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On the Volume of Sections of the Cube

On the Volume of Sections of the Cube Anal. Geom. Metr. Spaces 2021; 9:1–18 Research Article Open Access Grigory Ivanov and Igor Tsiutsiurupa* https://doi.org/10.1515/agms-2020-0103 Received April 17, 2020; accepted December 14, 2020 Abstract: We study the properties of the maximal volume k-dimensional sections of the n-dimensional cube [−1, 1] . We obtain a rst order necessary condition for a k-dimensional subspace to be a local maximizer of the volume of such sections, which we formulate in a geometric way. We estimate the length of the projection of a vector of the standard basis of R onto a k-dimensional subspace that maximizes the volume of the intersection. We nd the optimal upper bound on the volume of a planar section of the cube [−1, 1] , n ≥ 2. Keywords: Tight frame; section of cube; volume; Ball’s inequality MSC: 52A38, 49Q20, 52A40, 15A45 1 Introduction n n The problem of volume extrema of the intersection of the standard n-dimensional cube  = [−1, 1] with a k-dimensional linear subspace H has been studied intensively. The tight lower bound for all n ≥ k ≥ 1 was obtained by J. Vaaler [17], he showed that k n vol  ≤ vol ( \ H). k k A. Akopyan and R. Karasev [1] gave a new proof of Vaaler’s inequality in terms of waists. A deep generalization of Vaaler’s result for ` balls was made by M. Meyer and A. Pajor [15]. K. Ball in [3], using his celebrated version of the Brascamb-Lieb inequality, found the following upper bounds k/2 p n−k n k n k vol ( \ H) ≤ vol  and vol ( \ H) ≤ 2 vol  . (1.1) k k k k The leftmost inequality here is tight if and only if kjn (see [11]), and the rightmost one is tight whenever 2k ≥ n. Thus, if k does not divide n and 2k < n, the maximal volume of a section of  remains unknown. Before proving inequality (1.1), K. Ball obtained a particular case of it in [2], namely, the hyperplane case k = n− 1. Using the Fourier transform, he reduced the hyperplane case to a certain integral inequality. A sim- pler than original proof of this integral inequality was introduced later by F. Nazarov and A. Podkorytov [16]. Also, for the hyperplane case k = n − 1, the rightmost inequality in (1.1) was generalized to certain product measures which include Gaussian type measures by A. Koldobsky and H. König [12]. The Gaussian measure of sections of the cube was studied in [4] and [19]. A. Koldobsky [13] used the Fourier transform of a power of the radial function to express the volume of central hyperplane sections of star bodies in R and conrmed the Meyer and Pajor conjecture from [15] related to the volume of central sections of ` balls. This result was recently generalized by A. Eskenazis [7] (see also [6]). We refer the reader interested in the interaction between Convex Geometry and Fourier Analysis to [14]. In [9], a tight bound on the volume of a section of by a k-dimensional linear subspace was conjectured for all n > k ≥ 1. Grigory Ivanov: Institute of Science and Technology Austria (IST Austria), Kleusteneuburg, Austria; and Laboratory of Combi- natorial and Geometrical Structures, Moscow Institute of Physics and Technology, Moscow, Russia. E-mail: grimivanov@gmail.com *Corresponding Author: Igor Tsiutsiurupa: Laboratory of Combinatorial and Geometrical Structures, Moscow Institute of Physics and Technology, Moscow, Russia. E-mail: tsyutsyurupa.ik@phystech.edu Open Access. © 2021 Grigory Ivanov and Igor Tsiutsiurupa, published by De Gruyter. This work is licensed under the Creative Commons Attribution alone 4.0 License. 2 Ë Grigory Ivanov and Igor Tsiutsiurupa n n Conjecture 1. If the maximal volume of a section of the cube by a k-dimensional subspace of R is attained at a subspace H, then  \ H is an ane cube. k n Let C (n, k) vol  be the maximal volume of a section of  by a k-dimensional subspace L such that n n \ L is an ane cube. Conjecture 1 states that for any k-dimensional subspace H of R , one has n k vol ( \ H) ≤ C (n, k) vol  . k k n n A complete description of a k-dimensional subspace L of R such that the section  \ L is an ane cube of volume C (n, k) vol  is given in the following lemma. Lemma 1.1. The constant C (n, k) is given by l m j k n−kbn/kc k−(n−kbn/kc) n n C (n, k) = . (1.2) k k and is attained at the subspaces given by the following rule. 1. We partitionf1, 2, . . . , ng into k sets such that the cardinalities of any two sets dier by at most one. 2. Let fi , . . . , i g be one of the sets of the partition. Then, choosing arbitrary signs, we write the system of linear equations ±x[i ] = . . . = ±x[i ], where x[i] denotes the i-th coordinate of x in R . 3. Our subspace is the solution of the system of all equations written for each set of the partition at step (2). Since Lemma 1.1 was proven in [9], we provide a sketch of its proof in Appendix A. In this paper, we continue our study of maximizers of G(H) = vol ( \ H), H 2 Gr(n, k) with n > k > 1. (1.3) Using the approach of [10], which is described in detail below, we get a geometric rst order necessary con- dition for H to be a local maximizer of (1.3). Theorem 1.1. Let H be a local maximizer of (1.3), v be the projection of the i-th vector of the standard basis of n n R onto H, i 2 f1, 2, . . . , ng. Denote P =  \ H; we understand P as a k-dimensional polytope in H. Then 1. P = fx 2 H : jhx, v ij ≤ 1g . i=1 2. For every i 2 f1, 2, . . . , ng, v ≠ 0 and the intersection of P with the hyperplanefhx, v i = 1g is a facet of i i P. 3. For every i 2 f1, 2, . . . , ng, the line spanfv g intersects the boundary of P in the centroid of a facet of P. 4. Let F be a facet of P. Denote P = cof0, Fg. Then vol P 1 jv j k F i = , vol P 2 k where the summation is over all indices i 2 f1, 2, . . . , ng such that the line spanfv g intersects F in its centroid. One of the arguments used by K. Ball to prove the rightmost inequality in (1.1) is that the projection of a vector of the standard basis onto a maximizer of (1.3) for 2k ≥ n has length at least 2. We prove the following extension of this result. Theorem 1.2. Let n > k > 1 and H be a global maximizer of (1.3), v be the projection of a vector of the standard basis of R onto H. Then k k ≤ jvj ≤ . n + k n − k On the Volume of Sections of the Cube Ë 3 Using these results and some additional geometric observations, we prove the following. Theorem 1.3. Conjecture 1 is true for k = 2 and n ≥ 3. That is, for any two-dimensional subspace H  R the following inequality holds l mj k n n n 2 Area( \ H) ≤ C (n, 2) vol  = 4 . 2 2 n n This bound is optimal and is attained if and only if  \ H is a rectangle with the sides of lengths 2 and 2 . 2 Denitions and Preliminaries For a positive integer n, we refer to the set f1, 2, . . . , ng as [n]. The standard n-dimensional cube [−1, 1] n n is denoted by  . We use hp, xi to denote the standard inner product of vectors p and x in R . For vectors n n u, v 2 R , their tensor product (or, diadic product) is the linear operator on R dened as (u v)x = hu, xi v d n for every x 2 R . The linear hull of a subset S ofR is denoted by span S. For a k-dimensional linear subspace H ofR and a body K  H, we denote by vol K the k-dimensional volume of K. The two-dimensional volume 2 n of a body K  R is denoted by Area K. We denote the identity operator on a linear subspace H  R by I . If H = R , we use I for convenience. k k + For a non-zero vector v 2 R , we denote by H the ane hyperplanefx 2 R : hx, vi = 1g, and by H − k k and H the half-spacesfx 2 R : hx, vi ≤ 1g andfx 2 R : hx, vi ≥ −1g, respectively. It is convenient to identify a section of the cube with a convex polytope in R . Let fv , . . . , v g be the projections of the standard basis of R onto H. Clearly, \ H = fx 2 H : jhx, v ij ≤ 1g. i=1 n n That means that a section of  is determined by the set of vectorsfv g  H, which are the projections of the orthogonal basis. Such sets of vectors have several equivalent description and names. Denition 1. We will say that an ordered n-tuple of vectors fv , . . . , v g  H is a tight frame (or forms a 1 n tight frame) in a vector space H if v = I , (2.1) i i H where I is the identity operator in H and Aj is the restriction of an operator A onto H. We use Ω(n, k) to denote the set of all tight frames with n vectors in R . Denition 2. An n-tuple of vectors in a linear space H that spans H is called a frame in H. k n In the following trivial lemma we understand R  R as the subspace of vectors, whose last n − k n k n coordinates are zero. For convenience, we will considerfv g  R  R as k-dimensional vectors. Lemma 2.1. The following assertions are equivalent: k k 1. the vectorsfv , . . . , v g  R form a tight frame in R ; 1 n n k 2. there exists an orthonormal basisff , . . . , f g of R such that v is the orthogonal projection of f onto R , 1 n i i for any i 2 f1, 2, . . . , ng; k k 3. spanfv , . . . , v g = R and the Gram matrix Γ of vectorsfv , . . . , v g  R is the matrix of the projection 1 n 1 n operator from R onto the linear hull of the rows of the matrix M = (v , . . . , v ). 1 n 4. the k × n matrix M = (v , . . . , v ) is a sub-matrix of an orthogonal matrix of order n. 1 n 4 Ë Grigory Ivanov and Igor Tsiutsiurupa k k It follows that the tight frames in R are exactly the projections of orthonormal bases onto R . This ob- servation allows us to reformulate the problems in terms of tight frames and associated polytopes in R . Indeed, identifying H with R , we identify the projection of the standard basis onto H with a tight frame n + − fv , . . . , v g 2 Ω(n, k). Thus, we identify  \ H with the intersection of slabs of the form H \ H , 1 n v v i i i 2 f1, 2, . . . , ng. Vice versa, assertion (3) gives a way to reconstruct H from a given tight framefv , . . . , v g 1 n in R . Denition 3. We will say that an n-tuple S = fv , . . . , v g 2 R generates 1 n 1. the polytope + − (S) = H \ H , (2.2) v v i i i=1 which we call the section of the cube generated by S; 2. the matrix v v . We use A to denote this matrix. i i S i=1 To sum up, the global extrema of (1.3) coincide with that of F(S) = vol (S), where S 2 Ω(n, k) with n > k > 1. (2.3) To compare the local extrema of (1.3) and (2.3), we have to dene metrics on Ω(n, k) and on the Grassma- n k nian of k-dimensional subspaces of R . We endow the set of n-tuples of vectors in R with the metric 1/2 dist fv , . . . , v g,fu , . . . , u g = jv − u j . ( 1 n 1 n ) i i i=1 This denes the metric on the set of frames in R consisting of n vectors and on Ω(n, k). The standard metric on the Grassmanian is given by 0 H H Dist(H, H ) = P − P , H H wherek·k denotes the operator norm and P and P are the orthogonal projections onto the k-dimensional subspaces H and H , respectively. It was shown in [10] that the local extrema of (2.3) coincide with that of (1.3). However, we note that there is an ambiguity when we identify H with R . Any choice of orthonormal basis of H gives its own tight frame in R , all of them are isometric but dierent from each other. It is not a problem Ω(n,k) as there exists a one-to-one correspondence between Gr(n, k) and , where O(k) is the orthogonal group O(k) in dimension k. And, clearly, F(S ) = F(S ) whenever S = U(S ) for some U 2 O(k). 1 2 2 1 From now on, we will study properties of the maximizers of (2.3) and work with tight frames. 3 Operations on frames The following approach to problem (2.3) was proposed in [10] and used in [9] to study the properties of pro- jections of the standard cross-polytope. The main idea is to transform a given tight frame S into a new one S and compare the volumes of the sections of the cube generated by them. Since it is not very convenient to transform a given tight frame into another one, we add an intermediate step: we transform a tight frame S into a frame S, and then we transform S into a new tight frame S using a linear transformation. The main observation here is that we can always 0 0 ˜ ˜ transform any frame S = fv , . . . , v g into a tight frame S using a suitable linear transformation L: S = LS = 1 n fLv , . . . , Lv g. Equivalently, any non-degenerate centrally symmetric polytope in R is an ane image of a 1 n section of a high dimension cube. For a frame S in R , by denition put B = A = v v . v2S On the Volume of Sections of the Cube Ë 5 The operator B is well-dened as the condition span S = R implies that A is a positive denite operator. S S Clearly, for any frame S, B maps S to the tight frame B S: S S X X B v B v = B v v B = B A B = I . S S S S S S S k v2S v2S We obtain the following necessary and sucient condition for a tight frame to be a maximizer of (2.3). Lemma 3.1. The maximum of (2.3) is attained at a tight frame S 2 Ω(n, k) i for an arbitrary frame S in R inequality vol (S) 1 ≤ (3.1) vol (S) det A holds. ˜ ˜ ˜ ˜ Proof. For any frame S, we have that B S is a tight frame and vol (B S) = vol (S)/ det B . The maximum ˜ k ˜ k ˜ S S S ˜ ˜ of (2.3) is attained at a tight frame S i vol (B S) ≤ vol (S) for an arbitrary frame S. Hence k ˜ k ˜ ˜ vol (B S) vol (S) 1 vol (S) k ˜ S k k 1 ≥ = = det A vol (S) det B vol (S) vol (S) k ˜ k k Dividing by det A , we obtain inequality (3.1). ˜ ˜ Clearly, if S in the assertion of Lemma 3.1 is close to S, then the tight frame B S is close to S as well. Therefore, inequality (3.1) gives a necessary condition for local maximizers of (2.3). Let us illustrate how we will use it. Let S be an extremizer of (2.3), and T be a map from a subset of Ω(n, k) to the set of frames in R . In order to obtain properties of extremizers, we consider a composition of two operations: T ˜ S 0 S ! S ! S , where B is as dened above. For example, see Figure 1, where T is the operation of replacing a vector v of S by the origin. v v T : v → 0 B H H v v !(S) !(S ) !(S) −H −H v v Figure 1: A frame S  R consists of three vectors and (S) is a hexagon. A vector v 2 S is mapped to the origin, yielding ˜ ˜ ˜ the frame S. By construction, (S) is a parallelogram. Since the frame S = B S is a tight frame with only two nonzero vectors, these vectors are two orthogonal unit vectors. Hence (S ) is a square. Choosing a simple operation T, we may calculate the left-hand side of (3.1) in some geometric terms. We consider several simple operations: Scaling one or several vectors, mapping one vector to the origin, mapping one vector to another. On the other hand, the determinant in the right-hand side of (3.1) can be calculated for the operations listed above. In particular, the following rst-order approximation of the determinant was obtained by the rst author in [10, Theorem 1.2]. We provide a sketch of its proof in Appendix A. 6 Ë Grigory Ivanov and Igor Tsiutsiurupa Lemma 3.2. Let S = fv , . . . , v g  R be a tight frame and the n-tuple S be obtained from S by substitution 1 n v ! v + tx , where t 2 R, x 2 R , i 2 [n]. Then i i i i det A = 1 + t hx , v i + o(t). ˜ i i i=1 We state as lemmas several technical facts from linear algebra that will be used later. k k Lemma 3.3. Let A be a positive denite operator on R . For any u 2 R , we have −1/2 2 det (A ± u u) = 1 ± jA uj det A. Proof. We have −1/2 −1/2 det (A ± u u) = det A · det I ± A u A u . −1/2 −1/2 −1/2 Diagonalizing the operator I ± A u A u, we see that its determinant equals 1 ± A u . This completes the proof. For any n-tuple S of vectors of R with v 2 S, we use S \ v to denote the (n − 1)-tuple of vectors obtained from S by removing the rst occurrence of v in S. Lemma 3.4. Let S 2 Ω(n, k) and v 2 S be a vector such thatjvj < 1. Then S \ v is a frame in R and B is the S\v −1/2 k 2 k stretch of R by the factor 1 −jvj along spanfvg. In particular, for any u 2 R , we have B u ≥ juj. S\v Proof. Since A = I − v v > 0, we have that S \ v is a frame in R . Clearly, A stretches the space by S\v k S\v 2 −1/2 the factor 1 −jvj along spanfvg. Therefore, the operator B = A stretches the space by the factor S\v S\v −1/2 1 −jvj along the same direction. 4 Properties of a global maximizer Theorem 1.2 is formulated in terms of subspaces. For the sake of convenience, we introduce its equivalent reformulation in terms of tight frames. Theorem 4.1 (Frame version of Theorem 1.2). Let S 2 Ω(n, k) be a global maximizer of (2.3) for n > k > 1. Pick an arbitrary v 2 S. Then k k ≤ jvj ≤ . (4.1) n + k n − k Proof. The denition of tight frame implies that jpj = k. Hence there is a vector u 2 S such that p2S 2 2 juj ≤ k/n and a vector w 2 S such thatjwj ≥ k/n. We start with the rightmost inequality in (4.1). Let S be the n-tuple obtained from S by substitution u ! v. 2 k Sincejuj ≤ k/n < 1, we have A = I −u u +v v ≥ I −u u > 0. Hence S is a frame in R . By identity (2.2), k k ˜ ˜ inclusion (S)  (S) holds. Therefore, vol (S) ≤ vol (S). Using Lemma 3.1 and Lemma 3.3, we obtain k k 1 ≥ det A = 1 + B v 1 −juj . ˜ S\u 2 2 By Lemma 3.4, the right-hand side of this inequality is at least 1 +jvj 1 −juj . Therefore, juj k jvj ≤ ≤ . 1 −juj n − k Let us prove the leftmost inequality in (4.1). There is nothing to prove ifjvj ≥ k/n. Assume thatjvj < k/n. ˜ ˜ Let S be the n-tuple obtained from S by substitution v ! w. Since A = I−v v +u u > 0, S is a frame in R . S On the Volume of Sections of the Cube Ë 7 ˜ ˜ By identity (2.2), the inclusion (S)  (S) holds. Therefore, vol (S) ≤ vol (S). Using Lemma 3.1 and k k Lemma 3.3, we obtain 1 ≥ det A = 1 + B w 1 −jvj . S\v 2 2 Again, by Lemma 3.4, the right-hand side of this inequality is at least 1 +jwj 1 −jvj . It follows that jwj k jvj ≥ ≥ . 1 +jwj n + k This completes the proof. Clearly, Theorem 4.1 implies Theorem 1.2. These theorems can be sharpened in the case of sections by planes, i.e. k = 2. Lemma 4.1. Let S 2 Ω(n, 2) be a maximizer of (2.3) for k = 2 and n ≥ 3, let v 2 S. Then 2 2 ≤ jvj ≤ . (4.2) n + 1 n − 1 Proof. Lemma 1.1 provides an example of a two-dimensional subspace H satisfying Area  \ H = 4C (n, 2) = 4 dn/2ebn/2c. Therefore, the maximal area of a planar section of the cube  is at least 4 dn/2ebn/2c. Thus, we have Area(S) ≥ 4 dn/2ebn/2c. (4.3) Let us prove the leftmost inequality in (4.2). It is trivial if jvj ≥ 2/n. Assume that jvj < 2/n. Let S 2 Ω(n − 1, 2) be a maximizer of (2.3). By Ball’s inequality (1.1), we have Area(S ) ≤ 2(n − 1). (4.4) Consider S \ v. It is a frame by Lemma 3.4. Then, by Lemma 3.1 and Lemma 3.3, we get Area(S \ v) 1 1 ≤ p = p . Area(S ) det A 1 −jvj S\v By identity (2.2), we have (S)  (S \ v). By this and by inequalities (4.4) and (4.3), we get Area(S \ v) Area(S) 4 dn/2ebn/2c ≥ ≥ . 0 0 Area(S ) Area(S ) 2(n − 1) Combining the last two inequalities, we obtain Area(S ) n − 1 2 jvj ≥ 1 − ≥ 1 − p ≥ . Area(S) n + 1 2 dn/2ebn/2c We proceed with the rightmost inequality in (4.2). Let S 2 Ω(n + 1, 2) be a maximizer of (2.3). By Ball’s inequality (1.1), we have Area(S ) ≤ 2(n + 1). (4.5) We use S to denote the (n + 1)-tuple obtained from S by concatenating S with the vector v. Since S is a frame 2 2 in R , S is a frame in R as well. By Lemma 3.1 and Lemma 3.3, we get Area(S) 1 1 ≤ p = p . Area(S ) det A 1 +jvj By identity (2.2), we have (S) = (S). By this and by inequalities (4.5) and (4.3), we get Area(S) Area(S) 4 dn/2ebn/2c = ≥ . 0 0 Area(S ) Area(S ) 2(n + 1) 8 Ë Grigory Ivanov and Igor Tsiutsiurupa Combining the last two inequality, we obtain Area(S ) n + 1 2 jvj ≤ − 1 ≤ p − 1 ≤ . Area(S) n − 1 2 dn/2ebn/2c Remark 1. It is possible to sharpen inequality (4.1) for k > 2 and n > 2k using the same approach as in Lemma 4.1. The idea is to remove n mod k from or add n − (n mod k) vectors to a maximizer and compare the volume of a section of the cube generated by the new frame with the Ball bound (1.1). However, it doesn’t give a substantial improvement. Remark 2. As was pointed out to us by the anonymous reviewer, the results of this and the next section might be proven using the so-called shadow movement technique. Moreover, we expect that combining the result on convexity of the function related to the polar body of a symmetric shadow system [5] and suitable discrete substitutions, one could conrm Conjecture 1. 5 Local properties In this section, we prove some properties of the local maximizers of (2.3). We will perturb facets of (S) of a local maximizer S (that is, we will perturb the vectors of S in a specic way corresponding to a perturbation of some facets of the polytope (S)). To this end, we need to recall some general properties of polytopes connected to perturbations of a half-space supporting a polytope in its facet. 5.1 Properties of polytopes Recall that a point c is the centroid of a facet F of a polytope P  R if c = xdλ, (5.1) vol F k−1 where dλ is the standard Lebesgue measure on the hyperplane containing F. k + For a set W  R , we use P(W) to denote the polytopal set\ H . Let W be a set of pairwise distinct w2W w vectors such that • the set P(W) is a polytope; • for every w 2 W, the hyperplane H supports P(W) in a facet of P(W). That is, W is the set of scaled outer normals of P(W). Denote P = P(W). We x w 2 W and the facet F = P\H of P. Let c be the centroid of F. Transformation 1. We will “shift” a facet of a polytope parallel to itself. Let W be obtained from W by sub- w 0 0 0 stitution w ! w + h , where h 2 R. Denote P = P(W ). That is, the polytopal set P is obtained from P by jwj the shift of the half-space H by h in the direction of its outer normal. By the celebrated Minkowski existence and uniqueness theorem for convex polytopes (see, for example, [8, Theorem 18.2]), we have vol P − vol P = h vol F + o(h). (5.2) k k k−1 Transformation 2. We will rotate a facet around a codimension two subspace. Let u be a unit vector orthogo- nal to w. Dene c = H \ spanfwg and L = H \ (u + c ). Note that L is a codimension two ane w w u w w u subspace of R and an ane hyperplane in H . Clearly, for any non-zero t 2 R, L = H \ H = w u w w+tu fx 2 H : hx − c(w), ui = 0g . Thus, L divides F into two parts w u + − F = F \fx 2 H : hx − c(w), ui ≥ 0g and F = F \fx 2 H : hx − c(w), ui ≤ 0g w w On the Volume of Sections of the Cube Ë 9 H H H H H w w+u w w+u w w → w + u w → w + u h< 0 h> 0 ! ! P P Figure 2: Parallel shift of the facet F by the vector u = hw/jwj + − (one of the sets F or F is empty if c(w) 2̸ F). Let α 2 (−π/2, π/2) be the oriented angle between hyperplanes H and H such that α is positive for positive t. w w+tu Let W be obtained from W by substitution w ! w + tu, where t 2 R and u is a unit vector orthogonal 0 0 0 to w. Denote P = P(W ). Thus, for a suciently smalljtj, the polytopal set P is a polytope obtained from P by the rotation of the half-space H around the codimension two ane subspace L by some angle α = α(t). Clearly, in order to calculate the volume of P , we need to subtract from vol P the volume of the subset of P that is above H and to add to vol P the volume of the subset of P that is above H . Formally speaking, w+tu w denote 8 8 0 k + k + < < P \ R \ H for α ≥ 0 P\ R \ H for α ≥ 0 w w+tu + − Q = and Q = k + 0 k + : : P\ R \ H for α < 0 P \ R \ H for α < 0 w+tu w Then, we have (see Figure 3) 0 + − vol P − vol P = sign α vol Q − vol Q . (5.3) k k k k + − + − + There is a nice approximation for vol Q − vol Q . Let C (resp., C ) be the set swept out by F (resp., α α k k F ) while rotating around L by the angle α. By routine, +∞ vol C = jαj r vol F \ (L + ru) dr k α k−2 u 0 1 +∞ 0 Z Z @ A resp., vol C = jαj r vol F \ (L − ru) dr = −jαj r vol F \ (L + ru) dr . α u u k k−2 k−2 0 −∞ We claim that + + − − vol Q = vol C + o(α) and vol Q = vol C + o(α). (5.4) k k α k k α + + + This can be shown in the following way. If F is not a body in H , then vol Q = vol C = 0. As- k k α sume F is a body in H . Since a polytope is dened by a system of linear inequalities, there is a posi- tive constant b such that for any suciently small τ the Hausdor distance on H between Q \ H w+τu w+τu + d and C \ H is at most bα (recall that the Hausdor distance between A, B  R is d (A, B) = w+τu n α o H d d d inf ε > 0 : A  B + εB and B  A + εB , where B is the Euclidean unit ball). Note that the section of + + C by H is F rotated around L by some small angle β = β(τ). Therefore, there exist two homothets α w+τu u + + + F  H and F  H of F satisfying two properties: w w 1 2 + + + + • both the Hausdor distance between F and F and the Hausdor distance between F and F are at 1 2 ˜ ˜ most bα for a positive constant b; + + • the body swept out by F while rotating around L by the angle α is contained in Q and the body swept + + out by F contains Q . 2 10 Ë Grigory Ivanov and Igor Tsiutsiurupa + + By the rst property and a simple integration, the volumes of the bodies swept out by F and F coincide with 1 2 vol C up to the terms of the rst order of α. The second property implies the leftmost identity in (5.4). The k α rightmost one is obtained similarly. By identities (5.3) and (5.4), we obtain 0 + − vol P − vol P = α vol C − vol C + o(α) = α r vol F \ (L + ru) dr + o(α). k k k α k α k−2 u By this and by (5.1), we get vol P − vol P = αhc − c , ui vol F + o(α). k k k−1 Since w and u are orthogonal, we have juj juj α = arctan t = t + o(t). jwj jwj Finally, we obtain vol F 0 k−1 vol P − vol P = hc − c , ui t + o(t). (5.5) k k jwj H H H H H w w+tu w w+tu w F F L w u w → w + tu w → w + tu L − t> 0 t< 0 " " P P Figure 3: A rotation of the facet F around L 5.2 Local properties of sections of the cube Let S be a frame in R . For every v 2 S, we denote the set H \(S) by F . We say that v 2 S corresponds to a v v facet F of (S) if either F = F or F = −F . Clearly, if some vectors of S correspond to the same facet of (S), v v k k k then they are equal up to a sign. For a given frame S in R and u 2 R , a facet F of (S) and a vector u 2 R , we dene an F-substitution in the direction u as follows: • each vector v of S such that F  H is substituted by v + u; • each vector v of S such that −F  H is substituted by v − u; • all other vectors of S remain the same. In order to prove Theorem 1.1, we will use F-substitutions. At rst, we simplify the structure of a local maximizer. Lemma 5.1. Let S be a local maximizer of (2.3) and v 2 S. Then F is a facet of (S). Proof. Let K be a convex body in R , then its polar body is dened by n o y 2 R : hy, xi ≤ 1 for all x 2 K . On the Volume of Sections of the Cube Ë 11 Since (S) is the intersection of half-spaces of the formfhw, xi ≤ 1g with w 2 ±S, we have that cof±Sg is polar to (S) in span S = R . By the duality argument, it suces to prove that v 2 S is a vertex of the polytope cof±Sg . Assume that v is not a vertex of cof±Sg . Clearly, v 2 cof± S \ v g and v is not a vertex of the polytope cof± S \ v g . Therefore we have that ( ) ( ) k k spanfS\vg = span S = R . That is, S\v is a frame in R . Since B is a nondegenerate linear transformation, S\v B v is not a vertex of the polytope co ±B (S \ v) . By this and by the triangle inequality, there is a vertex u S\v S\v of cof±Sg such that u 2 S and B v < B u . S\v S\v Denote by S the n-tuple obtained from S by substitution v ! v + t(u − v), where t 2 (0, 1]. Since ˜ ˜ A ≥ A > 0, S is a frame in R . By the choice of u and identity (2.2), we have (S) = (S). Hence S\v vol (S) = vol (S). Lemma 3.1 implies that det A ≤ 1. k k ˜ On the other hand, by Lemma 3.3, we have det A = 1 + B (v + t(u − v)) 1 −jvj . S\v Inequality B v < B u implies that B v < B (v + t(u − v)) . By this and by Lemma 3.4, we conclude S\v S\v S\v S\v that det A > 1. This is a contradiction. Thus, v is a vertex of cof±Sg. The lemma is proven. As an immediate corollary of Lemma 5.1 and by the standard properties of polytopes, we have the follow- ing statement. Corollary 5.1. Let S be a local maximizer of (2.3) and v 2 S. Let S(t) be the n-tuple obtained from S by F - substitution in the direction tu with t 2 R and u 2 R . Then, for a suciently small jtj, S(t) is a frame, the ˜ ˜ vector v + ut corresponds to a facet of (S(t)). Moreover, vol (S(t)) is a smooth function of t at t = 0. In the following two lemmas, we will perturb a local maximizer by making F-substitutions. Geometrically speaking, making an F-substitution in the direction tu with u 2 R and t 2 R, we move the opposite facets F and −F of a local maximizer in a symmetric way. Thus, for a suciently small t, perturbations of the facets F and −F are independent. Lemma 5.2. Let S be a local maximizer of (2.3). Let v 2 S and d be the number of the vectors of S that corre- spond to F . Then vol F = djvj vol (S). (5.6) k−1 k jvj Proof. Denote by S the n-tuple obtained from S by F -substitution in the direction tv with t 2 R. Thus, we apply Transformation 1 to the facets ±F of (S). By Lemma 3.1, we have vol (S) 1 ≤ p . (5.7) vol (S) det A k ˜ By Lemma 3.2 and Corollary 5.1, both sides of this inequality are smooth as functions of t in a suciently small neighborhood of t = 0. Consider the Taylor expansions of both sides of inequality (5.7) as functions of t about t = 0. 2 2 By Lemma 3.3, det A = 1 + d(2t + t )jvj . Hence p = 1 − tdjvj + o(t). (5.8) det A + − Geometrically speaking, we shift the half-space H (resp., H ) by v v 1 1 t h = − = − + o(t) j(1 + t)vj jvj jvj in the directions of its outer normal. By this and by (5.2), we obtain 2t vol (S) − vol (S) = − vol F + o(t). (5.9) k k k−1 v jvj 12 Ë Grigory Ivanov and Igor Tsiutsiurupa Using identities (5.9) and (5.8) in (5.7) , we get 2t vol F v 2 k−1 1 − ≤ 1 − djvj t + o(t). jvj vol (S) Since S = S for t = 0 and the previous inequality holds for all t 2 (−ε, ε) for a suciently small ε, the coecients of t in both sides of the previous inequality coincide. That is, 2 vol F k−1 2 = djvj . jvj vol (S) This completes the proof. Lemma 5.3. Let S be a local maximizer of (2.3) and v 2 S. Then the line spanfvg intersects the hyperplane H in the centroid of the facet F . Proof. Denote the centroid of F by c and let c = spanfvg\H . Fix a unit vector u orthogonal to v. Denote by S v v v the n-tuple obtained from S by F -substitution in the direction tu with t 2 R. Thus, we apply Transformation 2 to the facets ±F of (S). By Lemma 3.1, we have vol (S) 1 ≤ p . (5.10) vol (S) det A k ˜ By Lemma 3.2 and Corollary 5.1, both sides of this inequality are smooth as functions of t in a su- ciently small neighborhood of t = 0. Consider the Taylor expansions of both sides of inequality (5.10) as functions of t about t = 0. By (5.5), we obtain vol (S) − vol (S) = Chc − c , ui t + o(t), k k where C = 2 vol F /jvj > 0. By Lemma 3.2, det A = 1 + o(t). Therefore, inequality (5.10) takes the k−1 ˜ following form 1 + hc − c , ui t + o(t) ≤ 1 + o(t). vol (S) Since S = S for t = 0 and the previous inequality holds for all t 2 (−ε, ε) for a suciently small ε, the coecients of t in both sides of the previous inequality coincide. That is, we conclude hc − c , ui = 0. Since c, c 2 H and the last identity holds for all unit vectors parallel to H , it follows that c = c . The v v v v lemma is proven. As a simple consequence of Lemma 5.3, we obtain the following result for the planar case. Theorem 5.1. Let S 2 Ω(n, 2) be a local maximizer of (2.3) for k = 2. Then, the polygon (S) is cyclic. That is, there is a circle that passes through all the vertices of (S). Proof. Denote the origin by o. Let ab be an edge of (S) and oh be the altitude of the triangle abo. By Lemma 5.3, h is the midpoint of ab. Hence, the triangle abo is isosceles and ao = bo. It follows that (S) is cyclic. We are ready to give a proof of Theorem 1.1. Proof of Theorem 1.1. Recall that any identication of H with R identies the projections of the standard basisfv , . . . , v g with a tight frame, denoted by S, that is a local maximizer of (2.3). 1 n Next, assertion 1 is trivial and holds for any section of the cube. Assertion 2 and assertion 3 are equivalent to Lemma 5.1 and Lemma 5.3, respectively. On the Volume of Sections of the Cube Ë 13 By Lemma 5.1, all vectors v 2 S such that span v intersects F correspond to F and have the same length that we denote byjvj. Then by Lemma 5.3, the span of each of these vectors intersects F in its centroid. Since the length of the altitude of the pyramid P is 1/jvj, we have 1 vol F k−1 vol P = . k F k jvj Hence assertion 4 follows from Lemma 5.2. 6 Proof of Theorem 1.3 We use the setting of tight frames developed in the previous sections to prove the theorem. More precisely, we use the obtained necessary conditions for a tight frame in R that maximizes (2.3) for n > k = 2 to prove that the section of the cube generated by the tight frame is a rectangle of area 4C (n, 2) = 4 dn/2ebn/2c p p with the sides of lengths 2 bn/2c and 2 dn/2e. First, let us introduce the notation. Let S = fv , . . . , v g 2 Ω(n, 2) be a global maximizer of (2.3) for 1 n k = 2 and n > 2. Clearly, (S) is a centrally symmetric polygon in R . The number of edges of (S) is denoted by 2f . Clearly, f ≤ n. By Theorem 5.1, the polygon (S) is cyclic; and we denote its circumradius by R. Let F , . . . F be the edges of (S) enumerated in clockwise direction (that is, edges F and F are opposite to 2f i i+f each other, i 2 [f]). We reenumerate the vectors of S in such a way that the vector v corresponds to the edge F for every i 2 [f]. The central angle subtended by the edge F is denoted by 2φ , i 2 [f ]. i i i Clearly, we have the following identities (see Figure 4): φ + ··· + φ = , (6.1) 1 f R cos φ = for all i 2 [f ], (6.2) jv j and Area(S) = R sin 2φ . (6.3) i=1 Also, we note here that l mj k n n Area(S) ≥ 4C (n, 2) = 4 . (6.4) 2 2 R v !(S) Figure 4: Notation for (S) There are several steps in the proof. We explain the main steps briey. In fact, we want to show that the number of edges of a local maximizer is 2f = 4. Using the discrete isoperimetric inequality (see below), we obtain an upper bound on Area(S) in terms of f . This upper bound yields the desired result for n ≥ 8 (the 14 Ë Grigory Ivanov and Igor Tsiutsiurupa bound is less than conjectured volume 4C (n, 2) for n ≥ 8). Finally, we deal with the lower-dimension cases using the necessary conditions obtained earlier. The discrete isoperimetric inequality for cyclic polygons says that among all cyclic f -gons with xed cir- cummradius there is a unique maximal area polygon – the regular f -gon. We will use a slightly more general form. Namely, xing one or several central angles of a cyclic polygon, its area is maximized when all other central angles are equal. In our notation xing the central angle φ and its vertically opposite, we have π π − 2φ 2 2 i R f sin ≥ R sin 2φ + (f − 1) sin ≥ 4C (n, 2). (6.5) f f − 1 This inequality immediately follows from the Jensen inequality and concavity of the sine function on [0, π]. 6.1 Step 1. Claim 1. The area of(S) such that f = 2 is at most 4C (n, 2). The bound is attained when(S) is a rectangle p p with the sides of lengths 2 dn/2e and 2 bn/2c. Proof. Since f = 2, the polygon (S) is an ane square. Hence the claim is an immediate consequence of Lemma 1.1. Thus, it suces to prove that f = 2 for any n > 2. 6.2 Step 2. Claim 2. For any n > 2 the following inequality holds n + 1 1 R ≤ . (6.6) 2 cos 2f Proof. Let φ be the smallest central angle. By identity (6.1), we have cos φ ≥ cos . Combining this with 1 1 2f the leftmost inequality in (4.2) and identity (6.2), we obtain 1 n + 1 1 R = ≤ . 2 2 2 jv j cos φ 2 cos 1 1 2f Claim 3. For any n > 2 the following inequality holds l mj k π 4 n n f tan ≥ . (6.7) 2f n + 1 2 2 Proof. By the discrete isoperimetric inequality (6.5), we have R f sin ≥ 4C (n, 2). Combining this with inequalities (6.6) and (6.4), we obtain l mj k sin 2 8 n n f ≥ C (n, 2) = . cos n + 1 n + 1 2 2 2f The claim follows. Claim 4. The following bounds on f hold: 1. f = 2 if n ≥ 8; 2. f ≤ 3 if n = 7; On the Volume of Sections of the Cube Ë 15 3. f ≤ 4 if n = 5. Proof. We consider the functions in the left- and right-hand sides of (6.7) as functions of f and n respectively. π 4 Set g(f) = f tan and h(n) = bn/2cdn/2e. Thus, inequality (6.7) takes the form g(f ) ≥ h(n). By routine 2f n+1 analysis, we have that g is strictly decreasing and h is increasing onfn 2 N : n ≥ 2g. The rst two assertions of the claim follows from this and the identity g(3) = h(7). Inequality f ≤ 4 for n = 5 follows from the direct computations of g(5), g(4) and h(5) (see Figure 5). f 2 3 4 5 n 5 6 7 p p p p p g(f) 2 3 4 2 − 1 5 5 − 2 5 h(n) 2 6/3 12/7 3 Figure 5: Some values of g and h Theorem 1.3 is proven for n ≥ 8. We proceed with the lower-dimensional cases. 6.3 Step 3. Claim 5. For n = 7, we have that f = 2. Proof. We showed that f ≤ 3 for n = 7. Assume that f = 3. We see that inequality (6.7) with such values turns into an identity. It follows that φ = φ = φ = π/6 and (S) is a regular hexagon. Hence the vectors of S 1 2 3 2 2 2 1 are of the same length. Since jvj = tr I = 2, we conclude thatjvj = 2/7 and R = = 14/3. 2 2 2 jv j cos π/6 v2S However, the volume of such a hexagon is strictly less than 4C (7, 2). We conclude that f = 2 for n = 7. Claim 6. For n 2 f3, 4, 6g, we have that f = 2. Proof. For n = 3, the statement is a simple exercise (see [18]). Conjecture 1 was conrmed in [11] for any n > k ≥ 1 such that kjn, in particular, for k = 2 and n 2 f4, 6g. Remark 3. The inequality on the area for n 2 f4, 6g and k = 2 is a special case of the leftmost inequality in (1.1) originally proved by K. Ball [3]. In [11], the equality cases in this Ball’s inequality are described. 6.4 Step 4. Claim 7. Let n = 5 and either f = 3 or f = 4. Then φ ≤ π/4 for every i 2 [f]. Proof. Assume that there is i 2 [f ] such that φ > π/4. Thus, cos φ < 1/ 2. Using identity (6.2) for i and for i i any j 2 [f], we have that r r cos φ jv j n + 1 3 j i = ≤ = , cos φ jv j n − 1 2 i j p p where the inequality follows from Lemma 4.1. Hence cos φ ≤ 3/2 cos φ < 3/2 and, therefore, φ > π/6. j i j This contradicts identity (6.1): π π π π = φ + ··· + φ > + (f − 1) > . 2 4 6 2 Thus, φ ≤ π/4 for every i 2 [f ]. Claim 8. Let n = 5 and either f = 3 or f = 4. Then φ ≥ π/10 for every i 2 [f]. i 16 Ë Grigory Ivanov and Igor Tsiutsiurupa Proof. Fix i 2 [f]. By identity (6.2) and Lemma 4.1, we have R ≤ . By inequality (6.5), we get 2 cos φ 3 π − 2φ sin 2φ + (f − 1) sin ≥ 4C (5, 2) = 4 6. cos φ f − 1 The function of φ in the left-hand side of this inequality is increasing on [0, π/2]. Since the inequality does not hold for φ = π/10, we conclude that φ is necessarily at least π/10. i i Claim 9. For n = 5, we have that f = 2. Proof. Assume that either f = 3 or f = 4. Denote by d the number of vectors in S that correspond to F . We i i want to rewrite the inequality of Lemma 5.2 using the circumradius and the center angle. Since the length of edge F is 2R sin φ and by identity (6.2), identity (5.6) takes the form i i 2 i 2R sin 2φ = Area(S). 2 2 R cos φ Set q(φ) = cos φ sin 2φ. Then for all i, j 2 [f], we have d q(φ ) i i = . d q(φ ) j i Clearly, there are i, j 2 [f ] such that d = 2 and d = 1. Therefore, q(φ )/q(φ ) = 2. By Claim 7 and Claim 8, i j i j we have that φ , φ 2 [π/10, π/4]. By simple computations, the maximum of q on the segment [π/10, π/4] i j p p is q(π/6) = 3 3/8 and the minimum is q(π/4) = 1/2. Hence max q(φ)/q(ψ) = 3 3/4 < 2 and we φ,ψ2[π/10,π/4] come to a contradiction. Thus, f = 2. Thus, we have proved that for a maximizer of (2.3), then f = 2. By Claim 1, the conjectured upper bound for the area of a planar section holds and also is tight. The proof of Theorem 1.3 is complete. Remark 4. We used the Ball inequality (1.1) to prove Theorem 1.3 for n 2 f3, 4, 6g. However, it can be done by using our approach without the Ball inequality. The proof is technical. Since it is not of great interest, we do not give a proof. A Sketches of proofs Sketch of the proof of Lemma 1.1. Let H be a k-dimensional subspace of R such that C (n, k) is attained. Denote P =  \ H. Since P is an ane k-dimensional cube, there are vectors fa , . . . , a g such that P = + − H \ H . a a i2[k] i i Letfv , . . . , v g be the projection of the vectors of the standard basis onto H. By the same arguments as 1 n in Lemma 5.1, the hyperplane H meets the polytope P in a facet of P for every i 2 [n]. Thus, v coincides i i with ±a for a proper sign and j 2 [k]. Or, equivalently, we partition [n] into k sets and H is the solution of a proper system of linear equations constructed as in (2) and (3), except we have not proved that (1) holds yet. Let us prove this assertion. Let d vectors of the standard basis of R project onto a pair ±a . Therefore, a i i k-tuple of vectorsf d a g is a tight frame. Identifying H with R and by the assertion (4) of Lemma 2.1, i i i2[k] 2 1 we conclude that a and a are orthogonal whenever i ≠ j. Therefore,ja j = and i j i n k vol  \ H = 2 d · . . . · d . (A.1) k 1 k Suppose d ≥ d + 2 for some i, j 2 [k]. Then d · d ≤ (d − 1)(d + 1). By this and by (A.1), we showed that i j i j i j (1) holds. It is easy to see that there are exactly n − kbn/kc of d ’s equaldn/ke and all others k − (n − kbn/kc) are equal tobn/kc. That is, C (n, k) is given by (1.2). This completes the proof. On the Volume of Sections of the Cube Ë 17 Sketch of the proof of Lemma 3.2. Recall that the cross product of k − 1 vectors fx , . . . , x g of R is the 1 k−1 vector x dened by hx, yi = det(x , . . . , x , y) for all y 2 R . 1 k−1 [n] For an ordered (k − 1)-tuple L = fi , . . . , i g 2 and a frame S = fv , . . . , v g, we use [v ] to denote 1 k−1 1 n L k−1 the cross product of v , . . . , v . i i 1 k−1 We claim the following property of the tight frames. k k Let S = fv , . . . , v g be a tight frame in R . Then the set of vectorsf[v ]g is a tight frame in R . [n] 1 n L L2 ( ) k−1 k n n We use Λ (R ) to denote the space of exterior k-forms on R . By assertion (2) of Lemma 2.1, there exists n n k an orthonormal basis ff g of R such that v is the orthogonal projection of f onto R , for any i 2 [n]. i 1 i i Then the (k − 1)-form v ^ ··· ^ v is the orthogonal projections of the (k − 1)-form f ^ ··· ^ f onto i i i i 1 k−1 1 k−1 k−1 k k−1 n [n] Λ (R )  Λ (R ), for any ordered (k − 1)-tuple L = fi , . . . , i g 2 . By Lemma 2.1 and since the 1 k−1 k−1 k−1 n (k − 1)-formsff ^ ···^ f g form an orthonormal basis of Λ (R ), we have that the set of [n] i i 1 k−1 fi ,...,i g2 1 k−1 ( ) k−1 k−1 k (k − 1)-forms fv ^ ··· ^ v g is a tight frame in Λ (R ). Finally, the Hodge star operator [n] i i 1 k−1 fi ,...,i g2 1 k−1 ( ) k−1 maps v ^···^ v to the cross product of vectors v , . . . , v . Since the Hodge star is an isometry, the set i i i i 1 k−1 1 k−1 of cross productsf[v ]g is a tight frame. The claim is proven. [n] L22 ( ) k−1 By linearity of the determinant, it is enough to prove the lemma for S = fv + tx, v , . . . , v g. Denote 1 2 n 0 0 v = v + tx and v = v , for 2 ≤ i ≤ n. 1 i 1 i By the Cauchy–Binet formula, we have 0 1 X X X 0 0 0 0 @ A det A = det v v = det v v . (A.2) i i i i [n] 1 i2Q Q2 ( ) By the properties of the Gram matrix, we have 0 0 0 0 det v v = det v , . . . , v . i i i i 1 k 1 k By this, by the denition of cross product and by identity (A.2), we obtain det A = 1 + 2t v , [v ] [v ], x + o(t). ˜ 1 Q\1 Q\1 [n] Q2 ,12Q ( ) [n] Sincehv , [v ]i = 0 for any J 2 such that 1 2 J, we have that the linear term of the Taylor expansion of 1 J k−1 det A equals X X 2t v , [v ] [v ], x = 2t hv , [v ]ih[v ], xi . 1 i L L Q\1 Q\1 [n] [n] Q2 ,12Q L2 ( ) ( ) k k−1 Sincef[v ]g [n] is a tight frame in R , we have that L2 ( ) k−1 hv , [v ]ih[v ], xi = hv , xi . i L L i [n] L2 ( ) k−1 Therefore, q q det A = det A + 2thv , xi + o(t) = 1 + thv , xi + o(t). ˜ S i i Acknowledgement: The authors acknowledge the support of the grant of the Russian Government N 075-15- 2019-1926. G.I. was supported also by the Swiss National Science Foundation grant 200021-179133. The authors are very grateful to the anonymous reviewer for valuable remarks. 18 Ë Grigory Ivanov and Igor Tsiutsiurupa References [1] Arseniy Akopyan, Alfredo Hubard, and Roman Karasev, Lower and upper bounds for the waists of dierent spaces, Topolog- ical Methods in Nonlinear Analysis 53 (2019), no. 2, 457–490. [2] Keith Ball, Cube slicing in R , Proceedings of the American Mathematical Society (1986), 465–473. [3] Keith Ball, Volumes of sections of cubes and related problems, Geometric aspects of functional analysis (1989), 251–260. [4] Franck Barthe, Olivier Guédon, Shahar Mendelson, and Assaf Naor, A probabilistic approach to the geometry of the ` -ball, The Annals of Probability 33 (2005), no. 2, 480–513. [5] Stefano Campi and Paolo Gronchi, On volume product inequalities for convex sets, Proceedings of the American Mathemat- ical Society 134 (2006), no. 8, 2393–2402. [6] Alexandros Eskenazis, Piotr Nayar, and Tomasz Tkocz, Gaussian mixtures: entropy and geometric inequalities, The Annals of Probability 46 (2018), no. 5, 2908–2945. [7] Alexandros Eskenazis, On Extremal Sections of Subspaces of L , Discrete & Computational Geometry (2019), 1–21. [8] Peter M. Gruber, Convex and discrete geometry, Springer Berlin Heidelberg, 2007. [9] Grigory Ivanov, On the volume of projections of cross-polytope, arXiv preprint arXiv:1808.09165 (2018). [10] Grigory Ivanov, Tight frames and related geometric problems, accepted for publication in Canadian Mathematical Bulletin. [11] Grigory Ivanov, On the volume of the John–Löwner ellipsoid, Discrete & Computational Geometry 63, 455–459 (2020). [12] Hermann König and Alexander Koldobsky, On the maximal measure of sections of the n-cube, Geometric Analysis, Mathe- matical Relativity, and Nonlinear Partial Dierential Equations, Contemp. Math 599 (2012), 123–155. [13] Alexander Koldobsky, An application of the Fourier transform to sections of star bodies, Israel Journal of Mathematics 106 (1998), no. 1, 157–164. [14] , Fourier analysis in convex geometry, no. 116, American Mathematical Soc., 2005. [15] Mathieu Meyer and Alain Pajor, Sections of the unit ball of ` , Journal of Functional Analysis 80 (1988), no. 1, 109–123. [16] Fedor L. Nazarov and Anatoliy N. Podkorytov, Ball, Haagerup, and distribution functions, Complex analysis, operators, and related topics, Springer, 2000, pp. 247–267. [17] Jerey Vaaler, A geometric inequality with applications to linear forms, Pacic Journal of Mathematics 83 (1979), no. 2, 543–553. [18] Chuanming Zong, The cube: A window to convex and discrete geometry, Cambridge Tracts in Mathematics, vol. 168, Cam- bridge University Press, 2006. [19] Artem Zvavitch, Gaussian measure of sections of dilates and translations of convex bodies, Advances in Applied Mathemat- ics 41 (2008), no. 2, 247–254. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Analysis and Geometry in Metric Spaces de Gruyter

On the Volume of Sections of the Cube

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10.1515/agms-2020-0103
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Anal. Geom. Metr. Spaces 2021; 9:1–18 Research Article Open Access Grigory Ivanov and Igor Tsiutsiurupa* https://doi.org/10.1515/agms-2020-0103 Received April 17, 2020; accepted December 14, 2020 Abstract: We study the properties of the maximal volume k-dimensional sections of the n-dimensional cube [−1, 1] . We obtain a rst order necessary condition for a k-dimensional subspace to be a local maximizer of the volume of such sections, which we formulate in a geometric way. We estimate the length of the projection of a vector of the standard basis of R onto a k-dimensional subspace that maximizes the volume of the intersection. We nd the optimal upper bound on the volume of a planar section of the cube [−1, 1] , n ≥ 2. Keywords: Tight frame; section of cube; volume; Ball’s inequality MSC: 52A38, 49Q20, 52A40, 15A45 1 Introduction n n The problem of volume extrema of the intersection of the standard n-dimensional cube  = [−1, 1] with a k-dimensional linear subspace H has been studied intensively. The tight lower bound for all n ≥ k ≥ 1 was obtained by J. Vaaler [17], he showed that k n vol  ≤ vol ( \ H). k k A. Akopyan and R. Karasev [1] gave a new proof of Vaaler’s inequality in terms of waists. A deep generalization of Vaaler’s result for ` balls was made by M. Meyer and A. Pajor [15]. K. Ball in [3], using his celebrated version of the Brascamb-Lieb inequality, found the following upper bounds k/2 p n−k n k n k vol ( \ H) ≤ vol  and vol ( \ H) ≤ 2 vol  . (1.1) k k k k The leftmost inequality here is tight if and only if kjn (see [11]), and the rightmost one is tight whenever 2k ≥ n. Thus, if k does not divide n and 2k < n, the maximal volume of a section of  remains unknown. Before proving inequality (1.1), K. Ball obtained a particular case of it in [2], namely, the hyperplane case k = n− 1. Using the Fourier transform, he reduced the hyperplane case to a certain integral inequality. A sim- pler than original proof of this integral inequality was introduced later by F. Nazarov and A. Podkorytov [16]. Also, for the hyperplane case k = n − 1, the rightmost inequality in (1.1) was generalized to certain product measures which include Gaussian type measures by A. Koldobsky and H. König [12]. The Gaussian measure of sections of the cube was studied in [4] and [19]. A. Koldobsky [13] used the Fourier transform of a power of the radial function to express the volume of central hyperplane sections of star bodies in R and conrmed the Meyer and Pajor conjecture from [15] related to the volume of central sections of ` balls. This result was recently generalized by A. Eskenazis [7] (see also [6]). We refer the reader interested in the interaction between Convex Geometry and Fourier Analysis to [14]. In [9], a tight bound on the volume of a section of by a k-dimensional linear subspace was conjectured for all n > k ≥ 1. Grigory Ivanov: Institute of Science and Technology Austria (IST Austria), Kleusteneuburg, Austria; and Laboratory of Combi- natorial and Geometrical Structures, Moscow Institute of Physics and Technology, Moscow, Russia. E-mail: grimivanov@gmail.com *Corresponding Author: Igor Tsiutsiurupa: Laboratory of Combinatorial and Geometrical Structures, Moscow Institute of Physics and Technology, Moscow, Russia. E-mail: tsyutsyurupa.ik@phystech.edu Open Access. © 2021 Grigory Ivanov and Igor Tsiutsiurupa, published by De Gruyter. This work is licensed under the Creative Commons Attribution alone 4.0 License. 2 Ë Grigory Ivanov and Igor Tsiutsiurupa n n Conjecture 1. If the maximal volume of a section of the cube by a k-dimensional subspace of R is attained at a subspace H, then  \ H is an ane cube. k n Let C (n, k) vol  be the maximal volume of a section of  by a k-dimensional subspace L such that n n \ L is an ane cube. Conjecture 1 states that for any k-dimensional subspace H of R , one has n k vol ( \ H) ≤ C (n, k) vol  . k k n n A complete description of a k-dimensional subspace L of R such that the section  \ L is an ane cube of volume C (n, k) vol  is given in the following lemma. Lemma 1.1. The constant C (n, k) is given by l m j k n−kbn/kc k−(n−kbn/kc) n n C (n, k) = . (1.2) k k and is attained at the subspaces given by the following rule. 1. We partitionf1, 2, . . . , ng into k sets such that the cardinalities of any two sets dier by at most one. 2. Let fi , . . . , i g be one of the sets of the partition. Then, choosing arbitrary signs, we write the system of linear equations ±x[i ] = . . . = ±x[i ], where x[i] denotes the i-th coordinate of x in R . 3. Our subspace is the solution of the system of all equations written for each set of the partition at step (2). Since Lemma 1.1 was proven in [9], we provide a sketch of its proof in Appendix A. In this paper, we continue our study of maximizers of G(H) = vol ( \ H), H 2 Gr(n, k) with n > k > 1. (1.3) Using the approach of [10], which is described in detail below, we get a geometric rst order necessary con- dition for H to be a local maximizer of (1.3). Theorem 1.1. Let H be a local maximizer of (1.3), v be the projection of the i-th vector of the standard basis of n n R onto H, i 2 f1, 2, . . . , ng. Denote P =  \ H; we understand P as a k-dimensional polytope in H. Then 1. P = fx 2 H : jhx, v ij ≤ 1g . i=1 2. For every i 2 f1, 2, . . . , ng, v ≠ 0 and the intersection of P with the hyperplanefhx, v i = 1g is a facet of i i P. 3. For every i 2 f1, 2, . . . , ng, the line spanfv g intersects the boundary of P in the centroid of a facet of P. 4. Let F be a facet of P. Denote P = cof0, Fg. Then vol P 1 jv j k F i = , vol P 2 k where the summation is over all indices i 2 f1, 2, . . . , ng such that the line spanfv g intersects F in its centroid. One of the arguments used by K. Ball to prove the rightmost inequality in (1.1) is that the projection of a vector of the standard basis onto a maximizer of (1.3) for 2k ≥ n has length at least 2. We prove the following extension of this result. Theorem 1.2. Let n > k > 1 and H be a global maximizer of (1.3), v be the projection of a vector of the standard basis of R onto H. Then k k ≤ jvj ≤ . n + k n − k On the Volume of Sections of the Cube Ë 3 Using these results and some additional geometric observations, we prove the following. Theorem 1.3. Conjecture 1 is true for k = 2 and n ≥ 3. That is, for any two-dimensional subspace H  R the following inequality holds l mj k n n n 2 Area( \ H) ≤ C (n, 2) vol  = 4 . 2 2 n n This bound is optimal and is attained if and only if  \ H is a rectangle with the sides of lengths 2 and 2 . 2 Denitions and Preliminaries For a positive integer n, we refer to the set f1, 2, . . . , ng as [n]. The standard n-dimensional cube [−1, 1] n n is denoted by  . We use hp, xi to denote the standard inner product of vectors p and x in R . For vectors n n u, v 2 R , their tensor product (or, diadic product) is the linear operator on R dened as (u v)x = hu, xi v d n for every x 2 R . The linear hull of a subset S ofR is denoted by span S. For a k-dimensional linear subspace H ofR and a body K  H, we denote by vol K the k-dimensional volume of K. The two-dimensional volume 2 n of a body K  R is denoted by Area K. We denote the identity operator on a linear subspace H  R by I . If H = R , we use I for convenience. k k + For a non-zero vector v 2 R , we denote by H the ane hyperplanefx 2 R : hx, vi = 1g, and by H − k k and H the half-spacesfx 2 R : hx, vi ≤ 1g andfx 2 R : hx, vi ≥ −1g, respectively. It is convenient to identify a section of the cube with a convex polytope in R . Let fv , . . . , v g be the projections of the standard basis of R onto H. Clearly, \ H = fx 2 H : jhx, v ij ≤ 1g. i=1 n n That means that a section of  is determined by the set of vectorsfv g  H, which are the projections of the orthogonal basis. Such sets of vectors have several equivalent description and names. Denition 1. We will say that an ordered n-tuple of vectors fv , . . . , v g  H is a tight frame (or forms a 1 n tight frame) in a vector space H if v = I , (2.1) i i H where I is the identity operator in H and Aj is the restriction of an operator A onto H. We use Ω(n, k) to denote the set of all tight frames with n vectors in R . Denition 2. An n-tuple of vectors in a linear space H that spans H is called a frame in H. k n In the following trivial lemma we understand R  R as the subspace of vectors, whose last n − k n k n coordinates are zero. For convenience, we will considerfv g  R  R as k-dimensional vectors. Lemma 2.1. The following assertions are equivalent: k k 1. the vectorsfv , . . . , v g  R form a tight frame in R ; 1 n n k 2. there exists an orthonormal basisff , . . . , f g of R such that v is the orthogonal projection of f onto R , 1 n i i for any i 2 f1, 2, . . . , ng; k k 3. spanfv , . . . , v g = R and the Gram matrix Γ of vectorsfv , . . . , v g  R is the matrix of the projection 1 n 1 n operator from R onto the linear hull of the rows of the matrix M = (v , . . . , v ). 1 n 4. the k × n matrix M = (v , . . . , v ) is a sub-matrix of an orthogonal matrix of order n. 1 n 4 Ë Grigory Ivanov and Igor Tsiutsiurupa k k It follows that the tight frames in R are exactly the projections of orthonormal bases onto R . This ob- servation allows us to reformulate the problems in terms of tight frames and associated polytopes in R . Indeed, identifying H with R , we identify the projection of the standard basis onto H with a tight frame n + − fv , . . . , v g 2 Ω(n, k). Thus, we identify  \ H with the intersection of slabs of the form H \ H , 1 n v v i i i 2 f1, 2, . . . , ng. Vice versa, assertion (3) gives a way to reconstruct H from a given tight framefv , . . . , v g 1 n in R . Denition 3. We will say that an n-tuple S = fv , . . . , v g 2 R generates 1 n 1. the polytope + − (S) = H \ H , (2.2) v v i i i=1 which we call the section of the cube generated by S; 2. the matrix v v . We use A to denote this matrix. i i S i=1 To sum up, the global extrema of (1.3) coincide with that of F(S) = vol (S), where S 2 Ω(n, k) with n > k > 1. (2.3) To compare the local extrema of (1.3) and (2.3), we have to dene metrics on Ω(n, k) and on the Grassma- n k nian of k-dimensional subspaces of R . We endow the set of n-tuples of vectors in R with the metric 1/2 dist fv , . . . , v g,fu , . . . , u g = jv − u j . ( 1 n 1 n ) i i i=1 This denes the metric on the set of frames in R consisting of n vectors and on Ω(n, k). The standard metric on the Grassmanian is given by 0 H H Dist(H, H ) = P − P , H H wherek·k denotes the operator norm and P and P are the orthogonal projections onto the k-dimensional subspaces H and H , respectively. It was shown in [10] that the local extrema of (2.3) coincide with that of (1.3). However, we note that there is an ambiguity when we identify H with R . Any choice of orthonormal basis of H gives its own tight frame in R , all of them are isometric but dierent from each other. It is not a problem Ω(n,k) as there exists a one-to-one correspondence between Gr(n, k) and , where O(k) is the orthogonal group O(k) in dimension k. And, clearly, F(S ) = F(S ) whenever S = U(S ) for some U 2 O(k). 1 2 2 1 From now on, we will study properties of the maximizers of (2.3) and work with tight frames. 3 Operations on frames The following approach to problem (2.3) was proposed in [10] and used in [9] to study the properties of pro- jections of the standard cross-polytope. The main idea is to transform a given tight frame S into a new one S and compare the volumes of the sections of the cube generated by them. Since it is not very convenient to transform a given tight frame into another one, we add an intermediate step: we transform a tight frame S into a frame S, and then we transform S into a new tight frame S using a linear transformation. The main observation here is that we can always 0 0 ˜ ˜ transform any frame S = fv , . . . , v g into a tight frame S using a suitable linear transformation L: S = LS = 1 n fLv , . . . , Lv g. Equivalently, any non-degenerate centrally symmetric polytope in R is an ane image of a 1 n section of a high dimension cube. For a frame S in R , by denition put B = A = v v . v2S On the Volume of Sections of the Cube Ë 5 The operator B is well-dened as the condition span S = R implies that A is a positive denite operator. S S Clearly, for any frame S, B maps S to the tight frame B S: S S X X B v B v = B v v B = B A B = I . S S S S S S S k v2S v2S We obtain the following necessary and sucient condition for a tight frame to be a maximizer of (2.3). Lemma 3.1. The maximum of (2.3) is attained at a tight frame S 2 Ω(n, k) i for an arbitrary frame S in R inequality vol (S) 1 ≤ (3.1) vol (S) det A holds. ˜ ˜ ˜ ˜ Proof. For any frame S, we have that B S is a tight frame and vol (B S) = vol (S)/ det B . The maximum ˜ k ˜ k ˜ S S S ˜ ˜ of (2.3) is attained at a tight frame S i vol (B S) ≤ vol (S) for an arbitrary frame S. Hence k ˜ k ˜ ˜ vol (B S) vol (S) 1 vol (S) k ˜ S k k 1 ≥ = = det A vol (S) det B vol (S) vol (S) k ˜ k k Dividing by det A , we obtain inequality (3.1). ˜ ˜ Clearly, if S in the assertion of Lemma 3.1 is close to S, then the tight frame B S is close to S as well. Therefore, inequality (3.1) gives a necessary condition for local maximizers of (2.3). Let us illustrate how we will use it. Let S be an extremizer of (2.3), and T be a map from a subset of Ω(n, k) to the set of frames in R . In order to obtain properties of extremizers, we consider a composition of two operations: T ˜ S 0 S ! S ! S , where B is as dened above. For example, see Figure 1, where T is the operation of replacing a vector v of S by the origin. v v T : v → 0 B H H v v !(S) !(S ) !(S) −H −H v v Figure 1: A frame S  R consists of three vectors and (S) is a hexagon. A vector v 2 S is mapped to the origin, yielding ˜ ˜ ˜ the frame S. By construction, (S) is a parallelogram. Since the frame S = B S is a tight frame with only two nonzero vectors, these vectors are two orthogonal unit vectors. Hence (S ) is a square. Choosing a simple operation T, we may calculate the left-hand side of (3.1) in some geometric terms. We consider several simple operations: Scaling one or several vectors, mapping one vector to the origin, mapping one vector to another. On the other hand, the determinant in the right-hand side of (3.1) can be calculated for the operations listed above. In particular, the following rst-order approximation of the determinant was obtained by the rst author in [10, Theorem 1.2]. We provide a sketch of its proof in Appendix A. 6 Ë Grigory Ivanov and Igor Tsiutsiurupa Lemma 3.2. Let S = fv , . . . , v g  R be a tight frame and the n-tuple S be obtained from S by substitution 1 n v ! v + tx , where t 2 R, x 2 R , i 2 [n]. Then i i i i det A = 1 + t hx , v i + o(t). ˜ i i i=1 We state as lemmas several technical facts from linear algebra that will be used later. k k Lemma 3.3. Let A be a positive denite operator on R . For any u 2 R , we have −1/2 2 det (A ± u u) = 1 ± jA uj det A. Proof. We have −1/2 −1/2 det (A ± u u) = det A · det I ± A u A u . −1/2 −1/2 −1/2 Diagonalizing the operator I ± A u A u, we see that its determinant equals 1 ± A u . This completes the proof. For any n-tuple S of vectors of R with v 2 S, we use S \ v to denote the (n − 1)-tuple of vectors obtained from S by removing the rst occurrence of v in S. Lemma 3.4. Let S 2 Ω(n, k) and v 2 S be a vector such thatjvj < 1. Then S \ v is a frame in R and B is the S\v −1/2 k 2 k stretch of R by the factor 1 −jvj along spanfvg. In particular, for any u 2 R , we have B u ≥ juj. S\v Proof. Since A = I − v v > 0, we have that S \ v is a frame in R . Clearly, A stretches the space by S\v k S\v 2 −1/2 the factor 1 −jvj along spanfvg. Therefore, the operator B = A stretches the space by the factor S\v S\v −1/2 1 −jvj along the same direction. 4 Properties of a global maximizer Theorem 1.2 is formulated in terms of subspaces. For the sake of convenience, we introduce its equivalent reformulation in terms of tight frames. Theorem 4.1 (Frame version of Theorem 1.2). Let S 2 Ω(n, k) be a global maximizer of (2.3) for n > k > 1. Pick an arbitrary v 2 S. Then k k ≤ jvj ≤ . (4.1) n + k n − k Proof. The denition of tight frame implies that jpj = k. Hence there is a vector u 2 S such that p2S 2 2 juj ≤ k/n and a vector w 2 S such thatjwj ≥ k/n. We start with the rightmost inequality in (4.1). Let S be the n-tuple obtained from S by substitution u ! v. 2 k Sincejuj ≤ k/n < 1, we have A = I −u u +v v ≥ I −u u > 0. Hence S is a frame in R . By identity (2.2), k k ˜ ˜ inclusion (S)  (S) holds. Therefore, vol (S) ≤ vol (S). Using Lemma 3.1 and Lemma 3.3, we obtain k k 1 ≥ det A = 1 + B v 1 −juj . ˜ S\u 2 2 By Lemma 3.4, the right-hand side of this inequality is at least 1 +jvj 1 −juj . Therefore, juj k jvj ≤ ≤ . 1 −juj n − k Let us prove the leftmost inequality in (4.1). There is nothing to prove ifjvj ≥ k/n. Assume thatjvj < k/n. ˜ ˜ Let S be the n-tuple obtained from S by substitution v ! w. Since A = I−v v +u u > 0, S is a frame in R . S On the Volume of Sections of the Cube Ë 7 ˜ ˜ By identity (2.2), the inclusion (S)  (S) holds. Therefore, vol (S) ≤ vol (S). Using Lemma 3.1 and k k Lemma 3.3, we obtain 1 ≥ det A = 1 + B w 1 −jvj . S\v 2 2 Again, by Lemma 3.4, the right-hand side of this inequality is at least 1 +jwj 1 −jvj . It follows that jwj k jvj ≥ ≥ . 1 +jwj n + k This completes the proof. Clearly, Theorem 4.1 implies Theorem 1.2. These theorems can be sharpened in the case of sections by planes, i.e. k = 2. Lemma 4.1. Let S 2 Ω(n, 2) be a maximizer of (2.3) for k = 2 and n ≥ 3, let v 2 S. Then 2 2 ≤ jvj ≤ . (4.2) n + 1 n − 1 Proof. Lemma 1.1 provides an example of a two-dimensional subspace H satisfying Area  \ H = 4C (n, 2) = 4 dn/2ebn/2c. Therefore, the maximal area of a planar section of the cube  is at least 4 dn/2ebn/2c. Thus, we have Area(S) ≥ 4 dn/2ebn/2c. (4.3) Let us prove the leftmost inequality in (4.2). It is trivial if jvj ≥ 2/n. Assume that jvj < 2/n. Let S 2 Ω(n − 1, 2) be a maximizer of (2.3). By Ball’s inequality (1.1), we have Area(S ) ≤ 2(n − 1). (4.4) Consider S \ v. It is a frame by Lemma 3.4. Then, by Lemma 3.1 and Lemma 3.3, we get Area(S \ v) 1 1 ≤ p = p . Area(S ) det A 1 −jvj S\v By identity (2.2), we have (S)  (S \ v). By this and by inequalities (4.4) and (4.3), we get Area(S \ v) Area(S) 4 dn/2ebn/2c ≥ ≥ . 0 0 Area(S ) Area(S ) 2(n − 1) Combining the last two inequalities, we obtain Area(S ) n − 1 2 jvj ≥ 1 − ≥ 1 − p ≥ . Area(S) n + 1 2 dn/2ebn/2c We proceed with the rightmost inequality in (4.2). Let S 2 Ω(n + 1, 2) be a maximizer of (2.3). By Ball’s inequality (1.1), we have Area(S ) ≤ 2(n + 1). (4.5) We use S to denote the (n + 1)-tuple obtained from S by concatenating S with the vector v. Since S is a frame 2 2 in R , S is a frame in R as well. By Lemma 3.1 and Lemma 3.3, we get Area(S) 1 1 ≤ p = p . Area(S ) det A 1 +jvj By identity (2.2), we have (S) = (S). By this and by inequalities (4.5) and (4.3), we get Area(S) Area(S) 4 dn/2ebn/2c = ≥ . 0 0 Area(S ) Area(S ) 2(n + 1) 8 Ë Grigory Ivanov and Igor Tsiutsiurupa Combining the last two inequality, we obtain Area(S ) n + 1 2 jvj ≤ − 1 ≤ p − 1 ≤ . Area(S) n − 1 2 dn/2ebn/2c Remark 1. It is possible to sharpen inequality (4.1) for k > 2 and n > 2k using the same approach as in Lemma 4.1. The idea is to remove n mod k from or add n − (n mod k) vectors to a maximizer and compare the volume of a section of the cube generated by the new frame with the Ball bound (1.1). However, it doesn’t give a substantial improvement. Remark 2. As was pointed out to us by the anonymous reviewer, the results of this and the next section might be proven using the so-called shadow movement technique. Moreover, we expect that combining the result on convexity of the function related to the polar body of a symmetric shadow system [5] and suitable discrete substitutions, one could conrm Conjecture 1. 5 Local properties In this section, we prove some properties of the local maximizers of (2.3). We will perturb facets of (S) of a local maximizer S (that is, we will perturb the vectors of S in a specic way corresponding to a perturbation of some facets of the polytope (S)). To this end, we need to recall some general properties of polytopes connected to perturbations of a half-space supporting a polytope in its facet. 5.1 Properties of polytopes Recall that a point c is the centroid of a facet F of a polytope P  R if c = xdλ, (5.1) vol F k−1 where dλ is the standard Lebesgue measure on the hyperplane containing F. k + For a set W  R , we use P(W) to denote the polytopal set\ H . Let W be a set of pairwise distinct w2W w vectors such that • the set P(W) is a polytope; • for every w 2 W, the hyperplane H supports P(W) in a facet of P(W). That is, W is the set of scaled outer normals of P(W). Denote P = P(W). We x w 2 W and the facet F = P\H of P. Let c be the centroid of F. Transformation 1. We will “shift” a facet of a polytope parallel to itself. Let W be obtained from W by sub- w 0 0 0 stitution w ! w + h , where h 2 R. Denote P = P(W ). That is, the polytopal set P is obtained from P by jwj the shift of the half-space H by h in the direction of its outer normal. By the celebrated Minkowski existence and uniqueness theorem for convex polytopes (see, for example, [8, Theorem 18.2]), we have vol P − vol P = h vol F + o(h). (5.2) k k k−1 Transformation 2. We will rotate a facet around a codimension two subspace. Let u be a unit vector orthogo- nal to w. Dene c = H \ spanfwg and L = H \ (u + c ). Note that L is a codimension two ane w w u w w u subspace of R and an ane hyperplane in H . Clearly, for any non-zero t 2 R, L = H \ H = w u w w+tu fx 2 H : hx − c(w), ui = 0g . Thus, L divides F into two parts w u + − F = F \fx 2 H : hx − c(w), ui ≥ 0g and F = F \fx 2 H : hx − c(w), ui ≤ 0g w w On the Volume of Sections of the Cube Ë 9 H H H H H w w+u w w+u w w → w + u w → w + u h< 0 h> 0 ! ! P P Figure 2: Parallel shift of the facet F by the vector u = hw/jwj + − (one of the sets F or F is empty if c(w) 2̸ F). Let α 2 (−π/2, π/2) be the oriented angle between hyperplanes H and H such that α is positive for positive t. w w+tu Let W be obtained from W by substitution w ! w + tu, where t 2 R and u is a unit vector orthogonal 0 0 0 to w. Denote P = P(W ). Thus, for a suciently smalljtj, the polytopal set P is a polytope obtained from P by the rotation of the half-space H around the codimension two ane subspace L by some angle α = α(t). Clearly, in order to calculate the volume of P , we need to subtract from vol P the volume of the subset of P that is above H and to add to vol P the volume of the subset of P that is above H . Formally speaking, w+tu w denote 8 8 0 k + k + < < P \ R \ H for α ≥ 0 P\ R \ H for α ≥ 0 w w+tu + − Q = and Q = k + 0 k + : : P\ R \ H for α < 0 P \ R \ H for α < 0 w+tu w Then, we have (see Figure 3) 0 + − vol P − vol P = sign α vol Q − vol Q . (5.3) k k k k + − + − + There is a nice approximation for vol Q − vol Q . Let C (resp., C ) be the set swept out by F (resp., α α k k F ) while rotating around L by the angle α. By routine, +∞ vol C = jαj r vol F \ (L + ru) dr k α k−2 u 0 1 +∞ 0 Z Z @ A resp., vol C = jαj r vol F \ (L − ru) dr = −jαj r vol F \ (L + ru) dr . α u u k k−2 k−2 0 −∞ We claim that + + − − vol Q = vol C + o(α) and vol Q = vol C + o(α). (5.4) k k α k k α + + + This can be shown in the following way. If F is not a body in H , then vol Q = vol C = 0. As- k k α sume F is a body in H . Since a polytope is dened by a system of linear inequalities, there is a posi- tive constant b such that for any suciently small τ the Hausdor distance on H between Q \ H w+τu w+τu + d and C \ H is at most bα (recall that the Hausdor distance between A, B  R is d (A, B) = w+τu n α o H d d d inf ε > 0 : A  B + εB and B  A + εB , where B is the Euclidean unit ball). Note that the section of + + C by H is F rotated around L by some small angle β = β(τ). Therefore, there exist two homothets α w+τu u + + + F  H and F  H of F satisfying two properties: w w 1 2 + + + + • both the Hausdor distance between F and F and the Hausdor distance between F and F are at 1 2 ˜ ˜ most bα for a positive constant b; + + • the body swept out by F while rotating around L by the angle α is contained in Q and the body swept + + out by F contains Q . 2 10 Ë Grigory Ivanov and Igor Tsiutsiurupa + + By the rst property and a simple integration, the volumes of the bodies swept out by F and F coincide with 1 2 vol C up to the terms of the rst order of α. The second property implies the leftmost identity in (5.4). The k α rightmost one is obtained similarly. By identities (5.3) and (5.4), we obtain 0 + − vol P − vol P = α vol C − vol C + o(α) = α r vol F \ (L + ru) dr + o(α). k k k α k α k−2 u By this and by (5.1), we get vol P − vol P = αhc − c , ui vol F + o(α). k k k−1 Since w and u are orthogonal, we have juj juj α = arctan t = t + o(t). jwj jwj Finally, we obtain vol F 0 k−1 vol P − vol P = hc − c , ui t + o(t). (5.5) k k jwj H H H H H w w+tu w w+tu w F F L w u w → w + tu w → w + tu L − t> 0 t< 0 " " P P Figure 3: A rotation of the facet F around L 5.2 Local properties of sections of the cube Let S be a frame in R . For every v 2 S, we denote the set H \(S) by F . We say that v 2 S corresponds to a v v facet F of (S) if either F = F or F = −F . Clearly, if some vectors of S correspond to the same facet of (S), v v k k k then they are equal up to a sign. For a given frame S in R and u 2 R , a facet F of (S) and a vector u 2 R , we dene an F-substitution in the direction u as follows: • each vector v of S such that F  H is substituted by v + u; • each vector v of S such that −F  H is substituted by v − u; • all other vectors of S remain the same. In order to prove Theorem 1.1, we will use F-substitutions. At rst, we simplify the structure of a local maximizer. Lemma 5.1. Let S be a local maximizer of (2.3) and v 2 S. Then F is a facet of (S). Proof. Let K be a convex body in R , then its polar body is dened by n o y 2 R : hy, xi ≤ 1 for all x 2 K . On the Volume of Sections of the Cube Ë 11 Since (S) is the intersection of half-spaces of the formfhw, xi ≤ 1g with w 2 ±S, we have that cof±Sg is polar to (S) in span S = R . By the duality argument, it suces to prove that v 2 S is a vertex of the polytope cof±Sg . Assume that v is not a vertex of cof±Sg . Clearly, v 2 cof± S \ v g and v is not a vertex of the polytope cof± S \ v g . Therefore we have that ( ) ( ) k k spanfS\vg = span S = R . That is, S\v is a frame in R . Since B is a nondegenerate linear transformation, S\v B v is not a vertex of the polytope co ±B (S \ v) . By this and by the triangle inequality, there is a vertex u S\v S\v of cof±Sg such that u 2 S and B v < B u . S\v S\v Denote by S the n-tuple obtained from S by substitution v ! v + t(u − v), where t 2 (0, 1]. Since ˜ ˜ A ≥ A > 0, S is a frame in R . By the choice of u and identity (2.2), we have (S) = (S). Hence S\v vol (S) = vol (S). Lemma 3.1 implies that det A ≤ 1. k k ˜ On the other hand, by Lemma 3.3, we have det A = 1 + B (v + t(u − v)) 1 −jvj . S\v Inequality B v < B u implies that B v < B (v + t(u − v)) . By this and by Lemma 3.4, we conclude S\v S\v S\v S\v that det A > 1. This is a contradiction. Thus, v is a vertex of cof±Sg. The lemma is proven. As an immediate corollary of Lemma 5.1 and by the standard properties of polytopes, we have the follow- ing statement. Corollary 5.1. Let S be a local maximizer of (2.3) and v 2 S. Let S(t) be the n-tuple obtained from S by F - substitution in the direction tu with t 2 R and u 2 R . Then, for a suciently small jtj, S(t) is a frame, the ˜ ˜ vector v + ut corresponds to a facet of (S(t)). Moreover, vol (S(t)) is a smooth function of t at t = 0. In the following two lemmas, we will perturb a local maximizer by making F-substitutions. Geometrically speaking, making an F-substitution in the direction tu with u 2 R and t 2 R, we move the opposite facets F and −F of a local maximizer in a symmetric way. Thus, for a suciently small t, perturbations of the facets F and −F are independent. Lemma 5.2. Let S be a local maximizer of (2.3). Let v 2 S and d be the number of the vectors of S that corre- spond to F . Then vol F = djvj vol (S). (5.6) k−1 k jvj Proof. Denote by S the n-tuple obtained from S by F -substitution in the direction tv with t 2 R. Thus, we apply Transformation 1 to the facets ±F of (S). By Lemma 3.1, we have vol (S) 1 ≤ p . (5.7) vol (S) det A k ˜ By Lemma 3.2 and Corollary 5.1, both sides of this inequality are smooth as functions of t in a suciently small neighborhood of t = 0. Consider the Taylor expansions of both sides of inequality (5.7) as functions of t about t = 0. 2 2 By Lemma 3.3, det A = 1 + d(2t + t )jvj . Hence p = 1 − tdjvj + o(t). (5.8) det A + − Geometrically speaking, we shift the half-space H (resp., H ) by v v 1 1 t h = − = − + o(t) j(1 + t)vj jvj jvj in the directions of its outer normal. By this and by (5.2), we obtain 2t vol (S) − vol (S) = − vol F + o(t). (5.9) k k k−1 v jvj 12 Ë Grigory Ivanov and Igor Tsiutsiurupa Using identities (5.9) and (5.8) in (5.7) , we get 2t vol F v 2 k−1 1 − ≤ 1 − djvj t + o(t). jvj vol (S) Since S = S for t = 0 and the previous inequality holds for all t 2 (−ε, ε) for a suciently small ε, the coecients of t in both sides of the previous inequality coincide. That is, 2 vol F k−1 2 = djvj . jvj vol (S) This completes the proof. Lemma 5.3. Let S be a local maximizer of (2.3) and v 2 S. Then the line spanfvg intersects the hyperplane H in the centroid of the facet F . Proof. Denote the centroid of F by c and let c = spanfvg\H . Fix a unit vector u orthogonal to v. Denote by S v v v the n-tuple obtained from S by F -substitution in the direction tu with t 2 R. Thus, we apply Transformation 2 to the facets ±F of (S). By Lemma 3.1, we have vol (S) 1 ≤ p . (5.10) vol (S) det A k ˜ By Lemma 3.2 and Corollary 5.1, both sides of this inequality are smooth as functions of t in a su- ciently small neighborhood of t = 0. Consider the Taylor expansions of both sides of inequality (5.10) as functions of t about t = 0. By (5.5), we obtain vol (S) − vol (S) = Chc − c , ui t + o(t), k k where C = 2 vol F /jvj > 0. By Lemma 3.2, det A = 1 + o(t). Therefore, inequality (5.10) takes the k−1 ˜ following form 1 + hc − c , ui t + o(t) ≤ 1 + o(t). vol (S) Since S = S for t = 0 and the previous inequality holds for all t 2 (−ε, ε) for a suciently small ε, the coecients of t in both sides of the previous inequality coincide. That is, we conclude hc − c , ui = 0. Since c, c 2 H and the last identity holds for all unit vectors parallel to H , it follows that c = c . The v v v v lemma is proven. As a simple consequence of Lemma 5.3, we obtain the following result for the planar case. Theorem 5.1. Let S 2 Ω(n, 2) be a local maximizer of (2.3) for k = 2. Then, the polygon (S) is cyclic. That is, there is a circle that passes through all the vertices of (S). Proof. Denote the origin by o. Let ab be an edge of (S) and oh be the altitude of the triangle abo. By Lemma 5.3, h is the midpoint of ab. Hence, the triangle abo is isosceles and ao = bo. It follows that (S) is cyclic. We are ready to give a proof of Theorem 1.1. Proof of Theorem 1.1. Recall that any identication of H with R identies the projections of the standard basisfv , . . . , v g with a tight frame, denoted by S, that is a local maximizer of (2.3). 1 n Next, assertion 1 is trivial and holds for any section of the cube. Assertion 2 and assertion 3 are equivalent to Lemma 5.1 and Lemma 5.3, respectively. On the Volume of Sections of the Cube Ë 13 By Lemma 5.1, all vectors v 2 S such that span v intersects F correspond to F and have the same length that we denote byjvj. Then by Lemma 5.3, the span of each of these vectors intersects F in its centroid. Since the length of the altitude of the pyramid P is 1/jvj, we have 1 vol F k−1 vol P = . k F k jvj Hence assertion 4 follows from Lemma 5.2. 6 Proof of Theorem 1.3 We use the setting of tight frames developed in the previous sections to prove the theorem. More precisely, we use the obtained necessary conditions for a tight frame in R that maximizes (2.3) for n > k = 2 to prove that the section of the cube generated by the tight frame is a rectangle of area 4C (n, 2) = 4 dn/2ebn/2c p p with the sides of lengths 2 bn/2c and 2 dn/2e. First, let us introduce the notation. Let S = fv , . . . , v g 2 Ω(n, 2) be a global maximizer of (2.3) for 1 n k = 2 and n > 2. Clearly, (S) is a centrally symmetric polygon in R . The number of edges of (S) is denoted by 2f . Clearly, f ≤ n. By Theorem 5.1, the polygon (S) is cyclic; and we denote its circumradius by R. Let F , . . . F be the edges of (S) enumerated in clockwise direction (that is, edges F and F are opposite to 2f i i+f each other, i 2 [f]). We reenumerate the vectors of S in such a way that the vector v corresponds to the edge F for every i 2 [f]. The central angle subtended by the edge F is denoted by 2φ , i 2 [f ]. i i i Clearly, we have the following identities (see Figure 4): φ + ··· + φ = , (6.1) 1 f R cos φ = for all i 2 [f ], (6.2) jv j and Area(S) = R sin 2φ . (6.3) i=1 Also, we note here that l mj k n n Area(S) ≥ 4C (n, 2) = 4 . (6.4) 2 2 R v !(S) Figure 4: Notation for (S) There are several steps in the proof. We explain the main steps briey. In fact, we want to show that the number of edges of a local maximizer is 2f = 4. Using the discrete isoperimetric inequality (see below), we obtain an upper bound on Area(S) in terms of f . This upper bound yields the desired result for n ≥ 8 (the 14 Ë Grigory Ivanov and Igor Tsiutsiurupa bound is less than conjectured volume 4C (n, 2) for n ≥ 8). Finally, we deal with the lower-dimension cases using the necessary conditions obtained earlier. The discrete isoperimetric inequality for cyclic polygons says that among all cyclic f -gons with xed cir- cummradius there is a unique maximal area polygon – the regular f -gon. We will use a slightly more general form. Namely, xing one or several central angles of a cyclic polygon, its area is maximized when all other central angles are equal. In our notation xing the central angle φ and its vertically opposite, we have π π − 2φ 2 2 i R f sin ≥ R sin 2φ + (f − 1) sin ≥ 4C (n, 2). (6.5) f f − 1 This inequality immediately follows from the Jensen inequality and concavity of the sine function on [0, π]. 6.1 Step 1. Claim 1. The area of(S) such that f = 2 is at most 4C (n, 2). The bound is attained when(S) is a rectangle p p with the sides of lengths 2 dn/2e and 2 bn/2c. Proof. Since f = 2, the polygon (S) is an ane square. Hence the claim is an immediate consequence of Lemma 1.1. Thus, it suces to prove that f = 2 for any n > 2. 6.2 Step 2. Claim 2. For any n > 2 the following inequality holds n + 1 1 R ≤ . (6.6) 2 cos 2f Proof. Let φ be the smallest central angle. By identity (6.1), we have cos φ ≥ cos . Combining this with 1 1 2f the leftmost inequality in (4.2) and identity (6.2), we obtain 1 n + 1 1 R = ≤ . 2 2 2 jv j cos φ 2 cos 1 1 2f Claim 3. For any n > 2 the following inequality holds l mj k π 4 n n f tan ≥ . (6.7) 2f n + 1 2 2 Proof. By the discrete isoperimetric inequality (6.5), we have R f sin ≥ 4C (n, 2). Combining this with inequalities (6.6) and (6.4), we obtain l mj k sin 2 8 n n f ≥ C (n, 2) = . cos n + 1 n + 1 2 2 2f The claim follows. Claim 4. The following bounds on f hold: 1. f = 2 if n ≥ 8; 2. f ≤ 3 if n = 7; On the Volume of Sections of the Cube Ë 15 3. f ≤ 4 if n = 5. Proof. We consider the functions in the left- and right-hand sides of (6.7) as functions of f and n respectively. π 4 Set g(f) = f tan and h(n) = bn/2cdn/2e. Thus, inequality (6.7) takes the form g(f ) ≥ h(n). By routine 2f n+1 analysis, we have that g is strictly decreasing and h is increasing onfn 2 N : n ≥ 2g. The rst two assertions of the claim follows from this and the identity g(3) = h(7). Inequality f ≤ 4 for n = 5 follows from the direct computations of g(5), g(4) and h(5) (see Figure 5). f 2 3 4 5 n 5 6 7 p p p p p g(f) 2 3 4 2 − 1 5 5 − 2 5 h(n) 2 6/3 12/7 3 Figure 5: Some values of g and h Theorem 1.3 is proven for n ≥ 8. We proceed with the lower-dimensional cases. 6.3 Step 3. Claim 5. For n = 7, we have that f = 2. Proof. We showed that f ≤ 3 for n = 7. Assume that f = 3. We see that inequality (6.7) with such values turns into an identity. It follows that φ = φ = φ = π/6 and (S) is a regular hexagon. Hence the vectors of S 1 2 3 2 2 2 1 are of the same length. Since jvj = tr I = 2, we conclude thatjvj = 2/7 and R = = 14/3. 2 2 2 jv j cos π/6 v2S However, the volume of such a hexagon is strictly less than 4C (7, 2). We conclude that f = 2 for n = 7. Claim 6. For n 2 f3, 4, 6g, we have that f = 2. Proof. For n = 3, the statement is a simple exercise (see [18]). Conjecture 1 was conrmed in [11] for any n > k ≥ 1 such that kjn, in particular, for k = 2 and n 2 f4, 6g. Remark 3. The inequality on the area for n 2 f4, 6g and k = 2 is a special case of the leftmost inequality in (1.1) originally proved by K. Ball [3]. In [11], the equality cases in this Ball’s inequality are described. 6.4 Step 4. Claim 7. Let n = 5 and either f = 3 or f = 4. Then φ ≤ π/4 for every i 2 [f]. Proof. Assume that there is i 2 [f ] such that φ > π/4. Thus, cos φ < 1/ 2. Using identity (6.2) for i and for i i any j 2 [f], we have that r r cos φ jv j n + 1 3 j i = ≤ = , cos φ jv j n − 1 2 i j p p where the inequality follows from Lemma 4.1. Hence cos φ ≤ 3/2 cos φ < 3/2 and, therefore, φ > π/6. j i j This contradicts identity (6.1): π π π π = φ + ··· + φ > + (f − 1) > . 2 4 6 2 Thus, φ ≤ π/4 for every i 2 [f ]. Claim 8. Let n = 5 and either f = 3 or f = 4. Then φ ≥ π/10 for every i 2 [f]. i 16 Ë Grigory Ivanov and Igor Tsiutsiurupa Proof. Fix i 2 [f]. By identity (6.2) and Lemma 4.1, we have R ≤ . By inequality (6.5), we get 2 cos φ 3 π − 2φ sin 2φ + (f − 1) sin ≥ 4C (5, 2) = 4 6. cos φ f − 1 The function of φ in the left-hand side of this inequality is increasing on [0, π/2]. Since the inequality does not hold for φ = π/10, we conclude that φ is necessarily at least π/10. i i Claim 9. For n = 5, we have that f = 2. Proof. Assume that either f = 3 or f = 4. Denote by d the number of vectors in S that correspond to F . We i i want to rewrite the inequality of Lemma 5.2 using the circumradius and the center angle. Since the length of edge F is 2R sin φ and by identity (6.2), identity (5.6) takes the form i i 2 i 2R sin 2φ = Area(S). 2 2 R cos φ Set q(φ) = cos φ sin 2φ. Then for all i, j 2 [f], we have d q(φ ) i i = . d q(φ ) j i Clearly, there are i, j 2 [f ] such that d = 2 and d = 1. Therefore, q(φ )/q(φ ) = 2. By Claim 7 and Claim 8, i j i j we have that φ , φ 2 [π/10, π/4]. By simple computations, the maximum of q on the segment [π/10, π/4] i j p p is q(π/6) = 3 3/8 and the minimum is q(π/4) = 1/2. Hence max q(φ)/q(ψ) = 3 3/4 < 2 and we φ,ψ2[π/10,π/4] come to a contradiction. Thus, f = 2. Thus, we have proved that for a maximizer of (2.3), then f = 2. By Claim 1, the conjectured upper bound for the area of a planar section holds and also is tight. The proof of Theorem 1.3 is complete. Remark 4. We used the Ball inequality (1.1) to prove Theorem 1.3 for n 2 f3, 4, 6g. However, it can be done by using our approach without the Ball inequality. The proof is technical. Since it is not of great interest, we do not give a proof. A Sketches of proofs Sketch of the proof of Lemma 1.1. Let H be a k-dimensional subspace of R such that C (n, k) is attained. Denote P =  \ H. Since P is an ane k-dimensional cube, there are vectors fa , . . . , a g such that P = + − H \ H . a a i2[k] i i Letfv , . . . , v g be the projection of the vectors of the standard basis onto H. By the same arguments as 1 n in Lemma 5.1, the hyperplane H meets the polytope P in a facet of P for every i 2 [n]. Thus, v coincides i i with ±a for a proper sign and j 2 [k]. Or, equivalently, we partition [n] into k sets and H is the solution of a proper system of linear equations constructed as in (2) and (3), except we have not proved that (1) holds yet. Let us prove this assertion. Let d vectors of the standard basis of R project onto a pair ±a . Therefore, a i i k-tuple of vectorsf d a g is a tight frame. Identifying H with R and by the assertion (4) of Lemma 2.1, i i i2[k] 2 1 we conclude that a and a are orthogonal whenever i ≠ j. Therefore,ja j = and i j i n k vol  \ H = 2 d · . . . · d . (A.1) k 1 k Suppose d ≥ d + 2 for some i, j 2 [k]. Then d · d ≤ (d − 1)(d + 1). By this and by (A.1), we showed that i j i j i j (1) holds. It is easy to see that there are exactly n − kbn/kc of d ’s equaldn/ke and all others k − (n − kbn/kc) are equal tobn/kc. That is, C (n, k) is given by (1.2). This completes the proof. On the Volume of Sections of the Cube Ë 17 Sketch of the proof of Lemma 3.2. Recall that the cross product of k − 1 vectors fx , . . . , x g of R is the 1 k−1 vector x dened by hx, yi = det(x , . . . , x , y) for all y 2 R . 1 k−1 [n] For an ordered (k − 1)-tuple L = fi , . . . , i g 2 and a frame S = fv , . . . , v g, we use [v ] to denote 1 k−1 1 n L k−1 the cross product of v , . . . , v . i i 1 k−1 We claim the following property of the tight frames. k k Let S = fv , . . . , v g be a tight frame in R . Then the set of vectorsf[v ]g is a tight frame in R . [n] 1 n L L2 ( ) k−1 k n n We use Λ (R ) to denote the space of exterior k-forms on R . By assertion (2) of Lemma 2.1, there exists n n k an orthonormal basis ff g of R such that v is the orthogonal projection of f onto R , for any i 2 [n]. i 1 i i Then the (k − 1)-form v ^ ··· ^ v is the orthogonal projections of the (k − 1)-form f ^ ··· ^ f onto i i i i 1 k−1 1 k−1 k−1 k k−1 n [n] Λ (R )  Λ (R ), for any ordered (k − 1)-tuple L = fi , . . . , i g 2 . By Lemma 2.1 and since the 1 k−1 k−1 k−1 n (k − 1)-formsff ^ ···^ f g form an orthonormal basis of Λ (R ), we have that the set of [n] i i 1 k−1 fi ,...,i g2 1 k−1 ( ) k−1 k−1 k (k − 1)-forms fv ^ ··· ^ v g is a tight frame in Λ (R ). Finally, the Hodge star operator [n] i i 1 k−1 fi ,...,i g2 1 k−1 ( ) k−1 maps v ^···^ v to the cross product of vectors v , . . . , v . Since the Hodge star is an isometry, the set i i i i 1 k−1 1 k−1 of cross productsf[v ]g is a tight frame. The claim is proven. [n] L22 ( ) k−1 By linearity of the determinant, it is enough to prove the lemma for S = fv + tx, v , . . . , v g. Denote 1 2 n 0 0 v = v + tx and v = v , for 2 ≤ i ≤ n. 1 i 1 i By the Cauchy–Binet formula, we have 0 1 X X X 0 0 0 0 @ A det A = det v v = det v v . (A.2) i i i i [n] 1 i2Q Q2 ( ) By the properties of the Gram matrix, we have 0 0 0 0 det v v = det v , . . . , v . i i i i 1 k 1 k By this, by the denition of cross product and by identity (A.2), we obtain det A = 1 + 2t v , [v ] [v ], x + o(t). ˜ 1 Q\1 Q\1 [n] Q2 ,12Q ( ) [n] Sincehv , [v ]i = 0 for any J 2 such that 1 2 J, we have that the linear term of the Taylor expansion of 1 J k−1 det A equals X X 2t v , [v ] [v ], x = 2t hv , [v ]ih[v ], xi . 1 i L L Q\1 Q\1 [n] [n] Q2 ,12Q L2 ( ) ( ) k k−1 Sincef[v ]g [n] is a tight frame in R , we have that L2 ( ) k−1 hv , [v ]ih[v ], xi = hv , xi . i L L i [n] L2 ( ) k−1 Therefore, q q det A = det A + 2thv , xi + o(t) = 1 + thv , xi + o(t). ˜ S i i Acknowledgement: The authors acknowledge the support of the grant of the Russian Government N 075-15- 2019-1926. G.I. was supported also by the Swiss National Science Foundation grant 200021-179133. The authors are very grateful to the anonymous reviewer for valuable remarks. 18 Ë Grigory Ivanov and Igor Tsiutsiurupa References [1] Arseniy Akopyan, Alfredo Hubard, and Roman Karasev, Lower and upper bounds for the waists of dierent spaces, Topolog- ical Methods in Nonlinear Analysis 53 (2019), no. 2, 457–490. [2] Keith Ball, Cube slicing in R , Proceedings of the American Mathematical Society (1986), 465–473. [3] Keith Ball, Volumes of sections of cubes and related problems, Geometric aspects of functional analysis (1989), 251–260. [4] Franck Barthe, Olivier Guédon, Shahar Mendelson, and Assaf Naor, A probabilistic approach to the geometry of the ` -ball, The Annals of Probability 33 (2005), no. 2, 480–513. [5] Stefano Campi and Paolo Gronchi, On volume product inequalities for convex sets, Proceedings of the American Mathemat- ical Society 134 (2006), no. 8, 2393–2402. 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Journal

Analysis and Geometry in Metric Spaces – de Gruyter

Published: Jan 1, 2021

Keywords: Tight frame; section of cube; volume; Ball’s inequality; 52A38; 49Q20; 52A40; 15A45

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