On the volume of sections of the cube

We study the properties of the maximal volume $k$-dimensional sections of the $n$-dimensional cube $[-1,1]^n$. We obtain a first 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 $\mathbb{R}^n$ onto a $k$-dimensional subspace that maximizes the volume of the intersection. We find the optimal upper bound on the volume of a planar section of the cube $[-1,1]^n,$ $n \geq 2.$


Introduction
The problem of volume extrema of the intersection of the standard -dimensional cube = [−1, 1] with a -dimensional linear subspace has been studied intensively. The tight lower bound for all ≥ ≥ 1 was obtained by J. Vaaler [Vaa79], he showed that vol ≤ vol ( ∩ ).
A. Akopyan and R. Karasev [AHK19] gave a new proof of Vaaler's inequality in terms of waists. A deep generalization of Vaaler's result for ℓ balls was made by M. Meyer and A. Pajor [MP88]. K. Ball in [Bal89], using his celebrated version of the Brascamb-Lieb inequality, found the following upper bounds The leftmost inequality here is tight if and only if | (see [Iva19]), and the rightmost one is tight whenever 2 ≥ . Thus, if does not divide and 2 < , the maximal volume of a section of remains unknown. For the hyperplane case = − 1, the rightmost inequality in (1.1) was generalized to certain product measures which include Gaussian type measures by A. Koldobsky and H. König [KK12].
In [Iva18a], a tight bound on the volume of a section of by a -dimensional linear subspace was conjectured for all > ≥ 1. Namely, let ( , )2 be the maximum volume of a section of by a -dimensional subspace such that ∩ is an affine cube.
Conjecture 1. The maximal volume of a section of the cube by a -dimensional subspace is attained on subspaces such that the section is an affine cube, i.e. We give a complete description of the set of -dimensional subspaces of R on which ( , ) is attained and satisfies identity (1.2) in Lemma 2.2.
Using the approach of [Iva18b], which is described in detail below, we get a geometric first order necessary condition for to be a local maximizer of (1.3).
where the summation is over all indices ∈ [ ] such that the line span{ } intersects 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 2 ≥ has length at least √ 2. We prove the following extension of this result.
Theorem 1.2. Let > ≥ 1 and be a global maximizer of (1.3), be the projection of a vector of the standard basis of R onto . Then Using these results and some additional geometric observations, we prove the following.
Theorem 1.3. Conjecture 1 is true for > = 2. That is, for any two-dimensional subspace ⊂ R the following inequality holds This bound is optimal and is attained if and only if ∩ is a rectangle with the sides of lengths 2 √︁ ⌈︀ 2 ⌉︀ and 2 √︁ ⌊︀ 2 ⌋︀ .

Definitions and Preliminaries
For a positive integer , we refer to the set {1, 2, . . . , } as [ ]. The standard -dimensional cube [−1, 1] is denoted by . We use ⟨ , ⟩ to denote the standard inner product of vectors and in R . For vectors , ∈ R , their tensor product (or, diadic product) is the linear operator on R defined as ( ⊗ ) = ⟨ , ⟩ for every ∈ R . The linear hull of a subset of R is denoted by span . For a -dimensional linear subspace of R and a body ⊂ , we denote by vol the -dimensional volume of . The two-dimensional volume of a body ⊂ R 2 is denoted by Area . We denote the identity operator on a linear subspace ⊂ R by . If = R , we use for convenience. For a non-zero vector ∈ R , we denote by the affine hyperplane { ∈ R : ⟨ , ⟩ = 1}, and by + and − the half-spaces { ∈ R : ⟨ , ⟩ ≤ 1} and { ∈ R : ⟨ , ⟩ ≥ −1}, respectively.
It is convenient to identify a section of the cube with a convex polytope in R . Let { 1 , . . . , } be the projections of the standard basis of R onto . Clearly, That means that a section of is determined by the set of vectors { } 1 ⊂ , which are the projections of the orthogonal basis. Such sets of vectors have several equivalent description and names.
Definition 1. We will say that an ordered -tuple of vectors { 1 , . . . , } ⊂ is a tight frame (or forms a tight frame) in a vector space if where is the identity operator in and | is the restriction of an operator onto . We use Ω( , ) to denote the set of all tight frames with vectors in R .
Definition 2. An -tuple of vectors in a linear space that spans is called a frame.
In the following trivial lemma we understand R ⊂ R as the subspace of vectors, whose last − coordinates are zero. For convenience, we will consider { } 1 ⊂ R ⊂ R as -dimensional vectors.
It follows that the tight frames in R are exactly the projections of orthonormal bases onto R . This observation allows us to reformulate the problems in terms of tight frames and associated polytopes in R . Indeed, identifying with R , we identify the projection of the standard basis onto with a tight frame { 1 , . . . , } ⊂ Ω( , ). Thus, we identify ∩ with the intersection of slabs of the form + ∩ − , ∈ [ ]. Vice versa, assertion (3) gives a way to reconstruct from a given tight frame { 1 , . . . , } in R .
Definition 3. We will say that an -tuple = { 1 , . . . , } ∈ R generates (1) the polytope which we call the section of the cube generated by ; (2) the matrix ∑︀ . We use to denote this matrix.
Moreover, it was shown in [Iva18b] that the local extrema of (2.3) coincide with that of (1.3). However, we note that there is an ambiguity when we identify with R . Any choice of orthonormal basis of gives its own tight frame in R , all of them are isometric but different from each other. It is not a problem as there exists a one-to-one correspondence between Gr( , ) and Ω( , ) O( ) , where O( ) is the orthogonal group in dimension . And, clearly, ( 1 ) = ( 2 ) whenever 2 = ( 1 ) for some ∈ O( ).
In the following lemma, we give a complete description of the set ℋ of -dimensional subspaces of R such that ∩ is an affine cube and vol ( ∩ ) = 2 ( , ) for ∈ ℋ. Since it was proven in [Iva18a], we provide a sketch of its proof in Appendix A.

Lemma 2.2. Constant
( , ) is given by (1.2) and is attained on the subspaces given by the following rule.
(1) We partition [ ] into sets such that the cardinalities of any two sets differ by at most one.
(2) Let { 1 , . . . , ℓ } be one of the sets of the partition. Then, choosing arbitrary signs, we write the system of linear equations where [ ] denotes the -th coordinate of in R .
(3) Our subspace is the solution of the system of all equations written for each set of the partition at step (2).
From now on, we will study properties of the maximizers of (2.3) and work with tight frames.

Operations on frames
The following approach to our problem was proposed in [Iva18b] and used in [Iva18a] to study the properties of projections of the standard cross-polytope.
The main idea is to transform a given tight frame into a new one ′ 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 into a frame˜, and then we transform˜into a new tight frame ′ using a linear transformation. The main observation here is that we can always transform any frame˜= { 1 , . . . , } into a tight frame ′ using a suitable linear transformation : ′ =˜= { 1 , . . . , }. Equivalently, any non-degenerate centrally symmetric polytope in R is an affine image of a section of a high dimension cube.
For a frame in R , by definition put The operator is well-defined as the condition span = R implies that is a positive definite operator. Clearly, maps any frame to a tight frame: We obtain the following necessary and sufficient 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 ∈ Ω( , ) iff for an arbitrary frame˜in R inequality Proof. For any frame˜, we have that˜˜is a tight frame and vol (˜˜) = vol (˜)/ det˜. The maximum of (2.3) is attained at a tight frame iff vol (˜˜) ≤ vol ( ) for an arbitrary frame˜. Hence ividing by √︀ det˜, we obtain inequality (3.1).
Clearly, if˜in the assertion of Lemma 3.1 is close to , then the tight frame˜˜is close to 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 be an extremizer of (2.3), and be a map from a subset of Ω( , ) to the set of frames in R . In order to obtain properties of extremizers, we consider a composition of two operations: where˜is as defined above. For example, see Figure 1, where is the operation of replacing a vector of by the origin.
v Figure 1. Here we map one vector to zero.
Choosing a simple operation , 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 first-order approximation of the determinant was obtained by the author in [Iva18b, Theorem 1.2]. We provide a sketch of its proof in Appendix A.
Lemma 3.2. Let = { 1 , . . . , } ⊂ R be a tight frame and the -tuple˜be obtained from We state as lemmas several technical facts from linear algebra that will be used later.
For any -tuple of vectors of R with ∈ , we use ∖ to denote the ( − 1)-tuple of vectors obtained from by removing the first occurrence of in .

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.
We start with the rightmost inequality in (4.1). Let˜be the -tuple obtained from by substitution → .
Clearly, Theorem 4.1 implies Theorem 1.2. These theorems can be sharpened in the planar case.
Since is a frame in R 2 ,˜is a frame in R 2 as well. By Lemma 3.1 and Lemma 3.3, we get By identity (2.2), we have ( ) = (˜). By this and by inequalities (4.5) and (4.3), we get Combining the last two inequality, we obtain Remark 1. It is possible to sharpen inequality (4.1) for > 2 and > 2 using the same approach as in Lemma 4.1. The idea is to remove mod from or add − ( mod ) 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.

Local properties
In this section, we prove some properties of the local maximizers of (2.3). We will perturb facets of ( ) of a local maximizer (that is, we will perturb the vectors of in a specific way corresponding to a perturbation of some facets of the polytope ( )). 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. • for every ∈ , the hyperplane supports ( ) in a facet of ( ). That is, is the set of scaled outer normals of ( ). Denote = ( ). We fix ∈ and the facet = ∩ of . Let be the centroid of . Transformation 1. We will "shift" a facet of a polytope parallel to itself. Let ′ be obtained from by substitution → +ℎ | | , where ℎ ∈ R. Denote ′ = ( ′ ). That is, the polytopal set ′ is obtained from by the shift of the half-space + by ℎ in the direction of its outer normal. By the celebrated Minkowski existence and uniqueness theorem for convex polytopes (see, for example, [Gru07, Theorem 18.2]), we have (one of the sets + or − is empty if ( ) / ∈ ). Let ∈ (− /2, /2) be the oriented angle between hyperplanes and + such that is positive for positive . Let ′ be obtained from by substitution → + , where ∈ R and is a unit vector orthogonal to . Denote ′ = ( ′ ). Thus, for a sufficiently small | |, the polytopal set ′ is a polytope obtained from by the rotation of the half-space + around the codimension two affine subspace by some angle = ( ). Clearly, in order to calculate the volume of ′ , we need to subtract from vol the volume of the subset of that is above + and to add to vol the volume of the subset of ′ that is above . Formally speaking, denote Then, we have (see Figure 3) There is a nice approximation for vol + − vol − . Let + (resp., − ) be the set swept out by + (resp., − ) while rotating around by the angle . By routine, We claim here without proof that vol + = vol + + ( ) and vol − = vol − + ( ).
By this and identity (5.3), we obtain By this and by (5.1), we get Since and are orthogonal, we have Finally, we obtain For every ∈ , we denote the set ∩ ( ) by . We say that ∈ corresponds to a facet of ( ) if either = or = − . Clearly, if some vectors of correspond to the same facet of ( ), then they are equal up to a sign. For a given frame in R and ∈ R , a facet of ( ) and a vector ∈ R , we define an -substitution in the direction as follows: • each vector of such that ⊂ is substituted by + ; • each vector of such that − ⊂ is substituted by − ; • all other vectors of remain the same. In order to prove Theorem 1.1, we will use -substitutions.
At first, we simplify the structure of a local maximizer.
Lemma 5.1. Let be a local maximizer of (2.3) and ∈ . Then is a facet of ( ).
Proof. Let be a convex body in R , then its polar body is defined by {︀ ∈ R : ⟨ , ⟩ ≤ 1 for all ∈ }︀ .
Since ( ) is the intersection of half-spaces of the form {⟨ , ⟩ ≤ 1} with ∈ ± , we have that co{± } is polar to ( ). By the duality argument, it suffices to prove that ∈ is a vertex of the polytope co {± } . Assume that is not a vertex of co {± } .
Clearly, ∈ co {± ( ∖ )} and is not a vertex of the polytope co {± ( ∖ )} . Therefore we have that span{ ∖ } = span = R . That is, ∖ is a frame in R . Since ∖ is a nondegenerate linear transformation, ∖ is not a vertex of the polytope co {︀ ± ∖ ( ∖ ) }︀ . By this and by the triangle inequality, there is a vertex of co{± } such that ∈ and | ∖ | < | ∖ |.
As an immediate corollary of Lemma 5.1 and by the standard properties of polytopes, we have the following statement.
Corollary 5.1. Let be a local maximizer of (2.3) and ∈ . Let˜be the -tuple obtained from by -substitution in the direction with ∈ R and ∈ R . Then, for a sufficiently small | |,˜is a frame, the vector + corresponds to a facet of (˜) and the polytopes ( ) and (˜) have the same combinatorial structure. Moreover, vol (˜) is a smooth function of at = 0.
In the following two lemmas, we will perturb a local maximizer by making -substitutions. Geometrically speaking, making an -substitution in the direction with ∈ R and ∈ R, we move the opposite facets and − of a local maximizer in a symmetric way. Thus, for a sufficiently small , perturbations of the facets and − are independent.
Lemma 5.2. Let be a local maximizer of (2.3). Let ∈ and be the number of the vectors of that correspond to . Then Proof. Denote by˜the -tuple obtained from by -substitution in the direction with ∈ R. Thus, we apply Transformation 1 to the facets ± of ( ). By Lemma 3.1, we have By Lemma 3.2 and Corollary 5.1, both sides of this inequality are smooth as functions of in a sufficiently small neighborhood of 0 = 0. Consider the Taylor expansions of both sides of inequality (5.6) as functions of about 0 = 0.
By Lemma 3.3, det˜= 1 + (2 + 2 )| | 2 . Hence Geometrically speaking, we shift the half-space + (resp., − ) by in the directions of its outer normal. By this and by (5.2), we obtain Using identities (5.8) and (5.7) in (5.6) , we get Since˜= for = 0 and the previous inequality holds for all ∈ (− , ) for a sufficiently small , the coefficients of in both sides of the previous inequality coincide. That is, This completes the proof.
Lemma 5.3. Let be a local maximizer of (2.3) and ∈ . Then the line span{ } intersects the hyperplane in the centroid of the facet .
Proof. Denote the centroid of by and let = span{ } ∩ . Fix a unit vector orthogonal to . Denote by˜the -tuple obtained from by -substitution in the direction with ∈ R. Thus, we apply Transformation 2 to the facets ± of ( ). By Lemma 3.1, we have Since˜= for = 0 and the previous inequality holds for all ∈ (− , ) for a sufficiently small , the coefficients of in both sides of the previous inequality coincide. That is, we conclude Since , ∈ and the last identity holds for all unit vectors parallel to , it follows that = . The lemma is proven.
As a simple consequence of Lemma 5.3, we obtain the following result for the planar case.
Theorem 5.1. Let ∈ Ω( , 2) be a local maximizer of (2.3) for = 2. Then, the polygon ( ) is cyclic. That is, there is a circle that passes through all the vertices of ( ).
Proof. Denote the origin by . Let be an edge of ( ) and ℎ be the altitude of the triangle . By Lemma 5.3, ℎ is the midpoint of . Hence, the triangle is isosceles and = . It follows that ( ) is cyclic.
We are ready to give a proof of Theorem 1.1.
Proof of Theorem 1.1. Recall that any identification of with R identifies the projections of the standard basis { 1 , . . . , } with a tight frame, denoted by , that is a local maximizer of (2.3).
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. By Lemma 5.1, all vectors ∈ such that span intersects correspond to and have the same length that we denote by | |. Then by Lemma 5.3, the span of each of these vectors intersects in its centroid. Since the length of the altitude of the pyramid is 1/| |, we have 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 2 that maximizes (2.3) for > = 2 to prove that the section of the cube generated by the tight frame is a rectangle of area 4 ( , 2) = 4 √︀ ⌈ /2⌉⌊ /2⌋ with the sides of lengths 2 √︀ ⌊ /2⌋ and 2 √︀ ⌈ /2⌉. First, let us introduce the notation. Let = { 1 , . . . , } ∈ Ω( , 2) be a global maximizer of (2.3) for = 2 and > 2. Clearly, ( ) is a centrally symmetric polygon in R 2 . The number of edges of ( ) is denoted by 2 . Clearly, ≤ . By Theorem 5.1, the polygon ( ) is cyclic; and we denote its circumradius by . Let 1 , . . . 2 be the edges of ( ) enumerated in clockwise direction (that is, edges and + are opposite to each other, ∈ [ ]). We reenumerate the vectors of in such a way that the vector corresponds to the edge for every ∈ [ ]. The central angle subtended by the edge is denoted by 2 , ∈ [ ]. Clearly, we have the following identities (see Figure 4) (6.1) 1 + · · · + = 2 , Also, we note here that There are several steps in the proof. We explain the main steps briefly. In fact, we want to show that the number of edges of a local maximizer is 2 = 4. Using the discrete isoperimetric inequality (see below), we obtain an upper bound on Area ( ) in terms of . This upper bound yields the desired result for ≥ 8 (the bound is less than conjectured volume 4 ( , 2) for ≥ 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 -gons with fixed circummradius there is a unique maximal area polygon -the regular -gon. We will use a slightly more general form. Namely, fixing one or several central angles of a cyclic polygon, its area is maximized when all other central angles are equal. For example, this follows from concavity of the sine function on [0, ]. In our notation fixing the central angle and its vertically opposite, we have 6.1.
Proof. Since = 2, the polygon ( ) is an affine square. Hence the claim is an immediate consequence of Lemma 2.2.
Thus, it suffices to prove that = 2 for any > 2.
Combining this with inequalities (6.6) and (6.4), we obtain sin The claim follows.
Remark 2. The inequality on the area for ∈ {4, 6} and = 2 is a special case of the leftmost inequality in (1.1) originally proved by K. Ball [Bal89]. In [Iva19], the equality cases in this Ball's inequality are described.
Proof. Fix ∈ [ ]. By identity (6.2) and Lemma 4.1, we have 2 ≤ 3 2 cos 2 . By inequality (6.5), we get 3 cos 2 The function of in the left-hand side of this inequality is increasing on [0, /2]. Since the inequality does not hold for = /10, we conclude that is necessarily at least /10.
Proof. Assume that either = 3 or = 4. Denote by the number of vectors in that correspond to . We want to rewrite the inequality of Lemma 5.2 using the circumradius and the center angle. Since the length of edge is 2 sin and by identity (6.2), identity (5.5) takes the form 2 2 sin 2 = 2 cos 2 Area ( ).
Thus, we have proved that for a maximizer of (2.3), then = 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 3. We used the Ball inequality (1.1) to prove Theorem 1.3 for ∈ {3, 4, 6}. 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.
Sketch of the proof of Lemma 2.2. Let be a -dimensional subspace of R such that ( , ) is attained. Denote = ∩ . Since is an affine -dimensional cube, there are vectors { 1 , . . . , } such that = ⋂︀ . . , } be the projection of the vectors of the standard basis onto . By the same arguments as in Lemma 5.1, the hyperplane meets the polytope in a facet of for every ∈ [ ]. Thus, coincides with ± for a proper sign and ∈ [ ]. Or, equivalently, we partition [ ] into sets and 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 vectors of the standard basis of R project onto a pair ± . Therefore, a -tuple of vectors { √ } ∈[ ] is a tight frame. Identifying with R and by the assertion (4) of Lemma 2.1, we conclude that and are orthogonal whenever ̸ = . Therefore, | | 2 = 1 and (A.1) vol ∩ = 2 √︀ 1 · . . . · . Suppose ≥ + 2 for some , ∈ [ ]. Then · ≤ ( − 1)( + 1). By this and by (A.1), we showed that (1) holds.