Research article

Covering cross-polytopes with smaller homothetic copies

  • Received: 30 November 2023 Revised: 08 January 2024 Accepted: 10 January 2024 Published: 11 January 2024
  • MSC : 52A20, 52C17, 52A15

  • Let Cn be an n-dimensional cross-polytope and Γp(Cn) be the smallest positive number γ such that Cn can be covered by p translates of γCn. We obtain better estimates of Γ2n(Cn) for small n and a universal upper bound of Γ2n(Cn) for all positive integers n.

    Citation: Feifei Chen, Shenghua Gao, Senlin Wu. Covering cross-polytopes with smaller homothetic copies[J]. AIMS Mathematics, 2024, 9(2): 4014-4020. doi: 10.3934/math.2024195

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  • Let Cn be an n-dimensional cross-polytope and Γp(Cn) be the smallest positive number γ such that Cn can be covered by p translates of γCn. We obtain better estimates of Γ2n(Cn) for small n and a universal upper bound of Γ2n(Cn) for all positive integers n.



    Let K be a convex body in Rn, i.e., a compact convex set having interior points. The set of convex bodies in Rn is denoted by Kn, and the set of convex bodies that are centrally symmetric is denoted by Cn. For each xRn and each λ>0, the set

    x+λK:={x+λyyK}

    is called a homothetic cop of K; when λ(0,1), it is called a smaller homothetic copy of K. For each KKn, we denote by c(K) the least number of translates of intK needed to cover K. Concerning the least upper bound of c(K) in Kn, there is a long-standing conjecture (see [1,2,3,4,5,6] for the origin, history, and classical known results concerning this conjecture):

    Conjecture 1. (Hadwiger's covering conjecture [4]) For each KKn, we have

    c(K)2n,

    and the equality holds if and only if K is a parallelotope.

    Although many people have conducted in-depth research, this conjecture is confirmed completely only for the planar case [7]. In [8], Chuanming Zong proposed a four-step program to attack this conjecture. In this program, it is important to estimate

    Γm(K):=inf{γ>0{cii[m]}Rn s.t. Ki[m](ci+γK)},

    i.e., Γm(K) is the smallest positive number γ such that K can be covered by m translates of γK. The map Γm(): Kn[0,1], KΓm(K) is called the covering functional with respect to m, where [m]:={iZ+1im}. Clearly, c(K)m if and only if Γm(K)<1. For each mZ+, Γm() is an affine invariant. More precisely,

    Γm(K)=Γm(T(K)), TAn,

    where An is the set of non-degenerate affine transformations from Rn to Rn.

    A compact convex set K is said to be an d-dimensional cross-polytope if there exist d linearly independent vectors v1,,vd such that

    K=conv{±v1,,±vd}.

    Clearly, any d-dimensional cross-polytope is the image of

    Cd={(α1,,αd)Rdi[d]|αi|1}

    under a non-degenerate affine transformation. Therefore, Γm(K)=Γm(Cd) holds for each pair of positive integers m and d. In a recent work [9], Xia Li et al. obtained some estimations of Γm(Cd) for large d. Moreover, they showed that, if PCn is a convex polytope with 2d vertices, then

    Γm(P)Γm(Cd), (1.1)

    which shows the importance of estimating Γm(Cd).

    It is well known that Γ2n([1,1]n)=1/2,n2. It is interesting to ask whether there exists a universal upper bound for Γ2n(Cn). In this paper, by using elementary yet interesting observations and refining techniques used in the recent works [9,10], we get better estimates of Γ2n(Cn). Based on this, we present the first nontrivial universal upper bound of Γ2n(Cn) for all positive n. By (1.1), results mentioned above yield also estimates of covering functionals of convex polytopes with few vertices.

    Throughout this paper, the dimension n of the underlying space is at least 3.

    For each n,kZ+, we put

    M(n,k)={(α1,,αn)Zni[n]|αi|k}.

    It is known that (cf. [11] or [12])

    #M(n,k)=ni=nk2ni(ni)(kni).

    Lemma 1. Let n,kZ+. If kn2, then

    (n+k)CnnCn+Sk,

    where

    Sk={(α1,,αn)Zni[n]|αi|=k}{o}.

    Moreover,

    #Sk=ki=12i(ni)(k1i1)+1.

    Proof. Let (α1,,αn) be an arbitrary point in (n+k)Cn. Then

    i[n]|αi|n+k.

    If i[n]|αi|n, then (α1,,αn)nCnnCn+Sk. Otherwise, there exists m[k] such that

    n+m1<i[n]|αi|n+m.

    On the one hand, since i[n](|αi||αi|)<n, we have i[n]|αi|m. Then there exist integers β1,,βn0 such that

    βi|αi|, i[n]  and  i[n]βi=m.

    Clearly, we have

    i[n]|αisgnαiβi|=i[n](|αi|βi)n.

    Therefore,

    (α1,,αn)=(α1sgn(α1)β1,,αnsgn(αn)βn)+(sgn(α1)β1,,sgn(αn)βn)nCn+Sm.

    On the other hand, set

    mi={|αi|, if |αi||αi|<12,|αi|+1, if |αi||αi|12,i[n]. (2.1)

    We have

    (α1,,αn)=(α1sgn(α1)m1,,αnsgn(αn)mn)+(sgn(α1)m1,,sgn(αn)mn).

    By the Triangle Inequality, we have

    ni[n]mi<i[n]|αi|i[n]mii[n]|αisgn(αi)mi|=i[n]||αi|mi|n2.

    Thus,

    m1++mn>n2k.

    Without loss of generality, assume that α1,,αn0, and

    α1,,αn01, αn0+1,,αn0[12,1), αn0+1,,αn[0,12).

    By (2.1), we have

    βiαimiαi, i[n0],βi=0<1=mi, i[n0][n0],βi=mi=0, i[n][n0].

    Then there exist integers m1,,mn such that

    βimimi, i[n]  and  i[n]mi=k.

    Set, for each i[n], fi(λ)=|αiλ|. Then fi is decreasing on [βi,αi]. We claim that

    fi(βi)fi(mi),i[n]. (2.2)

    The case when mi[βi,αi] is clear. If mi>αi, then mi=mi=αi+1 and 1/2αiαi<1. Thus

    fi(βi)fi(αi)=αiαi121(αiαi)=fi(αi+1)=fi(mi).

    Hence (2.2) holds as claimed. It follows that

    i[n]|αimi|=i[n]fi(mi)i[n]fi(βi)n.

    Therefore,

    (α1,,αn)=(α1m1,,αnmn)+(m1,,mn)nCn+Sk.

    Moreover,

    #Sk=#M2(n,k)#M2(n,k1)+1=ni=nk2ni(ni)(kni)ni=nk+12ni(ni)(k1ni)+1=2k(nnk)(kk)+2k1(nnk+1)(kk1)++(nn)(k0)2k1(nnk+1)(k1k1)(nn)(k10)+1=2k(nnk)+k1i=12i(nni)[(ki)(k1i)]+1=2k(nnk)(k1k1)+k1i=12i(nni)(k1i1)+1=ki=12i(nni)(k1i1)+1=ki=12i(ni)(k1i1)+1.

    For each nZ+, let k1(n) be the nonnegative integer satisfying

    k1(n)i=12i(ni)(k1(n)1i1)+12n<k1(n)+1i=12i(ni)(k1(n)i1)+1.

    It is easy to prove that k1(n)n2.

    Corollary 2. For each nZ+, we have

    Γ2n(Cn)nn+k1(n).

    Remark 3. It can be verified that

    ki=12i(ni)(k1i1)+12kki=1(ni)(k1i1)=2kki=1(nni)(k1i1)=2k[(nn1)(k10)++(nnk)(k1k1)]=2k(n+k1n1)2k(n+kn).

    For x(0,+), we define

    g(x)=2x(1+x)(1+x)xx.

    Clearly, g is strictly increasing on (0,+), and limx0+g(x)=1. For each t(1,+), let b(t) be the solution to the equation g(x)=t. Numerical calculation shows that b(2)0.205597. We can easily prove that, if k2(n) is the integer satisfying

    2k2(n)(n+k2(n)n)2n<2k2(n)+1(n+k2(n)+1n),

    then we have limnk2(n)n=b(2) [9,10]. It can be verified that

    limnk1(n)n>b(2).

    Therefore, the estimate in Corollary 2 is slightly better than that given by [9, Proposition 5] in the asymptotical sense, and it is much better for particular choices of small n. For example, we have k1(7)=2 and k2(7)=1. It follows that

    Γ128(C7)nn+k1(n)=77+20.78,

    which is better than Γ128(C7)nn+k2(n)=77+10.875 [9]. See Table 1 for more examples.

    Table 1.  Comparison of estimates of Γ2n(Cn).
    n k2(n) nn+k2(n) k1(n) nn+k1(n)
    7 1 0.875 2 0.778
    11 2 0.846 3 0.786
    16 3 0.842 4 0.8
    20 4 0.833 5 0.8
    25 5 0.833 6 0.806

     | Show Table
    DownLoad: CSV

    Theorem 4. For each n3, we have

    Γ2n(Cn)67.

    Proof. By numerical calculations, for each 3n49, we have Γ2n(Cn)67. Set c=b(2)0.02. Then for each n50, we have cnb(2)n1, which shows that (1+c)nn+b(2)n. Therefore,

    Γ2n(Cn)nn+b(2)nn(1+c)n0.8435.

    Thus, for each n3, we have

    Γ2n(Cn)max{67,0.8435}=67.

    By refining techniques used in the recent works [9,10], we get better estimates of Γ2n(Cn) and the first nontrivial universal upper bound of Γ2n(Cn). It is natural to find universal bounds of Γ2n(Bnp) for fixed p(1,), where Bnp is the closed unit ball of (Rn,p).

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors are supported by the [National Natural Science Foundation of China] grant numbers [12071444] and the [Fundamental Research Program of Shanxi Province of China] grant numbers [202103021223191].

    There is no conflicts of interest between all authors.



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