LARGE REGULAR SIMPLICES CONTAINED IN A HYPERCUBE
HIROSHI MAEHARA, IMRE Z. RUZSA, AND NORIHIDE TOKUSHIGE
ABSTRACT. We prove that the n-dimensional unit√ hypercube contains an n-dimensional regular simplex of edge length c n, where c > 0 is a constant independent of n.
Let ℓ∆n be the n-dimensional regular simplex of edge length ℓ, and let ℓQn be the n-dimensional hypercube of edge length ℓ. For simplicity, we omit ℓ if ℓ = 1, e.g., Qn denotes the unit hypercube. We are interested in the maximum edge length of a regular n-dimensional simplex contained in Qn.
Theorem. For every ε0 > 0 there is an N0 such that for every n > N0 one has ( ) 1 − ε √ 0 n ∆ ⊂ Q . 2 n n √ On the other hand, if ℓ∆n ⊂ Qn, then ℓ ≤ (n + 1)/2, which follows by comparing the√ circumscribed balls of ℓ∆n and Qn. (Recall that the circumra- dius of ∆n is n/(2n + 2).) This upper bound is reached iff there exists an Hadamard matrix of order n+1. Schoenberg [3] pointed out that this “read- ily established fact” went back to Coxeter, see√ also §4 of [1]. Our lower bound given by the theorem is approximately 1/ 2 of the upper bound.
Proof of Theorem. For a matrix (or a vector) A = (ai j), let us define its norm by ∥A∥ := maxi j |ai j|. Let Jn be the n × n all one matrix.
Lemma 1. Let A = (ai j) be an n × n real orthogonal matrix, and let c > 0 be a constant. If 1 ∥A∥ ≤ √ (1) c n
Date: November 18, 2008. 2000 Mathematics Subject Classification. Primary: 52C07 Secondary: 05B20. Key words and phrases. regular simplex, hypercube, Hadamard matrix. The first and the last authors were supported by MEXT Grant-in-Aid for Scientific Research (B) 20340022. Second author supported by Hungarian National Research Fund (OTKA), grants No. K 61908, K 72731. 1 2 HIROSHI MAEHARA, IMRE Z. RUZSA, AND NORIHIDE TOKUSHIGE and 1 ∥J A∥ ≤ , (2) n c √ then we have (c n/2)∆n ⊂ Qn.
Proof. Let pi = (ai1,...,ain) be√ the i-th row of the matrix A. Then the√ n n points p1,..., pn ∈ R form a 2∆n−1. By (1), we have ∥pi∥ ≤ 1/(c n) for all 1 ≤ i ≤ n. √ n Let g = (g1,...,√ gn) ∈ R be the barycenter of the above 2∆n−1, and let pn+1 := (1 − n + 1)g. Then√ a computation shows that the p + 1 points ∈ Rn ∆ p1,..., pn, pn+1√ form a 2√n. Moreover, it follows√ from (2)√ that ∥pn+1∥ = ∥g∥( n + 1−1) ≤ 1/(c n). Thus we have 2∆n ⊂ (2/(c n))Qn, as desired. ¤
Let us find orthogonal matrices satisfying (1) and (2). Let q be an odd prime power, and let Fq = {b0,...,bq−1} (b0 = 0) be the finite field of order q. Define a character χ : Fq → {0,±1} by χ(0) = 0, χ(x) = 1 if x is a square, and χ(x) = −1 if x is a nonsquare. Define an q × q matrix B = T (bi j) by setting bi j := χ(bi −b j). Then this matrix satisfies BB = qIq −Jq, BJq = JqB = O. (See pp. 202–203 in [2] for the proof and how to use this matrix to construct an Hadamard matrix of Paley type.) Finally we define an orthogonal q × q matrix Aq by 1 ( 1 ) A := √ B + √ J . q q q q
∥ ∥ ≤ √1 1 Then, it is easy to check that Aq satisfies Aq q + q and JqAq = Jq. Thus the matrix Aq satisfies (1) and (2) for c = 1 − o(1), and this verifies the theorem for the case when the dimension is an odd prime power.√ (By | − | ≤ using the fact that each entry ai j of Aq satisfies ai j √1/q 1/ q, instead of (1), we can remove the o(1), i.e., we actually get q/2∆q ⊂ Qq.) Now let q1,...,qr be distinct odd prime powers, and let n = q1 ···qr and ⊗ ··· ⊗ An := Aq1 Aqr . Then the matrix An is orthogonal with 1 r ( 1 ) ∥An∥ ≤ √ ∏ 1 + √ . (3) n i=1 qi ⊗···⊗ ⊗···⊗ Moreover, An satisfies JnAn = Jn because JnAn = (Jq1 Jqr )(Aq1 ⊗ ··· ⊗ Aqr ) = (Jq1 Aq1 ) (Jqr Aqr ) = Jn. We notice that r ( 1 ) 1 ) 1 ∏ 1 + √ ≤ ∏(1 + √ = ∑ √ =: g(n), (4) i=1 qi p|n p d|n d LARGE REGULAR SIMPLICES CONTAINED IN A HYPERCUBE 3 where the product in the middle term is taken for all primes p dividing n. Thus (3) gives (1) with c = 1/g(n), and Lemma 1 implies that √ ( n ) √ ∆n ⊂ Qn (5) g(n) 2 for every odd integer n.
Lemma 2. For every ε,δ > 0 there is an n0 with the following property. For every integer n > n0 there are odd integers n1,n2 such that 2n = n1 +n2, (1 − ε)n ≤ ni ≤ (1 + ε)n and g(ni) < 1 + δ for each i = 1,2. Proof. We will select an m, define q = ∏ p p≤m as the product of the primes up to m and select n1,n2 coprime to q. This guarantees that they are odd. First we average g(n) over integers coprime to q. Let (q,r) = 1. Write Ir for the set of integers { j ∈ [(1 − ε)n,(1 + ε)n] : j ≡ r (mod q)}. We have 1 ∑ (g( j) − 1) = ∑ √ Nd, j∈Ir d>1 d where Nd is the number of multiples of d in our set Ir. Clearly Nd = 0 if (d,q) > 1. If (d,q) = 1, then the multiples of d in this residue class form an arithmetic progression with difference qd, and we have the estimate 2εn N ≤ + 1. d qd
Furthermore Nd = 0 if d ≥ 2n. Our choice of q implies that whenever d > 1 and (d,q) = 1, then d > m, so we have ( ) 1 2εn ∑ (g( j) − 1) ≤ ∑ √ + 1 . qd j∈Ir m
Proof. Let p0, p1,..., ps be the vertices of ℓ∆s inside Qs, and let q0,q1,...,qt be the vertices of ℓ∆t inside Qt. We may assume that the origin is the centers of these regular√ simplices. Then the distance between pi and the origin is given by ℓ s/(2s + 2). We will construct ℓ∆s+t+1 with vertices u0,...,us,v0,...,vt as follows. Define ui for 0 ≤ i ≤ s and v j for 0 ≤ j ≤ t by s t s t ui = (pi,⃗0,x) ∈ R × R × R, v j = (⃗0,q j,0) ∈ R × R × R.
Choose x > 0 so that |ui − v j| = ℓ for all i, j. This can be done by solving s t |u − v |2 = ℓ2 + ℓ2 + x2 = ℓ2, i j 2s + 2 2t + 2 √ 1 1 1/2 which gives x = ℓ( 2s+2 + 2t+2 ) < ℓ/ s + 1 < 1. Namely, we have ui,v j ∈ Qs × Qt × [0,1] = Qs+t+1, for all i, j. ¤
We are ready to prove the theorem. Let ε0 > 0 be given. Set ε = ε0/2 and take δ > 0 so that √ 1 − ε = 1 − ε/(1 + δ). (7)
Plug these ε and δ into Lemma 2 to get k0 = k0(ε,δ) > 0 such that for every k > k0 there are k1,k2 satisfying 2k = k1 + k2, ki ≥ (1 − ε)k, and g(k ) < 1 + δ. Choose N ≥ 2k so that i 0 √0 √ (1 − ε) n − 1 > (1 − ε0) n (8) holds for all n > N0. LARGE REGULAR SIMPLICES CONTAINED IN A HYPERCUBE 5
Now, let n > N0 be given. First assume that n is odd, and write n = 2k+1. ∆ ⊂ Lemma 2 gives a decomposition 2k = k1 +k2. Then we have ℓi ki Qki for i = 1,2, where √ √ √ √ √ (5) ki (1 − ε)k (7) (1 − ε) k 1 − ε (8) 1 − ε0 ℓi = √ > √ = √ = n − 1 > n. g(ki) 2 (1 + δ) 2 2 2 2 √ 1−ε0 Applying Lemma 3 with s = k1,t = k2 and ℓ = 2 n, we have the desired result ℓ∆n ⊂ Qn. Next assume that n is even, and write n = 2k. Lemma 2 gives 2k = k1 +k2 and √ √ (3)(4) δ ∥ ∥ ≤ g√(ki) √1 + (7) 2√ 2 √ √1 Aki < = < =: . ki (1 − ε)k (1 − ε) n (1 − ε0) n c n Define an n × n orthogonal matrix C by ( ) A 0 C = k1 . 0 A √k2 ∥ ∥ ≤ ∥ ∥ ∥ ∥ ∥ ∥ Then we have C max Aki < 1√/(c n) and JnC = max Jki Aki = −ε √ ∆ 1 0 ∆ ⊂ 1. Thus, by Lemma 1, we have (c n/2) n = ( 2 n) n Qn. This completes the proof of the theorem. ¤
REFERENCES [1] M. Hudelson, V. Klee, D. Larman. Largest j-simplices in d-cubes: some relatives of the Hadamard maximum determinant problem. Linear Algebra Appl. 241/243 (1996) 519–598. [2] J. H. van Lint, R. M. Wilson. A course in combinatorics. Second edition. Cambridge University Press, Cambridge, 2001. [3] I. J. Schoenberg. Regular simplices and quadratic forms. Journal of the London Math- ematical Society 12 (1937) 48-55.
COLLEGEOF EDUCATION,RYUKYU UNIVERSITY,NISHIHARA,OKINAWA, 903- 0213 JAPAN E-mail address: [email protected]
ALFRED´ RENYIINSTITUTEOF´ MATHEMATICS,HUNGARIAN ACADEMYOF SCI- ENCES,BUDAPEST,PF. 127, H-1364 HUNGARY E-mail address: [email protected]
COLLEGEOF EDUCATION,RYUKYU UNIVERSITY,NISHIHARA,OKINAWA, 903- 0213 JAPAN E-mail address: [email protected]