Minimal volume product of three dimensional convex bodies with various discrete symmetries
MMinimal volume product of three dimensional convexbodies with various discrete symmetries
Hiroshi Iriyeh ∗ Masataka Shibata † July 20, 2020
Abstract
We give the sharp lower bound of the volume product of three dimensional convexbodies which are invariant under a discrete subgroup of O (3) in several cases. Wealso characterize the convex bodies with the minimal volume product in each case.In particular, this provides a new partial result of the non-symmetric version ofMahler’s conjecture in the three dimensional case. Let K be a convex body in R n , i.e, K is a compact convex set in R n with nonemptyinterior. Then K ⊂ R n is said to be centrally symmetric if it satisfies that K = − K .Denote by K n the set of all convex bodies in R n equipped with the Hausdorff metric andby K n the set of all K ∈ K n which are centrally symmetric.The interior of K ∈ K n is denoted by int K . For a point z ∈ int K , the polar body of K with respect to z is defined by K z = { y ∈ R n ; ( y − z ) · ( x − z ) ≤ x ∈ K } , where · denotes the standard inner product on R n . Then an affine invariant P ( K ) := min z ∈ int K | K | | K z | (1)is called volume product of K , where | K | denotes the n -dimensional volume of K in R n .It is known that for each K ∈ K n the minimum of (1) is attained at the unique point z on K , which is called Santaló point of K (see, e.g., [MP]). For a centrally symmetricconvex body K ∈ K n , the Santaló point of K is the origin o . See [Sc1, p.546] for furtherinformation. In the following, the polar of K with respect to o is denoted by K ◦ .Mahler conjecture [Ma57] states that for any K ∈ K n , P ( K ) ≥ n n ! (2) ∗ Graduate School of Science and Engineering, Ibaraki University,e-mail: [email protected] † Department of Mathematics, Tokyo Institute of Technology,e-mail: [email protected] a r X i v : . [ m a t h . M G ] J u l ould be hold. He proved it in the case where n = 2 ([Ma59]). The three dimensionalcase recently proved in [IS] (see also [FHMRZ] for a nice simple proof of an equipartitionresult used in [IS]). Although the case that n ≥ n -cube or, more generally,Hanner polytopes satisfy the equality in (2) and are predicted as the minimizers of P . Asfor non-symmetric bodies there is another well-known conjecture for the lower bound ofthe volume product as follows. Conjecture.
Any K ∈ K n satisfies that P ( K ) ≥ ( n + 1) n +1 ( n !) with equality if and only if K is a simplex.This was proved by Mahler for n = 2 [Ma59] (see also [Me]). Note that this conjectureremains open for n ≥
3, see e.g., [BF], [FMZ], [KR].On the other hand, Barthe and Fradelizi ([BF]) considered the sharp lower boundestimation of P ( K ) when K has many reflection symmetries, in other words, when K isinvariant under the action of a Coxeter group. It is worthwhile to extend the class of thegroups and to estimate the volume product of the G -invariant convex body K ⊂ R n frombelow for more general discrete subgroups G of the orthogonal group O ( n ). Recall that K n = { K ⊂ R n ; K is a convex body with o ∈ int K } . Problem.
Let G be a discrete subgroup of O ( n ). Denote by K n ( G ) the set of all convexbodies K ∈ K n which satisfies that K = g ( K ) for any g ∈ G . Then, consider theminimizing problem min K ∈K n ( G ) P ( K )and determine all the minimizers K ∈ K n ( G ).From this viewpoint, Mahler conjecture corresponds to the case G = { E, − E } ∼ = Z ( E is the identity matrix) and the non-symmetric case corresponds to G = { E } . Following[BF] for a subset A ⊂ R n we recall the subgroup of O ( n ) defined by O ( A ) = { g ∈ O ( n ); g ( A ) = A } . Similarly, we can define SO ( A ) ⊂ SO ( n ). For example, if P ⊂ R n is a regular polytopewith o as the centroid, then O ( P ) is a discrete subgroup of O ( n ). Indeed, Problem wassolved for G = O ( P ) for every regular polytope P ⊂ R n ([BF, Theorem 1 (i)]). Apartfrom the setting of [BF], we can settle the case G = SO ( P ) ⊂ SO ( n ) as well.In this paper, we shall focus on the three dimensional case of the above problem andconsider G -invariant convex bodies K ⊂ R for a discrete subgroup G of O (3). Note thatall the discrete subgroups of O (3) is well-known. Before we explain the details, we recallthe two dimensional case in which the problem has already been solved. For ‘ ∈ N , we put ξ = 2 π/‘ and R ‘ := cos ξ − sin ξ sin ξ cos ξ ! , V := − ! .
2p to conjugation, all the discrete subgroups of O (2) are C ‘ = h R ‘ i and D ‘ = h R ‘ , V i ( ‘ ∈ N ) . Note that K ( C ) = K and K ( C ) = K . We denote by ‘ the regular ‘ -gon with o asthe centroid. For simplicity, we denote := and (cid:3) := .The following is a summary of the known results. Theorem 1.1 ([Ma59, BF, BMMR]) . (i) P ( K ) ≥ P ( ) holds for K ∈ K ( C ) withequality if and only if K is the image of by an affine transformation of R . Thesame inequality holds for G = D . (ii) P ( K ) ≥ P ( (cid:3) ) holds for K ∈ K ( C ) with equality if and only if K is the image of (cid:3) by a linear transformation of R . The same inequality holds for G = D . (iii) Assume that ‘ ≥ . Then P ( K ) ≥ P ( ‘ ) holds for K ∈ K ( C ‘ ) with equality if andonly if K is a dilation or rotation of ‘ . The same inequality holds for G = D ‘ . In this subsection, we shall state known results and the main theorem in the case where n = 3. Let us first recall the list of all discrete subgroups of O (3) by using Schoenflies’notation. We equip R with the usual orthogonal xyz -axies and put R ‘ := cos ξ − sin ξ ξ cos ξ
00 0 1 , V := − , H := − , where ‘ ∈ N , ξ := 2 π/‘ . Here R ‘ , V , and H mean the rotation through the angle2 π/‘ about the z -axis, the reflection with respect to the zx -plane, and the reflection withrespect to the xy -plane, respectively. It is known that, up to conjugation, all the discretesubgroup of O (3) are classified as the seven infinite families C ‘ := h R ‘ i , C ‘h := h R ‘ , H i , C ‘v := h R ‘ , V i , S ‘ := h R ‘ H i ,D ‘ := h R ‘ , V H i , D ‘d := h R ‘ H, V i , D ‘h := h R ‘ , V, H i , and the following seven finite groups (see e.g., [CS, Table 3.1. (p. 36)]: T := { g ∈ SO (3); g = , T d := { g ∈ O (3); g = , T h := {± g ; g ∈ T } ,O := { g ∈ SO (3); g = } , O h := { g ∈ O (3); g = } = {± g ; g ∈ O } ,I := { g ∈ SO (3); g = } , I h := { g ∈ O (3); g = } = {± g ; g ∈ I } , where = , = , and = denote the regular tetrahedron (simplex), theregular octahedron, and the regular icosahedron with o as their centroids, respectively.Note that, for example, if we fix the configuration of , then T d is uniquely determined asthe subgroup of O (3) without any ambiguity of conjugations. In other words, any discretesubgroup conjugated to T d in O (3) is realized as { g ∈ O (3); g = } , where = k for some k ∈ O (3). Using the notation of [BF], T d = O ( ) ∼ = O ( ). We also note thatamong the above classification C ‘ , D ‘ , T, O, I are indeed subgroups of SO (3), and thatthe Santaló point of K ∈ K ( G ) is the origin o except only two subgroups G = C l or C lv ( l ∈ N ).Now we recall the known results of the three dimensional case. Note that a convexbody K ∈ K ( D h ) is exactly 1-unconditional and D h ∼ = ( Z ) as groups.3 heorem 1.2 ([SR], [M]) . For K ∈ K ( D h ) , we have P ( K ) ≥ P ( ) = 323 . The equality holds if and only if K is a -dimensional Hanner polytope ([R], [M]) , i.e., K is a dilate of the regular octahedron or the cube ◦ . Theorem 1.3 ([BF, Theorem 1]) . (i) P ( K ) ≥ P ( ) holds for K ∈ K ( T d ) . The equal-ity holds if and only if K is a dilate of the regular tetrahedron or ◦ . (ii) P ( K ) ≥ P ( ) holds for K ∈ K ( O h ) . The equality holds if and only if K is a dilateof the regular octahedron or the cube ◦ . (iii) P ( K ) ≥ P ( ) holds for K ∈ K ( I h ) . The equality holds if and only if K is a dilateof the regular icosahedron or the regular dodecahedron ◦ . (iv) Assume that ‘ ≥ . Then P ( K ) ≥ P ( P ‘ ) holds for K ∈ K ( D ‘h ) , where P ‘ is a ‘ -regular right prism defined by P ‘ := conv cos kξ sin kξ , cos kξ sin kξ − ; k = 0 , . . . , l − . Furthermore, the solution of Mahler’s conjecture for n = 3 [IS, Theorem 1] is corre-sponding to the case G = S ( ∼ = Z ). That immediately implies Corollary 1.4.
Assume that G = C h , T h , S , D d , or S . For K ∈ K ( G ) , we have P ( K ) ≥ P ( ) = 323 with equality if and only if K is a linear image of or ◦ which is invariant under thegroup G . The following is the main result of this paper.
Theorem 1.5. (i) P ( K ) ≥ P ( ) holds for K ∈ K ( T ) . The equality holds if and onlyif K is a dilate of the regular tetrahedron or ◦ . (ii) P ( K ) ≥ P ( ) holds for K ∈ K ( O ) . The equality holds if and only if K is a dilateof the regular octahedron or the cube ◦ . (iii) P ( K ) ≥ P ( ) holds for K ∈ K ( I ) . The equality holds if and only if K is a dilateof the regular icosahedron or the regular dodecahedron ◦ . (iv) Assume that ‘ ≥ . Then P ( K ) ≥ P ( P ‘ ) holds for K ∈ K ( C ‘h ) , where P ‘ is a ‘ -regular right prism. The equality holds if and only if K coincides with P ‘ or P ◦ ‘ up to a linear transformation in the abelian subgroup of GL (3 , R ) defined by G := a cos θ − a sin θ a sin θ a cos θ
00 0 b ; a, b > . he same inequality holds for K ∈ K ( D ‘ ) . The equality holds if and only if K coincides with P ‘ or P ◦ ‘ up to a linear transformation in G := a a
00 0 b ; a, b > . In particular, Theorem 1.5 (i) provides a new partial result of the non-symmetricversion of Mahler’s conjecture in the three dimensional case (cf. Theorem 1.3 (i)). Theresults from Theorem 1.2 to Theorem 1.5 are summarized as follows. ‘ · · · C ‘ { E } C C C C C · · · C ‘v C v C v C v C v C v C v · · · C ‘h C v • C h ◦ C h ◦ C h ◦ C h ◦ C h ◦ · · · S ‘ C v • S C h S C h • S · · · D ‘ C D ◦ D ◦ D ◦ D ◦ D ◦ · · · D ‘h C v ‡ D h † D h † D h † D h † D h † · · · D ‘d C h D d • D d D d D d D d · · ·◦ T † T d • T h ◦ O † O h ◦ I † I h † : The results in [BF] ‡ : The result in [SR] • : Results deduced from [IS] (The case where S ⊂ G and is a minimizer) ◦ : New results Gray: Duplicates The proof of results in this paper is based on a simple inequality [IS, Proposition 3.2]that gives rise to an effective method to estimate the volume product P ( K ) from below,which is a natural extension of the volume estimate by M. Meyer. He gave in [M] anelegant proof of the result by Saint-Raymond [SR] (see Theorem 1.2 above for n = 3) for1-unconditional bodies K ⊂ R n as follows. An 1-unconditional body K is determined bythe part K in the first octant: K := K ∩ { ( x , . . . , x n ) ∈ R n ; x ≥ , · · · , x n ≥ } . We denote the intersection of ∂K with each positive part of coordinate axes by P , . . . , P n ,respectively. For any point P in K , we can make the polytope which is the convex fullof vertices o , P , . . . , P n , and P . The volume of this polytope is less than or equal tothat of K . This inequality yields a test point contained in the dual (truncated) cone of K . Interchanging the role of K and the dual cone, we get another test point in K .By paring the two test points, we obtain the sharp lower bound estimate of P ( K ) for an1-unconditional bodies K .Meyer’s estimate for the above K can be generalized to the case of a convex truncatedcone and it was used in [BF]. Actually, n -dimensional convex bodies with many hyperplanesymmetries were treated. From the view point of the group action, that is the case ofCoxeter groups. In the argument in [BF], the hyperplane symmetries (reflections) areessential. For instance, for 1-unconditional body K , these symmetries yield that thepolar body K ◦ is also 1-unconditional.In general, K ∈ K ( G ) has a fundamental region ˆ K with respect to the discretegroup action by G . Then ˆ K is a convex truncated cone. If G ⊂ O ( n ) is not a Coxeter5roup, then the corresponding region ˆ K ◦ for the polar K ◦ of K is not necessarily convex.Nevertheless, by using the inequality [IS, Proposition 3.2], in some cases it is possible toobtain the sharp estimate of P ( K ) even for convex bodies K without enough hyperplanesymmetries. In this paper we focus on the three dimensional case.This paper is organized as follows. In Section 2, we review necessary facts about twodimensional bodies. Though all facts here may be well-known, we give short proofs forthe self-containedness. In Section 3, we give a detailed exposition of the “signed volumeestimate” for three dimensional G -invariant convex bodies. The principal estimate is theinequality in Lemma 3.10, which is frequently used in the arguments in Section 4. InSection 4, we prove the main result (Theorem 1.5). We apply the estimations arranged inSection 3 to the convex bodies in K ( G ) for each discrete subgroup G ⊂ O (3). By meansof the inequality in Lemma 3.10, the estimation is reduced to the two dimensional resultprepared in Sections 2 and 3. We also characterize the equality condition for each case inSection 5. In this section, we prepare necessary facts about 2-dimensional bodies. Although they areessentially proved in [BMMR], we give proofs of them for the sake of completeness. Theestimate of the volume product P ( K ) of a three dimensional convex body is eventuallydeduced to that case. In what follows, we denote by o , a , b , p , a ◦ , etc.the position vectors of points o, A, B, P, A ◦ , etc. in R or R , respectively. And a k b means that the two vectors a and b are parallel. We denote the positive hull and theconvex hull generated by points A , · · · , A m ∈ R n bypos( A , . . . , A m ) = { t a + · · · + t m a m ∈ R n ; t , . . . , t m ≥ } , conv( A , . . . , A m ) = { t a + · · · + t m a m ∈ R n ; t , . . . , t m ≥ , t + · · · + t m = 1 } . The following formula estimates the volume product of the intersection of a convex body K ⊂ R and a positive hull. Lemma 2.1.
Let K ⊂ R be a convex body. Assume that A, B ∈ ∂K and A ◦ , B ◦ ∈ ∂K ◦ with a · a ◦ = b · b ◦ = 1 . We put L := K ∩ pos( A, B ) and L ◦ := K ◦ ∩ pos( A ◦ , B ◦ ) . Thenwe have | L | | L ◦ | ≥ ( a − b ) · ( a ◦ − b ◦ ) / . y xLO AB y xL ◦ O A ◦ B ◦ roof. We put a = a a ! , b = b b ! , a ◦ = a ◦ a ◦ ! , b ◦ = b ◦ b ◦ ! . Since a · a ◦ = b · b ◦ = 1, we have a a ◦ + a a ◦ = 1 and b b ◦ + b b ◦ = 1. For any point P ( x, y ) in K , since the sum of the signed area of the triangle OAP and that of
OP B isless than or equal to | L | , we have12 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a a x y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 12 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x yb b (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | L | for any xy ! ∈ K, which means that 12 | L | − a + b a − b ! ∈ K ◦ . Similarly, we obtain 12 | L ◦ | − a ◦ + b ◦ a ◦ − b ◦ ! ∈ K. These test points yields that4 | L || L ◦ | ≥ ( a − b )( a ◦ − b ◦ ) + ( a − b )( a ◦ − b ◦ ) = ( a − b ) · ( a ◦ − b ◦ ) . Remark 2.2.
We can easily check that the above two test points contained in L ◦ and L ,respectively. Lemma 2.1 is closely related with the second proof of [BMMR, Lemma 7]. The next lemma is useful for, especially, the case of cyclically symmetric bodies in R . Lemma 2.3.
Under the same assumptions in Lemma 2.1, assume that B = R ( A ) and B ◦ = R ( A ◦ ) , where R denotes the rotation of angle ξ ∈ (0 , π ) . Then | L | | L ◦ | ≥ − cos ξ holds.Proof. We may assume that a = a a ! = ! , b = b b ! = cos ξ sin ξ ! , a ◦ = a ◦ a ◦ ! = a ◦ ! , b ◦ = b ◦ b ◦ ! = cos ξ − a ◦ sin ξ sin ξ + a ◦ cos ξ ! Then we obtain 4 | L || L ◦ | ≥ − (cos ξ − a ◦ sin ξ ) − cos ξ − (sin ξ ) a ◦ = 2(1 − cos ξ ).The following is the same equality condition as in the result [BF, Corollary 3] or[BMMR, Lemma 7]. Lemma 2.4.
Under the same assumptions in Lemma 2.3, if a k a ◦ and the equality of (3) holds, then either (i) or (ii) below holds. L = conv { o, A, C, B } and L ◦ = conv { o, A ◦ , B ◦ } , where c = ( a + b ) / (1 + cos ξ ) . (ii) L = conv { o, A, B } and L ◦ = conv { o, A ◦ , C ◦ , B ◦ } , where c ◦ = ( a ◦ + b ◦ ) / (1 + cos ξ ) .Proof. Under the setting of Lemma 2.3, the assumption a k a ◦ implies that a ◦ = 0.Hence, we can take the test points in the proof of Lemma 2.1 as c ◦ := 12 | L | − a + b a − b ! = 12 | L | sin ξ − cos ξ ! ∈ L ◦ , c := 12 | L ◦ | − a ◦ + b ◦ a ◦ − b ◦ ! = 12 | L ◦ | sin ξ − cos ξ ! ∈ L (see Remark 2.2). Then we have | L | ≥ the area of the quadrilateral oACB = 12 | L ◦ | (1 − cos ξ ) , | L ◦ | ≥ the area of the quadrilateral oA ◦ C ◦ B ◦ = 12 | L | (1 − cos ξ ) . Hence, the equality holds if and only if L and L ◦ coincide with the quadrilaterals oACB and oA ◦ C ◦ B ◦ , respectively. Here note that there exists a constant α such that c = 12 | L ◦ | sin ξ − cos ξ ! = sin( ξ/ | L ◦ | cos( ξ/ ξ/ ! = α a + b . Since the quadrilateral oACB is convex, we have α ≥
1. Similarly, we obtain c ◦ = α ◦ a ◦ + b ◦ , α ◦ ≥ . Case α = 1 ; In this case, L coincides with the triangle oAB and the dual face of theedge AB is the vertex C ◦ . That is, a · c ◦ = 1(= b · c ◦ ) holds and α ◦ = 21 + cos ξ Therefore, we obtain c ◦ = 21 + cos ξ a ◦ + b ◦ a ◦ + b ◦ ξ , which means that the condition (ii) holds. Case α > ; Since C is a vertex of L , its dual face is an edge of L ◦ . The edge containsthe segment A ◦ C ◦ or C ◦ B ◦ . From c · c ◦ = 1, we have c · a ◦ = 1 or c · b ◦ = 1. It followsthat α = 21 + cos ξ , c = a + b ξ . Then we have 1 = c · c ◦ = αα ◦ ξ ) = α ◦ . Hence, L ◦ coincides with the triangle oA ◦ B ◦ , that is, the condition (i) holds.8 Preliminaries for the three dimensional case
In this section, we recall the method of “signed volume estimate” introduced in [IS]. Theexposition here is simpler than that of [IS] and applicable to various truncated cones.Although the method can be extended to the higher dimensional case, from now on, weconcentrate on the three dimensional case.
Given a convex body K ∈ K whose interior contains the origin o , for any g ∈ O (3) ⊂ GL (3 , R ), we have ( gK ) ◦ = ( t g ) − K ◦ = gK ◦ . Hence, K ∈ K ( G ) implies that K ◦ ∈ K ( G ) for each subgroup G ⊂ O (3).Let C be an oriented piecewise C -curve on the boundary ∂K of K and r ( t ) (0 ≤ t ≤ C . Then we define a vector C ∈ R by C := 12 Z C r × d r = 12 Z r ( t ) × r ( t ) dt, which is independent of the choice of a parametrization of C . The norm of C is nothingbut the area of the ruled surface o ∗ C := { λ u ∈ R ; u ∈ C , ≤ λ ≤ } . If the curve C is on a plane in R passing through the origin o , then C is a normal vectorof the plane. For any x ∈ R , the signed volume of the solid x ∗ ( o ∗ C ) := { (1 − ν ) x + ν ξ ∈ R ; ξ ∈ o ∗ C , ≤ ν ≤ } is defined by 13 x · C = 16 Z det( x r ( t ) r ( t )) dt. Note that this quantity is essential for the signed volume estimate of truncated conesexplained later (see Lemma 3.8).Next, we examine the behavior of C under the action of O (3). For a rotation g ∈ SO (3),we have g C = g C . In general, this formula does not necessarily hold for g ∈ O (3). However, we obtain V C = − V C , H C = − H C , where V and H are the elements of O (3) defined in Section 1.3. Indeed, for a parametriza-tion r ( t ) of C , since V C = 12 Z ( V r ( t )) × ( V r ( t )) dt = 12 Z r ( t ) − r ( t ) r ( t ) × r ( t ) − r ( t ) r ( t ) dt, we get the first equality. The second one is also easily verified.We denote by −C the curve in R with the same image of the curve C but with theopposite direction. (Note that the notation −C here is different from the one used in [IS].)Then we obtain the formula −C = −C . .2 Curves on ∂K and their polars Let K ∈ K with o ∈ int K . For any two points A, B ∈ ∂K with A ∦ B , let us introducean oriented curve from A to B on the boundary ∂K defined by C ( A, B ) = C K ( A, B ) := ρ K ((1 − t ) a + t b )((1 − t ) a + t b ) : 0 ≤ t ≤ , where ρ K is the radial function of K defined by ρ K ( x ) := max { λ ≥ λ x ∈ K } for x ∈ R \ { o } . For any points A , . . . , A m ∈ ∂K with A i ∦ A i +1 , we denote by C K ( A , . . . , A m )the oriented curve on ∂K consists of successive oriented curves C ( A i , A i +1 ), i = 1 , . . . , m − C K ( A , . . . , A m ) := C ( A , A ) ∪ · · · ∪ C ( A m − , A m ) . In particular, if C K ( A , . . . , A m ) is a simple closed curve, that is, A m = A , then we denoteby S K ( A , . . . , A m − ) the part of ∂K enclosed by the curve such that the orientation ofthe part is compatible with that of the curve.Let K ∈ K with o ∈ int K and denote by µ K its Minkowski gauge. Note that ρ K ( x ) = 1 /µ K ( x ). Assume that ∂K is a C -hypersurface in R . From now on, weconsider the following class of convex bodies.ˇ K := n K ∈ K ; o ∈ int K, K is strongly convex, ∂K is of class C o , where K is said to be strongly convex if the Hessian matrix D ( µ K / x ) of C -function µ K / R is positive definite for each x ∈ R with | x | = 1. For a convex body K ∈ ˇ K ,we define a C -map Λ = Λ K : ∂K → ∂K ◦ byΛ( x ) = ∇ µ K ( x ) ( x ∈ ∂K ) . If K is strongly convex with the boundary ∂K of class C , then K ◦ is strongly convexwith ∂K ◦ of class C , and the map Λ : ∂K → ∂K ◦ is a C -diffeomorphism satisfying that x · Λ( x ) = 1 (see [Sc1, Section 1.7.2] and [IS, Section 3.1]). Note that the curve C ( A, B )is in the plane passing through the three points o , A , and B , but its image Λ( C ( A, B ))into ∂K ◦ is not necessarily contained in a plane of R . Lemma 3.1.
Let K ∈ ˇ K be a convex body and A, B ∈ ∂K be two points with A ∦ B .Let H ⊂ R be the plane passing through the three points o , A , B , and π H denotes theorthogonal projection onto H . Then π H ◦ Λ K = Λ K ∩ H holds on ∂ ( K ∩ H ) .Proof. It suffices to consider the case that the points
A, B are in the xy -plane, i.e., H = { x = ( x, y, z ) ∈ R ; z = 0 } . By definition, Λ K = ∇ µ K | ∂K is a C -diffeomorphism from ∂K to ∂K ◦ , where ∇ µ K = ( ∂ x µ K , ∂ y µ K , ∂ z µ K ). Setting L := K ∩ H , then L ⊂ H is strongly convex with its boundary of class C , o ∈ int K , and µ L = µ K | H . Since ∇ µ L = ( ∂ x ( µ K | H ) , ∂ y ( µ K | H )) : H → R and Λ L = ∇ µ L | ∂L : ∂L → Λ L ( ∂L ), we have π H ◦ Λ K | ∂L = ( ∂ x µ K , ∂ y µ K ) | ∂L = Λ L on ∂L .We next turn to the case of G -invariant convex bodies in ˇ K . For a subgroup G of O (3), let us introduce the classˇ K ( G ) := n K ∈ ˇ K ; gK = K for all g ∈ G o . For a convex body K in this class, the map Λ K behaves G -equivariantly as follows.10 emma 3.2. Let K ∈ ˇ K ( G ) . For any x ∈ ∂K and g ∈ G , we obtain g Λ K ( x ) = Λ K ( g x ) (4) Proof.
Let K ∈ ˇ K . We first suppose that x ∈ R \ { } and g ∈ O (3). By the definitionof µ K , we have µ K ( x ) = µ gK ( g x ). By differentiating it, ∇ µ K ( x ) = t g ∇ µ gK ( g x ) = g − ∇ µ gK ( g x )holds, because g ∈ O (3). Here, assume that K ∈ ˇ K ( G ), x ∈ ∂K , and g ∈ G . Then gK = K and g x ∈ ∂K hold. Since Λ K = ∇ µ K on ∂K , we obtain the equality (4).Not every G -invariant convex body K is equipped with the map Λ K . However, owingto the following approximation result, which is a special case of [Sc2, pp. 438], it sufficesto consider only the class ˇ K ( G ) for the purpose of this paper. Proposition 3.3 (Schneider) . Let G be a discrete subgroup of O (3) . Let K ∈ K ( G ) be a G -invariant convex body. Then, for any ε > there exists a G -invariant convexbody K (cid:15) ∈ ˇ K ( G ) having the property that δ ( K, K (cid:15) ) < ε , where δ denotes the Hausdorffdistance on K . In Section 4, we frequently use Proposition 3.3 for various discrete subgroups of O (3). Remark 3.4.
Note that Lemmas 3.1 and 3.2, and Proposition 3.3 hold for n -dimensionalcase by the same argument. Let K ∈ ˇ K . If we apply the method of signed volume estimate to K (see Section 3.4),then the lower bound estimate of the volume product P ( K ) is reduced to the estimationof some two dimensional situation. Here we prepare such an estimate. Proposition 3.5.
Let K ∈ ˇ K . For any two points A, B ∈ ∂K with a ∦ b , it holds that C ( A, B ) · Λ( C ( A, B )) ≥ ( a − b ) · Λ( a ) − Λ( b )4 . Moreover, if | a | = | b | , a k Λ( a ) , and b k Λ( b ) hold, then we obtain C ( A, B ) · Λ( C ( A, B )) ≥
12 (1 − cos ξ ) , where ξ is the angle between a and b .Proof. Let H ⊂ R be the plane that contains the curve C ( A, B ) and the origin o . ByLemma 3.1, we have o ∗ π H (Λ K ( C ( A, B ))) = o ∗ Λ K ∩ H ( C ( A, B ))= ( K ∩ H ) o H ∩ pos(Λ K ∩ H ( a ) , Λ K ∩ H ( b )) , where ( K ∩ H ) o H := { y ∈ H ; y · x ≤ x ∈ K ∩ H } . Moreover, if a k Λ( a ) and b k Λ( b ) hold, then we further obtain o ∗ π H (Λ K ( C ( A, B ))) = ( K ∩ H ) o H ∩ pos( a , b ).Thus, we can apply Lemmas 2.1–2.4 to the two dimensional body K ∩ H ⊂ H , and hencethe results hold. 11 roposition 3.6. Let K ∈ K . Let a , b ∈ ∂K with a ∦ b and | a | = | b | . Assume that a ◦ , b ◦ ∈ ∂K ◦ satisfies a · a ◦ = b · b ◦ = 1 , a k a ◦ , and b k b ◦ . Let H ⊂ R be the planepassing through the three points o , A , B , and π H denotes the orthogonal projection onto H . Then, setting L := K ∩ pos( A, B ) , L ◦ := π H ( K ◦ ) ∩ pos( A, B ) , we have | L | | L ◦ | ≥
12 (1 − cos ξ ) , where ξ is the angle between a and b . The equality holds if and only if either the following (i) or (ii) is satisfied :(i) L = conv { o, A, C, B } and L ◦ = conv { o, A ◦ , B ◦ } , where c = ( a + b ) / (1 + cos ξ ) . (ii) L = conv { o, A, B } and L ◦ = conv { o, A ◦ , C ◦ , B ◦ } , where c ◦ = ( a ◦ + b ◦ ) / (1 + cos ξ ) .Proof. The results are direct consequences of Lemmas 2.1–2.4.
Remark 3.7.
In the setting of Proposition 3.6, pos(
A, B ) = pos( A ◦ , B ◦ ) holds. Now we are in position to state the “signed volume estimate”, which is a natural gen-eralization of the standard volume estimate used in [M, I.2. Théorème] and [BF, Lemma11].
Lemma 3.8 ([IS, Proposition 3.2]) . Let K ∈ ˇ K . Let C be a piecewise C , oriented,simple closed curve on ∂K . Let S K ( C ) ⊂ ∂K be a piece of surface enclosed by the curve C such that the orientation of the surface is compatible with that of C . Then for any point x ∈ K , the inequality x · C ≤ | o ∗ S K ( C ) | holds. Remark 3.9.
Although in [IS, Proposition 3.2] the boundary ∂K is assumed to be C ∞ ,the proof works without any changes under the assumption that K ∈ ˇ K and the curve C is piecewise C . Lemma 3.10.
Under the same assumption as Lemma 3.8, the following inequality holds: | o ∗ S K ( C ) | | o ∗ S K ◦ (Λ( C )) | ≥ C · Λ( C ) . Proof.
By Lemma 3.8, we have x · C / ≤ | o ∗ S K ( C ) | for any x ∈ K . Hence we get atest vector C / (3 | o ∗ S K ( C ) | ) ∈ K ◦ . By assumption, the C -map Λ : ∂K → ∂K ◦ can bedefined. By applying Lemma 3.8 to a piece of surface S K ◦ (Λ( C )) ⊂ ∂K ◦ , we obtain C | o ∗ S K ( C ) | · Λ( C )3 ≤ | o ∗ S K ◦ (Λ( C )) | , as claimed. 12n Section 3.1, we defined the vector C ∈ R for a piecewise C -curve C ⊂ R by meansof line integral. Let K ∈ K be a convex body. If K has its boundary of class C , thena curve C ( A, B ) on ∂K is of class C and we get C ( A, B ) ∈ R . Note that the quantity C ( A, B ) can be defined for a more general convex body K ∈ K . Indeed, taking the curve C ( A, B ) is on the plane passing through o , A , and B into account, if C ( A, B ) is of class C , then C ( A, B ) = | o ∗ C ( A, B ) | a × b | a × b | (5)holds. On the other hand, since the quantity in the right hand side of (5) is also well-defined for a general curve C ( A, B ) which is not necessarily of class C on ∂K , we definethe vector C ( A, B ) ∈ R by the above formula in this general setting as well. Then we havethe following fact, which is necessary for determining the equality conditions of Theorem1.5 (see Section 5). Lemma 3.11 (cf. [IS, Lemma 6.2]) . Let K ∈ K and A , A , A ∈ ∂K . Assume that C ( A , A , A , A ) is a simple closed curve on ∂K , that is, S K ( A , A , A ) is a “triangle”on ∂K . Then, for any x ∈ K we have x · (cid:16) C ( A , A ) + C ( A , A ) + C ( A , A ) (cid:17) ≤ | o ∗ S K ( A , A , A ) | with equality if and only if o ∗ S K ( A , A , A ) = x ∗ ( o ∗ C ( A , A , A , A )) for some x ∈ S K ( A , A , A ) .Proof. For the inequality part, see [IS, Lemma 6.2].Assume that the equality holds. By the construction, if x / ∈ o ∗ S K ( A , A , A ), thenthe equality does not hold. Hence, x ∈ o ∗ S K ( A , A , A ). Then the left-hand side isnothing but the volume of the cone over o ∗ C ( A , A , A , A ) with the vertex x . Sincethe cone is contained in o ∗ S K ( A , A , A ), the assumption implies that the two solidscoincide and x ∈ S K ( A , A , A ). The converse is obvious. In this section, we give a proof of inequality part of Theorem 1.5 by case analysis. Westart with the case G = T . G = T Proposition 4.1.
The inequality P ( K ) ≥ P ( ) holds for any T -invariant convex body K ∈ K ( T ) .Proof. We put a = 1 √ , b = 1 √ − − , c = 1 √ − − , d = 1 √ − − ∈ S . is represented as = conv { A, B, C, D } , and we caneasily compute that P ( ) = 64 /
9. We denote by R A , R B , R C , and R D the rotationsthrough the angle 2 π/ oA , oB , oC , and oD , respectively. That is, R A = , R B = − − , R C = − − , R D = −
11 0 00 − . Under this setting, we see that T = h R A , R B , R C , R D i ⊂ SO (3) . Let K ∈ K ( T ). The volume product P is continuous with respect to the Hausdorffdistance on K . Therefore, by Proposition 3.3, it suffices to consider the case that K ∈ ˇ K ( T ). By a dilation of K , we may assume that A ∈ ∂K . Then, by the T -symmetry, B, C, D ∈ ∂K also holds. Putting ˜ K := o ∗ S K ( A, B, C ) and ˜ K ◦ := o ∗ Λ( S K ( A, B, C )),we have | K | = 4 | ˜ K | , | K ◦ | = 4 | ˜ K ◦ | . By applying Lemma 3.10 to the curve C = C K ( A, B, C, A ), we obtain916 | K | | K ◦ | ≥ | ˜ K | | ˜ K ◦ | =( C ( A, B ) + C ( B, C ) + C ( C, A )) · (Λ( C ( A, B )) + Λ( C ( B, C )) + Λ( C ( C, A ))) . (6)Next, let us compute the right-hand side of (6). Since C ( B, C ) = R D ( C ( A, B )) , C ( C, A ) = R D ( C ( A, B )) , (7)we have C ( A, B ) + C ( B, C ) + C ( C, A ) = ( E + R D + R D ) C ( A, B ) . Note that C ( A, B ) = |C ( A, B ) | t (0 , / √ , − / √
2) from (5). Hence,( E + R D + R D ) C ( A, B ) = |C ( A, B ) |√ −
11 1 − − − − = √ |C ( A, B ) | − . On the other hand, by Lemma 3.2, we have Λ( C ( B, C )) = R D (Λ( C ( A, B ))) and Λ( C ( C, A )) = R D (Λ( C ( A, B ))) from (7), so thatΛ( C ( A, B )) + Λ( C ( B, C )) + Λ( C ( C, A )) = ( E + R D + R D )Λ( C ( A, B ))14olds. Putting t ( x , x , x ) := Λ( C ( A, B )), we get C ( A, B ) · Λ( C ( A, B )) = 1 √ x − x ) |C ( A, B ) | . Since R B R C = − − , R B R C ( C ( A, B )) = C ( B, A ) , we obtain − x x x = Λ( C ( B, A )) = R B R C (Λ( C ( A, B ))) = R B R C Λ( C ( A, B )) = x − x − x , which implies that x = 0. Therefore,( E + R D + R D )Λ( C ( A, B )) = −
11 1 − − − x x = ( x − x ) − holds. Thus, the right-hand side of (6) is computed as √ |C ( A, B ) | − · ( x − x ) − = 3 √ x − x ) |C ( A, B ) | = 6 C ( A, B ) · Λ( C ( A, B )) . Finally, we estimate the right-hand side of the above equality from below. Putting t ( y , y , y ) := Λ( a ), we have a · Λ( a ) = 1 √ y + y + y ) = 1 , y y y = Λ( a ) = Λ( R A a ) = R A Λ( a ) = y y y , which implies that Λ( a ) = a . Similarly, Λ( b ) = b holds. By Proposition 3.5, we get C ( A, B ) · Λ( C ( A, B )) ≥
14 ( a − b ) · (Λ( a ) − Λ( b )) = 23 . Consequently, we obtain | K | | K ◦ | ≥
169 6 C ( A, B ) · Λ( C ( A, B )) ≥ , which completes the proof. G = O Proposition 4.2.
The inequality P ( K ) ≥ P ( ) holds for any O -invariant convex body K ∈ K ( O ) . roof. Let be the regular octahedron with vertices {± a , ± b , ± c } , where a := , b := , c := . Then, all the elements of O h are described as ± ± ± , ± ± ± , ± ± ± , ± ± ± , ± ± ± , ± ± ± . In particular, O = { g ∈ O h ; det g = 1 } . And we see that P ( ) = 32 /
3. Let K ∈ K ( O ).In order to examine the volume product of K , by Proposition 3.3, we may assume that K ∈ ˇ K ( O ). Putting ˜ K := O ∗ S K ( A, B, C ) and ˜ K ◦ := O ∗ Λ( S K ( A, B, C )), by the O -symmetry, we have | K | | K ◦ | = 64 | ˜ K | | ˜ K ◦ | . By Lemma 3.10, we obtain9 | ˜ K | | ˜ K ◦ | ≥ (cid:16) C ( A, B ) + C ( B, C ) + C ( C, A ) (cid:17) · (cid:16) Λ( C ( A, B )) + Λ( C ( B, C )) + Λ( C ( C, A )) (cid:17) . (8)We put R := , R A := −
10 1 0 , R B := − , where R is the rotation through the angle 2 π/ o and the point (1 , , R A and R B are rotations through the angle π/ oA and oB , respectively. Since C ( B, C ) = R ( C ( A, B )) , C ( C, A ) = R ( C ( A, B )) , the right-hand side of (8) becomes( E + R + R ) C ( A, B ) · ( E + R + R )Λ( C ( A, B )) . Note that C ( A, B ) = |C ( A, B ) | t (0 , , t ( x , x , x ) := Λ( C ( A, B )). Then,the right-hand side of (8) equals |C ( A, B ) | · x + x + x x + x + x x + x + x = 3 |C ( A, B ) | ( x + x + x ) . On the other hand, since RR A ( C ( A, B )) = R ( C ( A, C )) = C ( B, A ), we obtain x x − x = RR A Λ( C ( A, B )) = − Λ( C ( A, B )) = − x x x , x = − x . Thus, the right-hand side of (8) equals3 |C ( A, B ) | ( x + x + x ) = 3 |C ( A, B ) | x = 3 C ( A, B ) · Λ( C ( A, B )) . Since R A ( A ) = A and R B ( B ) = B hold, by a similar argument of the case G = T , weobtain Λ( a ) = a and Λ( b ) = b . Therefore, Proposition 3.5 yields that C ( A, B ) · Λ( C ( A, B )) ≥
14 ( a − b ) · (Λ( a ) − Λ( b )) = 12 , so that we obtain | K | | K ◦ | ≥ /
3, as claimed.
Remark 4.3.
In this case, the minimum of P is the same as the centrally symmetric case,that is, S = h R H i = h− E i ∼ = Z . However, the generator − E of S is not an elementof O . This means that Proposition 4.2 and [IS, Theorem 1] are independent results. G = I Proposition 4.4.
The inequality P ( K ) ≥ P ( ) holds for any I -invariant convex body K ∈ K ( I ) .Proof. Let be the regular icosahedron with the 12-vertices(0 , ± , ± φ ) , ( ± φ, , ± , ( ± , ± φ, , where φ = (1 + √ / a := φ , b := φ , c := φ . Let R be a rotation through the angle 2 π/ t (1 , , R = ∈ I, R ( A ) = B, R ( B ) = C, R ( C ) = A. By a simple calculation, we see that | | = 10(3 + √ / | ◦ | = 2(25 − √ K ∈ K ( I ). By Proposition 3.3, we may assume that K ∈ ˇ K ( I ). By a dilation, we canalso assume that A, B, C ∈ ∂K . Putting ˜ K := O ∗S K ( A, B, C ) , ˜ K ◦ := O ∗ Λ( S K ( A, B, C )),by the I -symmetry, we obtain | K | | K ◦ | = 400 | ˜ K | | ˜ K ◦ | . By Lemma 3.10, we get9 | ˜ K | | ˜ K ◦ | ≥ (cid:16) C ( A, B ) + C ( B, C ) + C ( C, A ) (cid:17) · (cid:16) Λ( C ( A, B )) + Λ( C ( B, C )) + Λ( C ( C, A )) (cid:17) . (9)Since C ( B, C ) = R ( C ( A, B )) , C ( C, A ) = R ( C ( A, B )) , the right-hand side of (9) equals( E + R + R ) C ( A, B ) · ( E + R + R )Λ( C ( A, B )) . C ( A, B ) k t (1 , φ , − φ ). Let R AB be the rotation through theangle π about the axis through o and the midpoint of the segment AB . Then R AB ∈ I and R AB Λ( C ( A, B )) = Λ( C ( B, A )) = − Λ( C ( A, B )) (10)hold. Note that the three vectors a × b = φ − φ , a − b = − φ φ − , a + b = φ φ + 1 are orthogonal to each other. Putting Λ( C ( A, B )) = y ( a × b ) + y ( a − b ) + y ( a + b ), by(10), we obtain − y ( a × b ) − y ( a − b ) + y ( a + b ) = − y ( a × b ) − y ( a − b ) − y ( a + b ) , so that y = 0. We put x t (1 , φ , − φ ) := C ( A, B ) for a nonzero real number x . Then, C ( A, B ) · Λ( C ( A, B )) = x φ − φ · { y ( a × b ) + y ( a − b ) } = 4 xy φ holds and the right-hand side of (9) becomes x ( E + R + R ) φ − φ · ( E + R + R ) y − φy φ y + y − φy + ( φ − y = 4 xy · = 12 xy = 3 φ C ( A, B ) · Λ( C ( A, B )) = 3(3 − √ C ( A, B ) · Λ( C ( A, B )) . We denote by R A the rotation through the angle 2 π/ o and A . Then R A ∈ I and R A ( a ) = a hold, so that we have Λ( a ) k a . Similarly, Λ( b ) k b holds. Since | a | = | b | , by Proposition 3.5, we obtain C ( A, B ) · Λ( C ( A, B )) ≥ − a · b | a | | b | ! = 12 − √ ! . Consequently, | K | | K ◦ | ≥
803 (5 − √ . Since the right-hand side equals the volume product of , it completes the proof. G = C ‘h Proposition 4.5.
Assume that ‘ ≥ . Then the inequality P ( K ) ≥ P ( P ‘ ) holds for any C ‘h -invariant convex body K ∈ K ( C ‘h ) .Proof. Let K ∈ K ( C ‘h ) for ‘ ≥
3. To prove the inequality for P , we may assume that K ∈ ˇ K ( C ‘h ) by Proposition 3.3. We put p := , a := , b := R ‘ a = cos ξ sin ξ , where ξ = 2 π/‘ . 18ecall that G is a subgroup of GL (3 , R ) defined in Section 1.3. Since gh = hg holds forevery g ∈ G and every h ∈ C ‘h , we see that gK ∈ ˇ K ( C ‘h ) for any g ∈ G . Thus, we canassume that P, A ∈ ∂K . Setting ˜ K := o ∗ S K ( P, A, B ) and ˜ K ◦ := o ∗ Λ( S K ( P, A, B )), bythe C ‘h -symmetry for K and K ◦ , we have | K | = 2 ‘ | ˜ K | , | K ◦ | = 2 ‘ | ˜ K ◦ | . By Lemma 3.10, we obtain9 | ˜ K | | ˜ K ◦ | ≥ (cid:16) C ( P, A ) + C ( A, B ) + C ( B, P ) (cid:17) · (cid:16) Λ( C ( P, A )) + Λ( C ( A, B )) + Λ( C ( B, P )) (cid:17) . (11)Since C ( B, P ) = R ‘ ( C ( A, P )), we obtain C ( B, P ) = R ‘ C ( A, P ) = − R ‘ C ( P, A ) . Since Λ( C ( B, P )) = Λ( R ‘ ( C ( A, P ))) = R ‘ (Λ( C ( A, P ))) by Lemma 3.2, we getΛ( C ( B, P )) = R ‘ Λ( C ( A, P )) = − R ‘ Λ( C ( P, A )) . It follows from the above two equalities that the right-hand side of (11) equals( E − R ‘ ) C ( P, A ) · ( E − R ‘ )Λ( C ( P, A )) + ( E − R ‘ ) C ( P, A ) · Λ( C ( A, B ))+ C ( A, B ) · ( E − R ‘ )Λ( C ( P, A )) + C ( A, B ) · Λ( C ( A, B )) . Here we denote these four terms by (I) , (II) , (III), and (IV), respectively.Let us start with the calculation of (III). For any x = t ( x , x , x ) ∈ R , we have( E − R ‘ ) x = (1 − cos ξ ) x + (sin ξ ) x − (sin ξ ) x + (1 − cos ξ ) x . On the other hand, by definition, C ( A, B ) k t (0 , ,
1) holds, which immediately impliesthat (III) = 0. Next, since Λ( C ( A, B )) = Λ( H ( C ( A, B ))) = H Λ( C ( A, B ))), we obtainΛ( C ( A, B )) = H (Λ( C ( A, B ))) = − H Λ( C ( A, B )) . Hence Λ( C ( A, B )) k t (0 , ,
1) holds, which means that (II) = 0. Moreover, we can put t (0 , x ,
0) := C ( P, A ) and t ( y , y , y ) := Λ( C ( P, A )) by their definitions. By the abovecalculation, we get(I) = (sin ξ ) x (1 − cos ξ ) x · (1 − cos ξ ) y + (sin ξ ) y − (sin ξ ) y + (1 − cos ξ ) y = 2(1 − cos ξ ) x y = 2(1 − cos ξ ) C ( P, A ) · Λ( C ( P, A )) . | K | | K ◦ | ≥ ‘ (cid:16) − cos ξ ) C ( P, A ) · Λ( C ( P, A )) + C ( A, B ) · Λ( C ( A, B )) (cid:17) . (12)Finally, we compute the right-hand side of (12). Since R ‘ ( P ) = P and p = t (0 , , R ‘ (Λ( P )) = Λ( P ) and Λ( p ) = p . From H ( A ) = A and B = R ‘ ( A ), we have H (Λ( A )) = Λ( A ) and Λ( B ) = R ‘ (Λ( A )). Thus, we can putΛ( a ) = z ◦ z ◦ , Λ( b ) = z ◦ cos ξ − z ◦ sin ξz ◦ sin ξ + z ◦ cos ξ for some z ◦ , z ◦ ∈ R . Since z ◦ = a · Λ( a ) = 1, Proposition 3.5 asserts that C ( P, A ) · Λ( C ( P, A )) ≥
14 ( p − a ) · (Λ( p ) − Λ( a )) = 12 , C ( A, B ) · Λ( C ( A, B )) ≥
14 ( a − b ) · (Λ( a ) − Λ( b )) = 12 (1 − cos ξ ) , so that | K | | K ◦ | ≥ ‘ − cos ξ ) . It is easy to check that the right-hand side equals the volume product of P ‘ . G = D ‘ Proposition 4.6.
Assume that ‘ ≥ . Then the inequality P ( K ) ≥ P ( P ‘ ) holds for any D ‘ -invariant convex body K ∈ K ( D ‘ ) .Proof. Let K ∈ K ( D ‘ ) for ‘ ≥
3. Recall that D ‘ = h R ‘ , V H i . To prove the inequality,we may assume that K ∈ ˇ K ( D ‘ ) by Proposition 3.3. We put p = , a = , q = − , b = cos ξ sin ξ . In a similar way as the proof of Proposition 4.5, by a linear transformation in G , we canassume A, B, P, Q ∈ ∂K . Let us consider the closed regions ˜ K := O ∗ S K ( P, A, B ) and˜ K ◦ := O ∗ Λ( S K ( P, A, B )). Since (
V H ) R − ‘ ( S K ( P, A, B )) = S K ( Q, B, A ) holds, we have | K | = 2 ‘ | ˜ K | , | K ◦ | = 2 ‘ | ˜ K ◦ | . Here, we show that ( E − R ‘ ) C ( P, A ) · Λ( C ( A, B )) = 0 . (13)For g ∈ D ‘ , we have g ( K ) = K and g ( ∂K ) = ∂K , so that g ( C K ( X, Y )) = C gK ( g ( X ) , g ( Y )) = C K ( g ( X ) , g ( Y ))for any x , y ∈ ∂K with x ∦ y . Since R ‘ , V H ∈ D ‘ , the above formula yields that R ‘ V H ( C ( B, A )) = C ( R ‘ V H ( B ) , R ‘ V H ( A )) = C ( A, B ) . C ( A, B )) = Λ( R ‘ V H ( C ( B, A ))) = R ‘ V H (Λ( C ( B, A ))) , so that Λ( C ( A, B )) = R ‘ V H (Λ( C ( B, A ))) = − R ‘ V H Λ( C ( A, B )) . Putting t ( x , x , x ) := Λ( C ( A, B )), we have x x x = cos ξ − sin ξ ξ cos ξ
00 0 1 − x x x = − (cos ξ ) x − (sin ξ ) x − (sin ξ ) x + (cos ξ ) x x . Thus ξ sin ξ sin ξ − cos ξ ! x x ! = ! (14)holds. On the other hand, since( E − R ‘ ) C ( P, A ) = − cos ξ sin ξ − sin ξ − cos ξ
00 0 0 |C ( P, A ) | = |C ( P, A ) | sin ξ − cos ξ , we obtain( E − R ‘ ) C ( P, A ) · Λ( C ( A, B )) = |C ( P, A ) | ((sin ξ ) x + (1 − cos ξ ) x ) = 0by (14), so that (13) is verified.Similarly as the case C ‘h , by the signed volume estimate, we obtain exactly the sameinequality as (12). Since R ‘ p = p and p · Λ( p ) = 1, we have Λ( p ) = R ‘ Λ( p ), so thatΛ( P ) = P holds. Similarly, V H ( A ) = A implies that Λ( A ) = A . Hence, Λ( B ) =Λ( R ‘ ( A )) = R ‘ (Λ( A )) = R ‘ ( A ) = B also holds. Thus, in the same way as in the case C ‘h , we can sharply estimate the right-hand side of (12) from below and obtain theconclusion. G = C h , T h , S , D d Proof of Corollary 1.4.
Let (cid:3) be an O h -invariant cube. By [IS, Theorem 1], (cid:3) is a mini-mizer of P on the set of S -invariant convex bodies. Under the matrix representations ofthese groups used throughout Section 4, S = h− E i ⊂ C h = h R , H i ⊂ T h ⊂ O h , S ⊂ S ⊂ D d ⊂ O h hold. Therefore, we have P ( (cid:3) ) = min K ∈K ( S ) P ( K ) ≤ min K ∈K ( C h ) P ( K ) ≤ min K ∈K ( T h ) P ( K ) ≤ min K ∈K ( O h ) P ( K ) ≤ P ( (cid:3) ) , P ( (cid:3) ) = min K ∈K ( S ) P ( K ) ≤ min K ∈K ( S ) P ( K ) ≤ min K ∈K ( D h ) P ( K ) ≤ min K ∈K ( O h ) P ( K ) ≤ P ( (cid:3) ) . In addition, [IS, Theorem 1] asserts that, if K is a minimizer of P on the set of S -invariantconvex bodies, then K or K ◦ is a parallelepiped. Thus, we obtain the conclusion.21 Proof of Theorem 1.5: equality conditions G = T Proposition 5.1.
Let K ∈ K ( T ) . If P ( K ) = P ( ) holds, then K is a dilate of or ◦ .Proof. Suppose that K ∈ K ( T ) satisfies P ( K ) = P ( ), i.e., K is a minimizer of P . Bythe approximation result (Proposition 3.3), there is a sequence { K m } m ∈ N ⊂ ˇ K ( T ) suchthat K m → K ( m → ∞ ) in the Hausdorff distance. By a dilation of K , we may assumethat A, B, C, D ∈ ∂K , where A, B, C, D are the points used in Section 4.1. In addition,normalizing K m by their dilations, we can assume A, B, C, D ∈ ∂K m . We denote thecurve C K m ( A, B ) by C m ( A, B ) for short. For each convex body K m , we can define themap Λ m : ∂K m → ∂K ◦ m . Let H be the plane through o , A , and B , and π H be theorthogonal projection to H . We put L m := o ∗ C m ( A, B ) ⊂ H, L ◦ m := o ∗ π H (Λ m ( C m ( A, B ))) ⊂ H. Then by the argument in the proof of Proposition 4.1, we have916 P ( K m ) ≥ | L m | | L ◦ m | ≥ . (15)Now we examine the two dimensional subsets L m and L ◦ m of H . By the T -symmetry,we have A ◦ := Λ m ( A ) = A and B ◦ := Λ m ( B ) = B , and L m = K m ∩ pos( A, B ) , L ◦ m = π H ( K ◦ m ) ∩ pos( A, B )hold. Putting L := K ∩ pos( A, B ) , L ◦ := π H ( K ◦ ) ∩ pos( A, B ) , then we obtain that L m → L and L ◦ m → L ◦ on H in the Hausdorff distance. Since K isa minimizer of P , taking m → ∞ , by (15), we have | L | | L ◦ | = 23 , which means that Proposition 3.6 holds with equality. Thus, either of the following twocases occurs.Case (i): L = conv { o, A, Z, B } and L ◦ = conv { o, A ◦ , B ◦ } , where z = 3( a + b ) / L = conv { o, A, B } and L ◦ = conv { o, A ◦ , Z ◦ , B ◦ } , where z ◦ = 3( a ◦ + b ◦ ) / Z ◦ ∈ π H ( K ◦ ), there exists s ∈ R such that z ◦ + s a × b ∈ K ◦ . We note that z ◦ k ( a ◦ + b ◦ ) / a + b ) /
2. Let R be the rotation through the angle π about the axisthrough o and the midpoint of A and B . Then, we see that R ∈ T and R ( z ◦ + s a × b ) = z ◦ − s a × b ∈ R ( K ◦ ) = K ◦ . Thus, the convexity of K ◦ implies Z ◦ ∈ K ◦ . Moreover, since o, A ◦ , B ◦ , Z ◦ ∈ pos( A, B ),we get L ◦ = conv { o, A ◦ , Z ◦ , B ◦ } ⊂ K ◦ ∩ pos( A, B ) ⊂ π H ( K ◦ ) ∩ pos( A, B ) = L ◦ , which implies that L ◦ = K ◦ ∩ pos( A, B ). It follows from A = A ◦ and B = B ◦ that L ◦ = o ∗ C K ◦ ( A ◦ , B ◦ ) , C K ◦ ( A ◦ , B ◦ ) = | L ◦ | a × b | a × b | . (16)Putting Y ◦ := R D Z ◦ and X ◦ := R D Z ◦ , we get o ∗ C K ◦ ( C ◦ , A ◦ ) = conv { o, C ◦ , Y ◦ , A ◦ } , o ∗ C K ◦ ( B ◦ , C ◦ ) = conv { o, B ◦ , X ◦ , C ◦ } . Now we consider the truncated convex cone ˆ K ◦ := o ∗ S K ◦ ( A ◦ , B ◦ , C ◦ ). For thesequence { K m } , by the proof of Proposition 4.1 (see also Lemma 3.11), we have43 | K m | ( C m ( A, B ) + C m ( B, C ) + C m ( C, A )) = 43 | K m | ( E + R D + R D ) C m ( A, B )= 4 | L m | | K m | ( E + R D + R D ) a × b | a × b | = 4 √ | L m | | K m | − ∈ K ◦ m . As m → ∞ , we obtain q ◦ := 4 √ | L | | K | − ∈ K ◦ . Moreover, it is easy to check that q ◦ ∈ ˆ K ◦ . By (16), we can calculate the volume of thesolid Q ◦ ∗ ( o ∗ C K ◦ ( A ◦ , B ◦ , C ◦ , A ◦ )) ⊂ ˆ K ◦ as13 q ◦ · (cid:16) C K ◦ ( A ◦ , B ◦ ) + C K ◦ ( B ◦ , C ◦ ) + C K ◦ ( C ◦ , A ◦ ) (cid:17) = 13 q ◦ · ( E + R D + R D ) | L ◦ | a × b | a × b | = 4 √ | L | | K | − · √ | L ◦ | − = 8 | L | | L ◦ | | K | = 169 | K | = | K ◦ | , which is exactly the volume of ˆ K ◦ . Hence, we can apply Lemma 3.11 with equalitycondition to get ˆ K ◦ = Q ◦ ∗ ( o ∗ C K ◦ ( A ◦ , B ◦ , C ◦ , A ◦ )). By the convexity, the segment Q ◦ A ◦ is on the boundary of ˆ K ◦ and that of K ◦ . Here, the endpoints of the three vectors a ◦ = 1 √ , y ◦ = 1 √ , z ◦ = 1 √ x + y + z ) / √ q ◦ = s t (1 , , −
1) ( s > a ◦ · ( q ◦ − a ◦ ) ≥
0, i.e., 1 √ · q ◦ = s √ ≥ , so that s ≥ √
3. On the other hand, a · q ◦ = s/ √ ≤ s = √ q ◦ = √ t (1 , , − q ◦ · a = q ◦ · b = q ◦ · c = 1. By the definition of the polar,we have K ⊂ { x ∈ R ; x · q ◦ ≤ } . Thus, we obtainconv { o, A, B, C } ⊂ K ∩ pos( A, B, C ) ⊂ { x ; x · q ◦ ≤ } ∩ pos( A, B, C ) = conv { o, A, B, C } , so that K ∩ pos( A, B, C ) = conv { o, A, B, C } . By the T -symmetry of K , we have K =conv { A, B, C, D } , which means that K is a dilate of .Finally, we consider Case (i). Then, we have L ◦ = conv { o, A ◦ , B ◦ } and o, A ◦ (= A ) , B ◦ (= B ) ∈ L ◦ . It then follows from the definition of L ◦ that L ◦ ⊂ K ◦ ∩ pos( A, B ) ⊂ π H ( K ◦ ) ∩ pos( A, B ) = L ◦ , so that K ◦ ∩ pos( A, B ) = L ◦ . By the similar argument exchanging K and K ◦ in Case(ii), we get the conclusion. G = O, I
Proposition 5.2. (i)
Let K ∈ K ( O ) . If P ( K ) = P ( ) holds, then K is a dilate ofor ◦ . (ii) Let K ∈ K ( I ) . If P ( K ) = P ( ) holds, then K is a dilate of or ◦ .Proof. We can show them similarly as Proposition 5.1, thus we omit their proofs. G = D ‘ Proposition 5.3.
Assume that ‘ ≥ . Let K ∈ K ( D ‘ ) . If P ( K ) = P ( P ‘ ) holds, then K coincides with P ‘ or P ◦ ‘ up to a linear transformation in G .Proof. We use the symbols p , a , q , and b defined in the proof of Proposition 4.6. ByProposition 3.3, there is a sequence { K m } m ∈ N ⊂ ˇ K ( D ‘ ) such that K m → K . Normalizing K m by a linear transformation in G , we can assume that P, A, B ∈ ∂K m for every m . Let π and π be the orthogonal projections to zx -plane and xy -plane, respectively. Let H be the plane through o, P, B , and π H be the orthogonal projection to H . For each K m ,by the same argument as in the proof of Proposition 5.1, we have P ◦ := Λ m ( P ) = P , A ◦ := Λ m ( A ) = A , and B ◦ := Λ m ( B ) = B , where Λ m : ∂K m → ∂K ◦ m . Moreover, we put L m := o ∗ C m ( P, A ) = K m ∩ pos( P, A ) ,L ◦ m := o ∗ π (Λ m ( C m ( P, A ))) = π ( K ◦ m ) ∩ pos( P, A ) ,M m := o ∗ C m ( A, B ) = K m ∩ pos( A, B ) ,M ◦ m := o ∗ π (Λ m ( C m ( A, B ))) = π ( K ◦ m ) ∩ pos( A, B ) . m → ∞ , these 2-dimensional sets converge to L := K ∩ pos( P, A ) , L ◦ := π ( K ◦ ) ∩ pos( P, A ) ,M := K ∩ pos( A, B ) , M ◦ := π ( K ◦ ) ∩ pos( A, B ) , respectively, in the Hausdorff distance. Since the inequality (12) holds for each K m , bytaking the limit m → ∞ , we obtain | K | | K ◦ | ≥ ‘ − cos ξ ) | L | | L ◦ | + | M | | M ◦ | ) . Since K is a minimizer of P , by the inequality in Proposition 3.6, we get | L | | L ◦ | = 12 , | M | | M ◦ | = 1 − cos ξ , which implies that Proposition 3.6 holds with equality for L and also for M . Hence, forthe 2-dimensional subsets L and L ◦ either of the following two cases happens.Case (A): L = conv { o, P, C, A } and L ◦ = conv { o, P ◦ , A ◦ } , where c = p + a .Case (B): L = conv { o, P, A } and L ◦ = conv { o, P ◦ , C ◦ , A ◦ } , where c ◦ = p ◦ + a ◦ .In addition, M and M ◦ are characterized as one of the following.Case (a): M = conv { o, A, D, B } and M ◦ = conv { o, A ◦ , B ◦ } , where d = ( a + b ) / (1 +cos ξ ).Case (b): M = conv { o, A, B } and M ◦ = conv { o, A ◦ , D ◦ , B ◦ } , where d ◦ = ( a ◦ + b ◦ ) / (1 +cos ξ ).Thus, it suffices to consider only four cases. Before that, we give some remarks. Remark 5.4. (i) In Case (B), since C ◦ ∈ L ◦ ⊂ π ( K ◦ ), there exists s ∈ R such that c ◦ + t (0 , s, ∈ K ◦ . In the following, we put c := c ◦ + t (0 , s, D ◦ ∈ M ◦ ⊂ π ( K ◦ ), there exists t ∈ R such that d ◦ + t (0 , , t ) ∈ K ◦ . In addition, we have R ‘ V H ( d ◦ + t (0 , , t )) ∈ K ◦ , so that sin ξ ξ t , sin ξ ξ − t ∈ K ◦ hold. The convexity of K ◦ implies d ◦ = sin ξ ξ ∈ K ◦ . Hence, M ◦ = conv { o, A ◦ , D ◦ , B ◦ } ⊂ K ◦ ∩ pos( A, B ) ⊂ π ( K ◦ ) ∩ pos( A, B ) = M ◦ holds, so that we obtain M ◦ = conv { o, A ◦ , D ◦ , B ◦ } = K ◦ ∩ pos( A, B ).25 ase (A) and (a):
Since
P, A ∈ K ◦ and L ◦ = conv { o, P ◦ , A ◦ } , by the convexity of K ◦ , we have L ◦ = conv { o, P ◦ , A ◦ } ⊂ K ◦ ∩ pos( P, A ) ⊂ π ( K ◦ ) ∩ pos( P, A ) = L ◦ , (17)that is, L ◦ = π ( K ◦ ) ∩ pos( P, A ) = K ◦ ∩ pos( P, A ). Similarly, M ◦ = K ◦ ∩ pos( A, B ) holds.By exchanging the roles of K and K ◦ , this case is reduced to Case (B) and (b) below. Case (A) and (b):
We put ˆ K ◦ := o ∗ S K ◦ ( P ◦ , A ◦ , B ◦ ) = K ◦ ∩ pos( P, A, B ). Since L ◦ = K ◦ ∩ pos( P, A ) from(17) and M ◦ = conv { o, A ◦ , D ◦ , B ◦ } from Remark 5.4 (ii), conv { o, P ◦ , A ◦ , D ◦ , B ◦ } ⊂ ˆ K ◦ holds. On the other hand, we have π ( ˆ K ◦ ) ⊂ L ◦ and π H ( ˆ K ◦ ) ⊂ R ‘ ( L ◦ ) for ‘ ≥ K ◦ ⊂ conv { o, P ◦ , A ◦ , D ◦ , B ◦ } for ‘ ≥
4. However, π ( ˆ K ◦ ) ⊂ L ◦ and π H ( ˆ K ◦ ) ⊂ R ‘ ( L ◦ ) do not hold for ‘ = 3. Therefore, we give another proof of ˆ K ◦ =conv { o, P ◦ , A ◦ , D ◦ , B ◦ } which is applicable for all ‘ ≥ K := o ∗ S K ( P, A, B ) = K ∩ pos( P, A, B ). In the same way asˆ K ◦ , we have conv { o, P, C, A, B, R ‘ ( C ) } ⊂ ˆ K . Then, it follows from the D ‘ -symmetries of K and K ◦ that2 ‘ | conv { o, P, C, A, B, R ‘ ( C ) }| ‘ | conv { o, P ◦ , A ◦ , D ◦ , B ◦ }| ≤ | K | | K ◦ | = 2 ‘ − cos ξ ) . (18)On the other hand, by an easy calculation, we have | conv { o, P, C, A, B, R ‘ ( C ) }| = (sin ξ ) / | conv { o, P ◦ , A ◦ , D ◦ , B ◦ }| = (sin ξ ) / ξ ). Hence, the inequality (18), in fact,holds with equality. This means that ˆ K ◦ = conv { o, P ◦ , A ◦ , D ◦ , B ◦ } .Again, by the D ‘ -symmetry, K ◦ is nothing but a regular ‘ -bipyramid such that thecross section of K ◦ with the xy -plane is the regular ‘ -gon with the vertices ( R ‘ ) j ( D ◦ )( j = 0 , . . . , ‘ − Case (B) and (a):
We put ˆ K ◦ := o ∗ S K ◦ ( P ◦ , A ◦ , B ◦ ) = K ◦ ∩ pos( P, A, B ). Then, by the same way as theabove case, conv { o, A ◦ , B ◦ , P ◦ , C ◦ , R ‘ ( C ◦ ) } ⊂ ˆ K ◦ holds. Again, by the same argument inthe above case, we obtain ˆ K ◦ = conv { o, A ◦ , B ◦ , P ◦ , C ◦ , R ‘ ( C ◦ ) } . By the D ‘ -symmetry, K ◦ coincides with an ‘ -regular right prism. Case (B) and (b):
We put ˆ K ◦ := o ∗ S K ◦ ( P ◦ , C , A ◦ , D ◦ , B ◦ , R ‘ ( C )). For each convex body K m ∈ ˇ K ( D ‘ ),we have2 ‘ | K m | (cid:16) C m ( P, A ) + C m ( A, B ) + C m ( B, P ) (cid:17) = 2 ‘ | K m | | L m | sin ξ − cos ξ + | M m | ∈ K ◦ m . As m → ∞ , we obtain q ◦ := 2 ‘ | K | | L | sin ξ − cos ξ + | M | ∈ K ◦ . Now, we calculate the (signed) volume of the solid˜ K ◦ := Q ◦ ∗ ( o ∗ C K ◦ ( P ◦ , C , A ◦ , D ◦ , B ◦ , R ‘ ( C ) , P ◦ )) ⊂ K ◦ .
26e put L ◦ := o ∗ C K ◦ ( P ◦ , C ) and L ◦ := o ∗ C K ◦ ( C , A ◦ ). Then we have C K ◦ ( P ◦ , C ) = | L ◦ |√ s − s , C K ◦ ( R ‘ ( C ) , P ◦ ) = | L ◦ |√ s s cos ξ + sin ξs sin ξ − cos ξ , C K ◦ ( C , A ◦ ) = | L ◦ |√ s − s , C K ◦ ( B ◦ , R ‘ ( C )) = | L ◦ |√ s sin ξ − cos ξs , C K ◦ ( A ◦ , B ◦ ) = | M ◦ | , | L ◦ | = | L ◦ | + | L ◦ |√ s . By a direct calculation, the signed volume of ˜ K ◦ equals q ◦ · | L ◦ | sin ξ − cos ξ + | M ◦ | + s | L ◦ |√ s − (1 − cos ξ )sin ξ = 2 ‘ | K | (2(1 − cos ξ ) | L | | L ◦ | + | M | | M ◦ | ) = ‘ | K | (1 − cos ξ ) = | K ◦ | ‘ . On the other hand, by the D ‘ -symmetry, we have R k‘ ( ˜ K ◦ ) ⊂ K ◦ (0 ≤ k ≤ ‘ −
1) andthe 3-dimensional Lebesgue measure of R k‘ ( ˜ K ◦ ) ∩ R k ‘ ( ˜ K ◦ ) is zero for 0 ≤ k = k ≤ ‘ − K ◦ is | K ◦ | / ‘ as calculated above, the sumof the signed volume of the subsets ˜ K ◦ , R ‘ ( ˜ K ◦ ) , . . . , R ‘ − ‘ ( ˜ K ◦ ) of K ◦ is nothing but thevolume of K ◦ . Consequently, we obtain˜ K ◦ ∪ R ‘ ( ˜ K ◦ ) ∪ · · · ∪ R ‘ − ‘ ( ˜ K ◦ ) = K ◦ , which implies that q ◦ ∈ ∂K ◦ . Thus, the segment Q ◦ A ◦ is on the boundary ∂K ◦ . Sincethe plane { x = 1 } contains the points C ◦ , A ◦ , and D ◦ , the first coordinate of q ◦ is greaterthan or equal to 1. On the other hand, since a · q ◦ ≤
1, the first coordinate of q ◦ equals1. Since | L | = 1 / | M | = (sin ξ ) /
2, and | K | = ( ‘ sin ξ ) /
2, we obtain q ◦ = ‘ | K | sin ξ − cos ξ sin ξ = − cos ξ sin ξ = d ◦ + . By the D ‘ -symmetry, K ◦ is an ‘ -regular right prism with 2 ‘ -vertices ( R ‘ ) k ( Q ◦ ), ( R ‘ ) k V H ( Q ◦ )(1 ≤ k ≤ ‘ ). G = C ‘h Proposition 5.5.
Assume that ‘ ≥ . Let K ∈ K ( C ‘h ) . If P ( K ) = P ( P ‘ ) , then K coincides with P ‘ or P ◦ ‘ up to a linear transformation in G .Proof. By a similar argument in the proof of Proposition 5.3, there exists a sequence { K m } m ∈ N ⊂ ˇ K ( C ‘h ) such that K m → K . We can assume that P, A, B ∈ ∂K m . For each K m , we have P ◦ := Λ m ( P ) = P . However, in this case, we note that A ◦ = A may not27old, where A ◦ := lim m →∞ Λ m ( A ). Since H ( A ) = A , we have H (Λ m ( A )) = Λ m ( A ). By A · Λ m ( A ) = 1, we can represent Λ m ( A ) asΛ m ( A ) = a m ( a m ∈ R ) . Here, let R θ be the rotation through the angle θ about the z -axis. Then, for each m ∈ N there exists θ m ∈ [0 , π ) such that Λ R θm K m ( A ) = . Taking a subsequence if necessary, we can assume θ m → θ for some θ ∈ R . Using R θ m K m and R θ K instead of K m and K , respectively, we may assume that Λ m ( A ) = A . Thus, wecan get the conclusion in a similar way to the proof of Proposition 5.3. References [BF] F. Barthe and M. Fradelizi,
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