ( see also pp.234-235 of [Be78] ).

###### Key words and phrases:

Isoparametric function, Morse-Bott function, exotic sphere, Eells-Kuiper
projective plane, $SC^{p}$-structure

###### 2010 Mathematics Subject Classification:

53C20, 57R60

The project is partially supported by the NSFC ( No.11071018 )

The second author is the corresponding author.

##  1\. Introduction

Let $N$ be a connected complete Riemannian manifold. A non-constant smooth
function $f$ on $N$ is called _transnormal_ , if there exists a smooth
function $b:\mathbb{R}\rightarrow\mathbb{R}$ such that $|\nabla f|^{2}=b(f),$
where $\nabla f$ is the gradient of $f$. If in addition, there exists a
continuous function $a:\mathbb{R}\rightarrow\mathbb{R}$ so that $\triangle
f=a(f),$ where $\triangle f$ is the Laplacian of $f$, then $f$ is called
_isoparametric_. Each regular level hypersurface is called an _isoparametric
hypersurface_ and the singular level set is called the _focal set_.

It was E. Cartan who first took up a systematical study of hypersurfaces with
constant principal curvatures in space forms, especially in unit spheres. Many
years later, a far-reaching generalization of Cartan's work on those
hypersurfaces in unit spheres was produced by M$\mathrm{\ddot{u}}$nzner in
[Mü80]. They together found that every hypersurface with constant principal
curvatures in a unit sphere determines an isoparametric function on the unit
sphere, and vice versa( see also [CR85], chapter 3 ). Thus, the set of level
hypersurfaces of an isoparametric function on a unit sphere consists of a
family of parallel hypersurfaces with constant principal curvatures, the so
called isoparametric foliation. Owing to Cartan and M$\mathrm{\ddot{u}}$nzner,
the classification of isoparametric hypersurfaces in a unit sphere has been
one of the most challenging problems in submanifold geometry. Up to now, the
classification has almost been completed. We refer to [CR85] for the
development of this subject. For the most recent progress and applications,
see for example [CCJ07], [Miy12] and [TY13].

In general Riemannian manifolds, a series of beautiful results, similar to the
case in a unit sphere has been proved or claimed by Wang in [Wa87]. Based on
Wang's work, Ge and Tang [GT12] made new contributions to the subject of
isoparametric functions on general Riemannian manifolds, especially on exotic
spheres. In a word, the results of [Wa87] and [GT12] show that the existence
of an isoparametric function ( even a transnormal function ) on a Riemannian
manifold $N$ may restrict strongly the geometry and topology of $N$. However,
whether a given manifold admits a metric and an isoparametric function is
still a problem. Actually, as posed in [GT12], it was not known that if any
exotic sphere with dimension more than $4$ admits a metric and an
isoparametric function with $2$ points as the focal set ( one of the main
results in [GT12] asserts that an exotic $4$-sphere( if exists ) admits no
properly transnormal functions ). Moreover, a transnormal function $f$ on
Gromoll-Meyer $7$-sphere with $2$ points as the focal set was constructed in
[GT12], however $f$ is not isoparametric.

One of our main aims is to develop a general way to construct metrics and
isoparametric functions on a given manifold. According to [Wa87], one can see
that a transnormal function on a complete Riemannian manifold is necessarily a
Morse-Bott function. As is well known that a Morse-Bott function is a
generalization of a Morse function, and it admits critical submanifolds
satisfying certain non-degenerate condition. As one of our prime results we
give a fundamental construction:

######  Theorem 1.1.

( A fundamental construction ) Let $N$ be a closed connected smooth manifold
and $f$ a Morse-Bott function on $N$ with the critical set $C(f)=M_{+}\sqcup
M_{-}$, where $M_{+}$ and $M_{-}$ are both closed connected submanifolds of
codimensions more than 1. Then there exists a metric $g$ on $N$ so that $f$ is
an isoparametric function. Moreover, the metric can be chosen so that the
critical submanifolds ${M}_{+}$ and ${M}_{-}$ are both totally geodesic.

Recall that a closed smooth $n$-manifold is called a _homotopy $n$-sphere_, if
it has the homotopy type of the unit sphere $S^{n}$. By using a theorem of S.
Smale, we have the following result as a consequence.

######  Corollary 1.1.

Every homotopy $n$-sphere with $n>4$ admits a metric and an isoparametric
function with 2 points as the focal set.

######  Remark 1.1.

Corollary 1.1 answers partially Problem 4.1 in [GT12].

Moreover, we can also construct metrics and isoparametric functions on
homotopy spheres and the Eells-Kuiper projective planes so that at least one
component of the critical set is not a single point.

Next, we intend to ask if our proof of Theorem 1.1 can be modified so that
each regular level hypersurface has more geometric properties by using a more
sagacious constructing method of the metric. As a partial negative solution to
this problem, we prove the following non-existence result:

######  Proposition 1.1.

Let $\Sigma^{n}$ be a homotopy sphere which admits a metric $g$ and an
isoparametric function $f$ with 2 points as the focal set. Suppose that the
level hypersurfaces are all totally umbilic. Then $\Sigma^{n}$ is
diffeomorphic to $S^{n}$.

By combining Proposition 1.1 with a topological argument, we obtain

######  Theorem 1.2.

Every odd dimensional exotic sphere ( a homotopy sphere not diffeomorphic to
the unit sphere ) admits no totally isoparametric functions with 2 points as
the focal set.

Recall that a _totally isoparametric_ function is an isoparametric function so
that each regular level hypersurface has constant principal curvatures, as
defined in [GTY11]. As mentioned before, an isoparametric function on a unit
sphere must be totally isoparametric.

Last but not least, we have both existence and non-existence results of
isoparametric functions on some homopoty spheres which also have
$SC^{p}$-property( see [Be78] for definition ).

The paper is organized as follows. In Section 2, we review some basic
definitions and give a proof to Theorem 1.1. Then we apply the construction in
Theorem 1.1 to homotopy spheres and the Eells-Kuiper projective planes in
Section 3. And in Section 4, we show Proposition 1.1 and Theorem 1.2. The last
section will be concerned with the existence and non-existence of
isoparametric functions on exotic spheres with $SC^{p}$-property. For
instance, Theorem 5.1 improves a beautiful result of Bérard-Bergery [BB77].

##  2\. A fundamental construction

This section is devoted to a complete proof of Theorem 1.1. Let's start by
recalling the definition of a Morse-Bott function. A smooth function $f$ on a
smooth manifold $N$ is called a _Morse-Bott function_ or a generalized Morse
function if: (i) the critical set defined by equation $\mathrm{d}f=0$ consists
of a family of smooth submanifolds, the so called critical submanifolds; (ii)
the Hessian $H_{f}$ is a non-degenerate quadratic form in the normal direction
of each critical submanifold to $N$.

In differential topology, it is well known that the topology of a manifold can
be analyzed by studying Morse-Bott functions on it. The following result is
essentially due to Wang [Wa87].

######  Proposition 2.1.

A transnormal function $f$ on a complete connected Riemannian manifold $N$ is
a Morse-Bott function. Moreover, the critical set $C(f)$ of $f$ coincides
exactly with the focal set of $f$.

###### Proof.

Combine Lemma 4 with Lemma 6 in [Wa87]. ∎

Owning to Proposition 2.1, it is natural to propose the following converse
problem.

######  Problem 2.1.

Let $f$ be a Morse-Bott function on a smooth manifold $N$. When does $N$ carry
a Riemannian metric $g$ so that $f$ is transnormal with respect to this $g$?

This problem motivates us to establish Theorem 2.2. To complete a proof to
Theorem 2.2, we still has a long way. Let's first investigate the topology of
a Morse-Bott function whose critical set has only two components.

######  Proposition 2.2.

Let $N$ be a connected closed manifold and $f:N\rightarrow\mathbb{R}$ a Morse-
Bott function so that $C(f)$ has two connected closed components, say $M_{+}$
and $M_{-}$. Then $N$ is diffeomorphic to the glued manifold

$D(M_{-})\sqcup M\sqcup_{\varphi}D(M_{+}),$

for some diffeomorphism $\varphi:\partial D(M_{-})\rightarrow\partial
D(M_{+})$, where $M=\partial D(M_{-})\times[0,1]$, $D(M_{+})$ and $D(M_{-})$
are respectively the normal disc bundles over $M_{+}$ and $M_{-}$.

###### Proof.

By the generalized Morse lemma and regular interval theorem( see [Hi76],
pp.149 and pp.153 ). ∎

######  Remark 2.1.

Duan and Rees [DR92] also considered the situation where the critical set of
the smooth map $f:N\rightarrow\mathbb{R}$ has only two components, both of
which are smooth submanifolds of $N$. Actually, they did not assume that the
critical set of $f$ are non-degenerate in any sense.

Next, we set about constructing the desired metric. The construction is a
modified version of A. Weinstein's construction( see, for example, [Be78],
pp.231-233 ).

######  Theorem 2.2.

Let $N$ be a closed smooth manifold, and $f:N\rightarrow\mathbb{R}$ a Morse-
Bott function with $C(f)=M_{+}\sqcup M_{-}$, where $M_{+}$ and $M_{-}$ are
both closed connected submanifolds of codimensions more than $1$. Then there
exists a Riemannian metric on $N$ so that $f$ is transnormal. In fact, the
metric can be chosen so that $M_{+}$ and $M_{-}$ are both totally geodesic.

###### Proof.

Without loss of generality, we can assume Im$f=[\alpha,\beta]$ and
$M_{+}=f^{-1}(\beta)$, $M_{-}=f^{-1}(\alpha)$. Our construction of the metric
is divided into three steps.

Step 1:

At first, by the generalized Morse lemma, there exist a vector bundle( a
tubular neighborhood ) over $M_{-}$ with some metric, say
$\xi^{-}=(\pi_{-},E_{-},M_{-})$ and an embedding
$\sigma_{-}:\mathcal{U}\rightarrow N$, where $\mathcal{U}\subset E_{-}$ is an
open neighborhood of the zero section, such that

(1) $\sigma_{-}|_{M_{-}}=\mathrm{id}$, where $M_{-}$ is identified with the
zero section of $\xi^{-}$ and $\sigma_{-}(\mathcal{U})$ is an open
neighborhood of $M_{-}$ in $N$. Notice here rank $\xi^{-}=$dim $N-$dim
$M_{-}\geq 2$.

(2) the composition
$\mathcal{U}\xrightarrow{\sigma_{-}}N\xrightarrow{f}\mathbb{R}$ is given by

$f(\sigma_{-}(p,v))=|v|^{2}+\alpha$

for any $(p,v)\in\pi_{-}^{-1}(p)\cap\mathcal{U}$ and $p\in M_{-}$.

Now, since $M_{-}$ is compact, we can choose $\epsilon>0$ so small that

$M_{-}(\epsilon):=\\{x\in
N|~{}f(x)\leq\alpha+\epsilon^{2}\\}\subset\sigma_{-}(\mathcal{U}).$

Thus, we have a disc bundle, $D_{\xi^{-}}(\epsilon):=\\{(p,v)\in E_{-}|p\in
M_{-},|v|\leq\epsilon\\}$ with $S_{\xi^{-}}(\epsilon):=\partial
D_{\xi^{-}}(\epsilon)$ so that
$\sigma_{-}(D_{\xi^{-}}(\epsilon))=M_{-}(\epsilon)$.

Next, for each dimension $d$, let us choose a smooth Riemannian metric $g^{d}$
on the disc $D^{d}(\epsilon):=\\{x\in\mathbb{R}^{d}|~{}|x|\leq\epsilon\\}$ as
$g^{d}=F(r)^{2}\mathrm{d}r^{2}+G(r)^{2}\mathrm{d}s_{d-1}^{2}$, where
$\mathrm{d}s_{d-1}^{2}$ is the canonical metric on $S^{d-1}(1)$, and $F,G$ are
smooth functions on $[0,\epsilon]$ with

$F(r)=\left\\{\begin{array}[]{cc}1,&\mathrm{for}~{}r\leq\frac{\epsilon}{2}~{};\\\
2r,&\mathrm{for}~{}r\geq\frac{3\epsilon}{4}~{},\end{array}\right.$

and

$G(r)=\left\\{\begin{array}[]{cc}r,&\mathrm{for}~{}r\leq\frac{\epsilon}{2}~{};\\\
1,&\mathrm{for}~{}r\geq\frac{3\epsilon}{4}~{}.\end{array}\right.$

The choice of the metric ensures the following two properties:

(1) $g^{d}$ is $O(d)$-invariant under the usual action of the orthogonal group
$O(d)$ with the origin $o\in\mathbb{R}^{d}$ fixed;

(2) $g^{d}$ is product near the boundary $\partial
D^{d}(\epsilon)=S^{d-1}(\epsilon)$. Moreover, $|\frac{\partial}{\partial
r}|=2r$ near the boundary, where $r=|x|$, and $\frac{\partial}{\partial r}$ is
the radius direction.

With this metric, each radius has the same length, say $\delta$. Furthermore,
it is not difficult to show each radius is a geodesic with arc length as
parameter in $D^{d}(\epsilon)$.

Let $h_{-}$ be any Riemannian metric on $M_{-}$. By a construction due to J.
Vilms [Vi70], there exists a Riemannian metric $h_{-}^{\prime}$ on
$D_{\xi^{-}}(\epsilon)$, such that the fibration
$\pi:(D_{\xi^{-}}(\epsilon),h_{-}^{\prime})\rightarrow(M_{-},h_{-})$ is a
Riemannian submersion with totally geodesic fibers isometric to
$(D^{d_{-}}(\epsilon),g^{d_{-}})$ with $d_{-}=$ rank $\xi^{-}$. In the case at
hand, this construction can be described as follows. Note that we have chosen
the Euclidean metric on $(\pi_{-},E_{-},M_{-})$, and let us choose a metric
linear connection on $(\pi_{-},E_{-},M_{-})$. This connection produces a
distribution $\mathcal{H}_{-}$ on the tangent bundle $TE_{-}$ of $E_{-}$,
complementary to the vertical distribution $\mathcal{V}_{-}$ ( kernel of
$\mathrm{d}\pi_{-}$ ), and the associated parallel transport is by linear
isometries. Then the desired metric $h_{-}^{\prime}$ on
$D_{\xi^{-}}(\epsilon)$ can be constructed in the following way: one takes the
vertical and horizontal distributions to be orthogonal, lifts $h_{-}$ from
$M_{-}$ to the horizontal distribution and then takes the metric $g^{d_{-}}$
on the fiber $D^{d_{-}}(\epsilon)$. Clearly parallel transport is still an
isometry for these metrics on the fibers, so that the fibers are totally
geodesic.

Notice that the metric $h_{-}^{\prime}$ is still product near the boundary
$S_{\xi^{-}}(\epsilon)$ of $D_{\xi^{-}}(\epsilon)$ and the zero section is
totally geodesic in $D_{\xi^{-}}(\epsilon)$. Moreover, any geodesic issuing
from a point in $S_{\xi^{-}}(\epsilon)$ and orthogonal to
$S_{\xi^{-}}(\epsilon)$ in $D_{\xi^{-}}(\epsilon)$ stays in the same fiber of
the fibration. In particular, all the geodesics orthogonal to the boundary
$S_{\xi^{-}}(\epsilon)$ in $D_{\xi^{-}}(\epsilon)$ have the same length
$2\delta$.

Now, by the diffeomorphism $\sigma_{-}:D_{\xi^{-}}(\epsilon)\rightarrow
M_{-}(\epsilon)$, we obtain an induced metric on $M_{-}(\epsilon)$, say
$h_{-}(\epsilon)$. On the other side, we can also construct an analogous
Riemannian metric, say $h_{+}(\epsilon)$, on $M_{+}(\epsilon):=\\{x\in
N|~{}f(x)\geq\beta-\epsilon^{2}\\}$ which is diffeomorphic to a disc bundle
$D_{\xi^{+}}(\epsilon)$ over $M_{+}$, where $\epsilon$ is small enough, the
same as before.

Step 2:

The task we face is to extend smoothly the metrics $h_{-}(\epsilon)$ and
$h_{+}(\epsilon)$ on $M_{-}(\epsilon)$ and $M_{+}(\epsilon)$ to the whole
manifold $N$ with appropriate properties. We are left to work on $M:=\\{x\in
N|~{}\alpha+\epsilon^{2}\leq f(x)\leq\beta-\epsilon^{2}\\}$, since
$N=M_{-}(\epsilon)\sqcup M\sqcup M_{+}(\epsilon)$.

To do it, by regular interval theorem( see [Hi76] ), $\partial
M_{-}(\epsilon)$ is diffeomorphic to $\partial M_{+}(\epsilon)$. More
precisely, there is a diffeomorphism $\tau:M\rightarrow
S_{\xi^{-}}(\epsilon)\times[\alpha+\epsilon^{2},\beta-\epsilon^{2}]$
satisfying

$\tau(f^{-1}(a))=S_{\xi^{-}}(\epsilon)\times\\{a\\},~{}\forall
a\in[\alpha+\epsilon^{2},\beta-\epsilon^{2}].$

Let us now choose a family $g_{t}$ of Riemannian metrics on
$S_{\xi^{-}}(\epsilon)$ , which are smooth in the parameter $t$ for
$t\in[\alpha+\epsilon^{2},\beta-\epsilon^{2}]$ and constant near
$\alpha+\epsilon^{2}$ and $\beta-\epsilon^{2}$, respectively. Moreover, we ask
that $g_{\alpha+\epsilon^{2}}$ is the metric induced from $h_{-}(\epsilon)$,
and $g_{\beta-\epsilon^{2}}$ is induced from $h_{+}(\epsilon)$ by the
diffeomorphism $\tau$.

Then a metric $h_{0}$ on
$S_{\xi^{-}}(\epsilon)\times[\alpha+\epsilon^{2},\beta-\epsilon^{2}]$ can be
constructed in such a way that the direct decomposition given by the
projections onto the two factors is orthogonal, the metric induced on each
$S_{\xi^{-}}(\epsilon)\times\\{t\\}$ is $g_{t}$, and the metric induced on
$\\{p\\}\times[\alpha+\epsilon^{2},\beta-\epsilon^{2}]$ is the canonical
metric with length $\beta-\alpha-2\epsilon^{2}$ on
$[\alpha+\epsilon^{2},\beta-\epsilon^{2}]$ for each point $p\in
S_{\xi^{-}}(\epsilon)$. It is not difficult to show that for any given point
$p\in S_{\xi^{-}}(\epsilon)$, the curve

$\gamma:[\alpha+\epsilon^{2},\beta-\epsilon^{2}]\rightarrow
S_{\xi^{-}}(\epsilon)\times[\alpha+\epsilon^{2},\beta-\epsilon^{2}],~{}t\mapsto(p,t),$

is a geodesic in
$S_{\xi^{-}}(\epsilon)\times[\alpha+\epsilon^{2},\beta-\epsilon^{2}]$.

Step 3:

At last, these metrics $h_{-}(\epsilon)$, $\tau^{*}h_{0}$ and
$h_{+}(\epsilon)$ on $M_{-}(\epsilon)$, $M$ and $M_{+}(\epsilon)$ respectively
can be glued together giving a desired metric $g_{N}$ on $N$. Observe that
$|\nabla f|\equiv 1$ on $M$. On the other hand, near the boundary of
$M_{-}(\epsilon)$, one has $|\nabla f|=|2r\nabla
r|=|\frac{1}{2r}\frac{\partial}{\partial r}|=1$, which also holds near the
boundary of $M_{+}(\epsilon)$. These properties guarantee that the glued
metric $g_{N}$ on $N$ is smooth. From this construction, it is obvious that
every geodesic starting orthogonally from any point in $M_{-}$($M_{+}$) must
arrive at $M_{+}$($M_{-}$) orthogonally with length
$2\delta+\beta-\alpha-2\epsilon^{2}$.

By a direct computation, we see

$|\nabla
f|^{2}=\left\\{\begin{array}[]{cc}\frac{4(f-\alpha)}{F^{2}(\sqrt{f-\alpha})}&\alpha\leq
f\leq\alpha+\epsilon^{2}\\\ 1&\alpha+\epsilon^{2}\leq
f\leq\beta-\epsilon^{2}\\\
\frac{4(\beta-f)}{F^{2}(\sqrt{\beta-f})}&\beta-\epsilon^{2}\leq
f\leq\beta,\end{array}\right.$

which means that $f$ is transnormal. Now, the proof is complete. ∎

######  Remark 2.2.

In general, $M_{+}$ and $M_{-}$ of a transnormal function $f$ are not
necessarily totally geodesic.

According to the proof of Theorem 2.2, we have the following enjoyable
observation.

######  Proposition 2.3.

The function $f$ is isoparametric on the domain $N-M$ with respect to the
metric constructed in Theorem 2.2.

###### Proof.

We only need to verify $f$ on
$M_{-}(\epsilon)=\sigma_{-}(D_{\xi^{-}}(\epsilon))$. Since we are working on a
Riemannian submersion with totally geodesic fibers, it follows that
$\triangle$ on $M_{-}(\epsilon)$ has a decomposition
$\triangle=\triangle^{\mathcal{H}}+\triangle^{\mathcal{V}}$, where
$\triangle^{\mathcal{H}}$ and $\triangle^{\mathcal{V}}$ are the horizontal and
vertical Laplacians on $M_{-}(\epsilon)$ with respect to the Riemannian
submersion $\pi\circ\sigma_{-}^{-1}:M_{-}(\epsilon)\rightarrow M_{-}$. In our
case, it is not difficult to see that $\triangle^{\mathcal{H}}f=0$. Thus, by a
direct computation,

$\triangle
f=\triangle^{\mathcal{V}}(r^{2})=\frac{2}{F^{2}}-\frac{2rF_{r}}{F^{3}}+\frac{2(\mathrm{dim}N-\mathrm{dim}M_{-}-1)rG_{r}}{F^{2}G},$

that is, $\triangle f$ is a continuous function of $f$ on $M_{-}(\epsilon)$.
The proof then follows. ∎

Up to now, we have constructed a metric $g_{N}$ on $N$ such that $f$ is
transnormal on whole $N$, and isoparametric near $M_{-}$ and $M_{+}$. In order
to prove Theorem 1.1, we have to deform the metric $g_{N}$ on $N$ so that $f$
is isoparametric. For this purpose, we prepare three lemmas.

######  Lemma 2.1 (Local version of Moser's theorem).

Let $Q^{n}=(0,1)^{n}$ and $\bar{Q}^{n}=[0,1]^{n}$ be the open and closed unit
cubes in Euclidean space, respectively. Let $f$, $g$ be two positive smooth
functions on $\bar{Q}^{n}$ such that $f=g$ near $\partial\bar{Q}^{n}$ and
$\int_{Q^{n}}(f-g)\mathrm{d}x=0$. Then there exists a diffeomorphism
$\psi:Q^{n}\rightarrow Q^{n}$ satisfying

(1) $g\circ\psi~{}\mathrm{det}\nabla\psi=f\;~{}in~{}Q^{n};$

(2) $\psi=\mathrm{id}$ near $\partial\bar{Q}^{n}$;

(3) $\psi$ is isotopic to $\mathrm{id}$.

In fact, the diffeomorphism $\psi$ can be decomposed as
$\psi=\psi_{n}\circ\psi_{n-1}\circ...\circ\psi_{1}:Q^{n}\rightarrow Q^{n}$,
and for any $1\leq s\leq n$,

$\psi_{s}:Q^{n}\rightarrow
Q^{n},(x_{1},x_{2},...,x_{n})\mapsto(x_{1},x_{2},...,x_{s-1},\upsilon_{s}(x_{1},x_{2},...,x_{n}),x_{s+1},...,x_{n})$

is a diffeomorphism such that $\frac{\partial\upsilon_{s}}{\partial x_{s}}>0$,
$\psi_{s}=\mathrm{id}$ near $\partial\bar{Q}^{n}$, and $\psi_{s}$ preserves
the line segments $\\{(x_{1},x_{2},...,x_{n})\in Q^{n}|x_{i}=c_{i},1\leq i\leq
n,i\neq s\\}$.

######  Lemma 2.2 (Global version of Moser's theorem).

Let $\tau$ and $\sigma$ be two volume elements on a compact connected manifold
$M$ with

$\int_{M}\tau=\int_{M}\sigma.$

Then there exists a diffeomorphism $\psi:M\rightarrow M$ so that $\psi$ is
isotopic to $\mathrm{id}:M\rightarrow M$ and satisfies $\psi^{*}\tau=\sigma$.

######  Remark 2.3.

For the proofs of Lemma 2.1 and Lemma 2.2, we refer to [Mo65] and [DM90].
However, we should indicate that Lemma 2 was not proved correctly in [Mo65],
and it was rectified in [DM90]. As Calabi pointed out that the global version
of Moser's theorem holds even for non-orientable manifolds if one uses the
concept of ``odd'' forms ( see [Mo65] ).

At last, we need a lemma for deforming metrics.

######  Lemma 2.3.

Let $\mathcal{S}^{n-1}$ be an $(n-1)$-dimensional smooth manifold and
$M^{n}=\mathcal{S}^{n-1}\times[\alpha_{0},\beta_{0}]$. Let $h$ be a metric on
$M$ defined by a smooth one-parameter family of metrics
$\\{h_{t}|~{}t\in[\alpha_{0},\beta_{0}]\\}$ on $\mathcal{S}$, i.e.
$h|_{(x,t)}=h_{t}+\mathrm{d}t^{2}$ for
$(x,t)\in\mathcal{S}^{n-1}\times[\alpha_{0},\beta_{0}]$. Define a new metric
$\tilde{h}$ on $M$ given by
$\\{\tilde{h}_{t}|~{}\tilde{h}_{t}=e^{2u}h_{t},~{}t\in[\alpha_{0},\beta_{0}]\\}$
depending only on a smooth function $u$ on $M$. Considering the smooth
projection function
$f:\mathcal{S}^{n-1}\times[\alpha_{0},\beta_{0}]\rightarrow\mathbb{R},(x,t)\mapsto
t$, we have

$\tilde{\triangle}f=\triangle f+(n-1)\frac{\partial u}{\partial t},$

where $\tilde{\triangle}$ and $\triangle$ are the Laplacians with respect to
$\tilde{h}$ and $h$ respectively.

###### Proof.

By a straightforward computation. ∎

After a long time of preparation, we are ready to prove theorem 1.1.

Proof of Theorem 1.1: Assume the same notations as in Theorem 2.2. By the
proof of Theorem 2.2, there exists a diffeomorphism
$\varphi:S_{\xi^{-}}(\epsilon)=\partial D_{\xi^{-}}(\epsilon)\rightarrow
S_{\xi^{+}}(\epsilon)=\partial D_{\xi^{+}}(\epsilon)$ such that $N$ is
diffeomorphic to the glued manifold $D_{\xi^{-}}(\epsilon)\sqcup
M^{\prime}\sqcup_{\varphi}D_{\xi^{+}}(\epsilon)$, where
$M^{\prime}=S_{\xi^{-}}(\epsilon)\times[\alpha+\epsilon^{2},\beta-\epsilon^{2}]$.
Additionally, we have constructed metrics $h_{-}^{\prime}$, $h_{+}^{\prime}$
and $h_{0}$ on $D_{\xi^{-}}(\epsilon)$, $D_{\xi^{+}}(\epsilon)$ and
$M^{\prime}$ respectively. As we asserted before, $f$ is isoparametric on
$D_{\xi^{-}}(\epsilon)$ and $D_{\xi^{+}}(\epsilon)$.

Let $\omega_{-}$ and $\omega_{+}$ be the volume elements of
$(S_{\xi^{-}}(\epsilon),h_{-}^{\prime}|_{S_{\xi^{-}}(\epsilon)})$ and
$(S_{\xi^{+}}(\epsilon),h_{+}^{\prime}|_{S_{\xi^{+}}(\epsilon)})$
respectively. Then
$(S_{\xi^{-}}(\epsilon),g_{\beta-\epsilon^{2}})=(S_{\xi^{-}}(\epsilon),\varphi^{*}(h_{+}^{\prime}|_{S_{\xi^{+}}(\epsilon)}))$
has the volume element $\varphi^{*}\omega_{+}$. By Lemma 2.2 ( Global version
of Moser's theorem ), there exists a diffeomorphism
$\psi:S_{\xi^{-}}(\epsilon)\rightarrow S_{\xi^{-}}(\epsilon)$, which is
isotopic to $\mathrm{id}$ and satisfies

$\psi^{*}(\varphi^{*}\omega_{+})=\lambda\omega_{-},$

where $\lambda=\frac{\int\varphi^{*}\omega+}{\int\omega_{-}}$. Defining
$\tilde{\varphi}:=\varphi\circ\psi$, we see that $\tilde{\varphi}$ is isotopic
to $\varphi$ and

$\tilde{\omega}=(\varphi\circ\psi)^{*}\omega_{+}=\psi^{*}(\varphi^{*}\omega_{+})=\lambda\omega_{-},$

where $\tilde{\omega}$ is the volume element of
$(S_{\xi^{-}}(\epsilon),\psi^{*}g_{\beta-\epsilon^{2}})$. Since
$\tilde{\varphi}$ is isotopic to $\varphi$, it follows that
$D_{\xi^{-}}(\epsilon)\sqcup
M^{\prime}\sqcup_{\tilde{\varphi}}D_{\xi^{+}}(\epsilon)$ is diffeomorphic to
$D_{\xi^{-}}(\epsilon)\sqcup M^{\prime}\sqcup_{\varphi}D_{\xi^{+}}(\epsilon).$

Consider now $\tilde{N}:=D_{\xi^{-}}(\epsilon)\sqcup
M^{\prime}\sqcup_{\tilde{\varphi}}D_{\xi^{+}}(\epsilon)$. Let
$\tilde{f}:\tilde{N}\rightarrow\mathbb{R}$ be a function such that
$\tilde{f}|_{D_{\xi^{\pm}}(\epsilon)}=f\circ\sigma_{\pm}|_{D_{\xi^{\pm}}(\epsilon)}$
and $\tilde{f}|_{M^{\prime}}=f\circ\tau^{-1}|_{M^{\prime}}$. It is obvious
that $\tilde{N}$ is diffeomorphic to $N$ and $\tilde{f}$ is a Morse-Bott
function with the critical submanifolds
$C(\tilde{f})=\tilde{M}_{+}\bigsqcup\tilde{M}_{-}$, where $\tilde{M}_{+}$ and
$\tilde{M}_{-}$ are diffeomorphic to $M_{+}$ and $M_{-}$, respectively. With
respect to the metrics constructed in Theorem 2.2, $\tilde{f}$ is
isoparametric on $(D_{\xi^{-}}(\epsilon),h_{-}^{\prime})$ and
$(D_{\xi^{+}}(\epsilon),h_{+}^{\prime})$. On $M^{\prime}$, choose a metric $g$
defined by a smooth family of metrics
$\\{g_{t}^{\prime}|~{}t\in[\alpha+\epsilon^{2},\beta-\epsilon^{2}]\\}$ on
$S_{\xi^{-}}(\epsilon)$ such that

$g_{\alpha+\epsilon^{2}}^{\prime}=g_{\alpha+\epsilon^{2}},~{}g_{\beta-\epsilon^{2}}^{\prime}=\psi^{*}g_{\beta-\epsilon^{2}}$

and $g_{t}^{\prime}$ is constant near $\alpha+\epsilon^{2}$ and
$\beta-\epsilon^{2}$. By lemma 2.3, let $u:M^{\prime}\rightarrow\mathbb{R}$ be
a smooth function to be determined and $\tilde{g}$ a new metric on
$M^{\prime}$ determined by the new family of metrics
$\\{\tilde{g}_{t}^{\prime}|~{}\tilde{g}_{t}^{\prime}=e^{2u}g_{t}^{\prime},t\in[\alpha+\epsilon^{2},\beta-\epsilon^{2}]\\}$,
then

$\tilde{\triangle}\tilde{f}=\triangle\tilde{f}+(n-1)\frac{\partial u}{\partial
t},$

where $n=$ dim $N$, $\tilde{\triangle}$ and $\triangle$ are Laplacians with
respect to $\tilde{g}$ and $g$, respectively.

Next let $h:[\alpha+\epsilon^{2},\beta-\epsilon^{2}]\rightarrow\mathbb{R}$ be
a smooth function with
$h(t)=\triangle\tilde{f}|_{S_{\xi^{-}}(\epsilon)\times\\{t\\}}$ for $t$ near
$\alpha+\epsilon^{2}$ and $\beta-\epsilon^{2}$. In order to find such an $u$
that $\tilde{f}$ is isoparametric on $(M^{\prime},\tilde{g})$, it is
sufficient to construct $u$ as the solution of the following equation

$\tilde{\triangle}\tilde{f}=h$

or equivalently

$h(t)=\triangle\tilde{f}+(n-1)\frac{\partial u}{\partial t}$

which implies that for any $(x,t)\in M^{\prime}$

$(n-1)(u(x,t)-u(x,\alpha+\epsilon^{2}))=\int_{\alpha+\epsilon^{2}}^{t}[h(s)-\triangle\tilde{f}(x,s)]\mathrm{d}s.$

On the other hand, by a direct computation, we have
$\triangle\tilde{f}=\frac{1}{2}\frac{\partial}{\partial
t}(\mathrm{log}~{}\mathrm{det}(g_{ij}))$, where
$g_{t}^{\prime}=\sum_{i,j=1}^{n}g_{ij}(x,t)\mathrm{d}x_{i}\mathrm{d}x_{j}$
with respect to local coordinates $(x_{1},x_{2},...,x_{n-1})$ around $x\in
S_{\xi^{-}}(\epsilon)$. It follows that

$\int_{\alpha+\epsilon^{2}}^{\beta-\epsilon^{2}}\triangle\tilde{f}(x,s)\mathrm{d}s=\mathrm{log}\frac{\sqrt{\mathrm{det}(g_{ij}(x,~{}\beta-\epsilon^{2}))}}{\sqrt{\mathrm{det}(g_{ij}(x,~{}\alpha+\epsilon^{2}))}}$

which is just equal to $\mathrm{log}\lambda$, independent of the choice of
$x\in S_{\xi^{-}}(\epsilon)$. Hence, an appropriate choice of the function $h$
satisfies the equation
$\int_{\alpha+\epsilon^{2}}^{\beta-\epsilon^{2}}[h(s)-\triangle\tilde{f}(\cdot,s)]\mathrm{d}s\equiv
0$. At last we get a solution

$u(x,t)=\frac{1}{n-1}\int_{\alpha+\epsilon^{2}}^{t}[h(s)-\triangle\tilde{f}(x,s)]\mathrm{d}s,$

for $(x,t)\in M^{\prime}$, such that
$u|_{S_{\xi^{-}}(\epsilon)\times\\{t\\}}\equiv 0$ for $t$ near
$\alpha+\epsilon^{2}$ and $\beta-\epsilon^{2}$. The argument above implies
that $\tilde{g}$ still has the product metric near the boundary of
$M^{\prime}$ as before. Therefore the three parts
$(D_{\xi_{-}}(\epsilon),h_{-}^{\prime})$,
$(D_{\xi_{+}}(\epsilon),h_{+}^{\prime})$ and $(M^{\prime},\tilde{g})$ can be
glued to gain a smooth metric $\tilde{g}_{\tilde{N}}$ on $\tilde{N}$ so that
$\tilde{f}$ is isoparametric.

The proof is now complete.$\hfill{\square}$

##  3\. Applications and examples

In this section, the general construction in Theorem 1.1 is applied to the
existence of isoparametric functions on various exotic spheres and Eells-
Kuiper projective planes.

First, we give a short

Proof of Corollary 1.1:

###### Proof.

A well known result of S. Smale ( ref. [Mil07], pp.128 ) states that there
exists a Morse function with 2 critical points on each homotopy sphere with
dimension more than 4. Thus, it follows from Theorem 1.1 that each homotopy
sphere with dimension more than 4 admits a metric and an isoparametric
function with 2 points as the focal set.

The proof is now complete. ∎

More generally, we also investigate the existence of isoparametric functions
on exotic spheres such that at least one component of the critical set is not
a single point.

Following J. Milnor, let $g:S^{m}\times S^{n}\rightarrow S^{m}\times S^{n}$ be
an orientation-preserving diffeomorphism. Then a smooth manifold
$M^{m+n+1}(g)$ depending on $g$ is obtained from disjoint spaces
$\mathbb{R}^{m+1}\times S^{n}$ and $S^{m}\times\mathbb{R}^{n+1}$ by matching
$(tx,y)$ and $(x^{\prime},t^{\prime}y^{\prime})$, where
$(x^{\prime},y^{\prime})=g(x,y)$ and $t^{\prime}=\frac{1}{t}$ for any $x\in
S^{m}$, $y\in S^{n}$ and $t\in(0,+\infty)$. In light of this construction, we
have the following observation.

######  Proposition 3.1.

Define a function $f$ on $M^{m+n+1}(g)$ by

$f:\mathbb{R}^{m+1}\times
S^{n}\rightarrow\mathbb{R},(X,y)\mapsto\frac{|X|^{2}}{1+|X|^{2}},$

and

$f:S^{m}\times\mathbb{R}^{n+1}\rightarrow\mathbb{R},(x,Y)\mapsto\frac{1}{1+|Y|^{2}}.$

$f$ is a Morse-Bott function with $C(f)=\\{0\\}\times S^{n}\sqcup
S^{m}\times\\{0\\}$. Moreover, $M^{m+n+1}(g)$ admits an isoparametric function
with focal set diffeomorphic to $S^{m}\sqcup S^{n}$ under some metric.

###### Proof.

It is obvious that $f$ is a well defined smooth function. Moreover, $f$ is a
Morse-Bott function with critical set $\\{0\\}\times S^{n}\sqcup
S^{m}\times\\{0\\}$. Therefore, $M^{m+n+1}(g)$ admits a metric and an
isoparametric function with focal set diffeomorphic to $S^{m}\sqcup S^{n}$
under some metric, by virtue of Theorem 1.1. ∎

In order to construct concrete examples, we have the following method by J.
Milnor [Mil59]. Starting with two smooth maps $f_{1}:S^{m}\rightarrow SO(n+1)$
and $f_{2}:S^{n}\rightarrow SO(m+1)$, a diffeomorphism $g:S^{m}\times
S^{n}\rightarrow S^{m}\times S^{n}$ is defined by

$g(x,y)=(f_{2}^{-1}(f_{1}(x)\cdot y)\cdot x,f_{1}(x)\cdot y)$

for $x\in S^{m}$ and $y\in S^{n}$. Denote $M(f_{1},f_{2}):=M(g)$. There is
also another way to construct manifolds, the so called Milnor pairing. The
Milnor pairing is a bilinear pairing given by

$\beta_{m,n}:\pi_{m}SO(n)\otimes\pi_{n}SO(m)\rightarrow\pi_{0}\mathrm{Diff^{+}}~{}S^{m+n}\rightarrow\Gamma_{m+n+1}=\Theta_{m+n+1},$

where $\mathrm{Diff^{+}}S^{m+n}$ is the group of all orientation preserving
diffeomorphisms from $S^{m+n}$ onto itself; $\Gamma_{l}$ denotes the group of
oriented diffeomorphism classes of twisted $l$-spheres, $\Theta_{l}$ denotes
the group of oriented homotopy $l$-spheres up to relation $h$-cobordent, and
in fact they are isomorphic for each $l$ ( [Mil07], pp.5 ). We remark that two
homotopy $l$-spheres, $l>4$, are $h$-cobordant if and only if they are
diffeomorphic( ref. [KM63] ). The following theorem of J. Milnor tells us how
the two constructing methods above are related.

######  Theorem 3.1 ([Mil07], pp. 218).

Given two smooth maps $f_{1}:S^{m}\rightarrow SO(n)\hookrightarrow SO(n+1)$
and $f_{2}:S^{n}\rightarrow SO(m)\hookrightarrow SO(m+1)$, the resulting
manifold $M(f_{1},f_{2})$ is diffeomorphic to $\beta_{m,~{}n}(f_{1},f_{2})$.

Applying Proposition 3.1 and Theorem 3.1 will lead to the following examples.

######  Example 3.1.

Each homotopy 7-sphere admits a metric and an isoparametric function with
focal set dffeomorphic to $S^{3}\sqcup S^{3}$, since $\Theta_{7}=$ Im
$\beta_{3,~{}3}$( see [Mil07], pp. 187 ).

######  Remark 3.1.

This example generalizes Proposition 4.1 in [GT12], where only Milnor
$7$-spheres ( those can be represented as $S^{3}$ bundles over $S^{4}$ ) are
considered. It is well known that not every homotopy $7$-sphere is a Milnor
sphere.

Recall that the group of homotopy spheres $\Theta_{n}$ has an important
subgroup $bP_{n+1}$ defined by Kervaire and Milnor in [KM63]. A homotopy
$n$-sphere $\Sigma^{n}$ represents an element of $bP^{n+1}$ if and only if
$\Sigma^{n}$ bounds a parallelizable manifold. To determine the group
$\Theta_{n}$, they did a systematic research of such subgroups. In particular,
they showed that $\Theta_{7}=bP_{8}$ and $bP_{4m+2}=\mathbb{Z}_{2}$ or $0$. As
in the literature, a Kervaire sphere $\Sigma^{4m+1}$ is the boundary of the
pluming of two copies of the unit disc bundle $D(TS^{2m+1})$ of the tangent
bundle of $S^{2m+1}$ and consequently it is a generator of $bP_{4m+2}$( see
[KM63], [HH67] and [MM80] ). For such homotopy spheres, we have the following
result.

######  Example 3.2.

For $m>0$, the Kervaire sphere $\Sigma^{4m+1}$ admits a metric and an
isoparametric function with focal set diffeomorphic to $S^{2m}\sqcup S^{2m}$.
The reason is that one can identify $\Sigma^{4m+1}$ with $M(f,f)$, where
$f:S^{2m}\rightarrow SO(2m)\hookrightarrow SO(2m+1)$ is the characteristic map
for the tangent bundle of $S^{2m+1}$( see [HH67] and [MM80] ).

Additionally, more examples concerning even dimensional homotopy sphere, as
well as the odd dimensional homotopy sphere which is not the boundary of any
parallelizable manifold are given.

######  Example 3.3.

According to D. L. Frank [Fr68], it is known that $\Theta_{8}=$ Im
$\beta_{3,~{}4}$, $2\Theta_{10}=$ Im $\beta_{3,~{}6}$. Hence, every homotopy
$8$-sphere admits a metric and an isoparametric function with focal set
diffeomorpic to $S^{3}\sqcup S^{4}$; every element in $2\Theta_{10}$ admits a
metric and an isoparametric function with focal set diffeomorphic to
$S^{3}\sqcup S^{6}$.

It is also proved by D. L. Frank in [Fr68] that there exists a homotopy
$15$-sphere $\Sigma^{15}$ which is contained in Im $\beta_{11,~{}3}$ but not
in $bP_{16}$. Therefore, this $\Sigma^{15}$ admits an isoparametric function
with focal set $S^{11}\sqcup S^{3}$ for certain metric, and it is not the
boundary of any parallelizable manifold.

In fact, Proposition 3.1 gives us more examples than we show here( see [Mil59]
and related references ).

At last, we will investigate the existence of isoparametric functions on
Eells-Kuiper projective planes. Recall that a closed manifold is called an
Eells-Kuiper projective plane if it admits a Morse function with three
critical points. Eells-Kuiper [EK1] obtained many remarkable results on such
manifolds. For instance, they showed that the integral cohomology ring of an
Eells-Kuiper projective plane $M$ is isomorphic to that of the real, complex,
quaternionic or Cayley projective plane. Consequently, the dimension of $M$
must be equal to $2,4,8$, or $16$. Moreover, for $m\in\\{1,4,8\\}$, an Eells-
Kuiper projective plane $M^{2m}$ is diffeomorphic to
$D(\xi)\sqcup_{\varphi}D^{2m}$ for some diffeomorphism
$\varphi:S(\xi)=\partial D(\xi)\rightarrow S^{2m-1}$, where $D(\xi)$ and
$S(\xi)$ are the associated disc bundle and sphere bundle for certain vector
bundle of rank $m$ over $S^{m}$. It follows that on every Eells-Kuiper
projective plane we can construct a Morse-Bott function whose critical set has
two components, a point and $S^{m}$. Therefore, by Theorem 1.1, we get the
following result.

######  Proposition 3.2.

For $m=4$ or $8$, each Eells-Kuiper projective plane $M^{2m}$ admits a metric
and an isoparametric function so that, one component of the focal set is a
single point and the other is diffeomorphic to $S^{m}$.

######  Remark 3.2.

On the projective planes $CP^{2}$, $HP^{2}$ and $CaP^{2}$ with Fubini-Study
metrics, there are isoparametric functions so that the focal sets have two
components, a single point, and $S^{2}$, $S^{4}$ and $S^{8}$, respectively. In
fact, these isoparametric foliations are all homogeneous. For $CP^{2}$ and
$HP^{2}$, one uses the Hopf fibrations to derive the desired conclusion. As
for $CaP^{2}$, one uses the cohomogenity one action of $\mathrm{Spin(9)}$ on
$F_{4}/\mathrm{Spin(9)}$ which is isometric to $CaP^{2}$.

##  4\. Non-existence results on exotic spheres

In this section, Proposition 1.1 and Theorem 1.2 are proved. Although any
exotic $n$-sphere with $n>4$ admits an isoparametric function with 2 points as
the focal set by Theorem 1.1, Proposition 1.1 means that regular level
hypersurfaces of such isoparametric functions on an exotic sphere can not have
strong extrinsic geometric properties. This is a way to distinguish exotic
spheres from standard spheres by geometry.

Proof of Proposition 1.1:

###### Proof.

Let $2\delta:=d(m_{+},m_{-})$, where $d$ is the distance function with respect
to the Riemannian metric $g$. The existence of thus isoparametric function $f$
implies $\Sigma^{n}$ can be decomposed as

$\Sigma^{n}=\mathrm{exp}D_{\delta}(T_{m_{-}}\Sigma^{n})\cup\mathrm{exp}D_{\delta}(T_{m_{+}}\Sigma^{n}),$

where $D_{\delta}(T_{m_{-}}\Sigma^{n})$ and $D_{\delta}(T_{m_{+}}\Sigma^{n})$
are the discs of radius $\delta$ at $T_{m_{-}}\Sigma^{n}$ and
$T_{m_{+}}\Sigma^{n}$ respectively. It follows that $\Sigma^{n}$ is
diffeomorphic to $D^{n}(1)\sqcup_{\eta}D^{n}(1)$, where
$\eta:S^{n-1}(1)\rightarrow S^{n-1}(1)$ is determined by the following diagram

$\begin{array}[]{ccc}S^{n-1}(1)&\hookrightarrow
D^{n}(1)\xrightarrow{\rho_{+}^{\delta}}T_{m_{+}}\Sigma^{n}\xrightarrow{\mathrm{exp}_{m_{+}}}&\Sigma^{n}\\\
\downarrow{\eta}&&\parallel\\\ S^{n-1}(1)&\hookrightarrow
D^{n}(1)\xrightarrow{\rho_{-}^{\delta}}T_{m_{-}}\Sigma^{n}\xrightarrow{\mathrm{exp}_{m_{-}}}&\Sigma^{n},\end{array}$

where $\rho_{+}^{\delta}$ and $\rho_{-}^{\delta}$ are the canonical linear
diffeomorphisms.

Since the regular hypersurfaces of $f$ are all totally umbilical, we can
assume the regular hypersurface $X_{t}:=\\{x\in\Sigma^{n}|~{}d(p,m_{-})=t\\}$
has constant principal curvature $\lambda(t)$ for any $t\in(0,2\delta)$.
According to Kowalski and Vanhecke ( [KV86], Theorem 12 ), we have

$\mathrm{exp}_{m_{-}}^{*}g=H(r)\sum_{i=1}^{n}(\mathrm{d}x_{i})^{2}+\frac{1-H(r)}{r^{2}}(\sum_{i=1}^{n}x_{i}\mathrm{d}x_{i})^{2},$

where
$H(r)=\mathrm{exp}\\{2\int_{0}^{r}(\lambda(t)-\frac{1}{t})\mathrm{d}t\\}$,
$r^{2}=\sum_{i=1}^{n}(x_{i})^{2}\in(0,\delta^{2}]$, and
$(x_{1},x_{2},...,x_{n})$ is the normal coordinate around $m_{-}$. Similarly,
near $m_{+}$,

$\mathrm{exp}_{m_{+}}^{*}g=H(r)\sum_{i=1}^{n}(\mathrm{d}y_{i})^{2}+\frac{1-H(r)}{r^{2}}(\sum_{i=1}^{n}y_{i}\mathrm{d}y_{i})^{2},$

where
$H(r)=\mathrm{exp}\\{2\int_{0}^{r}(\lambda(2\delta-t)-\frac{1}{t})\mathrm{d}t\\}$,
$r^{2}=\sum_{i=1}^{n}(y_{i})^{2}\in(0,\delta^{2}]$, and
$(y_{1},y_{2},...,y_{n})$ is the normal coordinate around $m_{+}$. By the
diagram introducing the diffeomorphism $\eta$ and the formulae of the metric
around $m_{+}$ and $m_{-}$, we conclude that $\eta:S^{n-1}(1)\rightarrow
S^{n-1}(1)$ is an isometry. It follows at once that $\Sigma^{n}$ is
diffeomorphic to $S^{n}(1)$, which completes the proof. ∎

At last, for odd dimensional homotopy spheres, we have Theorem 1.2 by a
topological argument.

Proof of Theorem 1.2:

###### Proof.

Let $\Sigma^{2n+1}$ be a homotopy sphere and
$f:\Sigma^{2n+1}\rightarrow\mathbb{R}$ a totally isoparametric function with
respect to a metric $g$ on $\Sigma^{2n+1}$. Then for any regular value $t$,
the regular hypersurface $M_{t}:=f^{-1}(t)$ has constant principal curvatures
in $(\Sigma^{2n+1},g)$. Let $m(t)$ be the number of distinct principal
curvatures of $M_{t}$ and assume $m(t)>1$ for some $t$. Then the tangent
bundle of $M_{t}$ has the decomposition as follows,

$TM_{t}=T_{1}\oplus T_{2}\oplus\cdot\cdot\cdot\oplus T_{m(t)},$

where $T_{1}$, $T_{2}$,…, $T_{m(t)}$ are the principal distributions. On the
other hand, we observe that each regular hypersurface $M_{t}$ is diffeomorphic
to $S^{2n}$, since $M_{t}$ is a geodesic hypersphere of each of two focal
points. Hence, the Euler characteristic classes satisfy

$0\neq e(TM_{t})=e(T_{1})e(T_{2})\cdot\cdot\cdot e(T_{m(t)}).$

However, the right side of the equality is equal to $0$, since
$\mathrm{rank}T_{k}<2n$, each class $e(T_{k})\in$
$H^{\mathrm{rank}T_{k}}(M_{t})=H^{\mathrm{rank}T_{k}}(S^{2n})$ vanishes, for
$k\in\\{1,2,...,m(t)\\}$. That is a contradiction, which implies that $m(t)=1$
and the regular hypersurface $M_{t}$ is totally umbilical for any regular
value $t$. The conclusion we hoped follows immediately from Proposition 1.1. ∎

######  Remark 4.1.

According to [HH67], there exists an exotic Kervaire sphere $\Sigma^{4m+1}$
which has a cohomogeneity one action. Consequently, by a pretty observation in
[GT12], $\Sigma^{4m+1}$ admits a totally isoparametric function $f$ under an
invariant metric. However, each component of the focal set of $f$ is not just
a point, but a smooth submanifold. Hence, the assumption on the focal set in
Theorem 1.2 is essential.

Due to Theorem 1.2, it is reasonable to ask

######  Problem 4.1.

Does there exist an even dimensional exotic sphere $\Sigma^{2n}(n>2)$ which
admits a metric and a totally isoparametric function with 2 points as the
focal set?

##  5\. Isoparametric functions and $SC^{p}$-property

In this section, both isoparametric functions and $SC^{p}$-structure are
investigated simultaneously. First, we recall the definition. Let $(M,g)$ be a
Riemannian manifold and $p$ be a point in $M$. If all the geodesics issued
from $p$ are simply closed geodesics with the same length, we say that $(M,g)$
satisfies the _$SC^{p}$ -property_ at $p\in M$.

For background knowledge and a systematical research, we refer to the classic
book [Be78].

Next, we give the following existence theorem which improves a beautiful
result of Bérard-Bergery [BB77]( see also pp.159, pp.234-235 in [Be78] ).

######  Theorem 5.1.

On any homotopy sphere $\Sigma^{2n}$ in $2\Theta_{2n}(n\geq 3)$, there exists
a Riemannian metric so that it possesses $SC^{p}$-property at two points, say
$m_{+}$ and $m_{-}$. Furthermore under the same metric, there exists an
isoparametric function $f$ on $\Sigma^{2n}$ with focal set
$C(f)=\\{m_{+},m_{-}\\}$. The later property means that $\Sigma^{2n}$ is
locally harmonic at both points ${m_{+}}$ and ${m_{-}}$.

###### Proof.

We first observe that, given any homotopy sphere $\Sigma^{2n}\in
2\Theta_{2n}$, there exists an orientation-preserving diffeomorphism
$\eta:S^{2n-1}\rightarrow S^{2n-1}$ such that

$\Sigma^{2n}=D^{2n}\sqcup_{\eta\circ\eta}D^{2n}.$

Let $(S^{2n-1},g_{0})$ be the standard unit sphere with volume element
$\omega_{0}$ and the standard antipodal map $\tau_{0}$. By the constructions
in Theorem 2.2 and Theorem 1.1, to prove Theorem 5.1, it suffices to find an
orientation-preserving diffeomorphism $\varphi:S^{2n-1}\rightarrow S^{2n-1}$
such that

(1) $\varphi$ is isotopic to $\eta\circ\eta$;

(2) $\varphi^{*}\omega_{0}=\omega_{0}$;

(3) $\varphi\circ\tau_{0}=\tau_{0}\circ\varphi$.

The conditions (1) and (2) imply the existence of a metric $g$ and an
isoparametric function $f$ with $C(f)=\\{m_{+},m_{-}\\}$, and condition (3)
guarantees that the metric $g$ satisfies the $SC^{p}$-property at $m_{+}$ and
$m_{-}$.

To construct the desired $\varphi$, let
$S_{+}:=\\{x=(x_{1},x_{2},...,x_{2n})\in S^{2n-1}|~{}x_{1}\geq 0\\}$, and
$S_{-}:=\\{x=(x_{1},x_{2},...,x_{2n})\in S^{2n-1}|~{}x_{1}\leq 0\\}$. First,
choose a diffeomorpism $\zeta:S^{2n-1}\rightarrow S^{2n-1}$, which is the
restriction of an orientation-preserving diffeomorphism from $D^{2n}$ to
$D^{2n}$, such that the composition $\eta^{\prime}:=\eta\circ\zeta$ satisfies
$\eta^{\prime}|_{S_{+}}=\mathrm{id}$ and $\eta^{\prime}$ is isotopic to
$\eta$. Thus we get two forms $(\eta^{\prime})^{*}\omega_{0}$ and $\omega_{0}$
on $S^{2n-1}$, which are equal on the domain $S_{+}$ and have equal integral
over the domain $S_{-}$. Applying Lemma 2.1 to the two forms, we have a
diffeomorphism $u:S^{2n-1}\rightarrow S^{2n-1}$ such that $u$ is isotopic to
$\mathrm{id}$, $u|_{S_{+}}=\mathrm{id}$, and
$u^{*}((\eta^{\prime})^{*}\omega_{0})=\omega_{0}$. Define a diffeomorphism
$\tilde{\eta}:=\eta^{\prime}\circ u$. It is clear that
$\tilde{\eta}|_{S_{+}}=\mathrm{id}$,
$(\tilde{\eta})^{*}\omega_{0}=\omega_{0}$.

Now, the diffeomorpism $\varphi$ is defined by the compositions
$\varphi:=\tau_{0}\circ\tilde{\eta}\circ\tau_{0}\circ\tilde{\eta}$. We are
left to verify that $\varphi$ possesses the required properties. For condition
(1), we note that $\tau_{0}$ is isotopic to $\mathrm{id}$ on the odd
dimensional sphere $S^{2n-1}$. While condition (2) is evident to verify. As
for condition (3), since $\tau_{0}\circ\tilde{\eta}\circ\tau_{0}$ is the
identity on $S_{-}$, it follows that $\tau_{0}\circ\tilde{\eta}\circ\tau_{0}$
commutes with $\tilde{\eta}$, and thus
$\varphi\circ\tau_{0}=\tau_{0}\circ\varphi$. It completes the proof. ∎

######  Remark 5.1.

The first half conclusion of Theorem 5.1 was proved by Bérard-Bergery [BB77] (
pp.237. see also pp.4, pp.234, pp.235 of [Be78]). The later half means that
every geodesic hypersphere centered at $m_{+}$ or $m_{-}$ is of constant mean
curvature. In particular, there exists at least one exotic sphere of dimension
$10$ which possesses these properties.

As for odd dimensional homotopy spheres, we have the following non-existence
result.

######  Theorem 5.2.

Let $\Sigma^{7}$ be a 7-dimensional homotopy sphere and
$f:\Sigma^{7}\rightarrow\mathbb{R}$ a transnormal function with the focal set
$C(f)=\\{m_{+},m_{-}\\}$ under some metric $g$. Suppose that $(\Sigma^{7},g)$
satisfies the $SC^{p}$-property at either $m_{+}$ or $m_{-}$. Then
$\Sigma^{7}$ is diffeomorphic to $S^{7}$ or the element $[14]$ in
$\Theta_{7}=\mathbb{Z}_{28}$.

###### Proof.

As in the proof of Proposition 1.1, let $2\delta:=d(m_{+},m_{-})$, where $d$
is the distance function with respect to the Riemannian metric $g$. Since $f$
is a transnormal function with $C(f)=\\{m_{+},m_{-}\\}$, $\Sigma^{7}$ has the
following decomposition,

$\Sigma^{n}=\mathrm{exp}D_{\delta}(T_{m_{-}}\Sigma^{n})\cup\mathrm{exp}D_{\delta}(T_{m_{+}}\Sigma^{n}),$

where $D_{\delta}(T_{m_{-}}\Sigma^{7})$ and $D_{\delta}(T_{m_{+}}\Sigma^{7})$
are the discs of radius $\delta$ at $T_{m_{-}}\Sigma^{7}$ and
$T_{m_{+}}\Sigma^{7}$ respectively. What's more, $\Sigma^{7}$ is diffeomorphic
to $D^{7}(1)\sqcup_{\sigma}D^{7}(1)$, where $\sigma:S^{6}(1)\rightarrow
S^{6}(1)$ is determined by the following diagram,

$\begin{array}[]{ccc}S^{6}(1)&\hookrightarrow
D^{7}(1)\xrightarrow{\rho_{+}^{\delta}}T_{m_{+}}\Sigma^{7}\xrightarrow{\mathrm{exp}_{m_{+}}}&\Sigma^{7}\\\
\downarrow{\sigma}&&\parallel\\\ S^{6}(1)&\hookrightarrow
D^{7}(1)\xrightarrow{\rho_{-}^{\delta}}T_{m_{-}}\Sigma^{7}\xrightarrow{\mathrm{exp}_{m_{-}}}&\Sigma^{7},\end{array}$

where $\rho_{+}^{\delta}$ and $\rho_{-}^{\delta}$ are the canonical linear
diffeomorphisms.

The assumption that $(\Sigma^{7},g)$ satisfies the $SC^{p}$-property at the
point $m_{+}$($m_{-}$) implies that it satisfies the $SC^{p}$-property at the
point $m_{-}$($m_{+}$). That is to say, the assumption can deduce that
$(\Sigma^{7},g)$ satisfies the $SC^{p}$-property at both of the 2 points
$m_{+}$ and $m_{-}$. Thus we obtain

$\sigma\circ\tau_{0}=\tau_{0}\circ\sigma.$

Define a map

$\begin{array}[]{ccc}\mathcal{A}&:&\Theta_{7}=\mathbb{Z}_{28}\rightarrow\Theta_{7}=\mathbb{Z}_{28}.\\\
&&D^{7}\sqcup_{\varphi}D^{7}\mapsto
D^{7}\sqcup_{\tau_{0}\circ\varphi\circ\tau_{0}}D^{7}\end{array}$

It is obvious that $\mathcal{A}$ is a well-defined homomorphism. According to
[ADPR], $\mathcal{A}$ is determined as follows,

$\begin{array}[]{ccc}\mathcal{A}&:&\Theta_{7}=\mathbb{Z}_{28}\rightarrow\Theta_{7}=\mathbb{Z}_{28}.\\\
&&[n]\mapsto[-n]\end{array}$

Now, we have the diffeomorphsims
$\mathcal{A}(\Sigma^{7})=\mathcal{A}(D^{7}\sqcup_{\sigma}D^{7})=D^{7}\sqcup_{\tau_{0}\circ\sigma\circ\tau_{0}}D^{7}=D^{7}\sqcup_{\sigma}D^{7}=\Sigma^{7}$.
Hence, it follows that $\Sigma^{7}=[0]$ or
$[14]\in\mathbb{Z}_{28}=\Theta_{7}$. ∎

######  Remark 5.2.

Similarly, let $\Sigma^{15}$ be a 15-dimensional homotopy sphere which bounds
a parallelizable manifold and $f:\Sigma^{15}\rightarrow\mathbb{R}$ a
transnormal function with the focal set $C(f)=\\{m_{+},m_{-}\\}$ under some
metric $g$. Suppose in addition that $(\Sigma^{15},g)$ satisfies the
$SC^{p}$-property at either the point $m_{+}$ or $m_{-}$. Then $\Sigma^{15}$
is diffeomorphic to $S^{15}$ or $\Sigma^{15}=[4064]\in
bP_{16}=\mathbb{Z}_{8128}$.

######  Remark 5.3.

Very recently, Tang and Zhang [TZ13] solved a problem of Bérard-Bergery and
Besse. That is, they showed that every Eells-Kuiper quaternionic projective
plane carries a Riemannian metric with $SC^{p}$ property for certain point
$p$. We don't know how to improve our Proposition 3.2. For instance, we don't
know whether there is a metric on every Eells-Kuiper quaternionic projective
plane with not only the property in Proposition 3.2, but also the
$SC^{p}$-property.

######  Acknowledgements .

The authors extend their heartfelt thanks to Professor Weiping Zhang for
providing the reference [DM90] and very useful comments on Moser's theorem.
Thanks are also due to Dr. Jianqian Ge for his good ideas to prove Proposition
1.1 and Theorem 1.2.

## References

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# Automated Proof of Mixed Trigonometric-polynomial Inequalities in the
Unbounded Case

Chen shiping  Sichuan Trade School, Yaan, 625107, Sichuan, China
<EMAIL_ADDRESS>[ Ge Xinyu  Chengdu Institute of Computer Application,
Chinese Academy of Sciences, Chengdu, 610000, Sichuan, China  University of
Chinese Academy of Sciences, Beijing, 100049, Beijing, China<EMAIL_ADDRESS>[

###### Abstract

Mixed trigonometric-polynomials frequently occur in applications in physics,
numerical analysis and engineering, the algorithm has been already proposed to
determine its sign on $(0,\frac{\pi}{2}]$. This paper proposes a procedure to
extend the interval to $(0,+\infty)$. Such generalization is by no means
trivial, for the process depends on boundedness of roots and square-free
factorization of mixed trigonometric-polynomials.

###### keywords:

Mixed trigonometric-polynomial; Boundedness of roots; Square-free
factorization; Automated proof of inequalities.

url]URL 1

url]URL 2

##  1 Introduction

Automatic proof of the transcendental inequalities and sign-deciding of
transcendental functions have always been hot and difficult problems in the
field of mathematical mechanization and automatic reasoning in recent years.
Among the transcendental functions, a class of functions named mixed
trigonometric-polynomial (denoted as Trigonometric-polynomial or MTP simply)
and defined by the formula $F(x)=\sum
a_{i}x^{p_{i}}\sin^{q_{i}}(\omega_{i}x)\cos^{r_{i}}(\omega_{i}x)$ attract more
and more scholars’ attention[1-8], which frequently occur in applications in
physics, numerical analysis and engineering, where $a_{i}\in\mathbb{R}$,
$p_{i},q_{i},r_{i}\in\mathbb{N}\cup\\{0\\}$, $\omega_{i}\in\mathbb{Q}$.
Moreover, an inequality of the form $F(x)>0$ is called mixed trigonometric
polynomial inequality (MTP inequality), where $F(x)$ is an MTP. Let $p$ be the
least common divisor of numerators of $\omega_{i}$s, and $q$ the greatest
common multiple of denominators, $y=\frac{p}{q}x$, then an MTP can be reduced
to another equivalent form $G(y)=g(y,\sin(y),\cos(y))=\sum
a_{i}y^{p_{i}}\sin^{q_{i}}(y)\cos^{r_{i}}(y)$, where
$g\in\mathbb{R}[x_{1},x_{2},x_{3}]$.

Ref[8] presents an algorithm for automated proof of MTP inequality and
deciding the sign of MTP. By Taylor expansion, the proving of the target
inequality is reduced to the verification of a series of polynomial
inequalities with only one variable, and then completed by algebraic
inequality-proving package such as BOTTEMA. The algorithm is complete and
experiments show that it is very effective for mixed trigonometric-polynomial
inequalities, furthermore, the procedure is ’readable’. However, the algorithm
limits the domain of inequality on $(0,\frac{\pi}{2}]$ because the Taylor
expansion of $\arctan(x)$ is convergent only on $[0,1]$, where it is assumed
that $x\geq 0$. Of course, the domain of the trigonometric-polynomial
inequalities can be extended to $(0,T)$ for $\forall T>0$ by the double angle
formulas of trigonometric function.

In this paper, we are to discuss how to decide the sign of mixed
trigonometric-polynomial $f(x)=f(x,\sin(x),\cos(x))$ on the unbounded interval
$(0,+\infty)$. The generalization is non-trivial, for it is necessary to
discuss the boundedness of MTP’s roots and the square-free decomposition of
MTPs to deal with the unbounded case.

Ref[9] discusses the boundedness of roots of the so-called trigonometric-
exponential polynomial with the form of $f(t)=\sum
e^{r_{j}t}(P_{1,j}(t)\cos(\omega_{j}t)+P_{2,j}(t)\sin(\omega_{j}t))$, and
presents a deciding procedure in two cases using the theory of semi-algebraic
sets , the first case is that the set of frequencies ${\omega_{j}}$ spans a
one-dimensional vector space over $\mathbb{Q}$ , the second is that the set of
frequencies ${\omega_{j}}$ spans a two-dimensional vector space over
$\mathbb{Q}$ and the polynomials $P_{1,j}(t)$ and $P_{2,j}(t)$ are all
constants. Ref[9]’s methods can be used to decide the boundedness of MTP’s
roots, while we will design a simpler and more convenient scheme for the
specific MTPs with the help of the Sylvester resultant.

Factorization of trigonometric functions is a classic field and the following
methods are frequently used. The first is the quotient ring
$Q[s,c]/<s^{2}+c^{2}-1>$, but it is not a unique factorization domain and so,
the factorization is not unique in general,furthermore,it is still needed to
decide whether each factor has multiple roots[10]. The second is using Tan-
half angle substitution, $\sin(x)=\frac{2t}{1+t^{2}}$,
$\cos(x)=\frac{1-t^{2}}{1+t^{2}}$, to transform the trigonometric polynomial
to a rational expression, where $t=\tan(\frac{x}{2})$ and $x\neq(2k+1)\pi$ for
$k\in\mathbb{N}$ [10-11]. The third is reducing the trigonometric functions to
polynomials in complex field $Q(I)[e,e^{-1}]$ by Euler Theorem[11], where $e$
denotes $e^{Ix}$. This scheme requires $g(e)\in Q(I)[e,e^{-1}]$ be even or odd
function of $e$. Besides, the above methods can deal with the trigonometric
functions with form $f(\sin(x),\cos(x))$ only, where
$f(x,y)\in\mathbb{R}[x,y]$ or $\mathbb{Q}[x,y]$. In this paper, we will
present a procedure to decompose the mixed trigonometric-polynomials with the
form $f(x,\sin(x),\cos(x))$, i.e. the monomials contain one variable and
trigonometric functions applied to the same variable, and the domain is
$(0,+\infty)$ or $(-\infty,+\infty)$ without excluding any special point.

The rest of the paper is organized as follows. Section 2 studies the positive
root boundedness of MTP. Section 3 proposes the scheme of square-free
factorization of MTP. Section 4 presents deciding procedure of the sign of MTP
on unbounded interval $(0,+\infty)$. We will conclude the paper in Section 5.

##  2 Decision of the Positive Root Boundedness of Mixed Trigonometric-
Polynomial

In this section, we aim to decide the positive root boundedness of mixed
trigonometric-polynomial $F(x)=f(x,\sin(x),\cos(x))$.

Algorithm 2.1 Decision of the positive root boundedness of MTP

Input: $F(x)=f(x,\sin(x),\cos(x))$, where $f\in\mathbb{R}_{alg}[x,y,z]$; Where
$\mathbb{R}_{alg}$ is the set of real algebraic numbers.

Output: whether the positive roots of $F(x)$ are bounded or not, if bounded,
output one upper bound.

1) $f\\_{lst}\leftarrow factor(square-free(f(x,s,c)))$;

2) To decide if positive roots of each function in $f\\_{lst}$ are bounded,
$f_{i}(x,s,c)\leftarrow f\\_{lst}[i]$; Write $f_{i}(x,s,c)$ as $f(x,s,c)$ for
convenience below.

2.1) if $degree(f,c)$ is odd, then

2.1.1)$f_{1}(x,s)\leftarrow square-free(res(f,s^{2}+c^{2}-1,c))$,

2.1.2) $f_{2}(x)\leftarrow square-free(res(f_{1},diff(f_{1},s),s))$;

2.1.3) $r_{0}\leftarrow$ the maximum positive root of $f_{2}(x)$. (
$r_{0}\leftarrow 0$ if $f_{2}(x)$ has no positive root );

2.1.4) Select $\forall r_{1}>r_{0}$, if $g(s)=f_{1}(r_{1},s)=0$ has no real
root, then the positive root of $f(x,\sin(x),\cos(x))$ is bounded and $r_{0}$
is an upper bound.

If $g(s)=f_{1}(r_{1},s)=0$ has at least one real root, then the positive root
of $f(x,\sin(x),\cos(x))$ is unbounded, which implies that the positive root
of $F(x)$ is unbounded. The algorithm ends.

2.2)Else (the case that $degree(f,c)$ is even)

2.2.1) Reduce $f(x,s,c)$ to $g(x,s)$ by the substitution $c^{2}=1-s^{2}$;
where we suppose that $degree(g,s)>0$, otherwise $g$ is a unary polynomial of
$x$ and its positive roots are bounded.

2.2.2) $g_{1}(x)\leftarrow res(s+1,g,s)\times res(s-1,g,s)\times
res(g,diff(g,s),s)$;

2.2.3) $r_{0}\leftarrow$ the maximum positive root of $g_{1}(x)$. (
$r_{0}\leftarrow 0$ if $g_{1}(x)$ has no positive root );

2.2.4) Select $\forall r_{1}>r_{0}$, if $g_{2}(s)=g(r_{1},s)$ has no real root
on $(-1,1)$, then the positive root of $g(x,\sin(x))$ is bounded, and $r_{0}$
is an upper bound.

If $g_{2}(s)=g(r_{1},s)$ has a least one real root on $(-1,1)$, then the
positive root of $g(x,\sin(x))$ is unbounded, i.e. the positive root of $F(x)$
is unbounded. The algorithm ends.

3) The maximum value of the positive roots bounds of all functions in
$f\\_lst$ is one bound of $F(x)$. The algorithm ends.

Here $res(f,g,x)$ denotes the Sylvester resultant of $f(x)$ and $g(x)$ about
$x$, similarly hereinafter.

Lemma 2.1[12] Suppose that $f(a,x)=a_{m}x^{m}+a_{m-1}x^{m-1}+\cdots+a_{0}$ is
a unary polynomial of real parameter coefficients, where $a$ represents
$a_{m},a_{m-1},\dots,a_{0}$. Denote $R(a)=res(f,f^{\prime},x)$, if $c_{1}$ and
$c_{2}$ are two points in the same connected branch of $R(a)\neq 0$ in
parameter space $R^{m+1}$, then $f(c_{1},x)$ and $f(c_{2},x)$ have the same
number of real roots.

Theorem 2.1 Algorithm 2.1 is correct.

###### Proof.

First, suppose that $degree(f,c)$ is odd.

If $\exists r_{1}>r_{0}$, $g(s)=f_{1}(r_{1},s)=0$ has no real root, then
$g_{r}(s)=f_{1}(r,s)=0$ has no real root for any $r>r_{0}$ by Lemma 2.1.
Specially, $\sin(r)$ is not a root of $g_{r}(s)$, so $f_{1}(r,\sin(r))\neq 0$,
which is to say that $res(f(r,\sin(r),c),\sin^{2}(r)+c^{2}-1,c)\neq 0$, so
$f(r,\sin(r),c)$ has no common root with $\sin^{2}(r)+c^{2}-1$, as
$\sin^{2}(r)+\cos^{2}(r)-1=0$, so $f(r,\sin(r),\cos(r))\neq 0$, hence the root
of $f(x,\sin(x),\cos(x))$ is not bigger than $r_{0}$, so its positive roots is
bounded.

If $\exists r_{1}>r_{0}$, $g(s)=f_{1}(r_{1},s)=0$ has at least one real root.
Suppose $T$ be an arbitrary given positive real number.

Let $c=0$ under the assumption that $f(x,s,c)=0$ and $s^{2}+c^{2}=1$, $s=1$ or
$-1$ obviously. If $f(x,1,0)$ or $f(x,-1,0)$ is identically equal to $0$, the
positive roots of $f(x,\sin(x),\cos(x))$ is unbounded clearly. Without loss of
generality, we can suppose that $f(x,1,0)$ and $f(x,-1,0)$ are non-constant
polynomials of $x$, their biggest positive roots are $r_{1}$ and $r_{2}$
respectively, then we have that $c\neq 0$ for $\forall
x>r_{3}=max\\{r_{1},r_{2}\\}$, i.e. $c$ has a constant sign for $x>r_{3}$.

By Lemma 2.1, for any $r>T_{0}=max\\{r_{3},T,r_{0}\\}$, where $r_{0}$ is the
maximum positive root of $f_{2}(x)$ in Algorithm 2.1, $g_{r}(s)=f_{1}(r,s)=0$
has at least one real root, denotes $s(r)$ as its smallest one, then by the
continuous dependence of the roots of a polynomial on its coefficients, we
know that $s(r)$ is continuous on $(T_{0},+\infty)$.

Recall $f_{1}(r,s(r))=0$ implies that $f(r,s(r),c)$ and $s(r)^{2}+c^{2}-1$
have common roots, denoted by $c_{0}$, by the assumption that $degree(f,c)$ is
odd we have that $c_{0}$ must be real-valued, so we can declare that $-1\leq
s(r)\leq 1$ due to $s(r)^{2}+c_{0}^{2}-1=0$.

Assume that $c>0$ for $x\in(r_{3},+\infty)$. For $\forall k\in\mathbb{N}$, the
value of $\sin(x)$ changes from $-1$ to $1$ continuously on
$[2k\pi-\frac{\pi}{2},2k\pi+\frac{\pi}{2}]$. Suppose $k$ is big enough such
that $2k\pi-\frac{\pi}{2}>T_{0}$, then $\sin(x)$ and $s(r)$ has at least one
intersection on $[2k\pi-\frac{\pi}{2},2k\pi+\frac{\pi}{2}]$, that is to say,
there exists $x_{0}\in[2k\pi-\frac{\pi}{2},2k\pi+\frac{\pi}{2}]$ such that
$\sin(x_{0})=s(x_{0})$, so $f_{1}(x_{0},\sin(x_{0}))=0$, and clearly
$\cos(x_{0})>0$. As
$f_{1}(x_{0},\sin(x_{0}))=res(f(x_{0},\sin(x_{0}),c)$,$\sin^{2}(x_{0})+c^{2}-1,c)=0$
implies $f(x_{0},\sin(x_{0}),c)$ has common root with
$\sin^{2}(x_{0})+c^{2}-1$, $\sin^{2}(x_{0})+c^{2}-1=0$ and $c>0$ derive
$c=\cos(x_{0})$, i.e. $f(x_{0},\sin(x_{0}),\cos(x_{0}))=0$, $x_{0}$ is a root
of $F(x)$, which is bigger than the given $T$, so the positive roots of $F(x)$
is unbounded.

Suppose that $c<0$ for $x\in(r_{0},+\infty)$. As $\sin(x)$ changes from $1$ to
$-1$ continuously on $[2k\pi+\frac{\pi}{2},2k\pi+\frac{3\pi}{2}]$, then for
$\forall k\in\mathbb{N}$ such that $2k\pi+\pi/2>k$, $\sin(x)$ and $s(r)$ have
at least one intersection on $[2k\pi+\frac{\pi}{2},2k\pi+\frac{3\pi}{2}]$,
that is to say, there exists
$x_{0}\in[2k\pi+\frac{\pi}{2},2k\pi+\frac{3\pi}{2}]$ such that
$\sin(x_{0})=s(x_{0})$, so $f_{1}(x_{0},\sin(x_{0}))=0$, obviously
$\cos(x_{0})<0$. $\cos(x_{0})$ is the common root of $f(x_{0},\sin(x_{0}),c)$
and $\sin(x_{0})^{2}+c^{2}-1$, i.e. $f(x_{0},\sin(x_{0}),\cos(x_{0}))=0$,
$x_{0}$ is a root of $F(x)$, which is bigger than the given $T$, so the
positive roots of $F(x)$ is unbounded.

Now suppose that $degree(f,c)$ is even, $g(x,s)=f(x,s,c)$ after the
substitution $c^{2}=1-s^{2}$. In this case, $F(x)=g(x,sin(x))$ and
$g\in\mathbb{R}_{alg}[x,y]$ obviously.

Let $A$ be a connected branch of
$\\{x\in\mathbb{R}|g_{1}(x)=res(s+1,g,s)\times res(s-1,g,s)\times
res(g,diff(g,s),s)\neq 0\\}$, which is an open interval, $(s+1)$ has constant
sign on $A$ and so does $(s-1)$, and for $\forall x_{1},x_{2}\in A$, by Lemma
2.1 $g(x_{1},s)$ and $g(x_{2},s)$ has the same number of real roots.

If $\exists r>r_{0}$, $g_{2}(s)=g(r,s)$ has a least one real root on $(-1,1)$,
then $\forall r\in(r_{0},+\infty)$, $g_{2}(s)=g(r,s)$ has at least one real
root on $(-1,1)$. From the previous reasoning we can obtain that for
$k\in\mathbb{N}$ such that $2k\pi-\pi/2>r_{0}$, $\exists
x_{0}\in[2k\pi-\pi/2,2k\pi+\pi/2]$ such that $g(x_{0},sin(x_{0}))=0$, which
implies that the positive root of $g(x,sin(x))$ is unbounded.

If $\exists r>r_{0}$, $g_{2}(s)=g(r,s)$ has no real root on $(-1,1)$, then
$\forall r\in(r_{0},+\infty)$, $g_{2}(s)=g(r,s)$ has no real root on $(-1,1)$,
so $g(r,sin(r))\neq 0$ for $\sin(r)\in(-1,1)$. If $\sin(r)=1$ or $-1$,
$r=2k\pi+\pi/2$ or $2k\pi-\pi/2$, as $g\in\mathbb{R}_{alg}[x,y]$, we can clare
that $g(2k\pi+\pi/2,1)\neq 0$ and $g(2k\pi-\pi/2,-1)\neq 0$. So
$g(r,sin(r))\neq 0$ for $\forall r>r_{0}$, i.e. $g(x,sin(x))$ has no real root
for $x>r_{0}$, $r_{0}$ is an upper bound of positive roots of
$F(x)=g(x,sin(x))$.

∎

To describe the running process of Algorithm 2.1, we present the following
examples.

Example 1 Decide whether the positive roots of
$F(x)=\frac{2}{3}x+x\cos(x)-\sin(x)$ is bounded.

Let $f(x,s,c)=\frac{2}{3}x+xc-s$, where $c$ denotes $\cos(x)$ and $s$ denotes
$\sin(x)$,

$f_{1}(x,s)=res(f,s^{2}+c^{2}-1,c)=x^{2}s^{2}-\frac{5}{9}x^{2}-\frac{4}{3}xs+s^{2}$,

$g(x)=res(f_{1},diff(f_{1},s),s)=-\frac{20}{9}x^{6}-\frac{56}{9}x^{4}-4x^{2}$,

$realroot(g(x),\frac{1}{10})=[[0,0]]$, which implies that $g(x)$ has one root
$0$.

$f_{2}(s)=subs(x=1,f_{1}(x))=2s^{2}-\frac{5}{9}-\frac{4}{3}s$,

$realroot(f_{2}(s),\frac{1}{10})=[[\frac{15}{16},1],[-\frac{5}{16},-\frac{1}{4}]]$,
$f_{2}$ has $2$ real roots, which implies that $g_{x}(s)=f_{1}(x,s)$ has $2$
real roots for $\forall x>0$.

As $f(x,1,0)=\frac{2}{3}x-1$, $f(x,-1,0)=\frac{2}{3}x+1$, whose roots are
$\frac{3}{2}$ and $-\frac{3}{2}$, so under the assumption that
$f(x,s,c)=\frac{2}{3}x+xc-s=0$ and $s^{2}+c^{2}=1$, if $x>\frac{3}{2}$, $c<0$
definitely.

Let $T$ be a arbitrary given positive real number,
$T_{0}=max\\{\frac{3}{2},T\\}$, let $k_{0}$ be a number such that
$2k_{0}\pi+\pi/2>T_{0}$, then for each natural number $k\geq k_{0}$, $\exists
x_{0}\in[2k\pi+\frac{\pi}{2},2k\pi+\frac{3\pi}{2}]$ such that
$F(x_{0})=f(x_{0},sin(x_{0}),cos(x_{0}))=0$. So $x_{0}$ is a positive root of
$F(x)$ and $x_{0}>T$, we get that the positive roots of $F(x)$ are unbounded.

Example 2 Decide whether the positive roots of
$F(x)=\frac{2}{3}x+\frac{1}{3}x\cos(x)-\sin(x)$ is bounded.

Let $f(x,s,c)=\frac{2}{3}x+\frac{1}{3}x\cdot c-s$,

$f_{1}(x,s)=res(f,s^{2}+c^{2}-1,c)=\frac{1}{9}x^{2}s^{2}+\frac{1}{3}x^{2}-\frac{4}{3}xs+s^{2}$,

$g(x)=res(f_{1},diff(f_{1},s),s)=\frac{4}{243}x^{6}+\frac{8}{81}x^{4}-\frac{4}{9}x^{2}$,

$realroot(g(x),\frac{1}{10})=[[-\frac{111}{64},-\frac{221}{128}],[0,0],[\frac{221}{128},\frac{111}{64}]]$,

$f_{2}(x)=subs(x=2,f_{1})=\frac{13}{9}s^{2}+\frac{4}{3}-\frac{8}{3}s$,

$realroot(f_{2}(x),\frac{1}{10})=[]$, which means that $f_{2}(x)$ has no real
roots and implies that $g_{x}(s)=f_{1}(x,s)$ has no real roots for $\forall
x>\frac{111}{64}$. Then for $\forall r>\frac{111}{64}$, $sin(r)$ is not the
root of $g_{r}(s)=f_{1}(r,s)$, which is to say,
$f_{1}(r,sin(r))=res(f(r,\sin(r),c),\sin^{2}(r)+c^{2}-1,c)\neq 0$. So,
$\cos(r)$ is not a root of $f(r,\sin(r),c)$, otherwise, $\cos(r)$ is the
common root of $f(r,\sin(r),c)$ and $\sin^{2}(r)+c^{2}-1$.

So, for $\forall r>\frac{111}{64}$, $r$ can not be the real roots of $F(x)$,
i.e. the real roots of $F(x)$ are not bigger than $\frac{111}{64}$.

##  3 Square-free factorization of Mixed Trigonometric polynomial

Euler Theorem shows that $\sin(x)=\frac{e^{Ix}-e^{-Ix}}{2I}$,
$\cos(x)=\frac{e^{Ix}+e^{-Ix}}{2}$. Hence, an MTP can be expressed as an
exponential polynomial in complex field $f(x,e^{Ix},e^{-Ix})$, where
$I^{2}=-1$.

Let $y=e^{Ix}$, then an MTP can be reduced to a Laurent polynomial in the form
of $f[x,y,y^{-1}]$. Denotes $\mathbb{LR}:=\mathbb{C}[x,y,y^{-1}]$, then for
any $P\in\mathbb{LR}$, there exists a $Q\in\mathbb{C}[x,y]$, $P=Q/y^{p}$,
$p\in\mathbb{Z}$. If $factor(Q)$ is a factorization of $Q$, we call
$factor(Q)/y^{p}$ or $y^{-p}factor(Q)$ a factorization of $P$.

By Lindemann Theorem, we have that

Lemma 3.1 For $\forall x\in\mathbb{C}$, there are at least one transcendental
number between $x,e^{Ix}$.

Adjust Theorem 6 of ref[13] slightly, we get the following lemma and its
proof.

Lemma 3.2 If $f_{1}(x,y),f_{2}(x,y)\in\mathbb{A}[x,y]$ are co-prime, then
$F_{1}(x)=f_{1}(x,e^{Ix})$ and $F_{2}(x)=f_{2}(x,e^{Ix})$ have no common roots
other than $0$.

###### Proof.

Suppose $F_{1}(x)$ and $F_{2}(x)$ have common root $x_{0}$ and $x_{0}\neq 0$.

Let $g(x)=res(f_{1}(x,y),f_{2}(x,y),y)$ then $g\in\mathbb{A}[x]$ and
$g(x_{0})=0$. As $f_{1}(x,y)$ and $f_{2}(x,y)$ are co-prime, then $g(x)$ can
not be identically zero, so $x_{0}$ is algebraic. Then
$h(y)=f_{1}(x_{0},y)\in\mathbb{A}[y]$, so
$h(e^{Ix_{0}})=f_{1}(x_{0},e^{Ix_{0}})=0$, which implies $e^{Ix_{0}}$ is
algebraic and contradicts Lemma 3.1. ∎

Theorem 3.1 If $f(x,y)\in\mathbb{A}[x,y]$ is irreducible, then
$F(x)=f(x,e^{Ix})$ has no multiple roots other than $0$.

###### Proof.

Let $f^{\prime}(x,y)\in\mathbb{A}[x,y]$ such that
$F^{\prime}(x)=f^{\prime}(x,e^{Ix})$, then
$f^{\prime}(x,y)=f^{\prime}_{x}+Iyf’_{y}$, so $degree(f^{\prime},x)\leq
degree(f,x)$, $degree(f^{\prime},y)\leq degree(f,y)$. As $f(x,y)$ is
irreducible, so $f(x,y)$ and $f^{\prime}(x,y)$ are co-prime. Then $F(x)$ and
$F^{\prime}(x)$ have no common roots other than $0$. We conclude that Theorem
3.1 holds. ∎

Corollary 3.1 If $f(x,y)\in\mathbb{A}[x,y]$ is square-free, then
$F(x)=f(x,e^{Ix})$ has no multiple roots other than $0$.

We extend the operation of complex conjugation to $\mathbb{LR}$ as follows.

Given $P=\sum\limits_{j=1}^{n}a_{j}(x)y^{v_{j}}\in\mathbb{LR}$, where
$a_{1},\cdots,a_{n}\in\mathbb{C}[x]$, define its conjugate to be
$con(P)=\sum^{n}_{j=1}\overline{a_{j}(x)}y^{-v_{j}}$, where
$\overline{a_{j}(x)}$ is the conjugate function of $a_{j}(x)$, and obviously,
$con(a_{j}(x))=\overline{a_{j}(x)}$.

In this section, we extend the coefficients of MTP to complex field. For
$P=\sum\limits_{j=1}^{n}a_{j}(x)y^{v_{j}}\in\mathbb{LR}$, $v\in\mathbb{Q}$,
denote
$LRhom[v](P)=P(x,e^{Ivx})=\sum\limits_{j=1}^{n}a_{j}(x)(\cos(v_{j}vx)+I\sin(v_{j}vx))$,
then $LRhom[v](P)$ is a mixed trigonometric-polynomial with coefficients in
complex field. $LRhom[1](P)$ is abbreviated as $LRhom(P)$.

For an MTP $F(x)=f(x,\sin(x),\cos(x))$, let
$P(x,y)=f(x,\frac{y-1/y}{2I},\frac{y+1/y}{2})\in\mathbb{LR}$, then
$LRhom(P)=F(x)$ obviously.

Lemma 3.3 For $P=\sum\limits_{j=1}^{n}a_{j}(x)y^{v_{j}}\in\mathbb{LR}$,
$LRhom[v](con(P))=\overline{LRhom[v](P)}$ for $\forall v\in\mathbb{Q}$.

###### Proof.

$LRhom[v](con(P))=LRhom[v](\sum^{n}_{j=1}\overline{a_{j}(x)}y^{-v_{j}})=\sum\limits_{j=1}^{n}\overline{a_{j}(x)}(\cos(-v_{j}vx)+I\sin(-v_{j}vx))=\sum\limits_{j=1}^{n}\overline{a_{j}(x)}(\cos(v_{j}vx)-I\sin(v_{j}vx))=\sum\limits_{j=1}^{n}\overline{a_{j}(x)(\cos(v_{j}vx)+I\sin(v_{j}vx))}=\overline{LRhom[v](P)}$
∎

By Lemma 3.3 we get that

Lemma 3.4 If $P=con(P)$, then $LRhom[v](P)$ is real-valued for $\forall
v\in\mathbb{Q}$.

Theorem 3.2 If $P=\sum_{j=1}^{n}a_{j}(x)y^{v_{j}}\in\mathbb{LR}$ such that
$P=con(P)$, $P$ can be factorized as $Cy^{p}P_{1}^{r_{1}}\cdots
P_{n}^{r_{n}}$, where $C\in\mathbb{C}$ and
$P_{1},\dots,P_{n}\in\mathbb{C}[x,y]$ are square-free and pairwisely co-prime,
then for each $i$,

1) there exist $C_{i}\in\mathbb{C}$, $p_{i}\in\mathbb{Z}$, such that
$P_{i}=C_{i}y^{p_{i}}con(P_{i})$;

2) $f_{i}(x)=LRhom(P_{i}Z^{-p_{i}/2}C_{i}^{-1/2})$ is real-valued or pure
imaginary, where $f_{i}$ has no multiple root other than $0$ for
$i=1,\dots,n$, $f_{i}$ and $f_{j}$ have no common root other than $0$ for
$i\neq j$; 3) $LRhom(P)=C_{0}f_{1}^{r_{1}}\cdots f_{n}^{r_{n}}$, where
$C_{0}=C(C_{1})^{r_{1}/2}\dots(C_{n})^{r_{n}/2}$.

###### Proof.

1) For $n=1$, $CZ^{p}(P_{1})^{r_{1}}=\overline{C}Z^{-p}con(P_{1})^{r_{1}}$,
thus $P_{1}=(\overline{C}/C)^{1/r_{1}}Z^{-2p/r_{1}}con(P_{1})$, the conclusion
holds.

Suppose the conclusion holds for $n=m-1$, that is to say, there exist
$C_{i},p_{i}$ such that $P_{i}=C_{i}Z^{p_{i}}con(P_{i})$ for $1<i<m$. Let
$n=m$,

$P=CZ^{p}(P_{1})^{r_{1}}\cdots(P_{m-1})^{r_{m-1}}P_{m}^{r_{m}}=CZ^{p}(C_{1})^{r_{1}}Z^{p_{1}r_{1}}con(P_{1})^{r_{1}}\cdots(C_{m-1})^{r_{m-1}}Z^{p_{m-1}r_{m-1}}con(P_{m-1})^{r_{m-1}}(P_{m})^{r_{m}}$

$=C_{0}Z^{p_{0}}con(P_{1})^{r_{1}}\dots con(P_{m-1})^{r_{m-1}}P_{m}^{r_{m}}$,
where $C_{0}=C(C_{1})^{r_{1}}\dots(C_{m-1})^{r_{m-1}}$,
$p_{0}=p+p_{1}r_{1}+\cdots+p_{m-1}r_{m-1}$.

On the other hand, $con(P)=\overline{C}Z^{-p}con(P_{1})^{r_{1}}\cdots
con(P_{m-1})^{r_{m-1}}con(P_{m})^{r_{m}}$.

So, by the assumption $P=con(P)$, we get that if $con(P_{1})\neq
0,\cdots,con(P_{m-1})\neq 0$, $P_{m}=C_{m}y^{p_{m}}con(P_{m})$, where
$C_{m}=(\overline{C}/C_{0})^{1/r_{m}}$, $p_{m}=(-p-p_{0})/r_{m}$. As
$con(P_{1}),\cdots,con(P_{m-1})$ have finite zeros at most, $P_{m}$ and
$con(P_{m})$ are both continuous, so $P_{m}=C_{m}y^{p_{m}}con(P_{m})$ holds.

2) $P_{i}=C_{i}Z^{p_{i}}con(P_{i})$ implies
$P_{i}^{2}=C_{i}Z^{p_{i}}con(P_{i})P_{i}$, so
$P_{i}^{2}C_{i}^{-1}Z^{-p_{i}}=con(P_{i})P_{i}$. Let
$Q_{i}=P_{i}Z^{-p_{i}/2}C_{i}^{-1/2}$, then $Q_{i}^{2}=con(P_{i})P_{i}$, so
that $Q_{i}^{2}=con(Q_{i}^{2})$.

Let $f_{i}(x)=LRhom[u_{1},\dots,u_{r};v_{1},\dots,v_{s}](Q_{i})$, then
$f_{i}(x)^{2}$ is real-valued due to Lemma 3.4, i.e. $f_{i}(x)$ is real-valued
or pure imaginary.

It is clearly that $f_{i}$ has no multiple root other than $0$ by Theorem 3.1,
$f_{i}$ and $f_{j}$ have no common root other than $0$ for $i\neq j$ by Lemma
3.2.

3) As $P=con(P)$, then $degree(P,y)=degree(P,y^{-1})$, denoted by $q$. It is
trivial that $p=-q$ and $degree(Q=P_{1}^{r_{1}}\cdots P_{n}^{r_{n}},y)=2q$.
Since $P_{i}=C_{i}y^{p_{i}}con(P_{i})$,
$p_{i}=degree(P_{i},y)=degree(con(P_{i}),y^{-1})$, hence
$p_{1}+\dots+p_{n}=degree(Q,y)$, i.e. $p=-q=-(p_{1}+\dots+p_{n})/2$.

So, $C_{0}f_{1}^{r_{1}}\cdots
f_{n}^{r_{n}}=C(C_{1})^{r_{1}/2}\dots(C_{n})^{r_{n}/2}(LRhom(P_{1}y^{-p_{1}/2}C_{1}^{-1/2}))^{r_{1}}\dots(LRhom(P_{n}y^{-p_{n}/2}C_{n}^{-1/2}))^{r_{n}}$

$=C(LRhom[u_{1},\dots,u_{r};v_{1},\dots,v_{s}](P_{1}y^{-\frac{p_{1}}{2}}))^{r_{1}}\dots(LRhom[u_{1},\dots,u_{r};v_{1},\dots,v_{s}](P_{n}y^{-\frac{p_{n}}{2}}))^{r_{n}}$

$=CLRhom[u_{1},\dots,u_{r};v_{1},\dots,v_{s}](y^{-\frac{p_{1}}{2}r_{1}-\dots-\frac{p_{n}}{2}r_{n}}P_{1}^{r_{1}}\dots
P_{n}^{r_{n}})$

$=LRhom[u_{1},\dots,u_{r};v_{1},\dots,v_{s}](Cy^{p}P_{1}^{r_{1}}\dots
P_{n}^{r_{n}})=LRhom[u_{1},\dots,u_{r};v_{1},\dots,v_{s}](P)$ ∎

Corollary 3.2 2.4 for reachability For each trigonometric-exponential
polynomial
$F(x)=f(x,e^{{u_{1}}x},\dots,e^{{u_{r}}x},\sin(v_{1}x),\dots,\sin(v_{s}x),\cos(v_{1}x),\dots,\cos(v_{s}x))$,
where
$f\in\mathbb{R}_{alg}[x,y_{1},\dots,y_{r},z_{1},\dots,z_{s},w_{1},\dots,w_{s}]$
there exists $C\in\mathbb{R}$ and real-valued trigonometric-exponential
polynomials $\\{f_{i}\\}$ such that $f_{i}$ has no multiple root other than
$0$ , $f_{i}$ and $f_{j}$ have no common root other than $0$ for $i\neq j$,
$F(x)=Cf_{1}(x)^{r_{1}}\dots f_{n}(x)^{r_{n}}$.

###### Proof.

Making substitution for F(x), $e^{{u_{i}}x}=y_{i}$ for $i=1,\dots,r$,
$\sin(v_{i}x)=\ frac{z_{i}-1/z_{i}}{2I}$, $\cos(v_{i}x)=\
frac{z_{i}+1/z_{i}}{2}$ for $i=1,\dots,s$, yields $P\in\mathbb{LR}$ such that
$LRrom[u_{1},\dots,u_{r};v_{1},\dots,v_{s}](P)=F(x)$.

By Theorem 2.4, there exists $C\in\mathbb{C}$ and trigonometric-exponential
polynomials $\\{f_{i}\\}$ such that $f_{i}$ has no multiple root other than
$0$ , $f_{i}$ and $f_{j}$ have no common root other than $0$ for $i\neq j$,
$F(x)=Cf_{1}(x)^{r_{1}}\dots f_{n}(x)^{r_{n}}$, each $f_{i}$ is real-valued or
pure imaginary.

If all $f_{i}$s are real-valued, then Corollary 2.4 holds. If $f_{i}$ is pure
imaginary, let $f_{i}^{\prime}=f_{i}/I$ and $C^{\prime}=CI^{r_{i}}$, then
$f_{i}^{\prime}$ is real-valued and $F(x)=C^{\prime}f_{1}(x)^{r_{1}}\dots
f_{i}^{\prime}(x)^{r_{i}}\dots f_{n}(x)^{r_{n}}$.

Now, $F(x)$ and all $f_{i}(x)(orf_{i}^{\prime}(x))$s are real-valued, so the
constant $C(orC^{\prime})$ must be a real number. That is to say the corollary
holds.

∎

Algorithm 3.1Square-free factorization of MTP

Input: an MTP $F(x)=f(x,\sin(x),\cos(x)),f\in\mathbb{R}_{alg}[x,y,z]$;

Output: $F(x)=C_{0}F_{1}(x)^{n_{1}}\cdots F_{m}(x)^{n_{m}}$; where
$C_{0}\in\mathbb{R}$ and $F_{i}(x)$ is a real-valued MTP and which has no
multiple roots other than $0$ for $i=1,\cdots,m$ , $F_{i}(x)$ and $F_{j}(x$)
have no common roots other than $0$ for $i\neq j$.

1) $P\leftarrow f(x,\frac{y-1/y}{2I},\frac{y+1/y}{2})$;

2) $P\\_lst\leftarrow factor(P)=Cy^{p}P_{1}^{r_{1}}\cdots P_{n}^{r_{n}}$;
where $C\in\mathbb{C}$, $P_{1},\cdots,P_{n}\in\mathbb{C}[x,y]$ are square-free
and pairwisely co-prime.

3) $F\leftarrow 1$; $C_{0}\leftarrow C$;

4) for $i$ form $1$ to $m$

4.1) $g\leftarrow P_{i}$;

4.2) $h\leftarrow g/con(g)$; where $h$ is of form $C_{i}y^{p_{i}}$,
$LRhom(g^{2}/h)=LRhom(g\times con(g))$ is real-valued;

4.3) $f_{i}\leftarrow LRhom(g/(C_{i}^{1/2}y^{p_{i}/2}))$; where $f_{i}$ is
real or pure imaginary;

4.4) if $f_{i}$ is pure imaginary, then $f_{i}\leftarrow f_{i}/I$,
$C_{0}\leftarrow C_{0}I^{r_{i}}$;

4.5) $F\leftarrow Ff_{i}^{r_{i}}$;

4.6) $C_{0}\leftarrow C_{0}(C_{i}^{1/2})^{r_{i}}$;

5) return $C_{0}F$.

Example 3 Decide whether $f(x)=\frac{2}{3}x+x\cos(x)-\sin(x)$ has multiple
roots.

Let $\cos(x)=\frac{e^{Ix}+e^{-Ix}}{2}$, $\sin(x)=\frac{e^{Ix}-e^{-Ix}}{2I}$,
$y=e^{Ix}$,

then $f=\frac{2x}{3}+x\frac{y+1/y}{2}-\frac{y-1/y}{2I}$,

$factor(f)=\frac{1}{6}\frac{4xy+3xy^{2}+3x+3Iy^{2}-3I}{y}$,
$4xy+3xy^{2}+3x+3Iy^{2}-3I$ is irreducible, which implies that $f(x)$ has no
multiple roots.

Example 4 Decide whether $f(x)=1-\sin^{3}(x)$ has multiple roots, and if so,
do factorization.

Let $\sin(x)=\frac{e^{Ix}-e^{-Ix}}{2I}$, $y=e^{Ix}$.

Then
$f(x)=g(y)=1-\frac{1}{8}I(y-\frac{1}{y})^{3}=\frac{-\frac{1}{8}I(y^{4}+2Iy^{3}-6y^{2}-2Iy+1)(y-I)^{2}}{y^{3}}=CP_{1}(P_{2})^{2}$,
where $C=-\frac{I}{8}$, $P_{1}=y^{4}+2Iy^{3}-6y^{2}-2Iy+1$, and $P_{2}=y-I$.

We get that $y=I$ is a multiple root of $g(x)$, and then
$\\{2k\pi+\frac{\pi}{2},k\in\mathbb{Z}\\}$ are multiple roots of $f(x)$.

As
$con(P_{1})=\frac{1}{y^{4}}-\frac{2I}{y^{3}}-\frac{6}{y^{2}}+\frac{2I}{y}+1$,
so $P_{1}=c_{1}y^{4}con(P_{1})$, where $c_{1}=1$.

Let

$\displaystyle g_{1}$
$\displaystyle=LRhom(P_{1}/(y^{4})^{1/2})=LRhom((y^{4}+2Iy^{3}-6y^{2}-2Iy+1)/y^{2})$
(1) $\displaystyle=LRhom(y^{2}+2Iy-6-2Iy^{-1}+y^{-2})$
$\displaystyle=e^{2Ix}+2e^{I\frac{\pi}{2}}e^{Ix}-6-2e^{I{\frac{\pi}{2}}}e^{-Ix}+e^{-2Ix}$
$\displaystyle=\cos(2x)+I\sin(2x)+2\cos(\frac{\pi}{2}+x)+2I\sin(\frac{\pi}{2}+x)-6-2\cos(\frac{\pi}{2}-x)$
$\displaystyle-2I\sin(\frac{\pi}{2}-x)+\cos(-2x)+I\sin(-2x)$
$\displaystyle=\cos(2x)+I\sin(2x)-2\sin(x)+2I\cos(x)-6-2\sin(x)-2I\cos(x)+\cos(2x)-I\sin(2x)$
$\displaystyle=2\cos(2x)-4\sin(x)-6$

As $con(P_{2})=\frac{1}{y}+I$, then $P_{2}=c_{2}ycon(P_{2})$, where
$c_{2}=-I=e^{-I\frac{\pi}{2}}$.

Let

$\displaystyle g_{2}$
$\displaystyle=LRhom(P_{2}/(-Iy)^{\frac{1}{2}})=LRhom((y-I)/(-Iy)^{\frac{1}{2}})$
(2)
$\displaystyle=(e^{Ix}-e^{I\frac{\pi}{2}})/e^{-I\frac{\pi}{4}+I\frac{x}{2}}$
$\displaystyle=e^{I\frac{x}{2}+I\frac{\pi}{4}}-e^{I\frac{3\pi}{4}-I\frac{x}{2}}$
$\displaystyle=\cos(\frac{x}{2}+\frac{\pi}{4})+I\sin(\frac{x}{2}+\frac{\pi}{4})-(\cos(\frac{3\pi}{4}-\frac{x}{2})+I\sin(\frac{3\pi}{4}-\frac{x}{2}))$
$\displaystyle=\cos(\frac{x}{2}+\frac{\pi}{4})+I\sin(\frac{x}{2}+\frac{\pi}{4})+\cos(\frac{x}{2}+\frac{\pi}{4})-I\sin(\frac{x}{2}+\frac{\pi}{4})$
$\displaystyle=2\cos(\frac{\pi}{4}+\frac{x}{2})$

Let $c_{0}=Cc_{1}^{1/2}(c_{2}^{1/2})^{2}=-\frac{I}{8}(-I)=-1/8$.

Then
$f(x)=c_{0}g_{1}(g_{2})^{2}=-\frac{1}{8}(2\cos(2x)-4\sin(x)-6)\cos^{2}(\frac{\pi}{4}+\frac{x}{2})=-\frac{1}{2}(\cos(2x)-2\sin(x)-3)(\cos(\frac{x}{2})-\sin(\frac{x}{2}))^{2}=-\frac{1}{2}f_{1}(f_{2})^{2}$,
where $f_{1}=\cos(2x)-2\sin(x)-3,f_{2}=\cos(\frac{x}{2})-\sin(\frac{x}{2})$.

##  4 Sign-deciding of Mixed Trigonometric Polynomial

If the positive roots of a square-free MTP are unbounded, it has no constant
sign on $(0,+\infty)$ definitely. So we need only to discuss the sign-deciding
of square-free MTP on a bounded interval.

Ref[8] has designed a complete algorithm $Deciding\\_arctan\\_polynomial$ for
deciding the sign of inverse tangent function polynomial $f(x,arctan(x))$ on
$(0,1)$. Each mixed trigonometric-polynomial can be transformed to a tangent
function polynomial $f(x,\tan(\frac{x}{2}))$. Let $y=\tan(\frac{x}{2})$, then
$x=2\arctan(y)$ and $f(x,\tan(\frac{x}{2}))=f(2\arctan(y),y)$, thus tangent
function polynomial is reduced to an inverse tangent function polynomial. So
it is trivial to decide the sign of an MTP on $(0,\frac{\pi}{2})$.(In general,
we don’t care whether the right end of the interval is open or closed.)

For $T>\frac{\pi}{2}$, to decide the sign of MTP on (0,T), the tan-half angle
substitutions $\cos(2t)=1-2\sin^{2}(t)$ and $\sin(2t)=2\sin(t)\cos(t)$ can
reduce $f(x)$ to another MTP $g(t)$ such that $0<t<\frac{\pi}{2}$.

Algorithm 4.1 $Decide\\_trigonometric\\_polynomial$

Input: $F(x)=f(x,\sin(x),\cos(x))$, where $f\in\mathbb{R}_{alg}[x,y,z]$ and is
square-free, a constant $T$;

Output: The sign of $F(x)$ on $(0,T)$.

1) $p\leftarrow[T/(\frac{\pi}{2})]+1$; where $[x]$ denotes the maximum integer
not bigger than $x$

2) $G(t)\leftarrow F(pt)$, and transform $G(t)$ to the form
$g(t,\sin(t),\cos(t))$;

3) $G(t)\leftarrow
g(t,\frac{2\tan(\frac{t}{2})}{1+\tan^{2}(\frac{t}{2})},\frac{1-\tan^{2}(\frac{t}{2})}{1+\tan^{2}(\frac{t}{2})})$;

4) $H(y)\leftarrow subs(\tan(\frac{t}{2})=y,t=2\arctan(y),G(t))$;

5) Return $Deciding\\_arctan\\_polynomial(H(y))$.

For a general MTP on $(0,+\infty)$, we design the following algorithm.

Algorithm 4.2 $Decide\\_general\\_trigonometric\\_polynomial$

Input: $F(x)=f(x,\sin(x),\cos(x))$, $f\in\mathbb{R}_{alg}[x,y,z]$;

Output: The sign of $F(x)$ on $(0,+\infty)$;

1) Run Algorithm 3.1 to factorize $F(x)$,
$F(x)=cf_{1}^{d_{1}}f_{2}^{d_{2}}\cdots f_{n}^{d_{n}}$; Where $c$ is a real
constant, $f_{1},f_{2},\cdots,f_{n}$ are real-valued MTP and has no multiple
roots other than $0$ for $i=1,\cdots,n$ , $f_{i}(x)$ and $f_{j}(x$) have no
common roots other than $0$ for $i\neq j$.

2) $p\leftarrow 0$, $q\leftarrow 0$; Where $p$ denotes the number of negative
terms, $q>0$ shows that the inequality is not strict.

3) $m\leftarrow 1$.

4) if $d_{m}$ is odd, then

Run Algorithm 2.1 to decide if the positive roots of $f_{m}$ are bounded,

4.1) If unbounded, then return $0$, the algorithm ends; where $0$ means that
$F(x)$ has no constant sign on $(0,+\infty)$.

4.2) If bounded and $T$ is a bound, then $s\leftarrow
Decide\\_trigonometric\\_polynomial(f_{m},T)$

If $s=-1$ then $p\leftarrow p+1$; where $s=-1$ implies that $f_{m}<0$ holds on
$(0,T)$.

If $s=0$ then return $0$,the algorithm ends; where $s=0$ implies that $f_{m}$
has no constant sign $(0,T)$.

elif $d_{m}$ is even, then

4.3) If the positive roots of $f_{m}$ are unbounded, then $q\leftarrow q+1$;

4.4) The positive roots of $f_{m}$ are bounded and $T$ is a bound, then
$s\leftarrow Decide\\_trigonometric\\_polynomial(f_{i},T)$,

If $s=0$ then $q\leftarrow q+1$.

5) $m\leftarrow m+1$, if $m<n$ then goto 4)

6) 6.1) If $c>0$, $p$ is even and $q=0$, then inequality $F(x)>0$ holds;

6.2) If $c>0$, $p$ is even and $q>0$, then inequality $F(x)\geq 0$ holds;

6.3) If $c>0$, $p$ is odd and $q>0$, then inequality $F(x)<0$ holds;

6.4) If $c>0$, $p$ is odd and $q>0$, then inequality $F(x)\leq 0$ holds;

For $c<0$, the conclusions can be drawn in the same manner.

Example 5 Decide the sign of $f(x)=\frac{2}{3}x+x\cos(x)-\sin(x)$ on
$(0,+\infty)$.

By Example 1, the positive roots of $f(x)$ are unbounded. By Example 3, $f(x)$
has no multiple root. So $f(x)$ has no constant sign on $(0,+\infty)$.

Example 6 Decide the sign of $f(x)=\frac{2}{3}x+\frac{1}{3}x\cos(x)-\sin(x)$
on $(0,+\infty)$.

By Example 2, the positive roots of $f(x)$ are less than $\frac{111}{64}$, so
we need only to discuss the problem on $(0,\frac{111}{64})$.

As $\frac{111}{64}>\frac{\pi}{2}$, let
$g(y)=f(2y)=\frac{4}{3}y+\frac{2}{3}y(1-2\sin(y)^{2})-2\sin(y)\cos(y)$, where
$0<y<\frac{111}{128}<\frac{\pi}{2}$.

Let $t=\tan(\frac{y}{2})$, then $0<t<1$, $\sin(y)=\frac{2t}{1+t^{2}}$,
$\cos(y)=\frac{1-t^{2}}{1+t^{2}}$, hence
$g(y)=\frac{2}{3}\frac{3y-2yt^{2}+3yt^{4}-6t+6t^{3}}{(1+t^{2})^{2}}$.

Now it is necessary only to determine the sign of
$g_{1}(y,t)=3y-2yt^{2}+3yt^{4}-6t+6t^{3}$. Obviously $y=2\arctan(t)$, so
$g_{1}(y,t)=3\arctan(t)-2\arctan(t)t^{2}+3\arctan(t)t^{4}-6t+6t^{3}$, denoted
as $G(t)$, the algorithm $Deciding\\_arctan\\_polynomial$ of Ref[8] can
declare that $G(t)>0$ holds on $(0,1)$, and so $g(y)=f(2y)>0$ on
$(0,\frac{\pi}{2})$, $f(x)>0$ on $(0,\pi)$. As $f(x)$ has constant sign on
$(\frac{111}{64},+\infty)$, we get that $f(x)>0$ holds on $(0,+\infty)$.

##  5 Conclusion

In this paper, we present algorithms to determine the boundedness of MTP’s
positive roots and to decompose MTPs without multiple roots, and then propose
a procedure to decide the sign of MTP on $(0,+\infty)$. Furthermore, the
algorithms presented are much useful for similar problems.

## References

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