MINKOWSKI LENGTH OF 3D LATTICE POLYTOPES

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1 MINKOWSKI LENGTH OF 3D LATTICE POLYTOPES OLIVIA BECKWITH, MATTHEW GRIMM, JENYA SOPRUNOVA, AND BRADLEY WEAVER Astrct. We study the Minkowski length L(P) of lttice polytope P, which is defined to e the lrgest numer of non-trivil primitive segments whose Minkowski sum lies in P. The Minkowski length represents the lrgest possile numer of fctors in fctoriztion of polynomils with exponent vectors in P, nd shows up in lower ounds for the minimum distnce of toric codes. In this pper we give polytime lgorithm for computing L(P) where P is 3D lttice polytope. We next study 3D lttice polytopes of Minkowski length 1. In prticulr, we show tht if Q, supolytope of P, is the Minkowski sum of L = L(P) lttice polytopes Q i, ech of Minkowski length 1, then the totl numer of interior lttice points of the polytopes Q 1,,Q L is t most 4. Both results extend previously known results for lttice polygons. Our methods differ sustntilly from those used in the two-dimensionl cse. Introduction Let P e convex lttice polytope in R n. Then P defines L F (P), vector spce over field F spnned y the monomils in P. Tht is, L F (P) = spn F {t m m P Z n }, where t m = t m 1 1 t mn n. In this pper we ddress the following question: Wht is the lrgest numer of fctors tht polynomil in L F (P) could hve? We lso study those fctors nd otin results regrding their Newton polytopes. Although these questions re interesting on its own, our motivtion comes from studying toric codes. The toric code C P, first introduced y Hnsen in [3], is defined y evluting the polynomils in L Fq (P) t ll the points t in the lgeric torus (F q )n. Tht is, C P is liner code whose codewords re the strings (f(t) t (F q )n ) for f L Fq (P). It is convenient to ssume tht P is contined in the squre [0,q 2] n, so tht ll the monomils in L Fq (P) re linerly independent over F q [8]. Thus C P hs lock length (q 1) n nd dimension equl to the numer of the lttice points in P. Note tht the weight of ech non-zero codeword in C P is the numer of points t (F q) n where the corresponding polynomil does not vnish. Therefore, the minimum distnce of C P (which is the minimum weight for liner codes) equls d(c P ) = (q 1) n mx Z(f), 0 f L Fq (P) where Z(f) is the numer of zeroes (i.e. points of vnishing) in (F q) n of f. For toric surfce codes, tht is, in the cse n = 2, Little nd Schenck in [6] used Hsse- Weyl ound nd the intersection theory on toric surfces to come up with the following generl ide: If q is sufficiently lrge, then polynomils f L Fq (P) with more solutely irreducile fctors will necessrily hve more zeroes in (F q) 2 ([6], Proposition 5.2). In 1991 Mthemtics Suject Clssifiction. 52B20, 94B27, 14G50. The uthors were supported y funds from NSF-REU Grnt DMS

2 2 OLIVIA BECKWITH, MATTHEW GRIMM, JENYA SOPRUNOVA, AND BRADLEY WEAVER [10] this ide ws expnded to produce explicit ounds for the minimum distnce of C P in terms of certin geometric invrint L(P), (full) Minkowski length of P, which ws introduced in tht pper. This invrint L(P) reflects the lrgest possile numer of solutely irreducile fctors polynomilf L Fq (P)cnhve. ApolytimelgorithmforcomputingL(P)forpolygons ws provided in [10]. In this pper, we extend this result to dimension 3 (Theorem 2.7, sed on Theorem 2.5). Moreover, [10] provides description of the fctoriztion f = f 1 f L(P) for f L Fq (P) with the lrgest numer of fctors: it turns out tht in such fctoriztion the Newton polygon P fi (which is the convex hull of the exponents of the monomils in f i ) is either primitive segment, unit simplex, or tringle with exctly 1 interior point. It is lso shown in [10] tht tringle with n interior point cn occur in such fctoriztion t most once. This implies tht the totl numer I of interior lttice points of P fi is t most 1. This result is essentil for estlishing ounds on the minimum distnce of toric surfce codes in [10]. This rgument is not directly extendle to dimension 3, s it does not seem fesile to otin description of the Newton polytopes P fi in dimension 3 (See [11] for some exmples). Nevertheless, in this pper we show tht in the 3D cse I 4 (Theorem 3.14). Our methods differ sustntilly from those used in the two-dimensionl cse. An initil version of our rgument relied hevily on the clssifiction of 3D Fno tetrhedr [4]. Although we were le to completely get rid of this dependency in our finl rgument, the clssifiction helped us significntly in our explortions. In [10] comintoril results out Minkowski length of polygons led to lower ounds on the minimum distnce of surfce toric codes. We hope tht our(entirely comintoril) pper will in the future led to similr ounds for 3D toric codes. 1. Minkowski Length of Lttice Polytopes Here we recll the definition of the (full) Minkowski length introduced in [10] s well s reproduce nd refine some results from tht pper using new methods which will lter e pplied to the 3D cse Minkowski sum. Let P nd Q e convex polytopes in R n. Their Minkowski sum is P +Q = {p+q R n p P, q Q}, which is gin convex polytope. Figure 1 shows the Minkowski sum of tringle nd squre P Q P +Q Figure 1. The Minkowski sum of two polygons

3 MINKOWSKI LENGTH OF 3D LATTICE POLYTOPES 3 Let f e Lurent polynomil in F q [t ±1 1,...,t ±1 n ]. Then its Newton polytope P f is the convex hull of the exponent vectors of the monomils ppering in f. Thus P f is lttice polytope s its vertices elong to the integer lttice Z n R n. Note tht if f,g F q [t ±1 1,...,t±1 n ] then the Newton polytope of their product P fg is the Minkowski sum P f +P g. A primitive segment E is lttice segment whose only lttice points re its endpoints. The difference of the endpoints is vector v E whose coordintes re reltively prime (v E is defined up to sign). A polytope which is the Minkowski sum of primitive segments is clled (lttice) zonotope. We sy tht two lttice polytopes re equl if they re the sme up to trnsltion. The utomorphism group of the lttice is the group of ffine unimodulr trnsformtions, denoted y AGL(n, Z), which consists of trnsltions y n integer vector nd liner trnsformtions in GL(n, Z). It is stndrd fct from lttice-point geometry tht ny two primitive segments in R n re AGL(n,Z)-equivlent ([7], Theorem II.1) Minkowski length. Let P e lttice polytope in R n. Definition 1.1. The (full) Minkowski length L = L(P) of lttice polytope P is the lrgest numer of primitive segments E 1,E 2,...,E L whose Minkowski sum is in P. Equivlently, L = L(P) is the lrgest numer of non-trivil lttice polytopes Q 1,...,Q L whose Minkowski sum is in P. Any collection of L = L(P) non-trivil lttice polytopes Q 1,...,Q L whose Minkowski sum is in P will e referred to s mximl (Minkowski) decomposition in P. The dimension of mximl decomposition is the dimension of the Minkowski sum Q 1 + +Q L. Exmple 1.2. In the figure elow, the first polygon, clled T 0, hs Minkowski length 1. For the second one, the Minkowski length is 2. Notice tht this tringle hs mny mximl decompositions: the sum of two horizontl segments, the sum of two verticl segments, the sum of two digonl segments, the sum of two stndrd 2-simplices, etc. For the lst polygon, the Minkowski length is 3, s there is prllelogrm inside tht is sum of three lttice segments. Clerly, L(P) is n AGL(n, Z)-invrint nd the summnds of every mximl decomposition in P re polytopes of Minkowski length 1. It does not seems fesile to descrie polytopes of Minkowski length 1 in generl. However, in dimension 2 such description is given in [10] nd we reproduce it here. Theorem 1.3. [10] Let P e convex lttice polygon in the plne with L(P) = 1. Then P is AGL(2, Z)-equivlent to primitive segment, the stndrd 2-simplex D or the tringle T 0 with vertices (1,0), (0,1) nd (2,2).

4 4 OLIVIA BECKWITH, MATTHEW GRIMM, JENYA SOPRUNOVA, AND BRADLEY WEAVER T 0 Figure 2. Polygons of Minkowski length 1 It is lso proved in [10] tht mximl decomposition Q P cn hve t most one summnd AGL(2,Z)-equivlent to T 0, nd if this is the cse, the remining summnds re [0,e 1 ], [0,e 2 ], nd [0,e 3 ], tht is, Q is AGL(2,Z)-equivlent to Q = T 0 + n 1 [0,e 1 ] + n 2 [0,e 2 ]+n 3 [0,e 1 +e 2 ]. Here e 1,e 2 re the stndrd sis vectors. We will recover this result (using new method tht will e lter pplied to the 3D cse) nd will lso show tht if tringle is summnd of mximl decomposition of polygon P, then the other summnds re either primitive segments, or exctly tht tringle. Tht is, if Q 1 nd Q 2 re tringles nd L(Q 1 +Q 2 ) = 2, then Q 1 = Q 2. This refinement will e importnt for our 3D discussion. Before we stte the result, we set nottion nd prove lemm which will lso e importnt for our future discussion in dimension 3. Let P e lttice polytope in R n. For ech segment whose endpoints re lttice points in P, consider its direction vector reduced modulo 3. Since v nd v define the sme segment, we identify such vectors. Using this equivlence reltion, we otin the set Z 3 P n 1 of equivlence clsses. Lemm 1.4. Let L(P) = L(Q) = 1 nd L(P + Q) = 2, where P nd Q re lttice polytopes in R n. Then if P nd Q ech hve segment of some clss, then those two segments re equl (re the sme up to trnsltion). If P hs t lest two segments of clss, then Q hs no segments from tht clss. Proof. If P ndqothhve lttice segments fromthesme equivlence clss, then, unless these segments re equl, their Minkowski sum contins segment of Minkowski length 3 (since either sum or difference of the direction vectors is multiple of 3). If P hs multiple segments from one clss, then these segments cnnot e trnsltes of ech other, s they would form prllelogrm in P of Minkowski length t lest 2. If Q hs segment from tht clss, it would e not trnslte of t lest one of the two segments in P nd we gin conclude L(P +Q) 3. Theorem 1.5. Let P e convex lttice polygon. If one of the summnds of mximl decomposition Q in P is AGL(2,Z)-equivlent to T 0, then Q is AGL(2,Z)-equivlent to Q = T 0 +n 1 [0,e 1 ]+n 2 [0,e 2 ]+n 3 [0,e 1 +e 2 ]. If one of the summnds of Q is AGL(2,Z)- equivlent to the stndrd 2-simplex, then the remining summnds tht re not primitive segments, re equl to. Proof. We hve four equivlence clsses in Z 3 P 1 : (1,1),(1, 1),(1,0),(0,1). Notice tht if nd Z 3 P 1 re linerly independent (tht is, ±), then they generte ll the clsses:, = {,,+, }.

5 MINKOWSKI LENGTH OF 3D LATTICE POLYTOPES 5 Now let one of the summnds in mximl decomposition in P e AGL(2,Z)-equivlent to T 0. Then we cn ssume this summnd is exctly T 0. The direction vectors of the lttice segments in T 0 re (1,0), (0,1), (1,1), (1, 1), (1,2), nd (2,1). The lst three re ll fromthe sme clss, so y Lemm 1.4 segments fromthis clss cnnot show up in other summnds. The first three re ll from distinct clsses, hence y the lemm only segments with direction vectors (1,0), (0,1), nd (1,1) cn show up in other summnds. Since ll four clsses re covered, we hve shown tht the direction vectors of lttice segments in other summnds cn only e (1,0), (0,1), nd (1,1). One cn use such segments to form tringle in four different wys. The result will e the stndrd 2-simplex nd its reflections. In ech of these four cses, it s esy to check tht the Minkowski sum of such tringle with T 0 is 3, which proves our first sttement. Next, let one of the summnds e equivlent to the stndrd 2-simplex, so we cn ssume it s exctly. The direction vectors (1,0),(0,1), nd ( 1,1) re ll from distinct clsses, hence if other summnds hve lttice segments from these clsses, they would hve to hve these direction vectors. If there is nother tringle in the mximl decomposition, it would hve to e equivlent to the stndrd 2-simplex, s T 0 is not possile y the ove rgument. The direction vectors would hve to elong to three distinct clsses, so t lest two of the sides would hve to hve direction vectors (1,0),(0,1), or ( 1,1). Here re eight tringles tht could e formed in this wy: The lst four hve segment with direction vector either ( 2,1) or ( 1,2), which re from the sme clss with (1,1), so the Minkowski sum of ny of these tringles with is t lest 3. For ll the remining tringles, except itself we esily check tht their sum with hs Minkowski length 3. Corollry 1.6. Let P,Q e lttice polytopes in R 3 with L(P + Q) = 2. Consider the intersection of plne π with ech P nd Q. If ech π P nd π Q contins lttice tringle, then these lttice tringles re the sme up to trnsltion.

6 6 OLIVIA BECKWITH, MATTHEW GRIMM, JENYA SOPRUNOVA, AND BRADLEY WEAVER Proof. Let u e primitive norml vector to π. Let A GL(3,Z) e mtrix whose lst row is u. (It is shown, for exmple, in [7], Theorem II.1, why such mtrix exists.) Then A mps π to the (x,y)-plne nd the result follows from the previous theorem. 2. Algorithm for Computing L(P) for 3D polytopes. It ws shown in [10] tht in the plne cse there lwys exists mximl decomposition in P of very simple form. Nmely, there exists mximl decomposition tht is equivlent to n 1 [0,e 1 ]+n 2 [0,e 2 ]+n 3 [0,e 1 +e 2 ] for some n 1,n 2,n 3 N. This fct ws used in [10] to uild n lgorithm for finding L(P). To extend this result to the 3D cse, we first mke definition. Definition 2.1. Let P R n e convex lttice polytope. Then the set of its mximl decompositions is prtilly ordered y inclusion. Tht is, we sy tht A+P 1 + +P k < B +Q 1 + +Q l if A+P 1 + +P k B +Q 1 + +Q l. Here A nd B re points in Z n. A mximl decomposition is clled smllest mximl decomposition if it is miniml with respect to this prtil order. Note tht smllest mximl decomposition is Minkowski sum of segments. Proposition 2.2. Let P R 2 e lttice polygon. Consider smllest mximl decomposition Z in P: P Z = A+n 1 E 1 + +n l E l. Then Are(E i +E j ) 1 for ny choice of 1 i,j l. Proof. Let v 1 nd v 2 e the primitive direction vectors of the segments E i nd E j nd ssume tht the re of the prllelogrm spnned y v 1 nd v 2 is t lest 2. Applying n AGL(2,Z) trnsformtion, we cn ssume tht A is the origin, v 1 = e 1 = (1,0) nd v 2 = (,), where 0 < nd > 1, which implies tht (1,1) Π = [0,e 1 ]+[0,(,)]. We show now tht there is lwys segment I of Minkowski length 2 tht lies strictly inside Π nd hence, we cn pss from Π to 2I nd get smller mximl decomposition. If oth nd re even then 2[0,(/2,/2)] is strictly inside of Π; if is odd nd is even then 2[0,((+1)/2,/2)] Π; if is even nd is odd, (1,1)+2[0,(/2,( 1)/2)] Π; if nd re oth odd, (1,1)+2[0,(( 1)/2),( 1)/2)] Π. Theorem 2.3. Let P R 2 e lttice polygon. If Z is smllest mximl decomposition in P, then it is AGL(2,Z)-equivlent to P Z = n 1 [0,e 1 ]+n 2 [0,e 2 ]+n 3 [0,e 1 +e 2 ]. Proof. Let P Z = n 1 E 1 + +n l E l with v 1,...,v l distinct primitive direction vectors of the segments E 1,...,E l. Applying n AGL(2,Z) trnsformtion we cn ssume tht v 1 = e 1. Next, since det(v 1,v 2 ) = ±1, we cn ssume tht v 2 = e 2. Then y the previous proposition, ny other v k is either (1,1) or (1, 1) s we cn lwys switch vector to its negtive. Notice tht these two vectors cnnot simultneously pper in smllest decomposition, s the sum of the corresponding segments would contin segment of Minkowski length 2. Finlly, these two remining cses re AGL(2, Z)-equivlent. We next tret the 3D cse.

7 MINKOWSKI LENGTH OF 3D LATTICE POLYTOPES 7 Proposition 2.4. Let P R 3 e lttice polytope. Consider smllest mximl decomposition Z in P P Z = A+n 1 E 1 + +n l E l. Then Vol(E i +E j +E k ) 2 for ny choice of 1 i,j,k l. Proof. Let v 1, v 2, nd v 3 e the primitive vectors tht go long the segments E i, E j, E k, nd ssume tht the volume of the prllelepiped spnned y v 1, v 2, nd v 3 is t lest 3. Applying n AGL(2, Z) trnsformtion (nd using Proposition 2.2), we cn ssume tht A is the origin, v 1 = e 1 = (1,0,0), v 2 = e 2 = (0,1,0), nd v 3 = (s,t,u), where 0 s t < u. The volume of the prllelepiped spnned y e 1,e 2,v 3 is u. If s = 0, then the re spnned y e 2 nd v 3 is u 3, which would contrdict the minimlity of Z. We next oserve tht the prllelepiped spnned y e 1,e 2, nd v 3 is defined y Π = { (x,y,z) R 3 0 z u, nd consider the following three cses. s u z x s u z +1, t u z y t u z +1 Cse 1. s u/2, t u/2 Using the description of Π ove, we cn esily check tht (1,1,2) nd (s,t,u 2) reothinπ. Hence prllelogrmwiththevertices (1,1,0),(1,1,2),(s,t,u 2), nd (s,t,u) is inside Π. This prllelogrm is Minkowski sum of three segments (1,1,0)+2[0,(0,0,1)]+[0,(s 1,t 1,u 2)] Π, whichcontrdictstheminimlityofz. Noticethtu 3ensuresthtthesegments involved in the decomposition re non-trivil. Cse 2. s > u/2, t > u/2 Then (2,2,2) nd (s 1,t 1,u 2) re in Π, so prllelogrm with the vertices (0,0,0),(2,2,2),(s 1,t 1,u 2),nd(s+1,t+1,u)isinsideΠ. Thisprllelogrm is Minkowski sum of three segments 2[0,(1,1,1)]+[0,(s 1,t 1,u 2)] Π, which contrdicts the minimlity of Z. Cse 3. s u/2, t > u/2 Then (1,2,2) nd (s,t 1,u 2) re in Π, so prllelogrm with the vertices (1,0,0),(1,2,2),(s,t 1,u 2), nd (s,t+1,u) is inside Π. This prllelogrm is Minkowski sum of three segments (1,0,0)+2[0,(0,1,1)]+[0,(s 1,t 1,u 2)] Π, nd we get the sme contrdiction gin. Theorem 2.5. Let P R 3 e lttice polytope. Let Z e smllest mximl decomposition in P, then it is AGL(2,Z)-equivlent to either n 1 [0,e 1 ]+n 2 [0,e 2 ]+n 3 [0,e 1 +e 2 +2e 3 ]+n 4 [0,e 1 +e 2 +e 3 ]+n 5 [0,e 1 +e 3 ]+n 6 [0,e 2 +e 3 ]+n 7 [0,e 3 ] or n 1 [0,e 1 ]+n 2 [0,e 2 ]+n 3 [0,e 3 ]+n 4 [0,e 1 +e 2 +e 3 ]+n 5 [0,e 1 ±e 2 ]+n 6 [0,e 1 +e 3 ]+n 7 [0,e 2 +e 3 ]. }

8 8 OLIVIA BECKWITH, MATTHEW GRIMM, JENYA SOPRUNOVA, AND BRADLEY WEAVER Proof. Assume first tht there re three segments in the mximl decomposition Z whose direction vectors generte prllelepiped of volume 2. We cn then ssume tht the first direction vector is e 1. By Proposition 2.2 we cn ssume tht the second vector is e 2. Next, we cn ssume tht the third direction vector v is of the form (s,t,u) where 0 s t < u. Since 2 = det(e 1,e 2,v), we know tht u = 2 nd the only options for the third vector re (0,1,2) nd (1,1,2). The first of these two options is impossile, s the sum of (0,1,2) nd (0,1,0) is not primitive, which contrdicts the minimlity of Z. We hve shown tht the third vector is (1,1,2) = e 1 +e 2 +2e 3. Let v = (,,c) e direction vector of some other segment in the mximl decomposition Z. We know tht det(e 1,e 2,v) 2, det(e 1,e 1 + e 2 + 2e 3,v) 2, nd det(e 2,e 1 +e 2 +2e 3,v) 2, which gives us the following restrints on the components of v: c 2 2, c 2 2, nd c 2. By flipping the direction vector v if necessry, we cn ssume tht c 0. If c = 0, then v = (1,1,0) or (1, 1,0). Both options re impossile s then the sum of v with (1,1,2) is (2,2,2) or (2,2,0), so we cn pss to smller mximl decomposition. If c = 1, then v = (0,0,1), (0,1,1), (1,0,1), (1,1,1). If c = 2, then v = (0,1,2), (1,0,2), (1,2,2), or (2,1,2). Adding either e 1 or e 2 to ech of these four vectors we cn get non-primitive vector, so none of these vectors occur in our mximl decomposition. We hve shown tht in the cse when there re three segments in the mximl decomposition Z tht generte prllelepiped of volume 2, then Z is AGL(2, Z)-equivlent to n 1 [0,e 1 ]+n 2 [0,e 2 ]+n 3 [0,e 1 +e 2 +2e 3 ]+n 4 [0,e 1 +e 2 +e 3 ]+n 5 [0,e 1 +e 3 ]+n 6 [0,e 2 +e 3 ]+n 7 [0,e 3 ]. Next, ssume tht ny three segments in the mximl decomposition Z generte prllelepiped of volume t most 1. If for ny three vectors the volume is zero, then we re in the plne cse nd we re done. Otherwise, we cn ssume tht first three vectors re e 1, e 2, nd e 3. Let v = (,,c) e ny other direction vector in the mximl decomposition Z. Then we hve 1, 1 nd c 1. By flipping the direction vector we cn ssume tht c 0. Here re the options for v tht we get, written in four lines: (1,1,0),(1, 1,0), (0,1,1),(0, 1,1), (1,0,1),( 1,0,1), (1,1,1),(1, 1,1),( 1,1,1),( 1, 1,1). Notice tht no two vectors from the sme line here cn occur in Z together s their sum is not primitive, which would contrdict the minimlity of Z. By flipping the direction of sis vectors, we cn ssume tht if ny of the four vectors in the lst line occur in Z, then it is (1,1,1). Let s ssume tht this is the cse nd (1,1,1) occurs in Z. We notice next ( 1,0,1) nd (0, 1,1) cn not occur in Z together s if we dd these two vectors together with (1,1,1), we get (0,0,3). We cn mke the sme oservtion out ( 1,0,1) nd (1, 1,0) nd then out (0, 1,1) nd (1, 1,0). This implies tht only one of (1, 1,0), (0, 1,1), nd ( 1,0,1) occurs in Z. By permuting e 1,e 2, nd e 3 we cn ssume tht the one tht occurs is (1, 1,0). In the cse when none of of the four vectors (1,1,1),(1, 1,1),( 1,1,1),( 1, 1,1) occur in Z, y pplying digonl chnge of sis with ±1 s on the min digonl (which will not chnge [0,e 1 ],[0,e 2 ], [0,e 3 ]), we cn turn ny pir of vectors from the set (1, 1,0),(0, 1,1),( 1,0,1) into corresponding vectors with positive entries. For exmple, mtrix with the digonl entries 1, 1,1 will turn ( 1,0,1) nd (0, 1,1)

9 MINKOWSKI LENGTH OF 3D LATTICE POLYTOPES 9 into (1,0,1) nd (0,1,1). Hence we will e le to get rid of ll the vectors with negtive entires except, possily, one. By permuting e 1, e 2, nd e 3, we cn ssume tht the vector withnegtive entryis(1, 1,0). Wehveshown thtifnythreesegments inz generte prllelepiped of volume t most 1, then Z is AGL(3,Z)-equivlent to either n 1 [0,e 1 ]+n 2 [0,e 2 ]+n 3 [0,e 3 ]+n 4 [0,e 1 +e 2 +e 3 ]+n 5 [0,e 1 +e 2 ]+n 6 [0,e 1 +e 3 ]+n 7 [0,e 2 +e 3 ] or n 1 [0,e 1 ]+n 2 [0,e 2 ]+n 3 [0,e 3 ]+n 4 [0,e 1 +e 2 +e 3 ]+n 5 [0,e 1 e 2 ]+n 6 [0,e 1 +e 3 ]+n 7 [0,e 2 +e 3 ]. Notice tht in 2D smllest mximl decomposition hs t most 3 distinct summnds; in 3D, s we hve just shown, such decomposition hs t most 7 distinct summnds. It turns out tht in dimension n smllest mximl decomposition hs t most 2 n 1 distinct summnds. Proposition 2.6. Let P R n e convex lttice polytope. Let Z e smllest mximl decomposition in Z. Then Z hs t most 2 n 1 distinct summnds. Proof. Reduce ll the summnds in Z modulo 2. Since the summnds re primitive segments, there will e 2 n 1 possiilities for reduced segment. If the numer of distinct segments in Z is t lest 2 n, we will hve two summnds tht re equl modulo 2. Then their sum is non-primitive, which contrdicts the minimlity of Z. Althoughweexpect thtthesumofthe2 n 1segmentswith0,1componentsmentioned in the proof of the ove proposition hs Minkowski length 2 n 1, we do not hve proof of this sttement, which would hve implied tht the ound of the proposition is shrp. Let lttice polytope P e descried y its fcets equtions. Then Brvinok s lgorithm [1, 5] counts the numer of lttice points in P in polynomil time. We will ssume tht the list P Z 3 of the lttice points in P R 3 is given, nd will explin how to find the Minkowski length of P in polynomil time in P Z 3. Theorem 2.7. Let P R 3 e lttice polytope with the given set of its lttice points P Z 3. Then the Minkowski length L(P) cn e found in polynomil time in P Z 3. Proof. This lgorithm relies on Theorem 2.5. We serch for ll possile decompositions of the form descried in the theorem. For every qudruple of points {A,B,C,D} P Z 3, where it is importnt which point goes first nd the order of the other three does not mtter, we check if [0,B A], [0,C A], nd [0,D A] generte prllelepiped of volume one or two. If the volume is one, these segments re equivlent to [0,e 1 ], [0,e 2 ], [0,e 3 ] nd we look for mximl decompositions equivlent to n 1 [0,e 1 ]+n 2 [0,e 2 ]+n 3 [0,e 3 ]+n 4 [0,e 1 +e 2 +e 3 ]+n 5 [0,e 1 ±e 2 ]+n 6 [0,e 1 +e 3 ]+n 7 [0,e 2 +e 3 ], tht is, mximl decompositions of the form n 1 E 1 +n 2 E 2 +n 3 E 3 +n 4 E 4 +n 5 E 5 +n 6 E 6 +n 7 E 7, where E 1 = [0,B A], E 2 = [0,C A], E 3 = [0,D A], E 4 = [0,B + C + D 3A], E 5 = [0,B +C 2A] or [0,B C], E 6 = [0,B +D 2A], E 7 = [0,C +D 2A]. If the volume is two, we check if the segments [0,B A], [0,C A], nd [0,D A] re primitive nd if ny two of them generte prllelogrm whose only lttice points re the vertices. If this is the cse, these three segments re equivlent to [0,e 1 ], [0,e 2 ], nd [0,e 1 + e 2 + 2e 3 ]. We then let E 1 = [0,B A], E 2 = [0,C A], E 3 = [0,D A],

10 10 OLIVIA BECKWITH, MATTHEW GRIMM, JENYA SOPRUNOVA, AND BRADLEY WEAVER E 4 = [0,(B+C+D 3A)/2], E 5 = [0,(B+D C 3A)/2], E 6 = [0,(C+D B 3A)/2], E 7 = [0,(D A B C)/2]. Next, for every 1 i 7, we find M i, the lrgest integer such tht there is some lttice point F in P with F + M i E i P. For ech 7-tuple of integers m = (n 1,...,n 7 ) where 0 n i M i, we check if some lttice trnslte of the zonotope Z m = n 1 E 1 + +n 7 E 7 is contined in P (we run through lttice points F in P to check if F +Z m is contined in P). For ll such zonotopes tht fit into P we look t n 1 + +n 7 nd find the mximl possile vlue N of this sum. Finlly, the lrgest such sum N over ll choices of {A,B,C,D} P Z 3 is L(P). Clerly, this lgorithm is polynomil in P Z 3. A group of REU students (In Brnett, Benjmin Fuln, nd Cndice Quinn) t Kent Stte University in Summer 2011 tried to generlize this lgorithm to dimension 4. Their first step ws to otin 4D version of Proposition 2.4. They showed tht if P R 4 is lttice polytope nd Z = A+n 1 E 1 + +n l E l is smllest mximl decomposition in P, then Vol(E i +E j +E k +E m ) 14 for ny choice of 1 i,j,k,m l, nd this ound is shrp. Unfortuntely, this ound is too high to otin description of smllest mximl decompositions in 4D, similr to the one of Proposition Lttice Polytopes of Minkowski Length 1 It ws shown in Theorem 1.6 of [10] tht if P Q = Q Q l is mximl decomposition of polygon P then t most one of Q i hs n integer lttice point in its interior, tht is, I(Q i ) 1. This fct ws crucil in [10] for estlishing ounds on the minimum distnce of the toric surfce code defined y P. We expect tht in order to extend these ounds to 3D codes, one needs to explore similr questions in dimension 3. As it ws mentioned ove, description of polytopes of Minkowski length 1 in dimension 3 does not seem fesile. We will insted reduce the lttice segments contined in 3D lttice polytope modulo 3, which will help us show tht if P is 3D polytope with L(P) = 1, then I(Q i ) is t most 4. Let P e lttice polytope in R 3 of Minkowski length 1. Then P hs t most 8 lttice points. Indeed, otherwise there would hve een two lttice points in P tht re congruent modulo 2, nd hence the segment connecting them would hve Minkowski length of t lest 2. For ech segment whose endpoints re lttice points in P, we consider its direction vector reduced modulo 3. Since v nd v define the sme segment, we identify such vectors. For thus defined modulo 3 segments there re 13 equivlence clsses: (1,1,1),(1,1, 1),(1,1,0),(1, 1,1),(1, 1, 1),(1, 1,0), (1,0,1),(1,0, 1),(1,0,0),(0,1,1),(0,1, 1),(0,1,0),(0,0,1), tht is, we re deling with the projective spce Z 3 P 2. Notice tht if,, nd c Z 3 P 2 re linerly independent (tht is,, nd c /, = {,,±}), they generte ll the clsses:,,c,+,,+c, c,+c, c,++c,+ c, +c, ++c. Let S e five-point lttice set contined in polytope of Minkowski length 1. There re ten lttice segments tht connect lttice points in S. We will clssify such sets S ccording to the numers of segments from distinct clsses in Z 3 P 2.

11 MINKOWSKI LENGTH OF 3D LATTICE POLYTOPES 11 Proposition 3.1. If L(P) = 1, then ny 5-point lttice set S in P is of one of the following types The segments re from clsses 4,2,2(+),2( ). Here 4 or 2 in front of segment s clss denotes its multiplicity, which is the numer of times it occurs mong the lttice segments in S The segments re from clsses 3,3,2(+),2( ). 3+(7) The segments re from clsses 3,,+,,c,+c, c,+ c. 2+2+(6) The segments re from clsses 2,2,+,,+c,+c,++c,c. (10) All ten lttice segments connecting points in S re from distinct clsses in Z 3 P 2. The elements,,c Z 3 P 2 in ech of the type descriptions re linerly independent. All types except for the lst one hve segments from clsses,,+,. Proof. We ssign direction to the segments y picking stndrd representtive from ech of the clsses. If two segments from the sme clss shre vertex, the rrows cnnot oth point to or wy from the vertex s in this cse the third side in the tringle is of Minkowski length t lest 3. We lso notice tht if two sides in tringle re from the sme clss, then the third one is lso from tht clss nd we get the tringle digrm elow. No other segment strting in one of these three vertices cn e of clss, so the only remining segment in S tht could e of clss, is the one connecting two remining points of S. Hence the lrgest numer of segments of the sme clss in S is 4. If we hve 4 segments of the sme clss we get the digrm elow. + + Figure We cll this type s there re 4 segments of one type nd 2 segments of ech of the three other types. Next, ssume we only hve 3 segments from clss. Then they would hve to form tringle. We could lso hve nother 3 segments of clss, forming tringle shring vertex with the first tringle. Then there re 3 segments of clss, 3 of clss, nd 2 of ech of + nd. We cll this type Assume next there is no other tringle. Connect one of the vertices of the tringle whose sides re of clss to fourth lttice point in S. Let this segment e of clss. The segment connecting the fourth lttice point to the fifth cnnot e from clsses,,+, or, s this would give either nother tringle or four segments of the sme

12 12 OLIVIA BECKWITH, MATTHEW GRIMM, JENYA SOPRUNOVA, AND BRADLEY WEAVER clss. Hence tht segment is of clss c, such tht the set {,,c} is linerly independent nd we get the digrm elow. We cll this type 3+(7). c ++c c +c + Figure 4. 3+(7) If there re no 3s ut there is 2, we get configurtion of type 2+2+(6). ++c +c c +c + Figure (6) Finlly, it is possile tht there re no repets mong clsses of segments. An exmple of this sitution is tetrhedron with the vertices (1,0,0),(0,1,0),(0,0,1),( 1, 1, 1) with one lttice point, the origin, strictly inside. We cll this type (10). Lemm 3.2. Let,,c,d Z 3 P 2 where nd c d. Then, c,d. Proof. If c, the conclusion is ovious, so we cn ssume tht,,c = Z 3 P 2. One of d,c +d,c d does not hve c in its expression in terms of,,c, hence it elongs to,. Proposition 3.3. Let P nd Q e 3D lttice polytopes of Minkowski length 1 with t lest five lttice points ech. If there exists 5-point lttice suset S of P of type or , then L(P +Q) 3. Proof. Pick ny 5-point lttice suset T of Q. Since in S we hve used up four clsses with multiplicities greter thn 1, y Lemm 1.4, T cnnot e of type (10), since the overll numer of clsses is 13. In S, we hve multiple segments of ech of the clsses,, +, for some, Z 3 P 2. Since T is not of type (10), we lso hve segments of clsses c, d, c+d, c d in T for some c,d Z 3 P 2, not necessrily with multiplicities. By Lemms 3.2 nd 1.4 we conclude L(P +Q) 3. Proposition 3.4. Let P nd Q e 3D lttice polytopes of Minkowski length 1. If there exist 3- nd 5-point lttice susets S nd T of P nd Q correspondingly, such tht S is tringle with sides of the sme clss nd Q is of type 3+(7), then L(P +Q) 3. In prticulr, if oth S nd T re of type 3+(7), then L(P +Q) 3.

13 MINKOWSKI LENGTH OF 3D LATTICE POLYTOPES 13 Proof. We first notice tht if lttice polytope contins lttice tringle with sides of the smeclss, thenthistringleisequivlent to T 0. Indeed, wecnesily mpthistringleto one in the (x,y)-plne y creting mtrix of determinnt 1 whose lst row is primitive vector orthogonl to the plne of the tringle. We know tht in the (x,y)-plne, up to the equivlence, there re only two tringles of length one, the unit tringle nd T 0. The unit tringle hs sides tht elong to three distinct clsses nd the sides of T 0 re ll from the sme clss. Hence the initil tringle is equivlent to T 0 nd, therefore, hs lttice point inside nd ll four points re in the sme plne. We cn ssume tht Q = T 0 nd the segments re of clsses,, +,, where = (1,2,0) nd = (1,0,0). We lso hve configurtion equivlent to T 0 in T. Let the multiple clss in this copy of T 0 e c. By Lemm 3.2, P nd Q shre segment. By Lemm 1.4, they cnnot shre segment of multiple clss, so they hve common lttice segment inside the tringles. We cn ssume tht the shred segment is of clss = (1,0,0). Let the equivlence etween the two copies of T 0 e relized y the integer mtrix A = ±1 p q 0 s t, 0 u v where sv ut = ±1. Let T 0 hve vertices ( 1,0,0),(0, 1,0), nd (1,1,0). If the (1,1)- entry of A equls 1, then A(T 0 ) T hs vertices ( 1,0,0), ( p, s, u), nd (p+1,s,u). Hence the (1/2,0,0), the hlf sum of the second nd third vertex, is in T nd oth S nd T contin segments connecting ( 1,0,0) to (1/2,0,0) whose Minkowski sum is 3. Next, ssume the (1,1)-entry of A is 1. Then the vertices of A(T 0 ) re (1,0,0), ( p, s, u), nd (p 1, s, u). Similrly to the ove, ( 1/2, 0, 0) is in T, ut the Minkowski sum of the segments [( 1,0,0),(1/2,0,0)] nd [( 1/2,0,0),(1,0,0)] is the segment [( 1/2,0,0),(5/2,0,0)], whose Minkowski length is 2, so we need to work with the fifth lttice point in T. Let s next ssume tht the segment connecting the fifth lttice point in T to the centrl lttice point in A(T 0 ) is of clss d. Then the remining three segments re of clsses d, d c, nd d+c. Clerly, one of these three segments in its expression in terms of,, nd c will not hve c, so it will hve to e of clss + or. We will ssume it is of clss, so it is shred with S nd its direction vector is (0,1,0). If this segment shres vertex with segment of clss in T, y Corollry 1.6 there would e shred tringle one of whose sides is of clss, which is impossile. Hence we cn ssume tht the segment with the direction vector (0,1,0) hs (p 1,s,u) s one of its endpoints. Then either (p 1,s+1,u) or (p 1,s 1,u) is the fifth lttice point in S nd its hlf sum with ( p, s, u), which is either ( 1/2,1/2,0) or (1/2,1/2,0), is in T. In oth cses, the Minkowski sum of T 0 with tringle in T whose vertices re ( 1/2,0,0), (1,0,0), nd either ( 1/2,1/2,0) or (1/2,1/2,0) is 3. Proposition 3.5. Let P nd Q e 3D lttice polytopes of Minkowski length 1. If there exist 5-point lttice susets S nd T of P nd Q of types 2+2+(6) nd 3+(7) correspondingly, then L(P +Q) 3. Proof. We ssume tht L(P +Q) = 2. Let the multiple clsses in S e nd nd the clss of multiplicity 3 in T e d. Let the tringle in T with ll sides of clss d e ABC nd the remining two lttice points in T e D nd E with AD of clss e nd BE of clss f, s depicted in the digrm elow.

14 14 OLIVIA BECKWITH, MATTHEW GRIMM, JENYA SOPRUNOVA, AND BRADLEY WEAVER C B d f E d d e A D By Lemm 3.2,, d,e nd, d,f. Since clsses nd cnnot occur in T nd clss d cnnot occur in S, segments of clsses + nd hve to pper in T, nd we cn ssume tht e = + nd f =. This is ecuse these two segments in T cnnot shre vertex, s the third side in the tringle formed y + nd would hve to e of clss either or. By Proposition 3.1, the segments connecting the lttice points in S re of clsses,, +,, + c, + c, + + c, nd c for some linerly independent,,c Z 3 P 2. Hence there re five options left for d: c, c, + c, c, +c. Notice tht in T we hve segments of clsses d++, d, d +, nd d+. Going through the five options for d, we oserve tht every time there re four lttice segments tht re shred etween S nd T. Two of them re + nd. The remining two for ech of the five cses re listed in the tle elow. d shred segments in S nd T c c (+) = (+c), c+( ) = (++c) c c (+) = (+c), c ( ) = (++c) + c + c ( ) = (+c),+ c+(+) = (++c) c c ( ) = c, c+(+) = (+c) +c +c ( ) = c, +c (+) = +c We hve checked tht there re lwys t lest four segments shred etween S nd T, with the extr condition tht in T none of these four segments is ED. In T, three of these segments cnnot ll hve E s n endpoint, s this would imply tht two of these segments in S lso shre n endpoint, so y Corollry 1.6 there is shred tringle, one of whose sides is d, which is impossile. Similrly, three of the shred segments cnnot ll hve D s n endpoint. Hence there re two possile scenrios, depicted in the digrm elow. The shred segments re mrked y +,,x, nd y. + y x + x y u u y + x + u x y

15 MINKOWSKI LENGTH OF 3D LATTICE POLYTOPES 15 In the first scenrio, the tringle formed y x nd y in S is shred, so its third side is lso shred, which leds to contrdiction. In the second scenrio, two tringles, one formed y + nd y, nd nother y nd x re shred, so their third side u ppers twice in S, which is impossile. Proposition 3.6. Let P nd Q e 3D lttice polytopes of Minkowski length 1. If there exist 5-point lttice susets S nd T of P nd Q correspondingly, of type 2+2+(6) ech, then L(P +Q) 3. Proof. We ssume tht L(P +Q) = 2. Let the nd e the clsses of lttice segments in S of multiplicity 2. Then S lso hs segments of clsses + nd. By Lemm 1.4 segments in T of multiplicity 2 cnnot e of clsses,, +,, so we cn ssume tht one of them is c nd the other is ++c, y switching direction vectors of,, nd c, if needed. Since we hve 13 clsses totl, S nd T overlp in t lest 3 segments. One of them is of clss +. We will show tht S nd T shre three segments tht form tringle one of whose sides is of clss either + or. Assume tht the segment of clss is lso shred. A segment of clss in T cnnot shre vertex with segment of clss + (then either the sum or difference of those clsses would lso e represented in S, ut clsses nd cnnot pper in S). Chnging direction of c nd/or oth nd, we cn ssume tht is s in the digrm elow. S T + c + ++c c ++c Let the clss of the third shred segment e x. Assume tht in S the segment of clss x shres vertex with. Then in T the segment of clss x would hve to shre vertex with s the only ones tht don t re c, +, nd ++c nd x cnnot e one of them. Hence nd x shre vertex in oth S nd T nd therefore y Corollry 1.6 S nd T shre tringle with sides, + c, nd +c. Hence x = + c nd the digrms for S nd T re elow. We see tht oths ndt hve tringles with c nd+c s sides, ut those tringles re not identicl, which contrdicts L(P +Q) = 2. Next, let next x shre vertex with + in S. If x does not shre vertex with + in T s well, then the options for x re + c nd + c. If x = + c then either x = c or x+ = ++c is in S, which is impossile. If x = + c, then either x++ = ++c or x (+) = c is in S, which is lso impossile. Hence tringle with se + is shred etween S nd T, which leds to the sme digrm nd the sme contrdiction s efore. It remins to consider the cse when is not shred etween S nd T. We cn lso ssume tht + c is not shred. (If it is shred replce S nd T in the ove

16 16 OLIVIA BECKWITH, MATTHEW GRIMM, JENYA SOPRUNOVA, AND BRADLEY WEAVER S T + c c +c +c c ++c + c ++c c rgument.) Then S nd T shre two segments oth of which hve the fifth point s n endpoint in oth S nd T. Then the tringle formed y these two segments is shred nd the se of tht tringle is +. Notice tht there is no room for other shred segments s they would hve to hve fifth point s vertex nd + is the only option for shred se. Let s denote one of the shred segments y d. We get the following digrm elow. d ++d d ++d + ++c c + c ++c We next serch for the expression of d in terms of, nd c so tht the only common segments etween S nd T re +, d, nd ++d. Below is the list of segments used in S nd T. S d d+ d+ d++ T 2c 2(++c) + + c d d c d+++c d++ We clerly hve d ±,±,±(+),±( ),±c,±(++c),±(+ c). All the remining options re lso very esy to get rid of. If d = ±+c,±+c,±( )+c, then d c +. If d = c, then d+ ppers in oth S nd T. If d = c, then d+ ppers in oth S nd T. If d = + c, then d++ ppers in oth S nd T.

17 MINKOWSKI LENGTH OF 3D LATTICE POLYTOPES 17 If d = c, then d+++c ppers in oth S nd T. If d = c, then d+ is of the sme clss s d c. Finlly, if d = c, then d+ is of the sme clss s d c. Every time we rrive t contrdiction, which proves the proposition. Proposition 3.7. Let P nd Q e two 3D lttice polytopes of Minkowski length 1. If oth P nd Q re of type (10) nd L(P + Q) = 2, then P nd Q re equl (the sme up to trnsltion). Proof. Since there re 13 clsses of segments totl, P nd Q shre t lest 7 segments. Among these shred segments we cn find three tht hve common endpoint in P. At lest two of these three shred segments hve common endpoint in Q. Hence P nd Q shre tringle ABC. Let the two remining lttice points in P e D nd E. At lest one of the two tetrhedr ABCD nd ABCE with se ABC, sy, ABCD, hs t lest two lterl edges tht re shred. Together with segment in the se these two edges form tringle with shred sides. At lest two of these three shred sides shre vertex in Q, so y Corollry 1.6 we get shred tringle. We hve shown tht P nd Q hve two pirs of shred tringles tht hve common edge, so P nd Q shre tetrhedron ABCD. Let the fifth vertices in P nd Q e E nd E. Since P nd Q shre t lest seven lttice segments, there is segment in P connecting one of A,B,C,D to E which is shred with Q. Let this segment e AE. The prllel segment in Q is of the form E X where X = A,B,C, or D. In ny of these cses, AE nd E X re djcent to shred segment AX, so tringles AEX nd AE X re trnsltes of ech other, which implies tht P nd Q re the sme up to trnsltion. Proposition 3.8. Let P nd Q e 3D lttice polytopes of Minkowski length 1. If there exist 5-point lttice susets S nd T of P nd Q of types 3+(7) nd (10) correspondingly, then L(P +Q) 3. Proof. We ssume tht L(P +Q) = 2. There re 13 clsses of segments totl, so S nd T hve to shre t lest 5 segments. Let S hve lttice points A, B, C, D, nd E, where ABC is tringle with sides of the sme clss nd D point inside tht tringle. If AE, BE, nd CE re ll shred with T, then two of them shre vertex in T nd hence one of the tringles ABE, ACE, or BCE is shred, which is impossile since sides of ABC cnnot e shred. Hence one of AE, BE, CE is not shred. Similrly, one of AD, BD, CD is not shred. This implies tht ED is shred, nd we cn ssume tht ADE is shred tringle, DB nd EC re shred, while DC nd EB re not shred. E E ++c +c c B +c ++c c A + D C A? + D

18 18 OLIVIA BECKWITH, MATTHEW GRIMM, JENYA SOPRUNOVA, AND BRADLEY WEAVER Then nd +c in T should not e djcent to ED (or either DC or EB would e shred). Hence one of, +c hs A s n endpoint in T. If it is, we hve either or in T; if it is +c, then either +c or is in T. Both of these options re impossile. Proposition 3.9. Let P nd Q e 3D lttice polytopes of Minkowski length 1. If there exist 5-point lttice susets S nd T of P nd Q of types 2+2+(6) nd (10) correspondingly, then L(P +Q) 3. Proof. We ssume tht L(P +Q) = 2. There re 13 clsses of segments totl, so S nd T hve to shre t lest 5 segments. Let S hve lttice points A, B, C, D, nd E, where AB nd CD re from the sme clss, nd BC nd AD re from the sme clss. At lest 3 segments strting t E re shred. In T t lest two of them shre vertex, hence S nd T shre tringle with vertex E. We cn ssume tht tht tringle is AEC. One of EB, ED is lso shred, let s ssume it s EB. This segment in T cnnot shre vertex with AE or EC (this would imply tht either AB or BC is shred), so it connects two remining vertices X nd Y. If ED is lso shred, sme would e true, ut there re only five vertices, so ED is not shred. Hence BD is shred. But it would hve to hve either X or Y s one of the vertices, hence EBD is shred, contrdiction. Theorem Let P, Q, R e three 3D lttice polytopes of Minkowski length 1 with t lest five lttice points ech. Then L(P +Q+R) 4. Proof. Let S,T, nd U e ny 5-point lttice susets of P, Q, nd R correspondingly. By the ove propositions, S, T, nd U re ll of type (10). If ny of the three polytopes hs more thn five points, then there re t lest 15 lttice segments connecting them. Hence there re multiple segments of the sme clss. Reducing the numer of points, we get sets S, T, nd U where t lest one of these lttice sets is not of type (10). Hence we cn ssume tht ech of P, Q, nd R hs exctly five lttice points nd is of type (10). Furthermore, ll three polytopes re trnsltes of ech other. It remins to show tht L(3P) 4. Let us ssume first tht P hs four lttice vertices A,B,C,D nd lttice point E inside. Let G e the centroid of P. Drw through G four plnes, ech prllel to one of the fcets of P. Ech of the plnes cuts off tetrhedron off of P, which is similr to P with coefficient of 3/4. These four tetrhedr cover P, so the interior lttice point E elongs to t lest one of them, sy, to the tetrhedron one of whose vertices is A. Then if we continue AE to the point of intersection F with the plne BCD, then AE/AF 3/4 nd 3AF 4AE, so in 3P we hve segment 3AF whose lttice length is t lest 4. We hve shown tht in this cse L(3P) 4. Next, let P of type (10) hve 5 lttice vertices nd no other lttice points. It ws shown in [9], Theorem 3.5 tht P is AGL(3,Z) equivlent to polytope with the vertices (0,0,0), (1,0,0), (0,1,0), (0,0,1), nd (1,,) where gcd(,) = 1 nd 0. The sum of the two segments, [0,,] nd [0,1,0] is prllelogrm of re. If 3 then y Lemm 1.7 of [10] the Minkowski length of this prllelogrm is t lest 3 nd hence L(2P) 3. We re left with two cses = = 1 nd = 1, = 2 (if = 0 then L(P) 1). In the second cse P is not of type (10) s (0,,) = (0,1, 1). It remins to del with the cse = = 1. Let I e the verticl segment of length 1 nd let J e the segment connecting the origin to (1,1,0). Then I + 1/2J P nd hence 2I + J 2P, so we gin hve L(2P) 3. Notice tht we hve lso proved the following Corollry.

19 MINKOWSKI LENGTH OF 3D LATTICE POLYTOPES 19 Corollry If P nd Q re 3D lttice polytopes of Minkowski length 1 with t lest 5 lttice points ech, then L(P +Q) 3 unless P nd Q re of type (10) nd re the sme up to trnsltion. Next exmple demonstrtes tht one could hve L(2P) = 2 for 3D lttice polytope P of type (10). Exmple Let P hve the vertices ( 1, 1, 1), (1,0,0), (0,1,0), nd (0,0,1). Then P is of type (10), L(P) = 1, nd L(2P) = 2. Proposition Let P nd Q e 3D lttice polytopes of Minkowski length 1. If P hs t lest 6 lttice points nd Q hs t lest 5, then L(P +Q) 3. Proof. Since P hs 6 lttice points, there re 15 lttice segments in P. Since there re 13 clsses totl, some of the segments will repet nd we will either hve two segments of the sme clss not shring endpoints, or three segments of the sme clss forming tringle. Hence we cn pick 5 points in P tht form lttice suset of type other thn (10), which implies L(P +Q) 3. Theorem Let Q 1 + +Q N P e mximl decomposition. Let I(Q i ) denote the numer of interior lttice points with respect to the 3D topology. Then the overll numer of interior lttice points N I = I(Q i ) 4. i=1 Furthermore, if more thn one Q i hs interior lttice points with respect to the 3D topology, we hve I 2. Proof. In order to hve n interior lttice point with respect to the 3D topology, Q i hs to hve t lest 5 lttice points. Hence, y Theorem 3.10, t most two of the Q i s could hve interior lttice points nd e three-dimensionl. Let s ssume tht this holds for two of the Q i s. If one of these two Q i s hs t lest 2 interior lttice points, we get contrdiction with Proposition Hence in this cse the totl numer of interior lttice points is t most 2. If only one of the Q i s hs interior lttice points with respect to the 3D topology, then there re t most 4 of them, s L(Q i ) = 1 nd Q i hs t most 8 lttice points totl. The following exmple shows tht there exists Minkowski length one 3D polytope tht hs 4 interior lttice points, so the ound of Theorem 3.14 is shrp. Exmple Consider simplex P with the vertices (0,0,0), (1,3,0), (0,2,3), nd (4,1,3). Then the interior lttice points of P re (1,2,1), (1,2,2), (1,1,1), nd (2,1,2). This cn e checked y hnd or using Polymke [2]. It is esy to verify tht there re no prllel lttice segments connecting lttice points in P, which implies tht L(P) = 1. An exmple of 3D lttice polytope of Minkowski length one with 8 lttice points (ll of them on the oundry) ws given in n MSRI-UP project directed y John Little [11]. The numer of lttice points in lttice polytope in the n-spce is t most 2 n. A group of REU students t Kent in Summer 2011 constructed lttice n-dimensionl Minkowski length one polytope with 2 n points. It would e interesting to see if there exists n n-dimensionl simplex tht hs 2 n lttice points nd Minkowski length one.

20 20 OLIVIA BECKWITH, MATTHEW GRIMM, JENYA SOPRUNOVA, AND BRADLEY WEAVER Acknowledgments. We re thnkful to the nonymous referee for pointing out gp in one of the rguments s well s for numerous corrections nd suggestions. References [1] Alexnder I. Brvinok, P olynomil time lgorithm for counting integrl points in polyhedr when the dimension is fixed, Mthemtics of Opertions Reserch, 19 (1994), [2] Ewgenij Gwrilow nd Michel Joswig, Polymke: frmework for nlyzing convex polytopes. Polytopescomintorics nd computtion. (Oerwolfch, 1997), 4373, DMV Sem., 29, Birkhuser, Bsel, [3] Hnsen, John P., Toric surfces nd error-correcting codes, Coding theory, cryptogrphy nd relted res (Gunjuto, 1998), Springer, Berlin, 2000, [4] Alexnder M. Ksprzyk, Toric Fno three-folds with terminl singulrities, Tohoku Mth. J. (2) 58 (2006), no. 1, [5] Köppe, Mtthis, A priml Brvinok lgorithm sed on irrtionl decompositions, SIAM Journl on Discrete Mthemtics, 21 (2007), [6] John Little nd Hl Schenck, Toric surfce codes nd Minkowski sums, SIAM J. Discrete Mth. 20 (2006), no. 4, (electronic). [7] Morris Newmn, Integrl Mtrices, Pure nd Applied Mthemtics, Volume 45, Acdemic Press (1972), ISBN-13: , 224 pges. [8] Diego Runo, On the prmeters of r-dimensionl toric codes, Finite Fields Appl. 13 (2007), no. 4, [9] Herert E. Scrf, Integrl polyhedr in three spce, Mthemtics of Opertions Reserch 10 (1985), no. 3, [10] Ivn Soprunov nd Jeny Soprunov, Toric surfce codes nd Minkowski length of polygons, SIAM J. Discrete Mth. 23 (2008/09), no. 1, [11] A. Gudinez; C. Outing; R. Veg, Indecomposle polyhedr nd toric codes, MSRI-UP 2009 Technicl report, Hrvey Mudd College, Deprtment of Mthemtics, 301 Pltt Boulevrd, Clremont, CA 91711, USA E-mil ddress: oeckwith@gmil.com Deprtment of Mthemtics, UCSD, 9500 Gilmn Dr., #0112, L Joll, CA USA E-mil ddress: mgrimm2@kent.edu Deprtment of Mthemtics, Kent Stte University, Summit Street, Kent, OH 44242, USA E-mil ddress: soprunov@mth.kent.edu URL: Grove City College, Deprtment of Mthemtics, 100 Cmpus Drive,Grove City PA 16127, USA E-mil ddress: weverr1@gcc.edu