aa r X i v : . [ m a t h . A T ] J u l INSCRIBING A REGULAR CROSSPOLYTOPE
R.N. KARASEV
Abstract.
In this paper we prove that it is possible to inscribe a regular crosspolytope(multidimensional octahedron) into a smooth convex body in R d , where d is an odd primepower. Some generalizations of this statement are also proved. Introduction
The problems of inscribing or outscribing a polytope of a given family to some smoothor convex line or surface have a long history. In [18] it was proved that a square canbe inscribed into any simple smooth closed curve in the plane (Schnirelmann’s theorem).In [7] it was proved that a cube can be outscribed about any convex body in R . In [2]it was shown that if a smooth closed hypersurface H ⊂ R d bounds a region with nonzeroEuler characteristic, then H contains vertices of a simplex, positively homothetic to anyprescribed simplex.The questions of inscribing and outsribing are widely discussed in the books [3, 12].More recent results and references can be found in [14, 15].The inscribing and outscribing problems are closely connected to the Knaster prob-lem [13] on the level surfaces of function on S d − . Conjecture 1 (Knaster’s problem for functions) . Let S d − be a unit sphere in R d . Supposethat the points x , . . . , x d ∈ S d − and a continuous function f : S d − → R are given. Thenthere exists a rotation with positive determinant ρ ∈ SO ( d ) such that f ( ρ ( x )) = f ( ρ ( x )) = · · · = f ( ρ ( x d )) . In [10, 5] some counterexamples to the Knaster problem were found, with some specialfunctions and point sets, the dimension d was required to be large. Still for some particularpoint sets { x , . . . , x d } the conjecture turns out to be true. In Section 3 we use a positivesolution to the Knaster problem, that is formulated in terms of the Euler class of somevector bundle.In the book [12] the problem of inscribing a regular octahedron into a convex body in R is formulated (Problem 11.5). For the case of d = 3 and a smooth convex body, thisproblem was solved in [14]. Here we prove a similar statement for larger dimensions. Mathematics Subject Classification.
Key words and phrases. inscribing, Knaster’s problem.The research of R.N. Karasev was supported by the President’s of Russian Federation grant MK-1005.2008.1, and partially supported by the Dynasty Foundation.
Definition 1.
Let an orthogonal cross in R d be d mutually orthogonal straight lines,passing through a single point o , the latter is called the center of the cross . Definition 2.
Let ( e , . . . , e d ) be some linear base in R d . The convex hull of the points e , − e , e , − e , . . . , e d , − e d ∈ R d and all its similar images (under transforms with positive determinant) are called a crosspoly-tope in R d . If the base is orthonormal, we call the crosspolytope regular . Definition 3.
A convex body K ⊂ R d is called non-angular , if none of the points o ∈ ∂K is a center of some orthogonal cross C such that C ∩ int K = ∅ .It is clear that smooth convex bodies are non-angular. Now we can formulate the mainresults. Theorem 1.
Let K ⊂ R d be a non-angular convex body, let d be an odd prime power.Then there exists a regular crosspolytope C ⊂ R d such that all its vertices are on ∂K . The same proof is used to establish the following fact.
Theorem 2.
Let H ⊂ R d be an image of some smooth embedding of S d − , where d = p k is an odd prime power. Let ( e , . . . , e d ) be some linear base in R d , let C be the convex hullof ( ± e , . . . , ± e d ) .Suppose that the group G = ( Z p ) k acts transitively on the vectors ( e , . . . , e d ) and thisaction is a restriction of some (special) orthogonal action of G on R d . Then there exists acrosspolytope C ′ ⊂ R d , similar to C , with all its vertices lying on H . It should be mentioned that the possibility to inscribe a regular crosspolytope into acentrally symmetric convex body follows from the positive solution of the Knaster problemfor the orthonormal base in [20], which is true for every dimension d .Theorem 2 for a regular crosspolytope in arbitrary dimension was announced in [4], butthe sketch of the proof was based on some two-dimensional reasoning. Though, the two-dimensional case (Schnirelmann’s theorem) was proved in [4] incorrectly, the main lemmaon the continuous dependence of the inscribed square on the smooth curve is false andit has simple counterexamples. In the plane [18] the parity of the number of inscribedsquares remains constant under the deformations of the curve. In the space of higherdimension there is an infinite variety of inscribed crosspolytopes, and the counting andparity argument does not work.In the case d = 2 the proof in this paper cannot be applied. In this case some largergroup (at least Z ) should be considered as the symmetry group.2. Reduction to a topological fact
Denote { , , . . . , n } = [ n ].First we outline the proof of Theorem 1 for strictly convex non-angular K , which alsoproves Theorem 2 for the case, when H is a boundary of some smooth strictly convex body K . NSCRIBING A REGULAR CROSSPOLYTOPE 3
Let R be the space of positively oriented orthonormal frames in R d . In the case ofTheorem 2 let R be the space of frames, that can be obtained from the given frame( e , . . . , e d ) by a special orthogonal transform. In both cases R is naturally identified with SO ( d ). Note that in Theorem 2 the lengths of the vectors are equal, so we consider themto be unit vectors.Consider a continuous map g from K × R to the linear space V = R d , defined as follows.For a point p ∈ K , a frame ( e , . . . , e d ) ∈ R and any i ∈ [ d ] put a i = max { a : p + ae i ∈ K } , b i = max { a : p − ae i ∈ K } . It is quite clear that for strictly convex K this map is continuous.Consider the coordinates s i = a i + b i , t i = a i − b i in V , and consider the one-dimensionalsubspace L ⊂ V , given by t = · · · = t d = 0 , s = · · · = s d . In the quotient space
V /L , let the d -dimensional linear hull of { t , . . . , t d } be U , and the d − { s , . . . , s d } be W . Now we have to prove that the quotientmap f : K × R → U ⊕ W maps some pair ( p, r ) ∈ int K × R to zero.3. Calculating the obstruction
In the sequel we consider cohomology mod p , the coefficients are omitted in the nota-tion. The necessary facts on algebraic topology and vector bundles can be found in [6, 16,17].Note that if K is non-angular (or smooth), the map f is nonzero at any pair ( p, r ) ∈ ∂K × R . In case of Theorem 1 we can consider some free transitive action of G = ( Z p ) k (where d = p k ) on R by permuting the vectors. In case of Theorem 2 the action of G on R is already given. Correspondingly, G acts on U and W by permuting the respectivecoordinates t i , s i .Thus the map f is an equivariant section of a G -bundle K × R × U × W → K × R over the G -space K × R and the obstruction for f being nonzero over the entire K × R is therelative Euler class (see [11, 8]), which resides in the equivariant cohomology H d − G ( K × R, ∂K × R ).The section f splits as a direct sum of s U ⊕ s W for the corresponding G -bundles. Underthis splitting the section s U is nonzero over ∂K × R and gives some relative Euler class in e ( s U ) ∈ H dG ( K × R, ∂K × R ). The section s W has some Euler class e ( s W ) ∈ H d − G ( K × R ) = H d − G ( R ). The Euler class of f is therefore e ( f ) = e ( s U ) e ( s W ) by the multiplicative rulefor the Euler class.It is clear from the K¨unneth formula that the cohomology H ∗ G ( K × R, ∂K × R ) equalsthe tensor product H ∗ ( K, ∂K ) ⊗ H ∗ G ( R ), in this particular case it is u × H ∗ G ( R ), where u is the d -dimensional generator of H ∗ ( K, ∂K ).Let us find e ( s U ) first. Note that if we deform K , leaving it non-angular, this classremains the same. Thus it is sufficient to consider K equal to the unit ball B . In this R.N. KARASEV case we fix the frame r in the pair ( p, r ) ∈ B × R and note that the section s U has theonly non-singular zero in the center of B × { r } . Hence the image of e ( s U ) under thenatural map H dG ( B × R, ∂B × R ) → H d ( B, ∂B ) gives the generator of H d ( B, ∂B ), hence e ( s U ) = u × ∈ H ∗ G ( K × R, ∂K × R ).Now note that the Euler class e ( s W ) coincides with the obstruction, which is used inestablishing a particular case of the Knaster problem in [19]. By definition W = R [ G ] / R asa representation of G , this representation has no trivial irreducible summands. Therefore(see [6, Ch. III § e ( W ) ∈ H d − G ( EG ) = H d − ( BG ) is nonzero. Consider the natural classifying G -map p : R → EG . In the paper [19] an action of G on SO ( d ) was considered, which isequivalent to the action of G on R , considered here. In [19, Proposition 3, p. 127] it wasshown that the cohomology map p ∗ : H mG ( EG ) → H mG ( SO ( d )) = H mG ( R ) is injective for all m < p k − p k − ), hence e ( s W ) = p ∗ e ( W ) = 0.Thus we have e ( f ) = u × e ( s W ) = 0 by the K¨unneth formula.4. Removing the limitation of strict convexity
Consider some sequence of strictly convex smooth bodies K n , that converges to K in theHausdorff metric. Each of K n has some inscribed crosspolytope C n . From the compactnessconsiderations, we can suppose that C n converges to some crosspolytope C , which can bedegenerate (consist of one point).Assume that C is degenerate. Again, from the compactness considerations, we mayassume that the guiding cross of C n tends to some cross X , centered at C ∈ ∂K . The cross X must have nonempty intersection with int K (from the non-angularity in Theorem 1 orsmoothness in Theorem 2) of some positive length ε . It means that some line of X , denoteit l , intersects int K by the open segment σ of length at least ε , and σ does not containthe center of X .Consider the guiding crosses X n of C n , close enough to X . Their corresponding lines l n intersect int K n by some open segments σ n , that tend to σ . It can be easily seen, that forlarge enough n the center of σ n does not coincide with the center of X n , so X n cannot be aguiding cross of an inscribed crosspolytope. That is a contradiction, so C is non-degenerate.It is not clear whether the non-angularity condition can be omitted in Theorem 1, sincethe crosspolytopes can degenerate in this case.5. The case of non-convex hypersurface
Let us prove Theorem 2 for arbitrary hypersurface H , which is an embedded sphere.We can smoothly deform H into an ellipsoid with distinct axes, let all the deformations H t be inside some ball B .Let us describe the configuration space of all crosspolytopes, similar (with positive de-terminant) to C , having center in B . They are parameterized by B × I × SO ( d ), where B is for centers, SO ( d ) is for rotations, and I = [ a, b ] is for homothety (0 < a < b ).It is clear that for small enough a and large enough b the crosspolytopes of sizes ≤ a and ≥ b cannot be inscribed into H t for any t ∈ [0 , L = ∂B × I × SO ( d ) S B × ∂I × SO ( d ). NSCRIBING A REGULAR CROSSPOLYTOPE 5
Let G = ( Z p ) k act on SO ( d ) as above. By the K¨unneth formula H ∗ G ( B × I × SO ( d ) , L ) = u × v × H ∗ G ( SO ( d )), where u is the generator of H d ( B, ∂B ), v is the generator of H ( I, ∂I ).Each surface H t can be considered as the (non-degenerate everywhere) zero set of somesmooth function f t , we can assume that f t depends on t continuously, since the homotopy H t is a restriction of some isotopy of the whole R d . Consider the map g t : B × I × SO ( d ) → R d , which maps the triple ( x, λ, ρ ) to( f t ( x + λρ ( e )) , f t ( x + λρ ( − e )) , . . . , f t ( x + λρ ( e d )) , f t ( x + λρ ( − e d ))) . The map g t commutes with the action of G on B × I × SO ( d ), and the action of G on R d by permuting the coordinates. Thus it makes sense to consider the Euler class e ( g t ) ∈ H dG ( B × I × SO ( d ) , L ). This class does not change under the deformations, nowwe calculate it for the case of H t being an ellipsoid. Similar to what is done in Section 3 wefind that in this case e ( g t ) = u × v × e ( W ) = 0. Hence the zero set of g t is nonempty, every H t has an inscribed crosspolytope. Moreover, similar to the results of [19], the family ofinscribed crosspolytopes of H has dimension at least ( d − d − Conclusion
The author thanks A.Yu. Volovikov for the discussion of these results. Some applicationof the above technique to the measure partition problem is given in [9].
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