The Hadwiger Conjecture for Unconditional Convex Bodies

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1 The Hadwiger Cojecture for Ucoditioal Covex Bodies Daiel Barg September 5, 206 Abstract We ivestigate the famous Hadwiger Cojecture, focusig o ucoditioal covex bodies. We establish some facts about coverig with homotethic copies for L p balls, ad prove a property of outer ormals to ucoditioal bodies. We also use a projectio method to prove a relatio betwee the bouded ad ubouded versios of the cojecture. Defiitios ad Backgroud Defiitio. A covex body (closed ad compact) i R is called ucoditioal if it is symmetric with respect to the coordiate hyperplaes. Defiitio 2. For covex bodies K ad T, let N(K, T ) be the miimum umber of traslates of T eeded to cover K. Cojecture (Hadwiger). For K a covex body i R, N(K, it(k)) 2. 2 L p balls 2. 2 traslates For B p the uit ball i the L p orm, we first show that N(B p, ( ) p B p ) 2, i.e. we ca achieve a homothety ratio of ( ) p with just 2 traslates. We start by determiig the coverig umber i(b ) =

2 N(B, it(b )) of the cross-polytope, B. First, it is impossible to cover it with less tha 2 traslates. If this was possible, we d have two vertices (poits of the form (0,..., ±,..., 0) ), say v ad v 2 satisfyig for some y R. But we have v y <, v 2 y < v y + v 2 y (v y) + (y v 2 ) = v v 2 = 2, so we eed at least 2 traslates. Let s suppose that traslates of the form ±λe i work, where e i are the stadard basis vectors ad λ (0, ]. We eed that x B i s.t. x λe i <. Sice x = i s.t. x i. Let λ =, the the traslate ± e i works, where the sig depeds o that of x i. The x e i = x i + x i = < if x i > 0 ad x + e i = x i + + x i = < if x i < 0 Thus, i(b ) = 2, ad, i particular, the homothety ratio still works with 2 traslates. For geeral p we may do a similar procedure. Assume traslates of the form ±λe i work. We eed that x Bp i s.t. x λe i p <. Sice x p =, i s.t. x p, so let us set λ = p. Assumig wlog that x i > 0, i.e. x i [ p, ], we have x p p p = x i p p + x i p = (x i p ) p + x p i. Differetiatig the above expressio with respect to x i we get p(x i p ) p px p i. But x i > x i p, so the orm is strictly decreasig o [ p, ]. Thus, x p p p p p = < I particular, the homothety ratio ( ) p N(B p, ( ) p B p ) 2. works with 2 traslates, so 2

3 2.2 2 traslates The above approach uses traslates that work for oe extreme of the L p balls (p = ). We ca also use the traslates that work for the other extreme: L, i.e. the cube. We kow that traslates i the directios of the diagoals give the coverig umber 2. Sice the L p balls are ucoditioal bodies, we may focus o just oe quadrat, so let s assume wlog that all coordiates are o-egative. Assume a traslate of the form ɛ = (ɛ,..., ɛ) covers the etire first quadrat of B p, for some ɛ > 0. To fid the best possible epsilo (correspodig to the smallest possible homothety ratio), we must fid the followig: max f(x) subject to g(x)=0, where x:x i 0 i f(x) = (x,..., x ) ɛ p p ad g(x) = (x,..., x ) p p. We solve this problem with the method of Lagrage multipliers, first cosiderig the boudary of B p itself (x i > 0 i), ad the cosiderig the boudary of the quadrat separately. f = µ g for some costat µ (p(x ɛ) p,..., p(x ɛ) p ) = µ(px p,..., px p ), where we have assumed that x i ɛ i (the fial result is the same for the case 0 < x i < ɛ).we get that (x i ɛ) p = µx p ɛ i, or x i = (assumig p > ). µ p We have that x p ɛ p = ( ) p = µ = ( ɛ p ) p x i = p. µ p The the correspodig distace (homothety ratio) is ( p,..., p ) ɛ p = p ( p ɛ) = ɛ p Now we examie the boudary of the quadrat. Suppose x i = 0 for some i. The f(x,..., x i, 0, x i+,..., x ) = ɛ p + j i(x j ɛ) p. We must ow maximize the sum j i(x j ɛ) p where (x j ) p =, but this is just the j eqi above Lagrage multiplier problem i dimesios, so the correspodig critical value is f(( ) p,..., 0,..., ( ) p ) = ɛ p + ( ɛ( ) p ) p. Notice that < ɛ( ) p < ɛ p ( ɛ( ) p ) p > ( ɛ p ) p (( ) p,..., 0,..., ( ) p ) ɛ) p > ( p,..., p ) ɛ p. 3

4 Sice the right had side correspods to the oly critical value give to us by the origial Lagrage problem, we kow that the absolute maximum must occur oe of the coordiate hyperplaes. We may cotiue to iductively assume that a greater distace occurs at a poit where some (ew) x j = 0, evetually gettig that the absolute maximum must occur at oe of the basis vectors e i. Let s assume wlog that it occurs at e. The the absolute maximum of f i the first quadrat is f(, 0,..., 0) = ( )ɛ p + ( ɛ) p We eed the right had side to satisfy ( )ɛ p + ( ɛ) p <. Notice that for p 2, ( )ɛ p + ( ɛ) p ( )ɛ 2 + ( ɛ) 2 = ɛ 2 2ɛ +. Tryig ɛ = (as a guess), we get that (x,..., x ) ɛ p ( ) p. I particular, the homothety ratio ( ) p works with 2 traslates for p 2, so N(B p, ( ) p B p ) 2. With the above (o-optimal) guess, we have gotte the same homothety ratio, but with more traslates. 2.3 Upper-Semi Cotiuity of N(K, λk) We have the followig, due to []: Theorem. Let L ad K be covex bodies such that d BM (L, K) < + ɛ λ ad N(K, λk) = A. The N(L, (λ + ɛ)l) A. We may use this theorem to fid a homothety ratio for large p. For p q, we have that q p p q p, i.e. d(bp, Bq ) p q. Thus d(b p, B ) p. We also have that N(B, 2 B ) = 2 (this is the best possible homothety ratio for 2 traslates). The, if we wat N(B p, λb p ) 2, it suffices to have p < + 2(λ log ) = 2λ, i.e. p >. For example, 2 log 2λ we may try to reproduce the above results by takig λ = ( ) p. The p > with. log p > log 2 log 2+ p log( ) log 2. Figure shows how this expressio grows 4

5 6 5 log 2 log Figure : For a fixed, it is eough to choose ay p greater tha the value of the above graph at to guaratee that 2 traslates work for a homothety ratio of ( ) p 3 Outer Normals Suppose K is a ucoditioal body. For x K, v a outer ormal at x meas (v, y) (v, x) y K. If K is strictly covex, the (v, y x) < 0 for y x i K. By ucoditioality, for i, the poit x i = (x, x 2,..., x i, x i, x i+,..., x ) K. (v, x i x) = (v, 2x i e i ) = 2x i v i < 0 x i v i > 0 for x i 0 x i 0 v i 0 v i = 0 x i = 0. Now, suppose K is smooth, so there is oly oe outer ormal v at each x K. We claim that the x i = 0 v i = 0. Suppose ot, so x, v, i such that x i = 0 but v i 0. y K, (v, y) (v, x) = (Proj e i (v), x) sice x i = 0. But (Proj e i (v), y) = (v, Proj e i (y)) (v, x). Therefore, Proj e i (v) is also a outer ormal to K at x, which cotradicts K beig smooth. Thus, we have proved the followig: Theorem 2. For K ucoditioal, smooth, ad strictly covex, for x K ad v the outer ormal to K at x, x i = 0 v i = 0. 5

6 4 Projectios 4. Bouded Bodies Give a (bouded) covex body K, we may assume that 0 K (does t chage coverig umber). Let π : R R be the orthogoal projectio of R oto R (where we thik of R as the orthogoal direct sum R R ad project parallel to R ). We hope to achieve that if N traslates of it(π(k)) are eough to cover π(k), the at most 2 traslates of it(k) per traslate of π(k) are eeded to cover K, i.e. the coverig umber of K is at most 2N. Cotiuig this argumet iductively (we may keep projectig dow to the dimesio below), we evetually reach R. This fial projectio must be a bouded segmet, which has a coverig umber of 2, so we get that N(K, it(k)) 2. Specifically, we cojecture that if y π(k) satisfies y x + it(π(k)) for some traslate x R, the z s.t. π(z) = y, z x ± ɛe + it(k) for some ɛ > 0, where e = (, 0,..., 0) Cojecture 2. The above cojecture works for ucoditioal bodies. Theorem 3 (Origially formulated by Boltyaski ad Solta, see [2]). Suppose K R is closed ad bouded. Suppose there are o lies parallel to R itersectig K at oly oe poit (perhaps we ca achieve this by rotatig K). The: i(k) 2i(π(K)) Proof. See Appedix; slight modificatio of the proof foud i [2]. 4.2 Equivalece of Bouded ad Ubouded Cojectures Suppose K R is a closed covex body. Let c(k) = N(K, it(k)), h(k) = mi { : K N (x i + λ i K) where x i R, λ i (0, ) i}. Clearly, N N i=0 c(k) h(k), ad by compactess they are equal i the case that K is bouded. We attempt to establish the equivalece of the followig two cojectures: Cojecture 3 (Hadwiger). For K bouded, c(k) = h(k) 2. 6

7 Cojecture 4. For K ubouded, if c(k) is fiite, the c(k) h(k) 2. Propositio. Cojecture 4 Cojecture 3. Proof. Suppose K R is bouded. Cosider the (ubouded) cylider K := K [0, ] R +. First, we claim that h( K) h(k), i particular that h( K) is fiite. For suppose that K (x i + λ i K), the clearly K i=0 (x i ɛe + + λ i K) ɛ > 0, where e+ is the ( + ) st stadard basis i=0 vector. This is because K projects exactly dow to K, except for the bottom face of the cylider. Thus, if x K satisfies x x i + λ i K, the k > 0, x + ke x i + λ i K. To cover the bottom face it suffices to shift all of our traslates dow ay amout. Now, by our assumptio h( K) 2 (+) = 2 2. Thus, for some x i ad λ i, K ( x i + λ i K). Each xi is of the form x i = x i ± ɛ i e + for x i R. Ay poit x K is also i K, with last coordiate 0. Clearly, if x x i + λ i K, the x xi + λ i K. Thus, K 2 (x i + λ i K), so h(k) 2. i=0 We cojecture that the reverse of the above propositio holds as well. 5 Ackowledgemets I would like to thak Boaz Slomka ad Beatrice-Hele Vritsiou for advisig me durig the summer, ad the NSF ad the Uiversity of Michiga for the opportuity. i=0 Refereces [] M. Naszodi. O the Costat of Homothety for Coverig a Covex Set with its Smaller Copies (99). [2] V. Boltyaski ad P. Solta. Combiatorial Geometry of Various Classes of Covex Sets (978). 7

8 A Proof of Theorem 3 Proof. Let p,..., p m be directios i R illumiatig π(k). Let e be the stadard basis i R. We claim that for big eough λ > 0, the directios p + λe,..., p m + λe, p λe,..., p m λe illumiate K. By our assumptio, for a K the lie l a R itersects K alog a segmet, call it I(a). Let u(a) be the top of this segmet, ad v(a) the bottom (where x is above y if x y = ke for k > 0). Let U = {u(a) : a K}, V = {v(a) : a K}. For a K \ (U V ), if a directio p R illumiates u(a), the it illumiates I(a) \ {v(a)}. Ideed, the illumiatig coditio guaratees that x a = u(a) + λp it(k) for some λ > 0. Thus by covexity the segmet s a betwee x a ad v(a) is i K, with every poit except v(a) a iterior poit. For b I(a) \ {v(a)}, the half-lie b + R + p must itersect s a at some poit i it(k). Thus, p illumiates b: x a s a b + R + p u(a) b v(a) Similarly, if a directio illumiates v(a), the it illumiates I(a) \ {v(a)}. Thus, it is eough to show that 2m directios are eeded to illumiate U V. We first illumiate the set U 0 V 0, where U 0 := U π ( (π(k)) ad V 0 := V π ( (π(k)). The directios p,..., p m illumiate π(k). By compactess, we ca fid closed sets F,..., F m such that F F m = π(k) ad directio p i illumiates F i. For ay x F i, we may agai defie u(x) ad v(x) as the top ad bottom poits of I(x), respectively, where I(x) is the itersectio of l x R with K. Let U i = {u(x) : x F i }, V i = {v(x) : x F i }. The U U m V V m = U 0 V 0. Our goal is to illumiate all of cl(u 0 V 0 ), ad use a compactess argumet to argue that oly a fiite umber of illumiatio vectors (i particular, 2m) actually suffices. 8

9 Let us first illumiate the poits v(x), for x π(k). Sice x F i is illumiated by p i, there exists some y it(π(k)) such that y = x + λp i for some λ > 0. By assumptio, K cotais the segmet I(y), ad y it(π(k)) meas that I(y) \ {v(y), u(y)} it(k). The quadrilateral with vertices u(y), u(x), v(x), v(y) is i K by covexity. Wlog let s say the loger diagoal d y is the segmet betwee u(y) ad v(x). Sice d(y) is the loger diagoal, (u(y), e ) > (c(x), e ), where c(x) = u(x)+v(x). Thus, there exists 2 a poit ũ(y) = u(y) δe it(k), with (u(y), e ) > (c(x), e ), for some δ > 0. Let d y be the segmet coectig ũ(y) ad v(x). Every poit of this segmet except v(x) is a iterior poit of K by covexity. Sice the half lie c(x) + R + p i ad d y are co-plaar, they must itersect at a poit c(x) + λ x p i it(k) for some λ x > 0: u(y) ũ(y) c(x) + R + p i v(y) d y d y u(x) c(x) v(x) Covexity implies that for 0 < λ λ x, c(x) + λp i it(k). I particular, for λ λ x, c(x) + λp i illumiates c(x), which implies that (c(x) v(x)) + λp i illumiates v(x). Equivaletly, the vector (c(x) v(x))+p λ i illumiates v(x) for λ λ x. c(x) v(x) = µe for some µ > 0, so we have show that for some k x > 0, p i + ke illumiates v(x) for k k x. We ca similarly illumiate all poits u(x), obtaiig that vectors of the form p i je illumiate u(x) for j j x > 0. Let us ow illumiate poits w cl(u 0 V 0 ) \ (U 0 V 0 ). By defiitio, these poits must be accumulatio poits of U 0 V 0. We first claim that such poits w must project to the boudary of π(k), that is, for w a accumulatio poit, π(w) (π(k)). Suppose, for a cotradictio, that π(w) it(π(k)), so there exists a ope ball B ɛ (π(w))) it(π(k), for some ɛ > 0. By cotiuity of the projectio fuctio π, π (B ɛ (π(w)) is a ope eighborhood of w. Moreover, this eighborhood caot cotai poits i U 0 V 0, else B ɛ (π(w)) would cotai a poit i (π(k)). But this meas 9

10 that w is a isolated poit of cl(u 0 V 0 ), cotradictig our assumptio. Thus, these accumulatio poits w ca be oe of two types. First, w may be equal to a poit u(x) or v(x), for some x (π(k)). I this case, we have already illumiated by the method above. Otherwise, w I(x)\{u(x), v(x)} for some x (π(k)). But, by our first argumet, the vectors illumiatig u(x) ad v(x) will the both illumiate w. We have show that for every poit w cl(u 0 V 0 ), vectors of the form p i ± ke illumiate w, for k k w, ad i m. Sice illumiatig directios illumiate a ope subset of K, for each such w there exists a eighborhood Ãw K such that (wlog) a vector of the form p i + k w e illumiates all of Ãw. Now cosider a poit a Ãw. If a = v(x) for some x (K), we kow that we may icrease k w without boud ad stil illumiate a, i.e. p i + ke illumiates a for k k w. Else, a is i the iterior of the segmet I(a). By the illumiatig coditio there exists a poit x a itk such that x a = a + λ(p i + k w e ), for some λ > 0. But the the segmet betwee x a ad u(a) is fully cotaied i K, with every poit except (possibly) u(a) a iterior poit. Thus, ay vector of the form p i + ke for k w k < illumiates a (ad all of Ãw): x a a u(a) p i + k w e v(a) Thus, we kow that for all w cl(u 0 V 0 ) there exists a eighborhood à w K covered by vectors of the form p i ± ke for k k w. The sets A w = Ãw cl(u 0 V 0 ) are ope i cl(u 0 V 0 ), ad thus form a ope cover. By compactess there exits a fiite subcover A w,..., A wn. Let ˆk = max {k w j }. j N The the 2m vectors p + ˆke,..., p m + ˆke, p ˆke,..., p m ˆke illumiate U 0 V 0. It remais to illumiate U V = U V \ cl(u 0 V 0 ), that is, poits w (U V ) such that π(w) it(π(k)). We shall illumiate cl(u V ) ad oce agai use a compactess argumet. Cosider a poit w cl(v ). If w cl(v 0 ), we have already illumiated w. Else, π(w) it(π(k)). There 0

11 is a vertical segmet I(w) K, with I(w) \ {u(w), v(w)} it(k), so the vector +e illumiates w. Moreover, e illumiates a ope eighborhood A w K aroud w, where A w may be chose small eough so that π(a w ) it(π(k)). K has affie dimesio, so ope sets i K must also be of dimesio. Choose ay vector p i illumiatig π(k). We claim that the plae Pw i cotaiig w ad parallel to spa{p i, e i } must cotai aother poit w A w, where w ca be writte as w = w + ap i ± be i, for a, b > 0. If = 2, this is obvious, as the ay plae Pw i is all of R 2. For 3, A w ad π(a w ) must both be of dimesio at least 2, so ay vertical plae P = spa{e, v} for v R itersects A w i a segmet. The poit c(w) = u(w)+v(w) it(k), so the segmet s 2 w coectig w to c(w) has all poits except w i it(k). But the for some k w > 0, the half lie w + R + (p i + k w e ) itersects it(s w ), so for k k w, the vector p i + ke illumiates w: w + R + (p i + k w e ) u(w ) w v(w ) s w u(w) c(w) w = v(w) e We have ow show that for each w V (similar argumet for w U \ U 0 ), that vectors of the form p i ± ke illumiate w, for k k w. By aother compactess argumet, we fid that there exists a costat k > 0 such that p + ke,..., p m + ke, p ke,..., p m ke illumiate cl(u V ). Fially, take λ = max{ˆk, k}, ad the the vectors p + λe,..., p m + λe, p λe,..., p m λe illumiate K, so the theorem is proved. p i