Symmetric Mahler's conjecture for the volume product in the three dimensional case

TL;DR

This paper proves Mahler's conjecture for the volume product of centrally symmetric convex bodies specifically in three dimensions and identifies the conditions for equality.

Contribution

It provides the first proof of Mahler's conjecture in three dimensions and characterizes the equality cases for symmetric convex bodies.

Findings

Mahler's conjecture holds in 3D for symmetric convex bodies.

The equality condition for the volume product is characterized.

The proof confirms the conjecture's validity in the three-dimensional case.

Abstract

In this paper, we prove Mahler's conjecture concerning the volume product of centrally symmetric convex bodies in $\mathbb{R}^n$ in the case where $n=3$. Furthermore, we determine the equality condition.

Equations

$$K^z=\left\{y\in {\mathbb{R}}^n;(y-z)\cdot (x-z)\le 1\text{ for any }x\in K\right\},$$

$$\mathcal{P}(K):=\underset{z\in \mathrm{int}K}{\min }|K||K^z|.$$

$$\mathcal{P}(K)=|K||K^{\circ }|\ge \frac{4^n}{n!}.$$

$$|K||K^{\circ }|\ge \frac{32}{3},$$

$$\mathcal{P}(K)=\underset{z\in \mathrm{int}K}{\min }|K||K^z|\ge \frac{(n+1{)}^{n+1}}{(n!{)}^2}.$$

$$\frac{c(\Sigma )}{c(B^{2n})}\le {\left(\frac{\mathrm{vol}(\Sigma )}{\mathrm{vol}(B^{2n})}\right)}^{1/n}$$

$$c_{\mathrm{HZ}}(K\times K^{\circ })=4.$$

$$\frac{4^n}{{\pi }^n}=\frac{c_{\mathrm{HZ}}(K\times K^{\circ }{)}^n}{{\pi }^n}\le \frac{\mathrm{vol}(K\times K^{\circ })}{{\pi }^n/n!}=\frac{|K||K^{\circ }|}{{\pi }^n/n!}.$$

$$\mathcal{F}=\frac{(F,G,H)}{\sqrt{F^2+G^2+H^2}}:\partial D\to S^2.$$

$$K_{++}$$

$$K_{-+}$$

$$K$$

$$K^{\circ }$$

$$|K_{++}|=|K_{-+}|=\frac{|K|}{4}.$$

$$K^{\circ }\subset [-1,1]\times [-1,1]\setminus \{\pm (1,1),\pm (-1,1)\}.$$

$$K_{++}^{\circ }$$

$$K_{-+}^{\circ }$$

$$\frac{1}{2}\left|\begin{array}{cc}1& b\\ u& v\end{array}\right|+\frac{1}{2}\left|\begin{array}{cc}u& v\\ c& 1\end{array}\right|\le |K_{++}^{\circ }|.$$

$$(1,1)\cdot (1-b,1-c)\le |K||K_{++}^{\circ }|,(-1,1)\cdot (-1-b,1+c)\le |K||K_{-+}^{\circ }|,$$

$$\mathrm{polygon}\left\{O,B,S_1,C\right\}\subset K_{++},\mathrm{polygon}\left\{O,C,S_2,-B\right\}\subset K_{-+},$$

$$|K_{++}^{\circ }|=\frac{2-b-c}{|K|},|K_{-+}^{\circ }|=\frac{2+b+c}{|K|}.$$

$$\left(1,\frac{2|K_{++}^{\circ }|-1+b}{1-c}\right).$$

$$K=\mathrm{conv}\{\pm S_1,\pm S_2\}=\mathrm{conv}\left\{\pm \frac{1}{1+b^2}\left(1-b,1+b\right),\pm \frac{1}{1+b^2}\left(-1-b,1-b\right)\right\}$$

$${\mathcal{C}}_K(P,Q):=\left\{{\rho }_K((1-t)P+tQ)((1-t)P+tQ);0\le t\le 1\right\}.$$

$$K^{\circ }=\left\{(u,v,w)\in {\mathbb{R}}^3;(u,v,w)\cdot (x,y,z)\le 1\text{ for all }(x,y,z)\in K\right\},$$

$$A_{\pm }:=\pm {\mathbb{R}}_{+}{e}_1\cap \partial K,B_{\pm }:=\pm {\mathbb{R}}_{+}{e}_2\cap \partial K,C_{\pm }:=\pm {\mathbb{R}}_{+}{e}_3\cap \partial K$$

$$\alpha$$

$$\beta$$

$$\gamma$$

$$O\ast {\mathcal{S}}_K(c)=\{\lambda u;u\in {\mathcal{S}}_K(c)\text{ and }0\le \lambda \le 1\}.$$

$$K_{+++}$$

$$K_{--+}$$

$$K_{++-}$$

$$K_{---}$$

Full text

1 Introduction

1.1 Mahler’s conjecture for the volume product

A convex body in $\mathbb{R}^{n}$ is a compact convex set in $\mathbb{R}^{n}$ with nonempty interior. Denote by $\mathcal{K}^{n}$ the set of all convex bodies in $\mathbb{R}^{n}$. A convex body $K\in\mathcal{K}^{n}$ is said to be centrally symmetric if it satisfies $K=-K$. We denote by $\mathcal{K}^{n}_{0}$ the set of all $K\in\mathcal{K}^{n}$ which are centrally symmetric.

Let $K\in\mathcal{K}^{n}$ be a convex body. The interior of $K$ is denoted by $\operatorname{int}K$. For $z\in\operatorname{int}K$, the polar body of $K$ with respect to $z$ is defined by

[TABLE]

where $\cdot$ denotes the standard inner product on $\mathbb{R}^{n}$. Denote by $|K|$ the $n$-dimensional volume of $K$ in $\mathbb{R}^{n}$. Then the volume product of $K$ is defined by

[TABLE]

Note that this quantity is an affine invariant, i.e., $\mathcal{P}(\mathcal{A}K)=\mathcal{P}(K)$ for any invertible affine map $\mathcal{A}:\mathbb{R}^{n}\to\mathbb{R}^{n}$. It is well-known that for each $K\in\mathcal{K}^{n}$ the minimum of (1) is attained at the unique point $z$ on $K$, which is called Santaló point of $K$ (see [Sc]). For a centrally symmetric convex body $K\in\mathcal{K}^{n}_{0}$, the Santaló point of $K$ is nothing but the origin $O$. In the following, the polar of $K$ with respect to $O$ is denoted by $K^{\circ}$.

The upper bound for $\mathcal{P}(K)$ is well-known as Blaschke-Santaló inequality (see [Sa]) and the bound is attained only for ellipsoids (see [MP]). In contrast, the sharp lower bound estimate is fairly difficult and it remains as a longstanding open problem since 1939 as follows.

Conjecture** (Symmetric Mahler’s conjecture; [Ma]).**

Any $K\in\mathcal{K}_{0}^{n}$ satisfies that

[TABLE]

This is trivial for $n=1$ and was proved by Mahler himself for $n=2$ [Ma0]. There are many alternative proofs and the characterization of the equality case for $n=2$ (see, e.g., [Me], [Sc]*Section 10.7, and references therein). However, the case where $n\geq 3$ is still widely open. We solve affirmatively symmetric Mahler’s conjecture in the three dimensional case.

Theorem 1.1**.**

For any three dimensional centrally symmetric convex body $K\in\mathcal{K}_{0}^{3}$, we have

[TABLE]

with equality if and only if either $K$ or $K^{\circ}$ is a parallelepiped.

Further, we exhibit known partial results about the conjecture. An asymptotic lower bound of the volume product was first given by Bourgain and Milman [BM] and the best known constant is due to Kuperberg [K]. Meanwhile, the conjecture itself has been proved for very restricted cases; for zonoids by Reisner [R, R2] (see also [GMR]) and for unconditional bodies by Saint-Raymond [SR] (see [Me2] for a short proof). Note that the equality of (2) is attained by an $n$-cube, for instance. In [NPRZ], the estimate (2) was confirmed for $K\in\mathcal{K}_{0}^{n}$ sufficiently close to the unit $n$-cube in the Banach-Mazur distance. For other partial results, see e.g., [BF], [FMZ], [LR], [RSW].

There is another question when $K\in\mathcal{K}^{n}$ is not necessarily assumed to be centrally symmetric.

Conjecture** (Non-symmetric Mahler’s conjecture).**

Any $K\in\mathcal{K}^{n}$ satisfies that

[TABLE]

This was proved by Mahler for $n=2$ (see [Ma0]). The alternative proof of Mahler’s result by Meyer [Me] is notable. Indeed, we were able to extract an important idea for attacking to the three dimensional symmetric case from [Me]. The idea, which is sketched in Section 1.3, improves the method of [Me2] for unconditional bodies. Note that non-symmetric Mahler’s conjecture remains open for $n\geq 3$, see e.g., [BF], [FMZ], [KR], [RSW].

1.2 An application to Viterbo’s conjecture

Recently, a surprising connection between symmetric Mahler’s conjecture and a conjecture in the field of symplectic geometry was discovered. In 2000, Viterbo posed in [V] an isoperimetric-type conjecture for symplectic capacities of convex bodies in $\mathbb{R}^{2n}$ with the standard symplectic structure $\omega_{0}$. A symplectic capacity $c$ is a symplectic invariant which assigns a non-negative real number to each of symplectic manifolds of dimension $2n$. The Hofer-Zehnder capacity $c_{\rm HZ}$ is one of the important symplectic capacities, which is related to Hamiltonian dynamics on symplectic manifolds. For details, see [AKO] or a foundational book [HZ].

Conjecture** (Viterbo [V]).**

For any symplectic capacity $c$ and any convex body $\Sigma\in\mathcal{K}^{2n}$,

[TABLE]

holds, where $B^{2n}$ denotes the $2n$-dimensional unit ball and $\mathrm{vol}(\Sigma)$ denotes the symplectic volume of $\Sigma$.

This conjecture is unsolved even in the case of $n=2$. Note that $c(B^{2n})=\pi$. In [AKO], Artstein-Avidan, Karasev, and Ostrover calculated the Hofer-Zehnder capacity of a convex domain $K\times K^{\circ}\subset\mathbb{R}^{2n}$ $(K\in\mathcal{K}_{0}^{n})$ as follows.

Theorem 1.2** ([AKO], Theorem 1.7).**

For any centrally symmetric convex body $K\in\mathcal{K}_{0}^{n}$, we have

[TABLE]

Using it, they gave the following remarkable observation that Viterbo’s conjecture implies symmetric Mahler’s conjecture (see [AKO]*Section 1).

[TABLE]

Conversely, if we assume that symmetric Mahler’s conjecture holds, then Theorem 1.2 implies the inequality (3) for the case that $c=c_{\rm HZ}$ and $\Sigma=K\times K^{\circ}$. Therefore, Theorem 1.1 immediately yields

Corollary 1.3**.**

Let $K\subset\mathbb{R}^{3}$ be a centrally symmetric convex body. Then the inequality (3) holds for $\Sigma=K\times K^{\circ}\subset(\mathbb{R}^{6},\omega_{0})$ with respect to the Hofer-Zehnder capacity $c_{\rm HZ}$.

Of course, due to the result of [Ma0], the inequality (3) also holds for $\Sigma=K\times K^{\circ}\subset(\mathbb{R}^{4},\omega_{0})$, where $K\in\mathcal{K}_{0}^{2}$, with respect to $c_{\rm HZ}$.

1.3 Organization of the paper and our method

In Section 2, we review the two dimensional case based on the argument in [Me2] with a slight modification. Let $K\in\mathcal{K}_{0}^{2}$. Let $B$ (resp. $C$) be the intersection of $\partial K$ and the positive part of $x$ (resp. $y$) axis. First, by using coordinate axises, we divide $K$ into four parts $\pm K_{++},\pm K_{-+}$. In [Me2], $K$ is supposed to be an unconditional body, which means that the four parts are all congruent and hence it suffices to consider $K_{++}$ in the first quadrant, and $K^{\circ}$ is also unconditional. Then, for any point $P\in K_{++}$, the area $|K_{++}|$ is estimated from below by the area of a quadrilateral $OBPC$, which gives a test point in $K^{\circ}_{++}$. The same observation on $K^{\circ}$ yields another test point in $K_{++}$. By paring these two points, we easily get $\mathcal{P}(K)\geq 8$. In the general case, we may assume that $|K_{++}|=|K_{-+}|$. Although $K^{\circ}$ is not any longer unconditional, we can find an appropriate division of $K^{\circ}$ by using the points $B^{\circ},C^{\circ}\in\partial K^{\circ}$ with $B^{\circ}\cdot B=1,C^{\circ}\cdot C=1$ (see e.g., [BMMR, Section 4]). Including the equality case we sketch a proof in Section 2.

From Section 3, we begin the proof of the three dimensional case. We first divide $K$ into eight parts $K_{\pm\pm\pm}$ as in the two dimensional case. However, compare to the two dimensional case, to make the corresponding decomposition of $K^{\circ}$ is not so simple. The first ingredient of our proof of Theorem 1.1 is to concentrate only on the class of convex bodies $K\in\mathcal{K}_{0}^{3}$ which are strongly convex and $C^{\infty}$ boundary. A smoothing approximation procedure (Proposition 5.4) guarantees that the problem is reduced to considering only this class. For a convex body $K$ in the class, we can define a diffeomorphism $\partial K\to\partial K^{\circ}$, which yields the corresponding eight pieces decomposition $K^{\circ}_{\pm\pm\pm}$ of $K^{\circ}$ from that of $K$. The precise setting is explained in Section 3.2.

The second ingredient is that only a few test points give a sharp estimate of $\mathcal{P}(K)$, provided $K$ has an additional symmetry. When $n=2$, the symmetry we need is $|K_{++}|=|K_{-+}|$, although this is not an essential assumption. When $n=3$, we have to control not only the volume of $K_{\pm\pm\pm}$ but also the areas of boundary faces of $K_{\pm\pm\pm}$ in the three coordinate planes. In Section 3.4, we give a sharp estimate of $\mathcal{P}(K)$ from below under the condition that $K$ has “good” symmetries. The condition is (10). Under this condition, the process of estimating $\mathcal{P}(K)$ is regarded as a direct generalization of the two dimensional case. A (signed) volume comparison inequality (Lemma 3.1) which yields test points on $K$ and $K^{\circ}$ is given in Section 3.3.

How can we release $K$ from the very strong condition (10)? Note that $\mathcal{P}(K)$ is invariant under linear transformations of $\mathbb{R}^{3}$. The third ingredient is to make the most of the freedom of the action of linear transformations. As a test case, in Appendix B, we prove Theorem 1.1 for $K\in\mathcal{K}^{3}_{0}$ equipped with an extra symmetry with respect to a hyperplane through $O$ (Proposition B.1). In this case, by a linear transformation $\mathcal{A}$, we can deform $K$ into $\mathcal{A}K$ which satisfies the condition (10) by means of the intermediate value theorem. However, for a general $K\in\mathcal{K}^{3}_{0}$ to find a linear transformation $\mathcal{A}$ such that $\mathcal{A}K$ satisfies the condition (10) is highly nontrivial.

Sections 4 and 5, which are the technical part of the present paper, are devoted to find an appropriate linear transformation $\mathcal{A}$ for each $K\in\mathcal{K}^{3}_{0}$ which is strongly convex with smooth boundary $\partial K$. We proceed as follows. First, we define $K(\theta,\phi,\psi)\in\mathcal{K}^{3}_{0}$ as the image of the initial $K$ under the action of $SO(3)$. (Here, $\theta,\phi,\psi$ mean rotation angles with $x,y,z$-axises, respectively.) Next, in Section 4.1 we define a linear transform $\mathcal{A}$ such that $\mathcal{A}K(\theta,\phi,\psi)$ satisfies the condition (11) (Proposition 4.2), which is a partial condition of (10). Finally, to get the full condition (10) for $\mathcal{A}K(\theta,\phi,\psi)$, we introduce in Section 4.2 three smooth functions $F(K),G(K),H(K)$ on a contractible region $D\subset\mathbb{R}^{3}$ with the coordinates $\theta,\phi,\psi$. These functions are defined by the volume of pieces of the eight part decomposition of $\mathcal{A}K(\theta,\phi,\psi)$ and the resulting two dimensional quantities. Then the problem to find $\mathcal{A}K$ which satisfies the desired condition (10) is reduced to the existence of a zero $(\theta,\phi,\psi)$ of the map $(F,G,H):D\to\mathbb{R}^{3}$. Here, if $(F,G,H)$ has no zero on $\partial D$, then we can define a map

[TABLE]

From Section 4.3 to 4.7, we discuss properties of the map $(F,G,H)$ in order to calculate the degree of the map $\mathscr{F}$. The centrally symmetric convex body $\mathcal{A}K(\theta,\phi,\psi)$ equips with additional symmetries under rotations by specific angles around $x,y,z$-axises. These induces certain symmetries of the functions $F,G$, and $H$ (Lemmas 4.6, 4.7, 4.9, and 4.11). The results are summarized in Proposition 4.13 and essential for the degree calculation in the next section.

In Section 5, we actually prove that $\deg\mathscr{F}\neq 0$ after a suitable perturbation of this map $\mathscr{F}$. This immediately implies the existence of a zero of the map $(F,G,H)$. Note that the smoothness of $\partial K$ also has enough merit to the calculation of $\deg\mathscr{F}$. The construction of the above perturbation of $\mathscr{F}$ is technical, so it is discussed in Appendix A. Anyway the perturbed $\mathscr{F}$ equips with $(\pm 1,0,0)$ as regular values. This fact is crucial for the calculation of $\deg\mathscr{F}$. Indeed, it enables us to reduce the calculation of $\deg\mathscr{F}$ to the calculation of the winding number of a map $\mathscr{G}:S^{1}\to S^{1}$ (Proposition 5.2), which is fairly accessible. We see that the winding number of $\mathscr{G}$ is odd. This reduction process is explained in Section 5.3 and the calculation of the winding number of the map $\mathscr{G}$ is carried out in Section 5.4. Consequently, the existence of a zero of the map $(F,G,H)$ ensures that $\mathcal{P}(K)\geq 32/3$ for any convex body $K\in\mathcal{K}_{0}^{3}$ which is strongly convex with $C^{\infty}$ boundary. Finally, a smoothing approximation theorem implies that the inequality holds for all $K\in\mathcal{K}_{0}^{3}$.

In Section 6, we determine the equality condition (Propositions 6.3 and 6.4). First, for a general $K\in\mathcal{K}^{3}_{0}$ with $\mathcal{P}(K)=32/3$, we find a linear transformation $\mathcal{A}$ such that $\mathcal{A}K$ satisfies the condition (10). Indeed, there exists a sequence $\{K_{n}\}_{n\in\mathbb{N}}$ such that $\lim_{n\rightarrow\infty}K_{n}=K$, each $K_{n}\in\mathcal{K}^{3}_{0}$ is strongly convex with smooth boundary. By the results in Sections 4 and 5, there exists $\{\mathcal{A}_{n}\}_{n\in\mathbb{N}}$ such that each $\mathcal{A}_{n}K_{n}$ satisfies the condition $\eqref{eq:3}$. As the limit of a subsequence, we get the required $\mathcal{A}$. Moreover, from $\mathcal{P}(K)=32/3$, we can show that $P_{i}(K^{\circ})=(Q_{i}(K))^{\circ}$ and $\lvert Q_{i}(K)\rvert\lvert P_{i}(K^{\circ})\rvert=8$ where $Q_{i}$ is the cross section with a coordinate plane and $P_{i}$ is the projection to the coordinate plane $(i=1,2,3)$. Thus, the result of the two dimensional case ([Me2], [R2]) implies that $Q_{i}(K)$ and $P_{i}(K^{\circ})$ are parallelograms. Next, we extend the sharp estimate in Section 3.4. Then we see $K$ is polyhedron with at most 20 vertices. Finally, by case analysis and calculating the dual faces of the vertices, we prove that either $K$ or $K^{\circ}$ is a parallelepiped.

2 The two dimensional case

In this section, we sketch the proof of Mahler’s theorem for the two dimensional case based on the argument in [Me2] (see also [BF, Section 2], [BMMR, Section 4]), because the equality case of Proposition 2.2 below will be used in Section 6. Although [Me2] treats only unconditional bodies, its slight modification yields the following

Theorem 2.1** ([Ma0], [R2]).**

Let $K\in\mathcal{K}_{0}^{2}$. Then $\mathcal{P}(K)\geq 8$ with equality if and only if $K$ is a parallelogram.

To prove it, for $K\in\mathcal{K}_{0}^{2}$, we divide $K$ into the following four pieces:

[TABLE]

Theorem 2.1 is an easy consequence of the following

Proposition 2.2**.**

Let $K\in\mathcal{K}^{2}_{0}$. Suppose that $\lvert K_{++}\rvert=\lvert K_{-+}\rvert=\lvert K\rvert/4$, $(1,0),(0,1)\in\partial K$. Then $\mathcal{P}(K)\geq 8$ holds. In addition, when $\mathcal{P}(K)=8$, there exists a constant $a\in(-1,1]$ such that

[TABLE]

Especially, $K$ and $K^{\circ}$ are squares, $\lvert K\rvert=4/(1+a^{2})$, and $\lvert K^{\circ}\rvert=2(1+a^{2})$.

Proof of Theorem 2.1.

Since the volume product $\mathcal{P}(K)$ is invariant under rotations of $K$ around the origin $O=(0,0)$, we may assume that $|K_{++}|=|K_{-+}|$ owing to the intermediate value theorem. By the assumption that $K$ is centrally symmetric, this means that

[TABLE]

The condition (5) is preserved under scaling by a diagonal matrix. Therefore, for any $K\in\mathcal{K}^{2}_{0}$, there exists a linear transformation $\mathcal{A}$ such that $\mathcal{A}K\in\mathcal{K}^{2}_{0}$ satisfies the assumptions of Proposition 2.2. Hence, $\mathcal{P}(K)=\mathcal{P}(\mathcal{A}K)\geq 8$. If $\mathcal{P}(K)=8$, then $\mathcal{A}K$ is a square. That is, $K$ is a parallelogram. ∎

Proof of Proposition 2.2.

We put $B:=(1,0),C:=(0,1)$. By the definition of the polar $K^{\circ}$, we have $\pm u=\pm B\cdot(u,v)\leq 1,\pm v=\pm C\cdot(u,v)\leq 1\text{ for any }(u,v)\in K^{\circ}$. Hence, $K^{\circ}\subset[-1,1]\times[-1,1]$. If $(1,1)\in K^{\circ}$, then we have $x+y\leq 1$ for any $(x,y)\in K$. The convexity of $K_{++}$, $K_{-+}$ and their area estimates easily yield $K=\operatorname{conv}\left\{\pm(1,0),\pm(0,1)\right\}$. This means (4) holds with $a=1$. Similarly, if $(-1,1)\in K^{\circ}$, then we get the same conclusion. Thus, since $K$ is centrally symmetric, it is sufficient to consider the case where

[TABLE]

Since $B,C\in\partial K$, by (6) and the definition of $K^{\circ}$, there exist points $B^{\circ}=(1,b),\,C^{\circ}=(c,1)\in K^{\circ}$ with $b,c\in(-1,1)$ satisfying that $B\cdot B^{\circ}=1$ and $C\cdot C^{\circ}=1$. (Note that these points are not necessarily unique in general.) The points $B^{\circ}$ and $C^{\circ}$ enable us to divide $K^{\circ}$ into the following:

[TABLE]

Now we shall find test points to estimate $\mathcal{P}(K)$. For any $P=(u,v)\in K^{\circ}$, the sum of the signed area of the triangle $OB^{\circ}P$ and that of the triangle $OPC^{\circ}$ is less than or equal to $|K^{\circ}_{++}|$, that is,

[TABLE]

This inequality with $b,c\in(-1,1)$ implies $S_{1}:=(1-b,1-c)/(2|K^{\circ}_{++}|)\in K_{++}$. A similar argument for the piece $K^{\circ}_{-+}$ yields $S_{2}:=(-1-b,1+c)/(2|K^{\circ}_{-+}|)\in K_{-+}$. Repeating the same arguments for $K_{++}$ and $K_{-+}$, we have two test points in $K^{\circ}$: $R_{1}:=2(1,1)/|K|,R_{2}:=2(-1,1)/|K|$. Since $R_{1}\cdot S_{1}\leq 1,\ R_{2}\cdot S_{2}\leq 1$, that is,

[TABLE]

we obtain $\mathcal{P}(K)=|K||K^{\circ}|=2|K|(|K^{\circ}_{++}|+|K^{\circ}_{-+}|)\geq 8$.

Hereafter, let us determine the equality condition. Now suppose that $\mathcal{P}(K)=8$. Since $K\supset\operatorname{conv}\{\pm(1,0),\pm(0,1)\}$, we have $\lvert K\rvert\geq 2$. If $\lvert K\rvert=2$, then we get $K^{\circ}=[-1,1]\times[-1,1]$, which contradicts to (6). Thus $\lvert K\rvert>2$ holds. We consider

[TABLE]

where $\operatorname{polygon}\left\{P_{1},\dots,P_{m}\right\}$ denotes the star-shaped polygon in $\mathbb{R}^{2}$ consists of successive vertices $P_{1},\dots,P_{m}\,(m\geq 3)$ in the counterclockwise order and the edges $P_{1}P_{2},\dots,P_{m-1}P_{m},P_{m}P_{1}$. Taking the volume of these polygons into account, we see that they coincide with $K_{++},K_{-+}$, respectively, and

[TABLE]

Consequently, we have $K=\operatorname{polygon}\left\{B,S_{1},C,S_{2},-B,-S_{1},-C,-S_{2}\right\}$. Similarly, $K^{\circ}=\operatorname{polygon}\left\{B^{\circ},R_{1},C^{\circ},R_{2},-B^{\circ},-R_{1},-C^{\circ},-R_{2}\right\}$.

Next, we consider the dual faces of the edges of $K$. Since the line segment $BS_{1}$ is a part of an edge of $K$, its dual face is a vertex of $K^{\circ}$, which is calculated as

[TABLE]

This point must be one of the elements of $\left\{\pm B^{\circ},\pm R_{1},\pm C^{\circ},\pm R_{2}\right\}$. Since $c\in(-1,1)$ and $\lvert K\rvert>2$, we have $\left(1,(2\lvert K^{\circ}_{++}\rvert-1+b)/(1-c)\right)=B^{\circ}=(1,b)$. Hence $\lvert K^{\circ}_{++}\rvert=(1-bc)/2$. From the same argument for $CS_{2}$, we have $\lvert K^{\circ}_{-+}\rvert=(1-bc)/2$. Moreover, $c=-b$ holds by (7). Then, we have $\lvert K^{\circ}_{++}\rvert=\lvert K^{\circ}_{-+}\rvert=2/\lvert K\rvert=(1+b^{2})/2$ and $S_{1}=(1-b,1+b)/(1+b^{2}),S_{2}=(-1-b,1-b)/(1+b^{2})$. Consequently, we obtain

[TABLE]

with $\lvert K\rvert=4/(1+b^{2})$. Then $K^{\circ}=\operatorname{conv}\left\{\pm(1,b),\pm(-b,1)\right\}$ and $\lvert K^{\circ}\rvert=2(1+b^{2})$. ∎

3 A sharp estimate for the three dimensional volume product

3.1 Preliminaries

Let $K\in\mathcal{K}^{n}$ with $O\in\operatorname{int}K$ and denote by $\mu_{K}$ its Minkowski gauge. Assume $\partial K$ is a $C^{\infty}$-hypersurface in $\mathbb{R}^{n}$. $K$ is called strongly convex if a $C^{\infty}$-function $\mu_{K}^{2}(x)/2$ on $\mathbb{R}^{n}$ is strongly convex, i.e., the Hessian matrix $D^{2}(\mu_{K}^{2}/2)(x)$ is positive definite for each $x$ with $|x|=1$. Then $\Lambda:=\nabla(\mu_{K}^{2}/2)=\nabla\mu_{K}:\partial K\to\partial K^{\circ}$ is a $C^{\infty}$-diffeomorphism. If $K$ is $C^{\infty}$ strongly convex, then $K^{\circ}$ is $C^{\infty}$ strongly convex (see [Sc, Section 1.7.2]).

3.2 Convex bodies in $\mathbb{R}^{3}$

From now on, we focus on centrally symmetric convex bodies in $\mathbb{R}^{3}$. Denote by $\hat{\mathcal{K}}$ the set of all centrally symmetric convex bodies $K\in\mathcal{K}^{3}_{0}$ which are strongly convex with smooth boundary $\partial K$. Throughout this paper, for a convex body $K\in$ $\mathcal{K}^{3}$ we fix the orientation on $K$ induced from the natural orientation of $\mathbb{R}^{3}$. From it the orientation of the boundary $\partial K$, that of any domain $S$ on $\partial K$ and that of $\partial S$ (that is, a closed curve on $\partial K$) are induced, respectively. For a point $P=(p_{1},p_{2},p_{3})$ in $\mathbb{R}^{3}$ we write $-P=(-p_{1},-p_{2},-p_{3})$. If $K\in\hat{\mathcal{K}}$ and $P\in K$, then $-P\in K$. Note that $\rho_{K}(P)P\in\partial K$ for any $P\in\mathbb{R}^{3}\setminus\{O\}$, where $\rho_{K}:=1/\mu_{K}$ is the radial function of $K$. For $K\in\hat{\mathcal{K}}$ and any distinct two points $P,Q\in\partial K$ with $P\not=-Q$, let us introduce an oriented curve from $P$ to $Q$ on the boundary $\partial K$ defined by

[TABLE]

We call it an oriented segment on $\partial K$. The polar body of $K\in\hat{\mathcal{K}}$ is given by

[TABLE]

which is also an element of $\hat{\mathcal{K}}$. Note that $O\in\operatorname{int}K^{\circ}$ and $(K^{\circ})^{\circ}=K$.

Let $K\in\hat{\mathcal{K}}$. First, we decompose the convex body $K$ into eight pieces. We choose an orthonormal basis $(e_{1},e_{2},e_{3})$ of $\mathbb{R}^{3}$ and take six points

[TABLE]

on the boundary $\partial K$, and consider six oriented segments on $\partial K$ as follows:

[TABLE]

Then another six segments $-\alpha,-\beta,-\gamma,-a,-b$, and $-c$ on $\partial K$ are automatically defined. We denote by $Q_{i}(K)\ (i=1,2,3)$ the sections of $K$ by the $yz$-plane, $zx$-plane, and $xy$-plane, respectively. Note that, for instance, in the $yz$-plane $Q_{1}(K)$ is a strongly convex set whose boundary consists of the successive arcs $\alpha,a,-\alpha,-a$. For smooth curves $c(t),d(t)\ (0\leq t\leq 1)$ in $\mathbb{R}^{3}$ such that the terminal point of $c$ coincides with the initial point of $d$, denote by $c\cup d$ the piecewise smooth curve which starts at $c(0)$ and ends at $d(1)$. Then closed curves $\alpha\cup a\cup-\alpha\cup-a$, $\beta\cup b\cup-\beta\cup-b$, and $\gamma\cup c\cup-\gamma\cup-c$ mutually do not intersect except for $A_{\pm},B_{\pm},C_{\pm}$.

For an oriented simple closed curve $c$ on $\partial K$ denote by $\mathcal{S}_{K}(c)\subset\partial K$ the piece of surface enclosed by $c$ with the orientation compatible with that of $c$. In this paper, we always equip $\mathcal{S}_{K}(c)$ with this orientation. We denote by $O*\mathcal{S}_{K}(c)$ the truncated cone over $\mathcal{S}_{K}(c)$:

[TABLE]

Using this notation, we divide $K$ into the following eight pieces:

[TABLE]

where for a curve $c\subset\mathbb{R}^{3}$, denote by $\widetilde{c}$ the curve which has the same image as $c$ with the reverse orientation, that is, $\widetilde{c}(t)=c(1-t)\ (0\leq t\leq 1)$.

Next, we decompose the polar body $K^{\circ}$ into eight pieces associated to the above decomposition of $K$. Using the map $\Lambda:\partial K\to\partial K^{\circ}$ defined in Section 3.1, we put six points and six curves on $\partial K^{\circ}$ as follows:

[TABLE]

From the definition of the polar body and the strong convexity of $K^{\circ}$, the point $A_{+}^{\circ}$ (resp. $B_{+}^{\circ},C_{+}^{\circ}$) is the unique point with the maximal $u$-coordinate (resp. $v$-coordinate, $w$-coordinate) among all the points in $K^{\circ}$.

Since $\Lambda:\partial K\to\partial K^{\circ}$ is a $C^{\infty}$-diffeomorphism, smooth closed curves $\alpha^{\circ}\cup a^{\circ}\cup-\alpha^{\circ}\cup-a^{\circ}$, $\beta^{\circ}\cup b^{\circ}\cup-\beta^{\circ}\cup-b^{\circ}$, and $\gamma^{\circ}\cup c^{\circ}\cup-\gamma^{\circ}\cup-c^{\circ}$ on $\partial K^{\circ}$ are mutually do not intersect except for $A^{\circ}_{\pm},B^{\circ}_{\pm},C^{\circ}_{\pm}$. Thus we can define a decomposition of $K^{\circ}$ into the following eight pieces

[TABLE]

which corresponds to the decomposition of $K$ described before.

Denote by $P_{i}\ (i=1,2,3)$ the orthogonal projections to the $vw$-plane, $wu$-plane and $uv$-plane, respectively. Since $K^{\circ}\subset\mathbb{R}^{3}$ is strongly convex, the projections $P_{i}(K^{\circ})\ (i=1,2,3)$ of $K^{\circ}$ are compact strongly convex domains in each plane.

Claim**.**

The boundaries $\partial(P_{i}(K^{\circ}))\ (i=1,2,3)$ coincide with smooth simple closed curves $P_{1}(\alpha^{\circ}\cup a^{\circ}\cup-\alpha^{\circ}\cup-a^{\circ})$, $P_{2}(\beta^{\circ}\cup b^{\circ}\cup-\beta^{\circ}\cup-b^{\circ})$, and $P_{3}(\gamma^{\circ}\cup c^{\circ}\cup-\gamma^{\circ}\cup-c^{\circ})$, respectively.

Proof.

Fix a point $Q\in\partial(Q_{1}(K))$. Since $Q$ lies on the set

[TABLE]

$Q$ is on one of the four segments in the right-hand side. If $Q\in\alpha$, then there exists $t\in[0,1]$ such that $Q=\alpha(t)$. By the definitions of $\alpha(t)$ and $\alpha^{\circ}(t)$, we have $\alpha(t)=Q_{1}(\alpha(t))=(0,\alpha_{2}(t),\alpha_{3}(t))$, $\alpha^{\circ}(t)=(\alpha^{\circ}_{1}(t),\alpha^{\circ}_{2}(t),\alpha^{\circ}_{3}(t))$, $P_{1}(\alpha^{\circ}(t))=(0,\alpha^{\circ}_{2}(t),\alpha^{\circ}_{3}(t))$, and

[TABLE]

Combining it with the conditions $\alpha(t)=Q_{1}(\alpha(t))\in\partial(Q_{1}(K))$ and $P_{1}(\alpha^{\circ}(t))\in P_{1}(K^{\circ})$, we find that $P_{1}(\alpha^{\circ}(t))\in\partial P_{1}(K^{\circ})$. Considering other cases similarly, we have

[TABLE]

On the other hand, for $P\in\partial(P_{1}(K^{\circ}))$, there exists a point $Q$ in $\partial(Q_{1}(K))=Q_{1}(\alpha\cup a\cup(-\alpha)\cup(-a))$ such that $Q\cdot P=1$. The point $Q$ is represented, for instance, by $\alpha(t)$ for some $t\in[0,1)$. By the definition, $\tilde{P}:=\alpha^{\circ}(t)=\Lambda(\alpha(t))$ is the unique point on $\partial K$ such that $Q\cdot\tilde{P}=1$. Moreover, by the strict convexity of $K^{\circ}$, we have $P_{1}(\tilde{P})=P$. Hence,

[TABLE]

also holds. ∎

3.3 A volume estimate of a piece of $K^{\circ}$ from below

The next task is to estimate $|K^{\circ}_{+++}|,|K^{\circ}_{-++}|,\ldots,|K^{\circ}_{+--}|$ from below. Consider $K^{\circ}_{+++}=O*\mathcal{S}_{K^{\circ}}(\alpha^{\circ}\cup\beta^{\circ}\cup\gamma^{\circ})$, for instance. The boundary $\partial K^{\circ}_{+++}$ is the union of the cone

[TABLE]

and $K^{\circ}_{+++}\cap\partial K^{\circ}$. For any point $P=(u,v,w)\in\mathbb{R}^{3}$, the signed volume of the cone

[TABLE]

is defined by

[TABLE]

where $\alpha^{\circ}(t)=(\alpha^{\circ}_{1}(t),\alpha^{\circ}_{2}(t),\alpha^{\circ}_{3}(t))$ and $|\cdot|$ denotes the determinant for a square matrix. Here we put

[TABLE]

then we have

[TABLE]

For simplicity, we put $\overline{\alpha^{\circ}}:=(\overline{\alpha^{\circ}_{1}},\overline{\alpha^{\circ}_{2}},\overline{\alpha^{\circ}_{3}})$. Similarly, we define $\overline{a^{\circ}},\overline{\beta^{\circ}},\overline{b^{\circ}},\overline{\gamma^{\circ}}$, and $\overline{c^{\circ}}$. Then the signed volume (8) of the cone $P*(O*(\alpha^{\circ}\cup\beta^{\circ}\cup\gamma^{\circ}))$ is represented as $(\overline{\alpha^{\circ}}+\overline{\beta^{\circ}}+\overline{\gamma^{\circ}})\cdot(u,v,w)/6$.

If the cone $K^{\circ}_{+++}$ is convex in $\mathbb{R}^{3}$ and $P\in K^{\circ}_{+++}$, then the quantity (8) coincides with the volume $|P*(O*(\alpha^{\circ}\cup\beta^{\circ}\cup\gamma^{\circ}))|$ and it is less than $|K^{\circ}_{+++}|$, which is the fact actually used in [Me2] and [BF]. In case $K^{\circ}_{+++}\subset\mathbb{R}^{3}$ is not convex or $P\in K^{\circ}$ is not in $K^{\circ}_{+++}$, the comparison of the signed volume (8) with $|K^{\circ}_{+++}|$ is a nontrivial problem. However, under the condition that $P$ lies in the convex set $K^{\circ}$ the following holds, which is a key lemma to estimate $\mathcal{P}(K)$.

Lemma 3.1**.**

Let $K^{\circ}\in\hat{\mathcal{K}}$. For any point $P=(u,v,w)\in K^{\circ}$, we have

[TABLE]

By the definition of polar, we obtain

[TABLE]

For the other seven parts similar estimates hold (see the table below).

Lemma 3.1 is a direct consequence of the following

Proposition 3.2**.**

Let $K\in\hat{\mathcal{K}}$. Let $c$ be a piecewise smooth, oriented, simple closed curve on $\partial K$. Let $\mathcal{S}_{K}(c)\subset\partial K$ be a piece of surface enclosed by the curve $c$. Then for any point $P\in K$, the inequality

[TABLE]

holds.

Proof.

Let $S$ be the boundary of the truncated cone $O*\mathcal{S}_{K}(c)$ and $S_{0}:=S\setminus\mathcal{S}_{K}(c)$. Then $S$ is a piecewise smooth surface. Let $\bm{r}$ be the position vector on $S$ and $\bm{n}$ the outward unit normal vector of $S$ at $\bm{r}$.

Since the vector $\bm{r}$ on $S_{0}$ is perpendicular to the normal vector $\bm{n}$ at the point, we have $\bm{r}\cdot\bm{n}=0$ on $S_{0}$. Hence, by the divergence theorem, $\int_{\mathcal{S}_{K}(c)}\bm{r}\cdot\bm{n}\,dA=\int_{O*\mathcal{S}_{K}(c)}\mathrm{div}\,\bm{r}\,dxdydz=3|O*\mathcal{S}_{K}(c)|$ holds, where $dA$ denotes the area element.

On the other hand, on $\mathcal{S}_{K}(c)$ we have $P\cdot\bm{n}\leq\bm{r}\cdot\bm{n}$ for any $P\in K$, because the support function $h_{K}(\bm{n}):=\max_{P\in K}P\cdot\bm{n}$ attains the maximum at $\bm{r}$. Integrating on $\mathcal{S}_{K}(c)$, we have

[TABLE]

Since the rotation of $P\times\bm{r}$ is calculated as $2P$, by Stokes’ formula, the left-hand side is equal to $(1/2)\int_{c}(P\times\bm{r})\cdot d\bm{r}$, which yields the result. ∎

To discuss the other seven parts of $K^{\circ}$ we need further preparations. For the curve $\widetilde{\alpha^{\circ}}$ on $\partial K^{\circ}$ defined by $\widetilde{\alpha^{\circ}}(t)=\alpha^{\circ}(1-t)\ (0\leq t\leq 1)$, we have

[TABLE]

for any $P=(u,v,w)\in\mathbb{R}^{3}$. Moreover, since

[TABLE]

we obtain

[TABLE]

We can do the same observation for $K^{\circ}_{-++}$, $K^{\circ}_{--+}$, and $K^{\circ}_{+-+}$. Since $K^{\circ}$ is centrally symmetric, the volume of $K^{\circ}_{++-}$, $K^{\circ}_{-+-}$, $K^{\circ}_{---}$, and $K^{\circ}_{+--}$ equals that of $K^{\circ}_{--+}$, $K^{\circ}_{+-+}$, $K^{\circ}_{+++}$, and $K^{\circ}_{-++}$, respectively.

Summarizing the above arguments, we obtain

[TABLE]

Furthermore, similar arguments for the convex body $K$ give the following results. Note that the calculation is easier than the case of the polar $K^{\circ}$, because the eight parts of $K$ are convex.

[TABLE]

In Section 3.4, these eight points $R_{i},S_{i}\ (i=1,2,3,4)$ will be effectively used as test points to estimate $\mathcal{P}(K)$ from below.

3.4 A sharp estimate

In this subsection, we give a sufficient condition of deducing inequality (2) (see the condition (10) below). In the following sections, for any $K\in\hat{\mathcal{K}}$ we carefully select a linear transformation $\mathcal{A}$ such that $\mathcal{A}K\in\hat{\mathcal{K}}$ satisfies this condition.

From the arguments in Section 3.3, we know that $R_{i}\in K^{\circ}$ and $S_{i}\in K$ $(i=1,2,3,4)$. By the definition of polar, we obtain $R_{i}\cdot S_{i}\leq 1$. In other words,

[TABLE]

Here, assume that

[TABLE]

We now check that this condition implies symmetric Mahler’s conjecture for the three dimensional case. Indeed, the condition (10) yields the volume of each eight part equals $|K|/8$, and hence equations (9) and (10) implies that

[TABLE]

It then follows from these inequalities that

[TABLE]

Let us examine the first term in the last line. Recalling the definition of $\overline{\alpha^{\circ}}=(\overline{\alpha^{\circ}_{1}},\overline{\alpha^{\circ}_{2}},\overline{\alpha^{\circ}_{3}})$, its $u$-component $\overline{\alpha^{\circ}_{1}}$ is twice the area of the convex region $O*P_{1}(\alpha^{\circ})$ in the $vw$-plane. Similarly, $\overline{a^{\circ}_{1}}$ is twice $|O*P_{1}(a^{\circ})|$. Since $K^{\circ}\in\hat{\mathcal{K}}$, the convex domain $P_{1}(K^{\circ})$ in the $vw$-plane is also centrally symmetric, and hence

[TABLE]

Again, since the curve $\alpha\subset K$ lies in the $yz$-plane, we have $\overline{\alpha}=(\overline{\alpha_{1}},0,0)$, where

[TABLE]

Here $\overline{\alpha_{1}}$ is nothing but twice the area of the convex region $O*\alpha$ in the $yz$-plane. From the assumption that $\overline{\alpha}=\overline{a}$ in the condition (10), we have $\overline{\alpha_{1}}=|Q_{1}(K)|/2$. Therefore, the first term $2\overline{\alpha}\cdot\left(\overline{\alpha^{\circ}}+\overline{a^{\circ}}\right)$ is equal to $|Q_{1}(K)|\,|P_{1}(K^{\circ})|$. Applying the same argument to the second and the third terms, we have

[TABLE]

and

[TABLE]

Consequently, we obtain

[TABLE]

It is well-known that convex domains $Q_{i}(K)$ and $P_{i}(K^{\circ})$ are polar bodies of each other (e.g. [GMR]*p. 274). By Mahler’s theorem [Ma0] (see also [R], [GMR], [MR], and Section 2), we have $|Q_{i}(K)|\,|P_{i}(K^{\circ})|\geq 4^{2}/2!=8\ (i=1,2,3)$. Thus

[TABLE]

holds. Hence we obtain

Proposition 3.3**.**

If $K\in\hat{\mathcal{K}}$ satisfies the condition (10), then $\mathcal{P}(K)\geq 32/3$.

4 The general case

The purpose of this and the next sections is to find a suitable linear transformation $\mathcal{A}$ for any centrally symmetric convex body $K$ in $\hat{\mathcal{K}}$ such that ${\mathcal{A}}K$ satisfies the condition (10). In other words, we will find a basis of $\mathbb{R}^{3}$ such that the hypothesis (10) is satisfied.

4.1 Linear transforms

In order to prove (2) for general $K\in\mathcal{K}^{3}_{0}$, we first consider some linear transform of $K\in\hat{\mathcal{K}}$. For this purpose, we introduce some notations. For $\theta,\phi,\psi\in\mathbb{R}$, $X(\theta)$, $Y(\phi)$, $Z(\psi)$ denote rotations in $\mathbb{R}^{3}$ as follows:

[TABLE]

Hereafter, we put $K(\theta,\phi,\psi):=X(\theta)Y(\phi)Z(\psi)K$.

In this section, we choose a linear transform $\mathcal{A}=\mathcal{A}(K)$ (depends on $K$) which satisfies

[TABLE]

where $\mathcal{A}K=\mathcal{A}(K)K$. We use the following spherical coordinates:

[TABLE]

First, we choose the angle $\Theta(K)\in(0,\pi)$ which satisfies that

[TABLE]

Since the left-hand side is increasing and the right-hand side is decreasing with respect to $\Theta(K)$, we see that $\Theta(K)$ is uniquely determined. Actually, we have the following

Lemma 4.1**.**

For each $K\in\hat{\mathcal{K}}$, there exists the unique $\Theta(K)\in(0,\pi)$ satisfying the relation (12). Moreover, $\Theta(\theta,\phi,\psi):=\Theta(K(\theta,\phi,\psi))$ is smooth with respect to $(\theta,\phi,\psi)\in\mathbb{R}^{3}$.

Proof.

We put

[TABLE]

Since $K\in\hat{\mathcal{K}}$, by the definition of $K(\theta,\phi,\psi)$, we see that $I$ is smooth in $\mathbb{R}^{4}$. On the other hand, since $\rho_{K}(P(\xi,\eta))>0$, we have

[TABLE]

By the intermediate value theorem, there exists the unique $\Theta(\theta,\phi,\psi)=\Theta(K(\theta,\phi,\psi))$ such that $I(\Theta(\theta,\phi,\psi),\theta,\phi,\psi)=0$. Moreover, by the implicit function theorem, we see that $\Theta(\theta,\phi,\psi)$ is smooth in $\mathbb{R}^{3}$. ∎

Next, we introduce the quantities $\Phi(K)\in(0,\pi)$ and $\Psi(K)\in(0,\pi)$ defined by

[TABLE]

Similarly as $\Theta(K)$, the above $\Phi(K)$ and $\Psi(K)$ are uniquely determined for each $K\in\hat{\mathcal{K}}$, and $\Phi(\theta,\phi,\psi):=\Phi(K(\theta,\phi,\psi))$ and $\Psi(\theta,\phi,\psi):=\Psi(K(\theta,\phi,\psi))$ are smooth functions on $\mathbb{R}^{3}$. Now, we define $\mathcal{A}=\mathcal{A}(K)$ as

[TABLE]

Proposition 4.2**.**

For every $K\in\hat{\mathcal{K}}$, the condition (11) holds for $\mathcal{A}K$ which is the image of $K$ by the above linear transform.

Proof.

Denote by $\chi_{A}$ the characteristic function:

[TABLE]

Then we have

[TABLE]

where we used the substitution $P=\mathcal{A}^{-1}\tilde{P}$, that is

[TABLE]

By using the polar coordinates $x=r\cos\xi$, $y=r\sin\xi\cos\eta$, $z=r\sin\xi\sin\eta$, we have

[TABLE]

Under the condition $r>0$ and $\sin\xi>0$, $z>0$ if and only if $0<\eta<\pi$. Thus we have

[TABLE]

Therefore, by Lemma 4.1, we have

[TABLE]

Next, since

[TABLE]

by the substitution $P=\mathcal{A}^{-1}\tilde{P}$, we have

[TABLE]

By using the above polar coordinates, if $y=z/\tan(\Theta(K))$ and $z>0$, then $\eta=\Theta(K)\in(0,\pi)$. Since

[TABLE]

we see that $x-z/(\sin(\Theta(K))\tan(\Psi(K)))>0$ if and only if $0<\xi<\Psi(K)$, and $x-z/(\sin(\Theta(K))\tan(\Psi(K)))<0$ if and only if $\Psi(K)<\xi<\pi$. Thus, by the definition of $\Psi(K)$, we obtain

[TABLE]

Finally, we show $|O*\gamma(\mathcal{A}K)|=|O*c(\mathcal{A}K)|$ similarly. Since

[TABLE]

by the substitution $P=\mathcal{A}^{-1}\tilde{P}$, we have

[TABLE]

By using the same polar coordinates, for the case $y>0$, $z=0$, we have $\eta=0$ and

[TABLE]

Hence, $x-y/\tan(\Phi(K))>0$ if and only if $0<\xi<\Phi(K)$, and $x-y/\tan(\Phi(K))<0$ if and only if $\Phi(K)<\xi<\pi$. Thus, by the definition of $\Phi(K)$, we obtain

[TABLE]

∎

4.2 Definition of a smooth map $(F,G,H)$

For simplicity, we denote $\mathcal{A}K:=\mathcal{A}(K)K$ and

[TABLE]

For every $K\in\hat{\mathcal{K}}$, we define real numbers $F,G,H$ by

[TABLE]

and introduce the following functions on $\mathbb{R}^{3}$:

[TABLE]

If we can find a zero $(\theta,\phi,\psi)$ of $(F,G,H)$, then ${\mathcal{A}K}(\theta,\phi,\psi)$ satisfies the condition (10) by Proposition 4.2. To show the existence of such a zero, we consider the following region in $\mathbb{R}^{3}$:

[TABLE]

Note that $\Theta$ is a smooth function introduced in the previous subsection and $0<\Theta(0,\phi,\psi)<\pi$. We divide the boundary $\partial D$ into the following six parts:

[TABLE]

Then, we have

[TABLE]

and $M_{i}\cap M_{j}$ is empty or a curve for each $i\not=j$. In the case where $(F,G,H)$ has no zeros on $\partial D$, we can use the degree of a map

[TABLE]

to find a zero in the interior of $D$. In order to calculate the degree of $\mathscr{F}$, in the rest of this section, we show the identities (17), (18), and (19) in Proposition 4.13 below.

4.3 Some formulas

We prepare some formulas.

Lemma 4.3**.**

For $K\in\hat{\mathcal{K}}$ and $\xi,\eta,\zeta\in\mathbb{R}$,

[TABLE]

Proof.

By the definition of $\mu_{K}$, we have

[TABLE]

By the definition of $\rho_{K}$, we get

[TABLE]

Moreover, for $P=P(\xi,\eta)\in S^{2}$, $X(-\zeta)P(\xi,\eta)=P(\xi,\eta-\zeta)$, which proves the first equation.

Next, by a direct calculation, we have $P(\xi,\eta\pm\pi)=-P(\pi-\xi,\eta)$. Since $K$ is centrally symmetric, we obtain

[TABLE]

∎

Lemma 4.4**.**

$\Theta(K)+\Theta(X(\pi-\Theta(K))K)=\pi.$**

Proof.

For $\eta_{0},\eta_{1}\in\mathbb{R}$, by Lemma 4.3, we have

[TABLE]

where we used the substitution $\tilde{\eta}=\eta+\pi-\Theta(K)$. Applying (13) with $(\eta_{0},\eta_{1})=(0,\Theta(K)),(\Theta(K),\pi)$ to the definition (12) of $\Theta(K)$, we obtain

[TABLE]

where we used Lemma 4.3 and the substitution $\tilde{\xi}=\pi-\xi$. Since $\pi-\Theta(K)\in(0,\pi)$, the uniqueness of $\Theta(X(\pi-\Theta(K))K))$ implies $\Theta(X(\pi-\Theta(K))K))=\pi-\Theta(K)$ from (12). ∎

4.4 Identities relating to the rotation $X(\pi-\Theta(K))$ and $X(\pi)$

Here we examine the behavior of the functions $F$, $G$, and $H$ under the rotation $X$ of $K$.

Lemma 4.5**.**

Put $L=X(\pi-\Theta(K))K$. Then

[TABLE]

Proof.

For simplicity, we put $\tilde{\Theta}=\Theta(L)$, $\Theta=\Theta(K)$, $\tilde{\Phi}=\Phi(L)$, $\Phi=\Phi(K)$, $\tilde{\Psi}=\Psi(L)$, $\Psi=\Psi(K)$. Note that $\tilde{\Theta}=\pi-\Theta$ by Lemma 4.4, which means the first formula.

By the definition of $\tilde{\Phi}$, we have

[TABLE]

By Lemma 4.3, we get

[TABLE]

and

[TABLE]

where we used the substitution $\tilde{\xi}=\pi-\xi$. Similarly,

[TABLE]

Consequently, we have

[TABLE]

By the uniqueness of $\Psi=\Psi(K)$, we see $\Psi=\pi-\tilde{\Phi}$.

Similarly, by the definition of $\tilde{\Psi}$,

[TABLE]

holds. By Lemma 4.3, we have

[TABLE]

Thus we get

[TABLE]

By the uniqueness of $\Phi=\Phi(K)$, we obtain $\tilde{\Psi}=\Phi$. ∎

Lemma 4.6**.**

[TABLE]

Proof.

We use the same notations as in Lemma 4.5. Putting $\tilde{\mathcal{A}}=\mathcal{A}(L)$, $\mathcal{A}=\mathcal{A}(K)$, we have

[TABLE]

By a direct calculation using Lemma 4.5,

[TABLE]

Thus, for any $\tilde{P}=(\tilde{x},\tilde{y},\tilde{z})\in\mathbb{R}^{3}$, we have

[TABLE]

By the substitution $x=\tilde{x}$, $y=\tilde{z}/\sin\Theta$, $z=-\tilde{y}\sin\Theta$,

[TABLE]

where the last equality comes from the fact that $\mathcal{A}K$ is centrally symmetric. Similarly, we obtain

[TABLE]

Since $d\tilde{y}d\tilde{z}=dydz$, we also obtain

[TABLE]

Therefore,

[TABLE]

∎

Using the above lemmas, we get identities relating to the rotation $X(\pi)$.

Lemma 4.7**.**

[TABLE]

[TABLE]

Proof.

Put $L=X(\pi-\Theta(K))K$. By Lemma 4.5, we have $\Theta(L)+\Theta(K)=\pi$. Thus,

[TABLE]

and so we can apply Lemmas 4.5 and 4.6 to $X(\pi)K=X(\pi-\Theta(L))L$. By Lemma 4.5, we have

[TABLE]

It follows from Lemma 4.6 that

[TABLE]

∎

4.5 Identities relating to the rotation $Y(\pi)$

Lemma 4.8**.**

Put $L=Y(\pi)K$. Then

[TABLE]

Proof.

For simplicity, we put $\tilde{\Theta}=\Theta(L)$, $\Theta=\Theta(K)$, $\tilde{\Phi}=\Phi(L)$, $\Phi=\Phi(K)$, $\tilde{\Psi}=\Psi(L)$, $\Psi=\Psi(K)$. We first claim that

[TABLE]

Actually, $Y(-\pi)P(\xi,\eta)=-P(\xi,\pi-\eta)$ holds by a direct calculation. Thus we have

[TABLE]

Since $K$ is centrally symmetric, the claim (14) holds.

By (14) and the definition of $\tilde{\Theta}$,

[TABLE]

By the substitution $\tilde{\eta}=\pi-\eta$,

[TABLE]

By the definition and uniqueness of $\Theta$, we get $\Theta=\pi-\tilde{\Theta}$. By the definition of $\tilde{\Phi}$,

[TABLE]

By Lemma 4.3, $\rho_{K}(P(\xi,\pi-\eta))=\rho_{K}(P(\pi-\xi,-\eta))$ holds. It then follows from (14) that

[TABLE]

where we used the substitution $\tilde{\xi}=\pi-\xi$. Therefore, by the definition of $\Phi$, $\Phi=\pi-\tilde{\Phi}$. Similarly, by the definition of $\tilde{\Psi}$,

[TABLE]

By (14) and $\Theta=\pi-\tilde{\Theta}$, we have $\rho_{L}^{2}(P(\xi,\tilde{\Theta}))=\rho_{K}^{2}(P(\xi,\pi-\tilde{\Theta}))=\rho_{K}^{2}(P(\xi,\Theta))$. Thus

[TABLE]

which means that $\tilde{\Psi}=\Psi$. ∎

Lemma 4.9**.**

[TABLE]

Proof.

We use the same notations as in Lemma 4.8. Putting $\tilde{\mathcal{A}}=\mathcal{A}(L)$, $\mathcal{A}=\mathcal{A}(K)$, we have

[TABLE]

Thus

[TABLE]

By Lemma 4.8 and a direct calculation,

[TABLE]

Thus, for any $\tilde{P}=(\tilde{x},\tilde{y},\tilde{z})\in\mathbb{R}^{3}$, we have

[TABLE]

where we used the substitution $x=-\tilde{x}$, $y=\tilde{y}$, $z=-\tilde{z}$. Similarly, we have

[TABLE]

Moreover, we get

[TABLE]

Therefore,

[TABLE]

∎

4.6 Identities relating to the rotation $Z(\pi)$

Lemma 4.10**.**

Put $L=Z(\pi)K$. Then

[TABLE]

Proof.

We put $\tilde{\Theta}=\Theta(L)$, $\Theta=\Theta(K)$, $\tilde{\Phi}=\Phi(L)$, $\Phi=\Phi(K)$, $\tilde{\Psi}=\Psi(L)$, $\Psi=\Psi(K)$. Since $Z(-\pi)P(\xi,\eta)=P(\pi-\xi,\pi-\eta)$, we have

[TABLE]

It follows from Lemma 4.3 that

[TABLE]

By (15) and the definition of $\tilde{\Theta}$, we have

[TABLE]

By the substitution $\tilde{\xi}=\pi-\xi$, $\tilde{\eta}=\pi-\eta$,

[TABLE]

By the definition of $\Theta$, we get $\Theta=\pi-\tilde{\Theta}$. Furthermore, we have

[TABLE]

By the definition of $\Phi$, we obtain $\Phi=\tilde{\Phi}$. Finally, the definition of $\tilde{\Psi}$ means

[TABLE]

By (15) and $\Theta=\pi-\tilde{\Theta}$, we obtain

[TABLE]

Thus

[TABLE]

Therefore, $\pi-\tilde{\Psi}=\Psi$. ∎

Lemma 4.11**.**

[TABLE]

Proof.

We use the same notations as in Lemma 4.10. Putting $\tilde{\mathcal{A}}=\mathcal{A}(L)$, $\mathcal{A}=\mathcal{A}(K)$, we have

[TABLE]

Then

[TABLE]

By Lemma 4.10 and a direct calculation,

[TABLE]

Thus, for any $\tilde{P}=(\tilde{x},\tilde{y},\tilde{z})\in\mathbb{R}^{3}$, we see

[TABLE]

where we used the substitution $x=-\tilde{x}$, $y=-\tilde{y}$, $z=\tilde{z}$. Similarly, we obtain

[TABLE]

Moreover, we have

[TABLE]

Therefore,

[TABLE]

∎

4.7 Key identities on $\partial D$

By using the lemmas prepared in the above subsections, we show Proposition 4.13 below, which states key identities on $\partial D$ for the degree calculation in the next section. To begin with, we introduce the map $\Gamma$ defined by

[TABLE]

Lemma 4.12**.**

$\Gamma$* satisfies the following properties:*

1. (i).

$\theta\mapsto\Gamma_{\psi}(\theta)$* is increasing.* 2. (ii).

$\Gamma_{\psi}\left([0,\pi-\Theta(0,0,\psi)]\right)=[\pi-\Theta(0,0,\psi),\pi]$. 3. (iii).

$\Gamma_{\psi},(\Gamma_{\psi})^{-1}$* is smooth with respect to $\theta$ and $\psi$.*

Proof.

By Lemma 4.1, $\Gamma_{\psi}(\theta)$ is smooth with respect to $\theta$ and $\psi$. We claim that

[TABLE]

If the inequality (16) holds, then the properties (i), (ii), and (iii) are easily verified. Indeed, (i) follows immediately. Note that $\Gamma_{\psi}(0)=\pi-\Theta(0,0,\psi)$. By Lemma 4.4, we have

[TABLE]

which proves (ii), because $\Gamma_{\psi}$ is increasing by (i). Moreover, $\Gamma_{\psi}^{-1}$ is smooth by the implicit function theorem.

Therefore, it suffices to prove (16). By the definition of $\Theta$,

[TABLE]

By Lemma 4.3, we have

[TABLE]

Thus, by the substitution $\tilde{\eta}=\eta-\theta$, we obtain

[TABLE]

By differentiating it with respect to $\theta$,

[TABLE]

which yields the inequality (16). ∎

Proposition 4.13**.**

Between $(0,\phi,\psi)\in M_{1}$ and $(\pi-\Theta(0,\phi,\psi),\phi,\psi)\in M_{2}$, the following formulas hold:

[TABLE]

Between $(\tilde{\theta},0,\psi)\in M_{3}$ and $(\theta,\pi,\psi)\in M_{4}$, the following formulas hold:

[TABLE]

where $\tilde{\theta}=(\Gamma_{\psi})^{-1}(\pi-\theta)\in[0,\pi-\Theta(0,0,\psi)]$. Between $(\theta,\pi-\phi,0)\in M_{5}$ and $(\theta,\phi,\pi)\in M_{6}$, the following formulas hold:

[TABLE]

Proof.

First, we show (17). We have

[TABLE]

Noting $\Theta(\theta,\phi,\psi)=\Theta(K(\theta,\phi,\psi))$, Lemma 4.6 implies

[TABLE]

Putting $\theta=0$, we get (17).

Next, let us consider the case $(\theta,\pi,\psi)\in M_{4}$. Since $X(\theta)Y(\pi)=Y(\pi)X(-\theta)$, $Y(\phi)Z(\pi)=Z(\pi)Y(-\phi)$, and $X(\theta)Z(\pi)=Z(\pi)X(-\theta)$, we have

[TABLE]

By Lemma 4.9, we have

[TABLE]

Since $F,G,H$ are $2\pi$-periodic with respect to $\theta$, we use Lemma 4.7 to obtain

[TABLE]

Here, by the definition of $M_{4}$, $0\leq\theta\leq\pi-\Theta(0,\pi,\psi)$ holds, and hence $\Theta(0,\pi,\psi)\leq\pi-\theta\leq\pi$. On the other hand, by Lemma 4.8, we have

[TABLE]

Thus $\pi-\Theta(0,0,\psi)\leq\pi-\theta\leq\pi$. For each $\pi-\theta\in[\pi-\Theta(0,0,\psi),\pi]$, putting $\tilde{\theta}=(\Gamma_{\psi})^{-1}(\pi-\theta)\in[0,\pi-\Theta(0,0,\psi)]$ (see Lemma 4.12), by using (20), we get

[TABLE]

which imply (18).

Finally, we consider the case $(\theta,\phi,\pi)\in M_{6}$. Note that

[TABLE]

Thus, by Lemmas 4.11 and 4.9, we obtain

[TABLE]

Since $K(0,\phi,\pi)=Z(\pi)Y(\pi)K(0,\pi-\phi,0)$, by the definition of $\Theta$, we have

[TABLE]

It means that $(\theta,\phi,\pi)\in M_{6}$ $(\theta\in[0,\pi-\Theta(0,\phi,\pi)])$ if and only if $(\theta,\pi-\phi,0)\in M_{5}$ $(\theta\in[0,\pi-\Theta(0,\pi-\phi,0)])$. Hence (19) is verified. ∎

5 Calculation of degree

5.1 Setting

We use new coordinates $(s,\phi,\psi)$ instead of $(\theta,\phi,\psi)$, where

[TABLE]

For the new coordinates $(s,\phi,\psi)$, we see that

[TABLE]

We define

[TABLE]

Note that

[TABLE]

and $(\hat{F},\hat{G},\hat{H})\in C^{\infty}(D,\mathbb{R}^{3})$.

Then $(\hat{F},\hat{G},\hat{H})|_{\partial D}$ is a triplet of continuous functions on $\partial D=M_{1}\cup\dots\cup M_{6}$, of class $C^{\infty}$ in the interior of each $M_{i}$ for $i=1,\dots,6$. Hence, $(\hat{F},\hat{G},\hat{H})|_{\partial D}$ can be regarded as a piecewise smooth vector field on $\partial D$. It follows from Proposition 4.13, (21), and (22) that $(\hat{F},\hat{G},\hat{H})$ satisfies the identities:

[TABLE]

for $\phi,\psi\in[0,\pi]$ and

[TABLE]

for $s\in[0,1],\psi\in[0,\pi]$ and

[TABLE]

for $s\in[0,1],\phi\in[0,\pi]$, where $T_{\psi}:[0,1]\to[0,1]$ is defined by

[TABLE]

By Lemma 4.12, $T_{\psi}(s)$ is smooth with respect to $\psi$ and $s$, decreasing with respect to $s$, $T_{\psi}(0)=1$, and $T_{\psi}(1)=0$.

As we mentioned in Section 4.2, if we can find a zero $(\theta,\phi,\psi)$ in $D$ of the vector field $(F,G,H)$, then the equations (10) in Section 3.4 are satisfied and the proof of 3-dimensional symmetric Mahler conjecture is completed. In Sections 5.3 and 5.4, we consider the case that $(F,G,H)$ has no zeros on the boundary $\partial D$. Now, assume that $(F,G,H)\not=0$ on $\partial D$, that is, $(\hat{F},\hat{G},\hat{H})\not=0$ on $\partial D$. Then we shall calculate the degree of

[TABLE]

5.2 Degree of a map

The degree of a map is one of the basic tools to find a zero of a vector field. Here we briefly review the necessary facts about the degree of maps (see e.g., [OR]).

Let $M,N$ be connected oriented smooth manifolds of dimension $n$. For a smooth map $f:M\to N$, the integer

[TABLE]

does not depend on the choice of the regular value $a\in N$, which is called the Brouwer-Kronecker degree of $f$ and denoted by $\deg(f)$. Since $\deg(f)$ has the property of the homotopy invariance, the notion can be defined for continuous maps. Note that since $\deg(f)$ can be described in terms of integration of differential forms [OR, Corollary 2.4], we can also use it when $M$ is piecewise smooth and $f:M\to N$ is continuous and smooth on the smooth part of $M$ as long as $f^{-1}(a)$ is contained in the smooth part of $M$. In our setting, $M=\partial D\subset\mathbb{R}^{3}$ and $f=\mathscr{F}:\partial D\to S^{2}$ is continuous, and smooth on the interior of $M_{i}\ (i=1,\ldots,6)$.

Another notion of degree we will need below is the Euclidean degree. Let $D_{0}\subset\mathbb{R}^{m+1}$ be a bounded open set. Let $g:D_{0}\cup\partial D_{0}\to\mathbb{R}^{m+1}$ be a smooth map. If $a\in\mathbb{R}^{m+1}\setminus g(\partial D_{0})$ is a regular value of $g|_{D_{0}}$, then $g^{-1}(a)$ is a finite set and

[TABLE]

is called the Euclidean degree of $f$. This is constant for all regular values contained in the same connected component of $\mathbb{R}^{m+1}\setminus g(\partial D_{0})$.

In our setting, the key idea for proving that $\deg(\mathscr{F})\neq 0$ is to reduce the calculation to the case of the Euclidean degree of $g=(\hat{G},\hat{H}):M_{1}\to\mathbb{R}^{2}$ (see Proposition 5.2). And the latter is reduced to the calculation of the winding number of it (see Proposition 5.3).

5.3 Reduction

For notational convenience, in Sections 5.3, 5.4, and Appendix A, we denote the above $\hat{F}$, $\hat{G}$, and $\hat{H}$ by $F$, $G$, and $H$, respectively, and consider

[TABLE]

In order to calculate $\deg\mathscr{F}$ by using (26), we first need to modify the map $\mathscr{F}$ so as to satisfy the property that $(\pm 1,0,0)$ are its regular values.

Proposition 5.1**.**

There exists a map $\tilde{\mathscr{F}}:\partial D\to S^{2}\subset\mathbb{R}^{3}$ satisfying the following properties (i)–(iv):

1. (i).

$\tilde{\mathscr{F}}$* is homotopic to $\mathscr{F}$. Especially, $\deg\tilde{\mathscr{F}}=\deg\mathscr{F}$ holds.* 2. (ii).

$\tilde{\mathscr{F}}(\partial M_{i})\not\ni(\pm 1,0,0)$* for $i=1,\dots,6$.* 3. (iii).

$(\pm 1,0,0)$* are regular values of $\tilde{\mathscr{F}}$.* 4. (iv).

$\tilde{\mathscr{F}}$* satisfies (23), (24), and (25).*

We give a proof of Proposition 5.1 in Appendix A. Owing to Proposition 5.1, in what follows we may assume that

[TABLE]

and $(\pm 1,0,0)$ are regular values of $\mathscr{F}$. Note that $\mathscr{F}(s,\phi,\psi)=(\pm 1,0,0)$ if and only if $G(s,\phi,\psi)=H(s,\phi,\psi)=0$ and $\operatorname{sgn}F(s,\phi,\psi)=\pm 1$. Hence, the inverse image of $(\pm 1,0,0)$ by $\mathscr{F}\rvert_{M_{1}}$ corresponds to the zeros of the vector field $(G,H)$ on $M_{1}$. By the condition (28), $(G,H)\not=(0,0)$ on $\partial M_{1}$ and the map

[TABLE]

is well-defined.

Proposition 5.2**.**

For the above maps $\mathscr{F}:\partial D\rightarrow S^{2}$ and $\mathscr{G}:\partial M_{1}\rightarrow S^{1}\subset\mathbb{R}^{2}$, we have

[TABLE]

where $w(\mathscr{G},(0,0))$ is the winding number.

Proof.

Since $(1,0,0)$ is a regular value of $\mathscr{F}$, by (26) we have

[TABLE]

Since the inverse image $\mathscr{F}^{-1}(1,0,0)$ is a finite set, we find out the parity of $\deg\mathscr{F}$ by counting the elements. For each $i=1,\ldots,6$, denote by $k_{i}$ the cardinality of $\mathscr{F}^{-1}(1,0,0)\cap\operatorname{int}M_{i}$, that is

[TABLE]

Similarly $k_{-}$ denotes the cardinality of $\mathscr{F}^{-1}(-1,0,0)\cap\operatorname{int}M_{1}$, that is

[TABLE]

By (28), we have $\deg\mathscr{F}=k_{1}+\dots+k_{6}\pmod{2}$.

If $P\in\mathscr{F}^{-1}(1,0,0)\cap M_{4}$, that is $P=(s,\pi,\psi)\in\mathscr{F}^{-1}(1,0,0)$ for some $s\in[0,1]$ and $\psi\in[0,\pi]$, then for $\tilde{P}=(T_{\psi}(s),0,\psi)\in M_{3}$ we have $\tilde{P}\in\mathscr{F}^{-1}(1,0,0)$ by (24). Since $(s,\pi,\psi)\mapsto(T_{\psi}(s),0,\psi)$ is a $C^{\infty}$-diffeomorphism from $\operatorname{int}M_{4}$ to $\operatorname{int}M_{3}$, we see $k_{4}\leq k_{3}$. Using the same argument for $P\in M_{3}\cap\mathscr{F}^{-1}(1,0,0)$, we obtain $k_{3}\leq k_{4}$, and hence $k_{3}=k_{4}$. Similarly, using the correspondence between $M_{5}$ and $M_{6}$ by (25), we get $k_{5}=k_{6}$. Furthermore, by (23) the set $\mathscr{F}^{-1}(1,0,0)\cap M_{2}$ exactly corresponds to $\mathscr{F}^{-1}(-1,0,0)\cap M_{1}$, which yields $k_{2}=k_{-}$. Consequently, we obtain

[TABLE]

and

[TABLE]

By (29) and (30), $k_{1}+k_{-}$ can be viewed as the number of zeros of the vector field $(G,H)$ on $M_{1}$. Note that $P\in\operatorname{int}M_{1}$ is a regular point of $(G,H)$ with $(G,H)(P)=(0,0)$ if and only if $P\in\operatorname{int}M_{1}$ is a regular point of $\mathscr{F}$ with $\mathscr{F}(P)=(\pm 1,0,0)$. Since $(G,H)\not=(0,0)$ on $\partial M_{1}$, $(0,0)$ is a regular value of $(G,H)$ because $(\pm 1,0,0)$ are regular values of $\mathscr{F}$. Thus, by (27) we have

[TABLE]

Moreover, it is known that

[TABLE]

where the right-hand side is the winding number of $(G,H):\partial M_{1}\rightarrow\mathbb{R}^{2}\setminus\{(0,0)\}$ around $(0,0)$. Since $(G,H):\partial M_{1}\rightarrow\mathbb{R}^{2}\setminus\{(0,0)\}$ is homotopic to

[TABLE]

we have $w((G,H),(0,0))=w(\mathscr{G},(0,0))$, which completes the proof. ∎

5.4 Calculation of $w(\mathscr{G},(0,0))$

Proposition 5.2 reduces the problem to the following

Proposition 5.3**.**

$w(\mathscr{G},(0,0))$* is odd.*

Proof.

To calculate the winding number, we represent each point $P_{t}$ on $\partial M_{1}$ as

[TABLE]

which parametrizes $\partial M_{1}$. Put

[TABLE]

Then $s$ is a piecewise smooth function of $t$. We have

[TABLE]

where $W(t):=\tilde{G}(P_{t})\tilde{H}^{\prime}(P_{t})-\tilde{G}^{\prime}(P_{t})\tilde{H}(P_{t}).$ First, by (23) and (24), we have

[TABLE]

Thus

[TABLE]

which yield that, for $3\pi\leq t\leq 4\pi$,

[TABLE]

So, we obtain

[TABLE]

where we used the substitution $4\pi-t=\tau-\pi$. On the other hand, for $\pi\leq t\leq 2\pi$, we have $\tilde{G}(t)=\tilde{G}(0,\pi,t-\pi)$, $\tilde{H}(t)=\tilde{H}(0,\pi,t-\pi)$. So, we obtain

[TABLE]

Hence

[TABLE]

Next, by (25), we have

[TABLE]

Thus

[TABLE]

which yield that, for $2\pi\leq t\leq 3\pi$,

[TABLE]

Therefore,

[TABLE]

where we used the substitution $\tau=t-2\pi$. On the other hand, for $0\leq t\leq\pi$, $\tilde{G}(P_{t})=\tilde{G}(0,t,0)$, $\tilde{H}(P_{t})=\tilde{H}(0,t,0)$. Thus,

[TABLE]

Hence,

[TABLE]

Consequently, we get

[TABLE]

Since $\tilde{G}(P_{t})=\cos s$, $\tilde{H}(P_{t})=\sin s$, and $W(t)=\frac{ds}{dt}$, we have

[TABLE]

where $s_{0}$ and $s_{1}$ are determined by

[TABLE]

respectively. By (31), we see

[TABLE]

This means that there exists $m\in\mathbb{Z}$ such that $s_{1}=s_{0}+(2m+1)\pi$. Hence

[TABLE]

which yields $w(\mathscr{G},(0,0))=2m+1$. ∎

5.5 Proof of the main theorem (inequality part)

Let us prove the inequality part of Theorem 1.1. A convex body $K\in\mathcal{K}_{0}^{3}$ can be approximated by an element of $\hat{\mathcal{K}}$ as follows (see e.g., [Mi, pp. 38], [Sc2, pp. 438]).

Proposition 5.4**.**

For any centrally symmetric convex body $K\in\mathcal{K}_{0}^{3}$ and any $\varepsilon>0$, there exists a convex body $L\in\mathcal{K}^{3}$ with the following properties:

• (a)

$\delta(K,L)<\varepsilon$,

• (b)

$L$* has $C^{\infty}$ support function and is strongly convex,*

• (c)

$L$* is also centrally symmetric,*

where $\delta$ denotes the Hausdorff distance on $\mathcal{K}^{3}$. In particular, $L\in\hat{\mathcal{K}}$.

Since the volume is continuous with respect to the Hausdorff distance, it suffices to show the inequality $\eqref{eq:1}$ for $K\in\hat{\mathcal{K}}$. Fix $K\in\hat{\mathcal{K}}$. We show the following

Claim**.**

$(\hat{F},\hat{G},\hat{H})$ has a zero on $D$.

Proof.

In the case where $(\hat{F},\hat{G},\hat{H})$ has a zero on $\partial D$, the claim trivially holds. If not,

[TABLE]

is well-defined and $\deg\mathscr{F}$ is odd by Propositions 5.2 and 5.3. By, for example, [OR]*pp. 157–158, Propositions 4.4 and 4.6, $(\hat{F},\hat{G},\hat{H})$ has a zero in $D$. ∎

We denote the zero of $(\hat{F},\hat{G},\hat{H})$ by $(s_{0},\phi_{0},\psi_{0})$. Putting $\theta_{0}=(\pi-\Theta(0,\phi_{0},\psi_{0}))s_{0}$, it means that $(F,G,H)=(0,0,0)$ at $(\theta_{0},\phi_{0},\psi_{0})$. Thus the convex body

[TABLE]

satisfies the key equality (10). Since $\mathcal{A}(X(\theta_{0})Y(\phi_{0})Z(\psi_{0})K)X(\theta_{0})Y(\phi_{0})Z(\psi_{0})$ is a linear transformation, by Proposition 3.3 we obtain

[TABLE]

Consequently, every $K\in\hat{\mathcal{K}}$ satisfies (2). Thus we complete the proof of the Mahler conjecture for any centrally symmetric convex body $K\in\mathcal{K}_{0}^{3}$.

6 The equality case

In this section, we show the equality part of Theorem 1.1. That is,

Theorem 6.1**.**

Let $K$ be a centrally symmetric convex body in $\mathbb{R}^{3}$ with $\mathcal{P}(K)=32/3$. Then, $K$ or $K^{\circ}$ is a parallelepiped.

6.1 Dual face

Let $K\in\mathcal{K}^{n}_{0}$ be a centrally symmetric convex polytope in $\mathbb{R}^{n}$. Then $K^{\circ}$ is also a centrally symmetric convex polytope. For a face $F$ of $K$, its dual face $F^{\circ}\subset K^{\circ}$ is defined by

[TABLE]

For a $k$-dimensional face $F$, the dimension of $F^{\circ}$ is $n-k-1$.

Assume $K\in\mathcal{K}^{2}_{0}$ is a polygon. If a face $F\subset K$ is a line segment which contains two points $(a,b)$ and $(c,d)$, then its dual face $F^{\circ}$ is a vertex of $K^{\circ}$. Putting $F^{\circ}=\{(\alpha,\beta)\}$, by the definition of $F^{\circ}$, we obtain that

[TABLE]

Hence,

[TABLE]

Assume $K\in\mathcal{K}^{3}_{0}$ is a polytope. If a $2$-dimensional face $F$ of $K$ contains three points $(x_{i},y_{i},z_{i})$  $(i=1,2,3)$ which do not on the same straight line, then its dual face $F^{\circ}$ is a vertex of $K^{\circ}$. Putting $F^{\circ}=\{(u,v,w)\}$, we similarly obtain

[TABLE]

Hence,

[TABLE]

6.2 Inequality

For a general convex body $K$, we need the inequality similarly as Proposition 3.2. Let $A_{i}$ $(i=1,2,3)$ be points with $A_{1}\cdot(A_{2}\times A_{3})>0$, and consider $C_{i,j}=\mathcal{C}_{K}(A_{i},A_{j})$ for $(i,j)=(1,2),(2,3),(3,1)$.

Lemma 6.2**.**

Assume $K\in\mathcal{K}^{3}_{0}$. Let $\mathcal{S}_{K}(C_{1,2}\cup C_{2,3}\cup C_{3,1})$ be a triangle on $\partial K$. For any $P\in K$, we have

[TABLE]

where $\overline{C_{i,j}}=(\overline{C_{i,j,1}},\overline{C_{i,j,2}},\overline{C_{i,j,3}})$ and $\overline{C_{i,j,k}}=\operatorname{sgn}((A_{i}\times A_{j})_{k})|O*P_{k}(C_{i,j})|$.

Proof.

Since (32) depends on $K$ continuously, by the Hausdorff approximation, we may assume $K\in\hat{\mathcal{K}}$. By Proposition 3.2, it is sufficient to show that

[TABLE]

The curve $C_{1,2}$ is parametrized by $l(t):=\rho_{K}((1-t)A_{1}+tA_{2})((1-t)A_{1}+tA_{2})$ $(t\in[0,1])$. By a direct calculation,

[TABLE]

Thus, we have

[TABLE]

Since the absolute value of $\left(\int_{0}^{1}l\times l^{\prime}\,dt\right)_{k}$ is equal to $2\lvert O*P_{k}(C_{1,2})\rvert$ and $\rho_{K}((1-t)A_{1}+tA_{2})>0$, we obtain (33). ∎

6.3 Linear transformation of $K\in\mathcal{K}_{0}^{3}$

To prove Theorem 6.1, for a general centrally symmetric convex body $K$, we need to find a transformation $\mathcal{A}$ such that $\mathcal{A}K$ satisfies (10).

Proposition 6.3**.**

Let $K\in\mathcal{K}_{0}^{3}$ be a convex body which satisfies $\mathcal{P}(K)=32/3$. Then there exists a linear transformation $\mathcal{B}\in GL(3)$ such that $L=\mathcal{B}K$ satisfies $\mathcal{P}(L)=32/3$, the condition (10), and $\pm(1,0,0),\pm(0,1,0),\pm(0,0,1)\in\partial L$.

Proof.

Fix $K\in\mathcal{K}^{3}_{0}$. By Proposition 5.4, there exists $(K_{m})_{m=1}^{\infty}\subset\hat{\mathcal{K}}$ such that $K_{m}\rightarrow K$ as $m\rightarrow\infty$. By the results in Sections 4 and 5, for each $m\in\mathbb{N}$, there exists a linear transformation $B_{m}=\mathcal{A}(K_{m}(\theta_{m},\phi_{m},\psi_{m}))X(\theta_{m})Y(\phi_{m})Z(\psi_{m})$ such that $B_{m}K_{m}$ satisfies the condition (10). Hence,

[TABLE]

Note that $(\theta_{m})_{m=1}^{\infty},(\phi_{m})_{m=1}^{\infty},(\psi_{m})_{m=1}^{\infty}\subset[0,\pi]$ are bounded sequences.

Claim**.**

$(\mathcal{A}(K_{m}))_{m=1}^{\infty}$ and $(\mathcal{A}(K_{m})^{-1})_{m=1}^{\infty}$ are bounded sequences.

Proof.

By the definition of $\mathcal{A}$ in Section 4.1, it is sufficient to show that

[TABLE]

are bounded, where $\Theta_{m}=\Theta(\theta_{m},\phi_{m},\psi_{m})$, $\Phi_{m}=\Phi(\theta_{m},\phi_{m},\psi_{m})$, $\Psi_{m}=\Psi(\theta_{m},\phi_{m},\psi_{m})$.

For simplicity, we put $\tilde{K}^{m}:=K_{m}(\theta_{m},\phi_{m},\psi_{m})$. Since $(K_{m})_{m=1}^{\infty}$ converges to $K$, there exists $r>1$ independent of $m$ such that

[TABLE]

where $B(0,r)$ is the open ball with center [math] and radius $r$. Since each $\tilde{K}^{m}$ is the image of $K_{m}$ by a rotation, the same inequality holds for $\tilde{K}^{m}$. Hence,

[TABLE]

Thus, by the definition (12) of $\Theta$, we have

[TABLE]

Hence,

[TABLE]

Since

[TABLE]

we obtain that $(1/\tan\Theta_{m})_{m=1}^{\infty}$ and $(1/\sin\Theta_{m})_{m=1}^{\infty}$ are bounded. Similarly, by the definitions of $\Phi$ and $\Psi$, we have

[TABLE]

Hence, the claim holds. ∎

Thus, taking subsequence if necessary, there exists a linear transformation $B\in GL(3)$ such that $\lim_{m\rightarrow\infty}B_{m}=B$ in $GL(3)$, and $\lim_{m\rightarrow\infty}B_{m}K_{m}=BK$ in the Hausdorff distance. Thus $\lim_{m\rightarrow\infty}\rho_{B_{m}K_{m}}=\rho_{BK}$ uniformly in $S^{2}\subset\mathbb{R}^{3}$. Since $(B_{m}K_{m})_{+++}$ and $\overline{\alpha}(B_{m}K_{m})$, etc. are expressed by the integral of $\rho_{B_{m}K_{m}}$ on closed subsets on $S^{2}$, taking as $m\rightarrow\infty$ in (LABEL:eq:52), we obtain that

[TABLE]

Moreover, since each $B_{m}K_{m}$ satisfies (10), we obtain

[TABLE]

as in Section 3.4. Since $\lim_{m\rightarrow\infty}(B_{m}K_{m})^{\circ}=(BK)^{\circ}$ in the Hausdorff distance,

[TABLE]

holds. By Theorem 2.1, $|Q_{i}(BK)|\,|P_{i}((BK)^{\circ})|\geq 8$ $(i=1,2,3)$. Since $\mathcal{P}(BK)=\mathcal{P}(K)=32/3$, we obtain $|Q_{i}(BK)|\,|P_{i}((BK)^{\circ})|=8$ $(i=1,2,3)$. If necessary, we choose a diagonal matrix $C$ to satisfy that $\pm(1,0,0),\pm(0,1,0),\pm(0,0,1)\in\partial(CBK)$. Then, $\mathcal{B}=CB$ satisfies the required conditions. ∎

6.4 The equality case

Proposition 6.4**.**

Let $L\in\mathcal{K}_{0}^{3}$ with $\mathcal{P}(L)=32/3$. Assume that $L$ satisfies the condition (10), $\pm(1,0,0),\pm(0,1,0),\pm(0,0,1)\in\partial L$, and $\lvert Q_{i}(L)\rvert\lvert P_{i}(L^{\circ})\rvert=8$ $(i=1,2,3)$. Then, $L$ or $L^{\circ}$ is a parallelepiped.

Proof.

For each $i=1,2,3$, we have $(Q_{i}(L))^{\circ}=P_{i}(L^{\circ})$ and $\lvert Q_{i}(L)\rvert\lvert P_{i}(L^{\circ})\rvert=8$. Moreover, (10) holds. Thus, we can apply Proposition 2.2 to each $Q_{i}(L)$. As a result, there exist $\alpha,\beta,\gamma\in(-1,1]$ such that

[TABLE]

Hence,

[TABLE]

Then

[TABLE]

It follows from Lemma 6.2 that

[TABLE]

for any $(x,y,z)\in L$. Hence, $(8/3\lvert L\rvert)(1/(1+\alpha^{2}),1/(1+\beta^{2}),1/(1+\gamma^{2}))\in L^{\circ}$ holds. Similarly, we obtain that

[TABLE]

On the other hand, since $(\pm 1,0,0),(0,\pm 1,0),(0,0,\pm 1)\in L$,

[TABLE]

hold for any $(u,v,w)\in L^{\circ}$. Hence,

[TABLE]

Now we determine the shapes of $L$ and $L^{\circ}$. Without loss of generalities, it suffices to consider the following four cases: (i) $\alpha,\beta,\gamma\in(-1,1)$,  (ii) $\alpha,\beta\in(-1,1)$, $\gamma=1$,  (iii) $\alpha=\beta=1$, $\gamma\in(-1,1)$,  (iv) $\alpha=\beta=\gamma=1$.

Case (i) $\alpha,\beta,\gamma\in(-1,1)$;

By the assumption $(0,0,1)\in\partial L$, we have $L^{\circ}\cap\{w=1\}\not=\emptyset$. By (35),

[TABLE]

hold. Thus, $L^{\circ}\cap\{w=1\}=\{(\beta,-\alpha,1)\}$. Similarly, we have

[TABLE]

By (38), we obtain

[TABLE]

Note that the points in (40) are mutually distinct, because $\alpha,\beta,\gamma\in(-1,1)$. Moreover, these six points are in different faces of the cube $[-1,1]^{3}$. We use these points to divide $L^{\circ}$.

The line segment connecting from $A^{\circ}_{+}$ to $B^{\circ}_{+}$ is on $\partial L^{\circ}$. Indeed, if $Q=\tau A^{\circ}_{+}+(1-\tau)B^{\circ}_{+}$ is an interior point of $L^{\circ}$ for some $\tau\in(0,1)$, then there exists a open ball $B(Q,\epsilon)$ with the center $Q$ and radius $\epsilon>0$ such that $B(Q,\epsilon)\subset L^{\circ}$. Thus $P_{3}(B(Q,\epsilon))\subset P_{3}(L^{\circ})$, and $P_{3}(Q)$ is an interior point of $P_{3}(L^{\circ})$. On the other hand, by (35), the line segment connecting from $(1,\gamma,0)$ to $(-\gamma,1,0)$ is a part of $\partial P_{3}(L^{\circ})$. It is a contradiction. Therefore, $\mathcal{C}_{L}(A^{\circ}_{+},B^{\circ}_{+})$ is the line segment connecting from $A^{\circ}_{+}$ to $B^{\circ}_{+}$. Similarly, we see that the segments connecting $C^{\circ}_{+}$, $A^{\circ}_{+}$, $B^{\circ}_{+}$ cyclically is a triangle on $\partial L^{\circ}$, because $(C^{\circ}_{+}\times A^{\circ}_{+})\cdot B^{\circ}_{+}=1+\alpha^{2}+\beta^{2}+\gamma^{2}>0$. We denote the triangle by $\mathcal{T}(C^{\circ}_{+},A^{\circ}_{+},B^{\circ}_{+})$. Put

[TABLE]

Then, by the symmetry of $L^{\circ}$, we have $\lvert L^{\circ}\rvert=2(\lvert L^{\circ}_{+++}\rvert+\lvert L^{\circ}_{-++}\rvert+\lvert L^{\circ}_{--+}\rvert+\lvert L^{\circ}_{+-+}\rvert)$. Applying Lemma 6.2 to $L^{\circ}_{+++}$, we have

[TABLE]

for any $(u,v,w)\in L^{\circ}$. Similarly, we obtain

[TABLE]

for any $(u,v,w)\in L^{\circ}.$ Therefore, the four points

[TABLE]

are contained in $L$. Combining the fact that $S_{+++}\in L$ with (36), we have

[TABLE]

Similarly, we obtain

[TABLE]

Thus,

[TABLE]

On the other hand, since the assumption (10) for $L$ means $\lvert L_{\pm\pm\pm}\rvert=\lvert L\rvert/8$, we have

[TABLE]

Thus, the above inequality holds with equality. Hence (41) holds with equality. That is, (36) for $S_{+++}$ holds with equality. Taking the proof of Proposition 3.2 into account, we have

[TABLE]

Hence $S_{+++}\in\partial L$. We put $S_{+++}=(s_{1},s_{2},s_{3})$ for simplicity.

Now, we consider the case $S_{+++}\not\in\{x=0\}$. Then

[TABLE]

is a part of a face of $L$. Its dual face is a vertex of $L^{\circ}$. Then the dual face is

[TABLE]

Comparing it with (39), we obtain

[TABLE]

Hence $-s_{3}+\alpha s_{2}+1=\beta s_{1}$. Then,

[TABLE]

On the other hand, the volume of $\operatorname{conv}\left\{O,A^{\circ}_{+},B^{\circ}_{+},C^{\circ}_{+}\right\}\subset L^{\circ}_{+++}$ is

[TABLE]

Hence,

[TABLE]

Next, we consider the case $S_{+++}\in\{x=0\}$. Since $S_{+++}\in\partial L$, we obtain $S_{+++}\in\partial Q_{1}(L)$. Since $S_{+++}$ is in the first octant, $S_{+++}$ is on the line segment connecting from $(0,1-\alpha,1+\alpha)/(1+\alpha^{2})$ to $(0,0,1)$, or on the line segment connecting from $(0,1,0)$ to $(0,1-\alpha,1+\alpha)/(1+\alpha^{2})$. For the former case, we have

[TABLE]

which yields $\beta s_{1}-\alpha s_{2}+s_{3}=1$, and $6\lvert L^{\circ}_{+++}\rvert=1+\alpha^{2}+\beta^{2}+\gamma^{2}$ holds as (42). Thus, we obtain (43) as well. For the latter case, we have

[TABLE]

Since $s_{1}=0$, we obtain $s_{2}+\alpha s_{3}=1$. Thus

[TABLE]

which yields (43) as well.

By repeating the above arguments for $L^{\circ}_{-++},L^{\circ}_{--+},L^{\circ}_{+-+}$, we obtain

[TABLE]

Thus, $L^{\circ}$ is a centrally symmetric octahedron. Then,

[TABLE]

is a parallelepiped.

Case (ii) $\alpha,\beta\in(-1,1)$, $\gamma=1$;

By (35), we have

[TABLE]

Thus, we obtain

[TABLE]

Hence, for any $v\in[-1,1]$, there exists $w(v)\in[-1,1]$ such that $(1,v,w(v))\in L^{\circ}$ and

[TABLE]

On the other hand, combining (35) with $\beta\in(-1,1)$, we have

[TABLE]

Consequently, we obtain $w(v)=-\beta$  $(v\in[-1,1])$ and

[TABLE]

Especially, we have $(1,-1,-\beta),(1,1,-\beta)\in L^{\circ}$. Combining (35) with $\alpha\in(-1,1)$,

[TABLE]

Hence $-\beta=\alpha=-\alpha$. Thus, $\alpha=\beta=0$. Especially, by (44), we obtain

[TABLE]

By the symmetry and convexity of $L^{\circ}$, we obtain

[TABLE]

Since $\alpha=\beta=0$, similarly as the above,

[TABLE]

which implies that $(0,0,1)\in L^{\circ}$. In summary,

[TABLE]

Now, since $\alpha=\beta=0$, by (37), we have

[TABLE]

Since

[TABLE]

we obtain

[TABLE]

Since $\mathcal{P}(L)=32/3$, we have

[TABLE]

On the other hand, by (45),

[TABLE]

holds. Therefore, $\lvert L^{\circ}\rvert=8/3$ and

[TABLE]

Hence, $L^{\circ}$ is a centrally symmetric octahedron and

[TABLE]

is a parallelepiped.

Case (iii) $\alpha=\beta=1$, $\gamma\in(-1,1)$;

By (35), we have

[TABLE]

Since $\gamma\in(-1,1)$, we see

[TABLE]

Hence

[TABLE]

By the symmetry of $L^{\circ}$, we obtain

[TABLE]

Similarly,

[TABLE]

Thus

[TABLE]

By the symmetry of $L^{\circ}$, we get

[TABLE]

By the convexity of $L^{\circ}$,

[TABLE]

On the other hand, we have $L^{\circ}\subset[-1,1]\times[-1,1]\times[-1,1]$ by (38). Combining it with $P_{3}(L^{\circ})=\operatorname{conv}\left\{\pm(1,\gamma,0),\pm(-\gamma,1,0)\right\}$, we obtain

[TABLE]

Thus,

[TABLE]

is a parallelepiped.

Case (iv) $\alpha=\beta=\gamma=1$;

By (35), we have

[TABLE]

Recall that $L^{\circ}\subset[-1,1]\times[-1,1]\times[-1,1]$. If $(1,1,1)\in L^{\circ}$, then we have

[TABLE]

for any $(x,y,z)\in L$. On the other hand, $(1,0,0),(0,1,0),(0,0,1)\in L$. By the definition of $L_{+++}$, we obtain

[TABLE]

Moreover, since

[TABLE]

$(-1,0,0),(0,1,0),(0,0,1)\in L$, and the definition of $L_{-++}$, we have

[TABLE]

By similar arguments for $L_{--+}$ and $L_{+-+}$, we obtain

[TABLE]

Thus

[TABLE]

is a parallelepiped. Similarly, if one of the eight points $(\pm 1,\pm 1,\pm 1)$ is in $L^{\circ}$, then $L$ is an octahedron and $L^{\circ}$ is a parallelepiped.

Thus, from now on we may assume that $(\pm 1,\pm 1,\pm 1)\not\in L^{\circ}$. Using the characterization of $Q_{i}(L)$ $(i=1,2,3)$, by Lemma 6.2, we have

[TABLE]

for any $(x,y,z)\in L$. Similar arguments about the other seven parts imply

[TABLE]

Next, by the characterization of $P_{i}(L^{\circ})$ $(i=1,2,3)$, there exists ${a}_{i},{b}_{i},{c}_{i}\in(-1,1)$  $(i=1,2)$ such that

[TABLE]

are in $L^{\circ}$. We consider three planes $H_{OA^{\circ}_{1}A^{\circ}_{2}}$, $H_{OB^{\circ}_{1}B^{\circ}_{2}}$, and $H_{OC^{\circ}_{1}C^{\circ}_{2}}$ determined by $O,A^{\circ}_{1},A^{\circ}_{2}$; $O,B^{\circ}_{1},B^{\circ}_{2}$; and $O,C^{\circ}_{1},C^{\circ}_{2}$, respectively. We divide $L^{\circ}$ by these planes into 8 pieces $L^{\circ}_{\pm\pm\pm}$. By the convexity of $L^{\circ}$, each line segment connecting any two points in $\{\pm A^{\circ}_{1},\pm A^{\circ}_{2},\pm B^{\circ}_{1},\pm B^{\circ}_{2},\pm C^{\circ}_{1},\pm C^{\circ}_{2}\}$ is in $L^{\circ}$. Thus $H_{OA^{\circ}_{1}A^{\circ}_{2}}$, $H_{OB^{\circ}_{1}B^{\circ}_{2}}$, and $H_{OC^{\circ}_{1}C^{\circ}_{2}}$ contain $\operatorname{conv}\left\{\pm A^{\circ}_{1},\pm A^{\circ}_{2}\right\}$, $\operatorname{conv}\left\{\pm B^{\circ}_{1},\pm B^{\circ}_{2}\right\}$, and $\operatorname{conv}\left\{\pm C^{\circ}_{1},\pm C^{\circ}_{2}\right\}$, respectively. We put

[TABLE]

Then $D^{\circ}$, $E^{\circ}$, and $F^{\circ}$ are in $L^{\circ}$. Applying Lemma 6.2 to $L^{\circ}$ with $C_{1,2}=\mathcal{C}_{L^{\circ}}(F^{\circ},D^{\circ})$, $C_{2,3}=\mathcal{C}_{L^{\circ}}(D^{\circ},E^{\circ})$, $C_{3,1}=\mathcal{C}_{L^{\circ}}(E^{\circ},F^{\circ})$, and $P=4(1,1,1)/3\lvert L\rvert\in L^{\circ}$, we have

[TABLE]

Similarly, we obtain that

[TABLE]

[TABLE]

Taking these sums, we obtain

[TABLE]

On the other hand, by the assumption, $\lvert L\rvert\lvert L^{\circ}\rvert=32/3$. Thus (47) holds with equality. Especially, $L^{\circ}_{+++}$ is the cone with the vertex $4(1,1,1)/3\lvert L\rvert$ over $O*(F^{\circ}B^{\circ}_{1}D^{\circ}C^{\circ}_{1}E^{\circ}A^{\circ}_{1}F^{\circ})$. It is not convex if $4/3\lvert L\rvert<1$. Thus $4/3\lvert L\rvert\geq 1$. It means that $(1,1,1)\in L^{\circ}$, which is a contradiction. Consequently, in the case $\alpha=\beta=\gamma=1$, $(\pm 1,\pm 1,\pm 1)\not\in L^{\circ}$ does not occur, and $L^{\circ}=[-1,1]\times[-1,1]\times[-1,1]$. Hence $L^{\circ}$ is a parallelepiped. ∎

Proof of Theorem 6.1.

Combining Proposition 6.3 and Proposition 6.4, the conclusion follows immediately. ∎

Appendix A Proof of Proposition 5.1

To prove Proposition 5.1, we have to modify the map $\mathscr{F}$ defined on $\partial D$. We consider $\partial D$ as a hexahedron, and deform $\mathscr{F}$ near each vertices of $\partial D$ and deform near each edges of $\partial D$. We introduce some auxiliary functions satisfy (23), (24), and (25). We first check that $G^{2}+H^{2}$ has same value at vertices.

Lemma A.1**.**

$(G^{2}+H^{2})$* takes same value at eight vertices on $\partial D$; $(0,0,0)$, $(0,\pi,0)$, $(0,\pi,\pi)$, $(0,0,\pi)$, $(1,0,0)$, $(1,\pi,0)$, $(1,\pi,\pi)$, $(1,0,\pi)$.*

Remark A.2**.**

More generally, if a vector field $(0,\check{G},\check{H})$ satisfies (23), (24), and (25), then $\check{G}^{2}+\check{H}^{2}$ takes same value at the eight vertices.

Proof of Lemma A.1.

By (23), we have

[TABLE]

By (24), we also obtain

[TABLE]

By (25),

[TABLE]

Thus we see the conclusion. ∎

To modify the vector field $(F,G,H)$ on $D\subset\mathbb{R}^{3}$, we introduce a cut-off function $\zeta(s,\phi,\psi)$. Let $\zeta:D\rightarrow\mathbb{R}$ be a smooth function such that, radially symmetric with respect to the origin, $0\leq\zeta\leq 1$, $\zeta(0,0,0)=1$, and $\operatorname{supp}\zeta$ is a small neighborhood of the origin. We put

[TABLE]

Lemma A.3**.**

$(0,G_{0},H_{0})$* satisfies (23), (24), and (25).*

Proof.

Since $\operatorname{supp}\zeta$ is a small neighborhood of the origin $(0,0,0)$, we see $\zeta(1,\phi,\psi)=\zeta(s,\pi,\psi)=\zeta(s,\phi,\pi)=0$. Combining them with the formulas of Lemma A.5 below, we can check that $(0,G_{0},H_{0})$ satisfies (23), (24), and (25) directly as follows:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

∎

Fix $\epsilon>0$. We also introduce cut-off functions $\zeta_{1}(s,\phi,\psi)$, $\zeta_{2}(s,\phi,\psi)$, and $\zeta_{3}(s,\phi,\psi)$ such that

[TABLE]

each $\operatorname{supp}\zeta_{i}$ $(i=1,2,3)$ is contained in $\epsilon/2$-neighborhood of $[\epsilon,1-\epsilon]\times\{0\}\times\{0\}$, $\{0\}\times[\epsilon,\pi-\epsilon]\times\{0\}$, and $\{0\}\times\{0\}\times[\epsilon,\pi-\epsilon]$ respectively, and $0\leq\zeta_{i}\leq 1$ $(i=1,2,3)$. We put

[TABLE]

where we choose the $\epsilon$ sufficiently small with $0<\epsilon<1/2$, and $\alpha_{i}$, $\beta_{i}$ $(i=1,2,3)$ are positive real numbers. Then, we can assume that $\operatorname{supp}G_{i}\cap\operatorname{supp}G_{j}=\emptyset$ and $\operatorname{supp}H_{i}\cap\operatorname{supp}H_{j}=\emptyset$ for $i\not=j$.

Lemma A.4**.**

For each $i=1,2,3$, the vector field $(0,G_{i},H_{i})$ on $D$ satisfies (23), (24), and (25).

Proof.

By the definition of $\zeta_{i}$, we have

[TABLE]

Combining them with the formulas of Lemma A.5 below, we can check that $(0,G_{1},H_{1})$ satisfies (23), (24), and (25) directly as follows:

[TABLE]

[TABLE]

Similarly, $(0,G_{i},H_{i})$ $(i=2,3)$ also satisfy (23), (24), and (25). ∎

Here, we show the lemma used in the above lemmas.

Lemma A.5**.**

$T_{0}^{-1}(s)=T_{\pi}(s)$, $T_{\pi}^{-1}(s)=T_{0}(s)$.

Proof.

Applying Lemma 4.5 to $K=K(\theta,\phi,\psi)$, we see that

[TABLE]

Combining (21) with the definition of $T_{\psi}$, we have

[TABLE]

By the definition of $\Gamma_{\psi}$,

[TABLE]

Put $\Theta_{0}:=\Theta(0,0,0)$ and $\Theta_{1}:=\Theta(0,\pi,0)$. Then, by (22) and (21),

[TABLE]

Thus, we obtain

[TABLE]

for $s,t\in[0,1]$, and such $T_{0}(s),T_{\pi}(t)\in[0,1]$ is uniquely determined, since $\Gamma_{\psi}:[0,\Theta(0,\pi,\psi)]\rightarrow[\pi-\Theta(0,0,\psi),\pi]$ is bijective by Lemma 4.12.

On substituting $(\theta,\phi,\psi)=(\Theta_{0}T_{\pi}(t),0,\pi)$ into (48), we have

[TABLE]

By (50) and Lemmas 4.7 and 4.10, the first term of the left-hand side is

[TABLE]

By (50),

[TABLE]

Consequently, we get

[TABLE]

By (49) with $s=T_{\pi}(t)$, the uniqueness asserts that $t=T_{0}(T_{\pi}(t))$. ∎

Proof of Proposition 5.1.

For

[TABLE]

we consider the following conditions:

• (C1)

$\check{\mathscr{F}}$ is homotopic to $\mathscr{F}$.

• (C2)

$\check{F}^{2}+\check{G}^{2}+\check{H}^{2}\not=0$ on $\partial D$.

• (C3)

$(\check{F},\check{G},\check{H})$ satisfies the identities (23), (24), and (25).

• (C4)

$\check{G}^{2}+\check{H}^{2}\not=0$ at the vertices of $\partial D$.

• (C5)

$\check{G}^{2}+\check{H}^{2}\not=0$ on $\partial M_{i}$ $(i=1,\dots,6)$.

We shall construct $\check{\mathscr{F}}$ such that the conditions (C1)–(C5) hold and $(\pm 1,0,0)\in S^{2}$ are regular values of $\check{\mathscr{F}}$.

First, we put

[TABLE]

where we take sufficiently small $\alpha_{0}>0$ if $(G^{2}+H^{2})(0,0,0)=0$, take $\alpha_{0}=0$ if $(G^{2}+H^{2})(0,0,0)\not=0$. Then

[TABLE]

By Lemma A.3, $(0,G_{0},H_{0})$ satisfies (23), (24), and (25). By (51) and Remark A.2, (C4) holds for $(F,\check{G}_{0},\check{H}_{0})$.

Since $F^{2}+G^{2}+H^{2}\not=0$ on $\partial D$ by the assumption, for sufficiently small $\alpha_{0}$, $F^{2}+(G+\alpha_{0}G_{0})^{2}+(H+\alpha_{0}H_{0})^{2}\not=0$ on $\partial D$. Thus, $\check{\mathscr{F}}_{0}$ satisfies the conditions (C1) and (C2). By Lemma A.3, $\check{\mathscr{F}}_{0}$ satisfies the condition (C3). Consequently, (C1)–(C4) hold for $\check{\mathscr{F}}_{0}$.

Second, by continuity, there exist $\delta_{0}>0$ and $\delta_{1}>0$ such that $\sqrt{\check{G}_{0}^{2}+\check{H}_{0}^{2}}\geq\delta_{1}$ on $\delta_{0}$-neighborhood of the vertices of $D$. For $\alpha_{i}$ and $\beta_{i}$ above with $\sqrt{\alpha_{i}^{2}+\beta_{i}^{2}}\leq\delta_{1}/4$, on $\delta_{0}$-neighborhood of the vertices, we see $\lvert(G_{i},H_{i})\rvert\leq\sqrt{\alpha_{i}^{2}+\beta_{i}^{2}}\leq\delta_{1}/4$. Thus

[TABLE]

on $\delta_{0}$-neighborhood of the vertices.

Thus, the points on $\partial M_{i}$ with $(\check{G}_{0}+G_{1}+G_{2}+G_{3},\check{H}_{0}+H_{1}+H_{2}+H_{3})=(0,0)$ are contained in the following 12 lines: $[\delta_{0}/2,1-\delta_{0}/2]\times\{0,\pi\}\times\{0,\pi\}$, $\{0,1\}\times[\delta_{0}/2,\pi-\delta_{0}/2]\times\{0,\pi\}$, $\{0,1\}\times\{0,\pi\}\times[\delta_{0}/2,\pi-\delta_{0}/2]$. Moreover, on the 12 lines, $(G_{i},H_{i})$ $(i=1,2,3)$ are constants. Consider the image $U$ of $(G+\alpha_{0}G_{0},H+\alpha_{0}H_{0})$ on the 12 lines. Since $U$ is the union of 12 curves of class $C^{\infty}$, the $2$-dimensional Lebesgue measure of $U$ is [math]. Therefore, there exists $\alpha_{i},\beta_{i}$ such that $|\alpha_{i}|$ and $|\beta_{i}|$ are small, and

[TABLE]

for $i=1,2,3$. In this setting, on the 12 lines, we obtain

[TABLE]

Thus, we put $\check{G}_{1}:=\check{G}_{0}+G_{1}+G_{2}+G_{3}$, $\check{H}_{1}:=\check{H}_{0}+H_{1}+H_{2}+H_{3}$, and

[TABLE]

for sufficiently small $|\alpha_{i}|$ and $|\beta_{i}|$. Then $\check{\mathscr{F}}_{1}$ satisfies (C1)–(C5).

Next, we will construct a deformation of $\check{\mathscr{F}}_{1}$ which satisfies that $(\pm 1,0,0)$ are its regular values on $M_{1}\cup M_{2}$.

Since $(\check{G}_{1},\check{H}_{1})\not=(0,0)$ on $\partial M_{1}$, there exists $\delta_{2}>0$ such that

[TABLE]

where $V\subset M_{1}$ is some small neighborhood of $\partial M_{1}$. By Sard’s theorem, there exist positive real numbers $\alpha_{4},\beta_{4}$ with $\sqrt{\alpha_{4}^{2}+\beta_{4}^{2}}\leq\delta_{2}/2$ such that $(\alpha_{4},\beta_{4})$ is a regular value of $(\check{G}_{1},\check{H}_{1})$ on $M_{1}$, and $(-\beta_{4},\alpha_{4})$ is a regular value of $(\check{G}_{1},\check{H}_{1})$ on $M_{2}$. We note that, since $|\alpha_{4}|$ and $|\beta_{4}|$ are small, $F\not=0$ at the regular points $(\check{G}_{1},\check{H}_{1})^{-1}(\alpha_{4},\beta_{4})$ on $M_{1}$ and $(\check{G}_{1},\check{H}_{1})^{-1}(-\beta_{4},\alpha_{4})$ on $M_{2}$. Let $\zeta_{4}(\phi,\psi)\in C_{0}^{\infty}(\operatorname{int}M_{1},[0,1])$ be a cut-off function such that $\zeta_{4}=1$ on $M_{1}\setminus V$. We put

[TABLE]

Then $(0,G_{4},H_{4})$ satisfies (23), (24), and (25). Thus, we define $\check{G}_{2}:=\check{G}_{1}-G_{4}$, $\check{H}_{2}:=\check{H}_{1}-H_{4}$, and

[TABLE]

Then, $\check{\mathscr{F}}_{2}$ satisfies (C1)–(C5). On $V(\subset M_{1})$, we obtain

[TABLE]

So, there exist no zeros of $(\check{G}_{2},\check{H}_{2})$ on $V$. On $M_{1}\setminus V$, we have

[TABLE]

Thus, by the definition of $(\alpha_{4},\beta_{4})$, $(0,0)$ is a regular value of $(\check{G}_{2},\check{H}_{2})$ on $M_{1}\setminus V$, and $F\not=0$ at the regular values. Therefore, $(\pm 1,0,0)$ are regular values of $\check{\mathscr{F}}_{2}$ on $M_{1}$. Similarly, we obtain that $(\pm 1,0,0)$ are regular values of $\check{\mathscr{F}}_{2}$ on $M_{2}$.

Moreover, we can modify $\check{\mathscr{F}}_{2}$ similarly with respect to $M_{3},\dots,M_{6}$. Then the modified $\check{\mathscr{F}}$ satisfies the conditions (i)–(iv) in Proposition 5.1. ∎

Appendix B The case with a hyperplane symmetry

Although, for a general $K\in\hat{\mathcal{K}}$, it is very hard to find a suitable linear transformation $\mathcal{A}$ such that ${\mathcal{A}}K$ satisfies the condition (10), it is not so difficult to do so if $K$ has additional symmetries. In this appendix, let us consider the case where $K\in\hat{\mathcal{K}}$ is symmetric with respect to a plane. We start with the following simple observation.

Example**.**

Consider the case that $K$ is symmetric with respect to the $yz$-plane, $zx$-plane, and $xy$-plane (more generally, its image $\mathcal{A}K$ by a linear transformation $\mathcal{A}$ is symmetric with respect to these three planes). In this case, the condition (10) automatically holds, and hence we have $\mathcal{P}(K)\geq 32/3$. This fact is the three dimensional case of the result by Saint-Raymond [SR].

From now on, suppose that $K\in\hat{\mathcal{K}}$ is symmetric with respect to a plane. Since the volume product is invariant under linear transformations of $K$, we may assume that the hyperplane of symmetry is the $yz$-plane. In this case, the area of $O*\beta$ and that of $O*(-b)$ coincide. Taking the signs into consideration, $\overline{\beta}=\overline{b}$ holds. Similarly, $\overline{\gamma}=\overline{c}$ holds. Concerning the volume of the eight parts of $K$, we have

[TABLE]

Consequently, the condition (10) holds, provided $K$ satisfies

[TABLE]

For a given convex body $K\in\hat{\mathcal{K}}$, we consider the image of $K$ by the following linear transformation:

[TABLE]

For $K\in\hat{\mathcal{K}}$, we introduced vectors $\overline{\alpha}=(\alpha_{1},\alpha_{2},\alpha_{3})$, etc., in Section 3.3. From now on, to clarify the dependence on the convex body, we denote the vector $\overline{\alpha}$ for $A(\varphi)X(\theta)K\in\hat{\mathcal{K}}$ by $\overline{\alpha}(A(\varphi)X(\theta)K)$, etc. Similarly, for the image $A(\varphi)X(\theta)K\in\hat{\mathcal{K}}$ of a convex body $K$, we denote by $(A(\varphi)X(\theta)K)_{+++},\dots$ the decomposition in Section 3.2.

First, since $A(\varphi)$ and $X(\theta)$ keep the symmetry of $K$ with respect to the $yz$-plane, we have

[TABLE]

for any $\varphi,\theta$ $\in\mathbb{R}$. Next, for a given $\theta\in\mathbb{R}$ we can choose $\varphi=\varphi(\theta)$ $\in\mathbb{R}$ to satisfy

[TABLE]

see Proposition 4.2 in the special case that $\Phi(A(\varphi)X(\theta)K)=\Psi(A(\varphi)X(\theta)K)=\pi/2$, $|(A(\varphi)X(\theta)K)_{+++}|=|(A(\varphi)X(\theta)K)_{-++}|$, $|(A(\varphi)X(\theta)K)_{--+}|=|(A(\varphi)X(\theta)K)_{+-+}|$ for $X(\theta)K$. Note that such $\varphi$ is uniquely determined for each $\theta$, and $\varphi$ depends on $\theta$ continuously. Moreover, we can easily see that there exists some $\theta_{0}$ such that $\varphi(\theta_{0})=0$. In this setting, we obtain $\varphi(\theta_{0}+\pi/2)=0$ and

[TABLE]

Next, we put

[TABLE]

Since $\varphi$ is continuous with respect to $\theta$, $F$ is continuous on $\mathbb{R}$. By (53), we have $F(\theta_{0}+\pi/2)=-F(\theta_{0})$. By using the intermediate value theorem, there exists $\theta_{1}\in[\theta_{0},\theta_{0}+\pi/2]$ such that $F(\theta_{1})=0$, which yields

[TABLE]

Consequently, $A(\varphi(\theta_{1}))X(\theta_{1})K\in\hat{\mathcal{K}}$ satisfies the condition (52), and hence (10). Since $A(\varphi(\theta_{1}))X(\theta_{1})$ is a linear transformation of $\mathbb{R}^{3}$, we obtain

[TABLE]

Thus, we have proved the following

Proposition B.1**.**

Let $K\in\hat{\mathcal{K}}$ be a three dimensional centrally symmetric convex body which is symmetric with respect to a plane. Then $\mathcal{P}(K)\geq 32/3$ holds.

Acknowledgements

The authors would like to thank the anonymous referees for their careful reading and valuable comments, especially, for pointing out references [Me2] and [BMMR], for pointing out a simplification of the proof of Proposition 3.2, and for their suggestions about the notations of the eight parts decomposition of convex bodies $K,K^{\circ}$ and the curves on their boundaries. Including them their valuable feedback has resulted in considerable improvements of exposition of the paper.