Markov complexity of monomial curves
aa r X i v : . [ m a t h . A C ] N ov MARKOV COMPLEXITY OF MONOMIAL CURVES
HARA CHARALAMBOUS, APOSTOLOS THOMA, MARIUS VLADOIU
Abstract.
Let A = { a , . . . , a n } ⊂ N m . We give an algebraic characterizationof the universal Markov basis of the toric ideal I A . We show that the Markovcomplexity of A = { n , n , n } is equal to two if I A is complete intersection andequal to three otherwise, answering a question posed by Santos and Sturmfels.We prove that for any r ≥ A ( r ) .Moreover, we prove that for any integer l there exist integers n , n , n such thatthe Graver complexity of A is greater than l . Introduction
Let k be a field, n, m ∈ N , A = { a , . . . , a n } ⊂ N m and A ∈ M m × n ( N ) bethe matrix whose columns are the vectors of A . We let L ( A ) := Ker Z ( A ) be thecorresponding sublattice of Z n and denote by I A the corresponding toric ideal of A in k [ x , . . . , x n ]. We recall that I A is generated by all binomials of the form x u − x w where u − w ∈ L ( A ).A Markov basis of A is a finite subset M of L ( A ) such that whenever w , u ∈ N n and w − u ∈ L ( A ) (i.e. A w = A u ), there exists a subset { v i : i = 1 , . . . , s } of M that connects w to u . This means that ( w − P pi =1 v i ) ∈ N n for all 1 ≤ p ≤ s and w − u = P si =1 v i . A Markov basis M of A is minimal if no subset of M is aMarkov basis of A . For a vector u ∈ L ( A ) we let u + , u − be the unique vectors in N n such that u = u + − u − . If M is a minimal Markov basis of A then a classicalresult of Diaconis and Sturmfels states that the set { x u + − x u − : u ∈ M} is aminimal generating set of I A , see [6, Theorem 3.1]. The universal Markov basis of A , which we denote by M ( A ), is the union of all minimal Markov bases of A , wherewe identify elements that differ by a sign, see [9, Definition 3.1]. The intersection ofall minimal Markov bases of A via the same identification, is called the indispensablesubset of the universal Markov basis M ( A ) and is denoted by S ( A ). The Graverbasis of A , G ( A ), is the subset of L ( A ) whose elements have no proper conformaldecomposition , i.e. u ∈ L ( A ) is in G ( A ) if there is no other v ∈ L ( A ) such that v + ≤ u + and v − ≤ u − , see [14, Section 4]. The Graver basis of A is always afinite set and contains the universal Markov basis of A , see [14, Section 7]. Thusthe following inclusions hold: S ( A ) ⊆ M ( A ) ⊆ G ( A ) . Mathematics Subject Classification.
Key words and phrases.
Toric ideals, Markov basis, Graver basis, Lawrence liftings.The third author was partially supported from grant PN–II–ID–PCE–2011–3–1023, nr.247/2011, awarded by UEFISCDI. n [4] a description was given for the elements of S ( A ) and M ( A ) that had a ge-ometrical flavour: it considered the various fibers of A in N n and the connectedcomponents of certain graphs. It did not examine the problem from a strict alge-braic point of view such as conformality. This point of view is seen in [9], but onlyfor the elements of S ( A ) from the side of sufficiency. In [9], the authors show thatany vector with no proper semiconformal decomposition is necessarily in S ( A ), see[9, Lemma 3.10]. In this paper we attempt to give the complete algebraic char-acterization for the elements of S ( A ) and M ( A ). This is done in Section 1. InProposition 1.1 we prove that the condition of [9, Lemma 3.10] is not only sufficientbut also necessary. Next, to give the algebraic characterization of the vectors in M ( A ), we introduce the notion of a proper strongly semiconformal decomposition and prove that the nonzero vectors with no proper strongly semiconformal decom-position are precisely the vectors of M ( A ), see Proposition 1.4. The relationshipbetween these decompositions is given in Lemma 1.2. Schematically the followingimplications hold:proper conformal ⇒ proper strongly semiconformal ⇒ proper semiconformal . In Example 1.3 we show that these implications are the best one could hope. Itis important to note that the definitions of conformal and semiconformal decom-positions involve exactly two summands. The natural and easy generalization todecompositions involving l summands, l ≥
2, does not produce anything new: suchdecompositions lead to a conformal and semiconformal decomposition with exactlytwo summands. This fact stands in contrast to the definition of a strongly semi-conformal decomposition. As is shown in Example 1.5, a vector may have a properstrongly semiconformal decomposition into l vectors with l >
2, but not a properstrongly semiconformal decomposition into exactly 2 vectors. There are howevercertain classes of integer configurations for which the notion of proper strongly semi-conformality into l vectors with l ≥ A as we show in Lemmas 2.3 and 3.2. Another class is given by the Lawrenceliftings of monomial curves in A , as follows from Theorems 2.6 and 3.4. We alsonote that incidence matrices of graphs have this property, see [12, Propositions 4.3and 4.8].For A ∈ M m × n ( N ) as above and r ≥
2, the r –th Lawrence lifting of A is denotedby A ( r ) and is the ( rm + n ) × rn matrix A ( r ) = r − times z }| { A A
0. . .0 0 AI n I n · · · I n , see [13]. We write L ( A ( r ) ) for Ker Z ( A ( r ) ), denote by A ( r ) the matrix A ( r ) , and identifyan element of L ( A ( r ) ) with an r × n matrix: each row of this matrix corresponds to n element of L ( A ) and the sum of its rows is zero. The type of an element of L ( A ( r ) )is the number of nonzero rows of this matrix. The Markov complexity , m ( A ), is thelargest type of any vector in the universal Markov basis of A ( r ) as r varies. The Graver complexity of A , g ( A ), is the largest type of any vector in the Graver basisof A ( r ) , as r varies. We note that the study of A ( r ) , for A ∈ M m × n ( N ) was motivatedby consideration of hierarchical models in Algebraic Statistics, see [13]. Aoki andTakemura, in [3], while studying Markov bases for certain contingency tables withzero two-way marginal totals, gave the first examples of matrices with finite Markovcomplexity, see [3, Theorem 4]. In [13, Theorem 1], Santos and Sturmfels provedthat m ( A ) is bounded above by the Graver complexity of A , g ( A ), and since thelatter one is finite, m ( A ) is also finite. In fact, g ( A ) is the maximum 1-norm of anyelement in the Graver basis of the Graver basis of A , [13, Theorem 3]. Up to now,no formula for m ( A ) is known in general and there are only a few classes of toricideals for which m ( A ) has been computed, see [2, 3, 13]. In this paper we compute m ( A ), when A is a monomial curve in A . This answers a question posed by Santosand Sturmfels in [13], see Example 6 of that paper. We succeed in computing m ( A )by applying the results of Section 1.The topic of monomial curves has been the subject of extensive research ever sinceHerzog in [8] studied such configurations. We recall that a monomial curve in the d -dimensional affine space A d is defined as the curve { ( t n , . . . , t n d ) : t ∈ k } , where n , . . . , n d are positive integers such that gcd( n , . . . , n d ) = 1. In this paper wewrite A = { n , n , n } ⊂ Z > and implicitly assume that gcd( n , n , n d ) = 1. In [8,Theorem 3.8], it was shown that the toric ideal I A is either a complete intersectionor if not, then it is minimally generated by exactly three binomials. We deal witheach case separately. In Section 2, we consider the case when I A is not a completeintersection. We show that the universal Markov basis of A ( r ) is unique and computeits cardinality, see Theorem 2.6. We also show in the same Theorem 2.6 that m ( A ) =3. In Section 3, we consider the case when I A is a complete intersection. As beforewe show that the universal Markov basis of A ( r ) is unique, compute its cardinality,and show that m ( A ) = 3, see Theorem 3.4. We note that Herzog in [8] describesall possible minimal generating sets of I A in either case, an essential tool to ourstudy. To be more precise, with the notation of [8], for i ∈ { , , } we consider c i to be the smallest element of Z > such that there exist integers r ij , r ik ∈ N with { i, j, k } = { , , } , and with the property that c i n i = r ij n j + r ik n k . What determineswhether I A is a complete intersection or not is whether there are i, j ∈ { , , } suchthat r ij = 0. If r ij > i, j = 1 , , I A is minimally generated byexactly three binomials and in this case, I A has a unique minimal generating setwhich is explicitly described in [8, Proposition 3.2, Proposition 3.3]. If there exist i, j ∈ { , , } such that r ij = 0 then I A is a complete intersection and has no uniqueminimal binomial generating set. In this case the universal Markov basis of A isexplicitly described in [8, Proposition 3.5].As mentioned above, in Theorems 2.6 and 3.4, the uniqueness of the minimalMarkov basis of A ( r ) for r ≥ A in A . Wewant to further dwell on this fact. It is well known that A (2) has a unique minimal arkov basis for all A ⊂ N m , see [14, Theorem 7.1]. Moroever for r ≥
3, theuniqueness of the minimal Markov bases for A ( r ) is noted for some classes of toricideals, see [3, 10, 13], see also related [10, Conjecture 3.8]. It is thus of interest tonote that Lawrence liftings of monomial curves in A have unique minimal Markovbasis, although the curves themselves may not.In Section 4 we compare m ( A ) with g ( A ) when A = { n , n , n } is a monomialcurve in A . To find g ( A ) we have to compute the Graver basis of the Graverbasis of A , which is difficult since there is no good description of the Graver bases ofmonomial curves in A . In Theorem 4.2 we show that there is no integer that boundsfrom above the Graver complexities of all monomial curves and thus the differencebetween m ( A ) and g ( A ) can be made arbitrarily large. Furthermore we pose somequestions that were motivated and partly based on the extensive computations thatwere done with 4ti2, [1]. 1. Universal Markov basis
Let A = { a , . . . , a n } ⊂ N m and A ∈ M m × n ( N ), L ( A ) ⊂ Z n , I A ⊂ k [ x , . . . , x n ]be the corresponding matrix, lattice and toric ideal of A respectively. We note thatthe only invertible element of N A is and L ( A ) ∩ N n = { } .Let u , w , w ∈ L ( A ) be such that u = w + w . We say that the above sum isa conformal decomposition of u and write u = w + c w if u + = w +1 + w +2 and u − = w − + w − . If both w , w are nonzero, we call such a decomposition proper .The Graver basis of A , G ( A ), consists of the nonzero vectors in L ( A ) for whichthere is no proper conformal decomposition and is a finite set, see for example [14,Algorithm 7.2].The notion of a semiconformal decomposition was introduced in [9, Definition 3.9].Let u , v , w ∈ L ( A ). We say that u = v + sc w is a semiconformal decomposition of u if u = v + w and v ( i ) > w ( i ) ≥ w ( i ) < v ( i ) ≤ ≤ i ≤ n . Here v ( i ) denotes the i th coordinate of the vector v . We callthe decomposition proper if both v , w are nonzero. It is easy to see that u = v + sc w if and only if u + ≥ v + and u − ≥ w − . We remark that cannot be written as thesemiconformal sum of two nonzero vectors since L ( A ) ∩ N n = { } . When writing asemiconformal decomposition of u it is necessary to specify the order of the vectorsadded. A semiconformal decomposition of u for which the order of the vectors canbe reversed is a conformal decomposition, that isif u = v + sc w and u = w + sc v then u = v + c w . We note that a semiconformal decomposition of u gives rise to a semiconformaldecomposition of − u and vice versa, by simply reversing the order of the summands: u = v + sc w ⇔ − u = ( − w ) + sc ( − v ) . Let u ∈ L ( A ). The fiber F u is the set { t ∈ N n : u + − t ∈ L ( A ) } . F u is a finite set, seefor example [5, Proposition 2.3]. Next we show that lack of a proper semiconformaldecomposition is not only a sufficient condition for an element to be in S ( A ) as wasshown in [9, Lemma 3.10], but it is also a necessary condition. roposition 1.1. The indispensable part S ( A ) of the universal Markov basis con-sists of all nonzero vectors in L ( A ) which have no proper semiconformal decompo-sition.Proof. We only need to show that if u ∈ S ( A ) then u has no proper semiconformaldecomposition. Suppose that u = v + sc w for some nonzero vectors v , w ∈ L ( A ).Since u ∈ S ( A ) the binomial x u + − x u − belongs to all minimal system of generators of I A . It follows that F u consists of exactly two elements, u + and u − , see [4, Corollary2.10]. On the other hand u = v + sc w implies that u + − v = u − + w ∈ N n . Moreover A ( u + − v ) = A u + and thus u + − v is in the fiber of u + . This implies that either u + − v = u + and thus v = or u + − v = u − which then implies that w = , acontradiction in either case. (cid:3) Let u , u , . . . , u l ∈ L ( A ), l ≥
2. We say that u = ssc u + · · · + u l , is a stronglysemiconformal decomposition if u = u + · · · + u l and the following conditionsare satisfied: u + > u +1 and u + > ( i − X j =1 u j ) + u + i for all i = 2 , . . . , l. When l = 2, we simply write u = u + ssc u . Note that u = u + ssc u implies that u + > u +1 and u − > u − . We say that the decomposition is proper if all u , . . . , u l arenonzero. We remark that if u = ssc u + · · · + u l is proper then u + , u + − u , . . . , u + − P li =1 u i = u − ∈ N n and thus are distinct elements of F u . In the following lemmawe show the implications amongst the three types of decompositions we definedabove. It is immediate that a conformal decomposition is also a semiconformaldecomposition. Lemma 1.2.
Let u ∈ L ( A ) be a nonzero vector. Then the following hold: (i) If u has a proper conformal decomposition then u has a proper strongly semi-conformal decomposition, (ii) If u has a proper strongly semiconformal decomposition then u has a propersemiconformal decomposition.Proof. For (i) note that if u = v + c w then u + = v + + w + and u − = v − + w − . If u + = v + then w + = and thus w = , a contradiction. Similarly, one shows that u − = w − and thus u = v + ssc w .In order to prove (ii), assume that u admits a proper strongly semiconformaldecomposition: there exists l ≥ u , . . . , u l ∈ L ( A ) \ { } such that u = ssc u + · · · + u l . In particular, u + > u +1 . We let v = u + · · · + u l . We will show that u = u + sc v . Indeed, since A ( u + − u ) = A u + and u + − u = u + − u +1 + u − > it follows that u + − u belongs to the fiber of u + . Moreover, u + = u + − u since u = . On the other hand, if u + − u = u − then we obtain that u = u and thus u + = u +1 , a contradiction. Therefore, the fiber F u contains at least three differentvectors: u + , u + − u , u − . Hence we have the following expression x u + − x u − = x u + − u +1 ( x u +1 − x u − ) + x t ( x u + − u − t − x u − − t ) , here x t is the monomial with the property that gcd( x u + − u , x u − ) = x t . The lastcondition implies that there exists a nonzero vector w ∈ L ( A ) such that w + = u + − u − t and w − = u − − t . It is easy to see that u = u + sc w . On the otherhand we have u = u + v and thus v = w , which implies our claim. (cid:3) The reverse implications from Lemma 1.2 do not follow as the next example shows.
Example 1.3.
Let A = { , , } . The toric ideal I A is a complete intersectionand there are exactly two minimal Markov bases of A : { ( − , , , (4 , , − } and { ( − , , , (1 , , − } , see Proposition 3.1. Thus M ( A ) = { ( − , , , , − , , − } while S ( A ) = { ( − , , } . Moreover, computation with 4ti2 [1] showthat the elements of G ( A ) are u = (0 , , − u = (3 , − , u = (4 , , − u = (1 , , − u = (7 , − , − u = (11 , , − u = (1 , − , u = (2 , − , u = u + ssc u while u = u + c u and that even though u = u + sc u the strong semiconformality does not hold: u = u + ssc u .If u ∈ L ( A ) we associate an A -degree to u and the binomial B = x u + − x u − ∈ I A as follows: deg A ( u ) = deg A ( B ) = n X i =1 u + ( i ) a i . We note that by [4, Proposition 2.2] u is in a minimal Markov basis of A and x u + − x u − is part of a minimal generating set of I A if and only if x u + − x u − is notin the ideal generated by the binomials of I A of strictly smaller A -degrees. Proposition 1.4.
The universal Markov basis M ( A ) of A consists of all nonzerovectors in L ( A ) with no proper strongly semiconformal decomposition.Proof. Let u ∈ M ( A ) and suppose that u = ssc u + · · · + u l . We consider thebinomials B = x u + − x u − and B i = x u + i − x u − i for i = 1 , . . . , l . It is immediate that B = x u + − u +1 B + x u + − u − u +2 B + · · · + x u + − P l − i =1 u i − u + l B l . The strongly semiconformality assumption implies that the coefficients in the aboveexpression are all non-constant monomials. Since B i ∈ I A for i = 1 , . . . , l it followsthat B is in the ideal generated by the binomials of I A of strictly smaller A -degreethan B . By [7, Section 1.3], B can not be part of any minimal system of generatorsof I A , a contradiction.Suppose now that u ∈ L ( A ) \ { } and that u
6∈ M ( A ). It follows that B = x u + − x u − is in the ideal generated by the binomials of I A of strictly smaller A -degree than B . By [5, Proposition 3.11], there are monomials x t i = 1 and binomials B i = x u i + − x u i − for i = 1 , . . . , l where u i ∈ L ( A ) such that B = x t B + · · · + x t l B l ,x t x u + = x u + , x t i x u + i = x t i − x u − i − , for i = 2 , . . . , l, and x t l x u − l = x u − . Note that the binomials B i in this expression need not be distinct. Thus u = u + · · · + u l and t = u + − u +1 , t i = u + − ( i − X j =1 u j ) − u + i , i = 2 , . . . , l . ince t i > for i = 1 , . . . , l it follows that u = ssc u + u + · · · + u l . (cid:3) The following example shows that it is necessary to define the strongly semicon-formality in terms of l vectors where l ≥ Example 1.5.
Let A = { (2 , , (2 , , (0 , , (1 , , (3 , } be a subset of N . Thecorresponding matrix A ∈ M × ( N ) is A = ! . Using 4ti2 [1] and [4, Theorem 2.6] we get that M ( A ) = { (1 , − , − , , , (0 , , , , − , (0 , , , − , , (1 , , , − , } . We consider the vector u = (2 , , , − , − ∈ L ( A ). We note that u + = (2 , , , , u − = (0 , , , ,
1) and deg A ( u ) = (4 , F u = { (2 , , , , , (1 , , , , , (0 , , , , , (0 , , , , , (0 , , , , } . We denote the elements of F u by v , . . . , v written in the above order. It is straight-forward that u = ssc u + u + u , where u = v − v , u = v − v , u = v − v ∈ L ( A ). Thus according toTheorem 1.4, u / ∈ M ( A ). It is interesting to note that u does not have a properstrongly semiconformal decomposition into two vectors. Indeed, if u = w + ssc w then u + − w = u − + w ∈ F u and cannot equal v = u + nor v = u − . Thus u + − w is either v , v or v . This implies that w = v − v i and w = v i − v foran integer i ∈ { , , } . But for i ∈ { , } we have w − = v = u − , while if i = 4 then w +1 = v = u + , which leads to a contradiction in either case since u = w + ssc w .Next, for r ≥ A ( r ) of A . The following remarkallows us to restrict the discussion of semiconformal decompositions of elements of L ( A ( r ) ) to semiconformal decompositions of elements of maximum type r . Remark 1.6.
Let r ≥ v ∈ L ( A ( r ) ) be a nonzero vector of type s < r and withoutloss of generality assume that the nonzero rows of v are the first s rows. Note that s ≥
2. We define w ∈ L ( A ( s ) ) as the vector whose i th row is just the i th row of v for i = 1 , . . . , s . Then v admits a proper semiconformal decomposition if and onlyif w ∈ L ( A ( s ) ) admits a proper semiconformal decomposition. The equivalence isimmediate since the unique semiconformal decomposition of is into zero vectors.The above discussion holds for proper strongly semiconformal decompositions v aswell.Let v = ( v i ) ∈ L ( A ( r ) ). If z = [ v − v . . . ] T and w = [ + v v . . . v r ] T , itis clear that v = z + w and that z and w are also in L ( A ( r ) ). The following lemmacomments on the semiconformality of the decomposition of v into these two vectors. Lemma 1.7.
Let v ∈ L ( A ( r ) ) , z , w be as above. If v has type greater than or equalto and v = z + sc w is proper then v = z + ssc w . Similarly if v = w + sc z is properthen v = w + ssc z . roof. Suppose that v = z + sc w is proper semiconformal. Assume by contradictionthat v is not the strongly semiconformal sum of z and w . It follows that v + = z + or v − = w − . If v + = z + then we obtain that v + i = for all i ≥ v i ∈ L ( A ) we have v i = for all i ≥
3. Therefore the type of v is less than three, acontradiction. Hence v − = w − and consequently v − = . This implies that v = and thus z = , a contradiction. The assertion about the second decomposition isproved similarly. (cid:3) Markov complexity for monomial curves in A which are notcomplete intersections In this section we study Lawrence liftings of monomial curves in A whose corre-sponding toric ideals are not complete intersections. Suppose that A = { n , n , n }⊂ Z > is a monomial curve such that I A is not a complete intersection ideal. For1 ≤ i ≤ c i be the smallest element of Z > such that c i n i = r ij n j + r ik n k , r ij , r ik ∈ Z > with { i, j, k } = { , , } . Below, we recall the description of the uniqueMarkov basis of A given in [8, Proposition 3.2, Proposition 3.3]. Theorem 2.1.
Let A = { n , n , n } be a set of positive integers with gcd( n , n , n ) =1 , and with the property that I A is not a complete intersection ideal. Let u =( − c , r , r ) , u = ( r , − c , r ) , u = ( r , r , − c ) . Then A has a unique mini-mal Markov basis, M ( A ) = { u , u , u } and u + u + u = . It follows immediately from Theorem 2.1 that M ( A ) = S ( A ). Another way tosee that is applying [11, Remark 4.4.3], since I A is a generic lattice ideal.To any vector u ∈ Z we assign a sign-pattern: we put + if the coordinate of u ispositive, − if the coordinate of u is negative and 0 if the corresponding coordinateis 0. When we don’t know exactly the sign of the coordinate we use the symbol ∗ . For example the vector (0 , , −
3) has the sign-pattern 0 + − while the elements u , u , u of M ( A ) have sign-patterns − + +, + − + and + + − . We note thatwhen α, β ∈ Z > the vectors α ( − u ), β u and α ( − u ) + β u have sign-patterns+ − − , + − + and + − ∗ . By looking at the sign patterns, it is immediate that α ( − u ) + β u = α ( − u ) + ssc β u . We generalize and isolate this remark. Remark 2.2.
Let i = j ∈ { , , } and α, β ∈ Z > . Then α ( − u i ) + β u j = α ( − u i ) + ssc β u j .The geometry of the plane implies the following lemma. Lemma 2.3.
Let A = { n , n , n } be such that I A is not a complete intersection.Let = v ∈ L ( A ) . Then either v = α ( ± u i ) for an α ∈ Z > and i ∈ { , , } or v = α ( − u i ) + ssc β u j for some α, β ∈ Z > and i = j ∈ { , , } .Proof. Since u + u + u = and L ( A ) = Z u + Z u + Z u it follows thatrank Z ( L ( A )) = 2. Therefore the three vectors u , u , u define a complete pointedpolyhedral fan of the plane, span R ( u , u ). This polyhedral fan has thirteen non-empty faces. Other than the vertex { } , six of the faces are one-dimensional: R + u , R + ( − u ), R + u , R + ( − u ), R + u and R + ( − u ). The rest are two-dimensional: + {− u , u } , R + {− u , u } , R + {− u , u } , R + {− u , u } , R + {− u , u } and R + {− u , u } .Let v ∈ L ( A ) be a nonzero vector. Since v belongs to the polyhedral fan then itfollows at once that v belongs either to one of the six one-dimensional cones, thatis v = α ( ± u i ) for some α ∈ Z > and i ∈ { , , } , or v belongs to the interior ofone of the six two-dimensional cones, that is α ( − u i ) + β u j for some α, β ∈ Z > and i = j ∈ { , , } . It is immediate, by Remark 2.2, that this sum determines a properstrongly semiconformal decomposition of v . (cid:3) In the following lemma we consider a subset T of L ( A ( r ) ). The elements of T havetype 2 and 3. In Theorem 2.6 we will see that T equals the universal Markov basisof A ( r ) . Lemma 2.4.
Let A = { n , n , n } be such that I A is not a complete intersection. Let r ≥ and let T be the subset of L ( A ( r ) ) containing all vectors of type whose nonzerorows are of the form u , − u , with u ∈ G ( A ) and all vectors of type whose nonzerorows are permutations of u , u , u . Then T ⊂ S ( A ( r ) ) . Moreover | T | = k (cid:16) r (cid:17) +6 (cid:16) r (cid:17) ,where k is the cardinality of the Graver basis of A .Proof. We will show that T ⊂ S ( A ( r ) ), the last part of the conclusion being imme-diate. By Remark 1.6 we may assume that r = 3. We first show that the elementof type 2 u − u0 , u ∈ G ( A )is indispensable. Suppose that is not. Then by Proposition 1.1 it admits a propersemiconformal decomposition, which is of the following form due to Remark 1.6 u − u0 = v − v + sc v − v . Thus u = v + sc v and − u = − v + sc ( − v ). But u = v + sc v also implies that − u = − v + sc v and thus u = v + c v . Thus u is not in the Graver basis of A ,a contradiction. Therefore all elements of type 2 whose nonzero rows belong to theGraver basis of A are indispensable. Next we prove that the element of type 3 u u u is indispensable. Arguing by contradiction and applying again Proposition 1.1, itwould imply that the vector has the following proper semiconformal decomposition u u u = v v v + sc w w w . Hence we have u i = v i + sc w i for i = 1 , ,
3. Since u i ∈ S ( A ) it follows that either v i or w i is zero. It follows that one of the vectors [ v v v ] T or [ w w w ] T as at least two zero rows, and thus it must be zero. This is a contradiction andconsequently all elements of type 3 with the rows being permutations of u , u , u are indispensable. (cid:3) In the next lemma we show that a class of elements of L ( A ( r ) ) is not part of M ( A ( r ) ). Lemma 2.5.
Let A = { n , n , n } be such that I A is not a complete intersection.Let r ≥ and T ⊂ S ( A ( r ) ) be as in Lemma 2.4. Suppose that for ≤ i ≤ r , α i , β i ∈ N , = v = α ( − u ) + β u ... α s ( − u ) + β s u β s +1 ( − u ) + α s +1 u ... β r ( − u ) + α r u ∈ L ( A ( r ) ) \ T .
Then v has a proper strongly semiconformal decomposition into summands of L ( A ( r ) ) where one of them has type .Proof. First suppose that v is of a type 2. Since v T it follows that its nonzerorows do not belong to G ( A ), and thus v admits a proper conformal decomposition.Applying Lemma 1.2(i) it follows that v has also a proper strongly semiconformaldecomposition. Next suppose that the type of v is greater than or equal to 3. ByRemark 1.6 we can assume that the type of v is exactly r where r ≥
3. We denotethe row vectors of v by v , . . . , v r . We notice that 1 ≤ s < r . Indeed, if s = 0 or s = r then we obtain that ( P ri =1 α i ) u − ( P ri =1 β i ) u = . Since u , u are linearlyindependent this implies that P ri =1 α i = P ri =1 β i = 0. Therefore we have α i = β i = 0for all i , which leads to v = , a contradiction.Suppose that s = 1. Since the row vectors of v add up to zero and u , u arelinearly independent, we obtain that α = P i ≥ α i and β = P i ≥ β i . Note that α = 0 or β = 0, otherwise we would have v = , a contradiction. If α = 0 then α i = 0 for all i . Let z = [ u − u . . . ] T . Then v = z + ssc ( v − z ). The case β = 0 is similar. Furthermore, if α , β = we notice that v can be decomposedas the sum of two vectors z , w of type 2 and r − z = α ( − u ) + β u β ( − u ) + α u ... , w = ( P i ≥ α i )( − u ) + ( P i ≥ β i ) u β ( − u ) + α u ... β r ( − u ) + α r u . The sign-patterns of α ( − u ) + β u and ( P i ≥ α i )( − u ) + ( P i ≥ β i ) u are of theform + − ∗ . Thus v = z + sc w or v = w + sc z . Applying Lemma 1.7 it follows that v = z + ssc w or v = w + ssc z .We remark that if s = r −
1, then a similar argument holds. Suppose now that2 ≤ s ≤ r −
2. Since v ∈ L ( A ( r ) ) and u , u are linearly independent we have the ollowing relations s X i =1 α i = r X j = s +1 α j and s X i =1 β i = r X j = s +1 β j . We may assume for the rest of the proof that s ≤ r − s , otherwise we replace v by − v . Since s ≤ r − s there exist i, j with 1 ≤ i ≤ s and s + 1 ≤ j ≤ r such that α i ≥ α j . For simplicity of notation we may assume that i = 1 and j = r . Firstsuppose that α = α r . If β ≤ β r then v = α ( − u ) + β u ... β r − ( − u ) + α r − u ( β r − β )( − u ) + sc α ( − u ) + β u ... β ( − u ) + α r u , otherwise v = α ( − u ) + β r u ... β r ( − u ) + α r u + sc ( β − β r ) u α ( − u ) + β u ... β r − ( − u ) + α r − u . In both situations we may apply Lemma 1.7 and we obtain that v has a properstrongly semiconformal decomposition. Next we examine what happens when α >α r . If β ≥ β r then v can be written as a sum of two vectors z , w of types 2 and r − z = α r ( − u ) + β r u ... β r ( − u ) + α r u , w = ( α − α r )( − u ) + ( β − β r ) u α ( − u ) + β u ... β r − ( − u ) + α r − u . Since the type of v is r , the type of z is 2. Moreover since α − α r > β − β r ≥ α − α r )( − u )+( β − β r ) u is + −∗ , the same as the sign-patternof α r ( − u ) + β r u . Therefore the two vectors add up semiconformally either in thisorder or the the reversed. Hence v = z + sc w or v = w + sc z and applying Lemma 1.7we see that the sums are also strongly semiconformal. The last case to consider iswhen β < β r . If α r + β > v can be decomposed as the semiconformal sumof w , z where w = ( α − α r )( − u ) α ( − u ) + β u ... β r − ( − u ) + α r − u ( β r − β )( − u ) , z = α r ( − u ) + β u ... β ( − u ) + α r u . ndeed, the the sign-patterns of the first and last row vector of each summand forcethe sum to be semiconformal. We show that the sum is also strongly semiconformal.Indeed, if v − = z − then v − k = and thus v k = for all k with 2 ≤ k ≤ r −
1, acontradiction since the type of v is bigger than 2. On the other hand, if v + = w + then v +1 = w +1 and the contradiction follows from the sign-patterns of w and z which are + − ∗ . So, v can be written as a strongly semiconformal sum.Finally suppose that α r + β = 0, that is α r = β = 0. Let j ∈ { s + 1 , . . . , r } besuch that α j >
0. We may assume that j = r −
1. Then v = ( α − − u ) α ( − u ) + β u ... β r − ( − u ) + ( α r − − u β r ( − u ) + ssc − u ... u . (cid:3) It follows from Lemma 2.3 that for an arbitrary v ∈ L ( A ( r ) ), whose row vectorsare v , . . . , v r , we can associate in a unique way a vector ( α, α ′ , β, β ′ , γ, γ ′ ) ∈ N ,which we denote by λ ( v ), such that α is the coefficient of u in P i =1 v i , α ′ thecoefficient of − u , β the coefficient of u , β ′ the coefficient of − u and so on. Onthe other hand since P ri =1 v i = we obtain that( α − α ′ ) u + ( β − β ′ ) u + ( γ − γ ′ ) u = 0 . By replacing in the equation above u = − u − u and by the linear independenceof u , u we see that α − α ′ = β − β ′ = γ − γ ′ . Theorem 2.6.
Let A = { n , n , n } be such that I A is not a complete intersection.Then m ( A ) , the Markov complexity of A , is 3. Moreover, for any r ≥ we have M ( A ( r ) ) = S ( A ( r ) ) and the cardinality of M ( A ( r ) ) is k (cid:16) r (cid:17) + 6 (cid:16) r (cid:17) , where k is thecardinality of the Graver basis of A .Proof. We let T be the subset of elements of S ( A ( r ) ) described in Lemma 2.4. Wewill show that every nonzero vector v ∈ L ( A ( r ) ) \ T admits a proper stronglysemiconformal decomposition. This will imply via Proposition 1.4 that M ( A ( r ) ) ⊂ T and thus S ( A ( r ) ) = M ( A ( r ) ) = T . Let v ∈ L ( A ( r ) ) \ T be a nonzero vector suchthat λ ( v ) = ( α, α ′ , β, β ′ , γ, γ ′ ). We have two main cases: αβγ = 0 and αβγ = 0.Assume first that αβγ = 0. It follows from Lemma 2.3 that there exist threedifferent rows of v , and we may assume that are the first three, such that in theirunique semiconformal decomposition we have a positive multiple of u , u , u . With-out loss of generality, we may assume that for 1 ≤ i ≤ u i appears in row i , thatis v i = a i ( − u j i ) + sc b i u i where j i = i and a i , b i ∈ N with b i = 0. We provenow that v = z + ssc w , where w = ( u u u . . . ) T and z = v − w . Notefirst that z is nonzero since v T . For 1 ≤ i ≤
3, the i th row of z is equal to a i ( − u j i ) + ( b i − u i , while all other rows of z are equal to the corresponding rowsof v . By the sign-patterns it is easy to see that the first three rows of v have a emiconformal decomposition: a i ( − u j i ) + b i u i = ( a i ( − u j i ) + ( b i − u i ) + sc u i , ≤ i ≤ . When i = 1, the sign-pattern of u is − + + and thus ( a ( − u j ) + b u ) + > ( a ( − u j ) + ( b − u ) + . Therefore v + > z + and v = z + sc w . Finally, assumeby contradiction that the decomposition is not strongly semiconformal. Then wenecessarily have v − = w − . Since w (1) = u (1) < z (1) = 0. On theother hand z = a ( − u j ) + sc ( b − u and since j ∈ { , } then the sign-patternof z is − ∗ ∗ if a = 0 or b = 1. Thus z (1) = 0 implies a = 0 and b = 1 andso z = . Similarly we have that z = z = . Since w k = for all k ≥ v − = w − it follows that v − k = for all k ≥
4. Therefore v k = v + k for all k ≥ v k ∈ L ( A ) we obtain v k = 0 for all k ≥
4. This implies z k = for all k ≥ v = w ∈ T , a contradiction. Thus we have provedthat v = z + ssc w , as desired. We should note now that if α ′ β ′ γ ′ = 0 then a similarargument shows that v = w ′ + ssc z ′ , where w ′ = − w .In the second case we have αβγ = 0 and from the above remark we may alsoassume that α ′ β ′ γ ′ = 0. Without loss of generality we let α = 0. If α ′ = 0 thenit follows from Lemma 2.3 that v is of the form described in Lemma 2.5 and weare done. Otherwise α ′ = 0 and since α ′ β ′ γ ′ = 0 we have β ′ = 0 or γ ′ = 0. Weanalyze just the case β ′ = 0, the other one being analogous. It follows from thedefinition of λ ( v ) that α ′ ( − u ) + β u + γ u + γ ′ ( − u ) = and using the relation − u = u + u we get from the linear independence of u and u that α ′ + β = 0.Thus, since α ′ , β ≥ α ′ = β = 0, a contradiction. Therefore we obtain that T = M ( A ( r ) ) and thus m ( A ) = 3. (cid:3) Remark 2.7.
In [9], a lower bound for m ( A ) is given by Ho¸sten and Sullivant for A ⊂ N m : it equals the maximum 1-norm of the elements in the Graver basis of S ( A ), see [9, Theorem 3.11]. By looking at the explicit description of the elementsof S ( A ) as given in Theorem 2.1, it is easy to see that the Graver basis of S ( A )contains exactly one element: (1 , , m ( A ).3. Markov complexity for monomial curves in A which arecomplete intersections In this section we study Lawrence liftings of monomial curves in A whose cor-responding toric ideals are complete intersections. Suppose that A = { n , n , n }⊂ Z > is a monomial curve such that I A is a complete intersection ideal. For1 ≤ i ≤ c i be the smallest element of Z > such that c i n i = r ij n j + r ik n k , r ij , r ik ∈ N with { i, j, k } = { , , } . In [8, Proposition 3.4], it was shown that either(0 , − c , c ) ∈ M ( A ) or ( c , , − c ) ∈ M ( A ) or ( − c , c , ∈ M ( A ). Below, werecall the description of the universal Markov basis of A given in [8, Proposition 3.5]when (0 , − c , c ) ∈ M ( A ). Proposition 3.1.
Let A = { n , n , n } be a set of positive integers such that gcd( n , n , n ) = 1 , I A is a complete intersection and (0 , − c , c ) ∈ M ( A ) . Let = ( − c , r , r ) and u = (0 , − c , c ) . The universal Markov basis of A is M ( A ) = { u , d · u + u : −⌊ r c ⌋ ≤ d ≤ ⌊ r c ⌋} , For the rest of this section we will assume that (0 , − c , c ) ∈ M ( A ), the othertwo cases being similar. Let us note that the sign patterns of u and u are − + +and 0 − +. Lemma 3.2.
Let A = { n , n , n } be such that I A is a complete intersection and (0 , − c , c ) ∈ M ( A ) . If = v ∈ L ( A ) then there are unique α, β ∈ N such that v = α ( − u ) + sc β ( ± u ) or v = β ( ± u ) + sc α u .Proof. Since rank Z ( L ( A )) = 2, if v = then there are a, b ∈ Z such that v = a u + b u for some integers a, b . It is easy to see how to write v in the desiredform. (cid:3) We note that the semiconformal decomposition of Lemma 3.2 might not bestrongly semiconformal as is the case for ( c + 1)( − u ) + sc u . Similarly, one canshow that any of the semiconformal sums from Lemma 3.2 with α, β ∈ Z > maynot be in general strongly semiconformal. In the next lemma we identify certainelements which are not part of the universal Markov basis of A ( r ) . Lemma 3.3.
Let A = { n , n , n } be such that I A is a complete intersection and (0 , − c , c ) ∈ M ( A ) . Let r ≥ . Suppose that = v ∈ L ( A ( r ) ) is of the followingform v = α ( − u ) + β u ... α s ( − u ) + β s u β s +1 ( − u ) + α s +1 u ... β r ( − u ) + α r u , where ≤ s ≤ r − and α i , β i ∈ Z > for ≤ i ≤ r . Then v admits a properstrongly semiconformal decomposition into two vectors, where one of the summandsis a vector of type two.Proof. Note first that our hypotheses imply that the type of v is r . Since v ∈ L ( A ( r ) ),the sum of the rows of v is zero. Since u , u are linearly independent we have(1) s X i =1 α i = r X j = s +1 α j and s X i =1 β i = r X j = s +1 β j . We will analyze two cases. For the first case we assume that there exist integers i, j corresponding to nonzero rows of v with 1 ≤ i ≤ s and s + 1 ≤ j ≤ r such thateither a) α i ≤ α j and β i ≤ β j or b) α i ≥ α j and β i ≥ β j . Without loss of generalitywe may assume that i = 1, j = r . Let us first suppose that α ≤ α r while β ≤ β r . hen v = z + w , where z = α ( − u ) + β u ... β ( − u ) + α u , w = α ( − u ) + β u ... β r − ( − u ) + α r − u ( β r − β )( − u ) + ( α r − α ) u . We will show that either v = z + ssc w or v = w + ssc z . Indeed, note first that bythe assumptions z is a type two vector and w is nonzero since v is of type r ≥ β ( − u ) + α u is − + ∗ , for the r th row of v we have either v r = z r + sc w r or v r = w r + sc z r . Indeed, since β r − β ≥ α r − α ≥ w r corresponding to thecases: α = α r and β = β r , α = α r and β < β r , α < α r and β = β r , α < α r and β < β r . More precisely, the corresponding four sign-patterns of w r are: 000,0 + − , − + + and − + ∗ and one can notice that z r and w r add up semiconformallyin this order or the reversed. This implies that v is the semiconformal sum of z and w in this order or the reversed one. Applying now Lemma 1.7 we obtain that v isthe strongly semiconformal sum of z and w . Next let us suppose that α ≥ α r while β ≥ β r . Thus v = z + w , where z = α r ( − u ) + β r u ... β r ( − u ) + α r u , w = ( α − α r )( − u ) + ( β − β r ) u α ( − u ) + β u ... β r − ( − u ) + α r − u . Since α r β r >
0, it follows that the sign-pattern of z is + ∗ − . The four possiblesign-patterns of w are 000, + − − , 0 + − and + ∗ − . As before one can concludethat either v = z + ssc w or v = w + ssc z .For the second case the following holds for all integers i, j corresponding to nonzerorows of v with 1 ≤ i ≤ s and s + 1 ≤ j ≤ r : either α i < α j and β i > β j or α i > α j and β i < β j . Consider i = 1 and j = r . Let us first suppose that α < α r and β > β r . We will show that v = z + ssc w , where z = α ( − u ) + β r u ... β r ( − u ) + α u , w = ( β − β r ) u α ( − u ) + β u ... β r − ( − u ) + α r − u ( α r − α ) u . Indeed, we notice first that v = z + sc w , since the sign-patterns of z and z r are+ − ∗ and − + ∗ , while the sign patterns of w and w r are 0 − + and − + +.Moreover the sign patterns of z r and w r show also that v − r = w − r and implicitly v − = w − . In addition, z = · · · = z r − = imply that v + = z + , otherwise we wouldhave v = · · · = v r − = and thus the type of v is two, a contradiction. Thereforewe also have v + = z + and v − = w − and thus v = z + ssc w , as desired. Next we onsider the case where α > α r and β < β r . Then v = w + sc z , where w = ( α − α r )( − u ) α ( − u ) + β u ... β r − ( − u ) + α r − u ( β r − β )( − u ) , z = α r ( − u ) + β u ... β ( − u ) + α r u , since the the sign-patterns of w and w r are + −− and 0+ − , while the sign patternsof z and z r are + − ∗ and − + ∗ . An argument similar to the previous one showsthat the sum is also strongly semiconformal, that is v = w + ssc z . (cid:3) We are ready to prove the main result of this section.
Theorem 3.4.
Let A = { n , n , n } be such that I A is a complete intersection.Then m ( A ) , the Markov complexity of A , is . Moreover, for any r ≥ we have M ( A ( r ) ) = S ( A ( r ) ) and the cardinality of M ( A ( r ) ) is k (cid:16) r (cid:17) , where k is the cardinalityof the Graver basis of A .Proof. We denote by T the set of type two vectors from L ( A ( r ) ) whose nonzero rowsare u , − u , where u ∈ G ( A ). If r = 2 then S ( A (2) ) = M ( A (2) ) = G ( A (2) ) = T , by[14, Theorem 7.1], hence the conclusion follows immediately. Hence we may assumethat r ≥
3. In general we have T ⊆ S ( A ( r ) ) ⊆ M ( A ( r ) ) for any r ≥
2, so it remainsto prove only that M ( A ( r ) ) ⊆ T to get the desired conclusion. The latter will followvia Proposition 1.4 if we show that any nonzero vector v ∈ L ( A ( r ) ) \ T has a properstrongly semiconformal decomposition.Without loss of generality we may assume that we are in the setting of Proposi-tion 3.1. Let v ∈ L ( A ( r ) ) \ T be a nonzero vector. We may assume via Remark 1.6that the type of v is r . By Lemma 3.2 we know that for each row vector v i of v either v i = α i ( − u ) + sc β i ( ± u ) or v i = β i ( ± u ) + sc α i u for α i , β i ∈ N with α i + β i >
0. We have three cases to analyze:(a) α i β i = 0 for all i = 1 , . . . , r ,(b) ∃ i ∈ { , . . . , r } such that α i = 0,(c) ∃ i ∈ { , . . . , r } such that β i = 0.For case (a) we apply Lemma 3.3 and we are done. For case (b) let i ∈ { , . . . , r } be such that α i = 0, and since α i + β i > β i >
0. Since the type of v is r ,by Lemma 3.2, v is of the form given in Lemma 3.3, the only difference being thatsome of the coefficients α i , β i might be zero, (not simultaneously). If i ≤ s then itfollows from (1) that there exists j with s + 1 ≤ j ≤ r such that β j = 0. Let z bethe type two vector with i -th row u and j -th row − u and let w = v − z . Then w is nonzero since the type of v is r ≥
3. Moreover by the sign patterns it followsthat v = z + sc w . That the sum is also strongly semiconformal follows either byLemma 1.7 if β i = 1 or by noticing that v + i = z + i and v − i = w − i if β i >
1, whichimplies v + = z + and v − = w − . For case (c) a similar argument as in case (b)holds. (cid:3) e note that in the case of a monomial curve A in A whose corresponding toricideal is a complete intersection, S ( A ) consists of exactly one vector and the Graverbasis of S ( A ) is { } . Thus in this case the bound given in [9, Theorem 3.11] is 0,and it is strictly smaller than m ( A ).4. Graver complexity of monomial curves in A In [13, Theorem 3], it was shown that g ( A ), the Graver complexity of A , is themaximum 1-norm of any element in the Graver basis of the Graver basis of A .However computing the Graver basis of the Graver basis of A is computationallychallenging to say the least. In this section we give a lower bound for the Gravercomplexity of a monomial curve A in A . This shows that in general the upper boundfor Markov complexity is rather crude: given any k ∈ N , one can find appropriateconfiguration A = { n , n , n } so that the g ( A ) ≥ k , while m ( A ) ≤
3. First we showthat in order to compute the Graver complexity of a monomial curve in A , one canassume that the configuration consists of vectors that are pairwise relatively prime. Proposition 4.1.
Let A = { n , n , n } such that ( n , n , n ) = 1 . For ≤ i < j ≤ we let d ij = ( n i , n j ) and consider A red = { n /d d , n /d d , n /d d } . Then g ( A ) = g ( A red ) .Proof. Let d ∈ N divide both n , n and consider A ′ = { n /d, n /d, n } . It is clearthat ( a , a , a ) ∈ L ( A ) if and only if ( a , a , a /d ) ∈ L ( A ′ ). This implies that thereis a one-to-one correspondence between the elements of the Graver bases of A and A red . Moreover it is clear that if there is an element of type t in the Graver basis of A ( r ) then there is an element of type t in the Graver basis of A ( r ) red and vice versa.Hence g ( A ) = g ( A red ). (cid:3) To prove the lower bound of the next theorem, we create an element whose typeis the desired lower bound in the appropriate Lawrence lifting.
Theorem 4.2.
Let A = { n , n , n } such that ( n , n , n ) = 1 and d ij = ( n i , n j ) forall i = j . Then g ( A ) ≥ n d d + n d d + n d d . In particular, if n , n , n are pairwise prime then g ( A ) ≥ n + n + n .Proof. By Proposition 4.1 g ( A ) = g ( A red ) and thus we may assume that d ij = 1 forall i = j . We note that v = ( n , − n , v = ( n , , − n ) and v = (0 , n , − n )are clearly in the Graver basis of A . Let k = n + n + n and consider the r × n matrix w with row vectors w i for i = 1 , . . . , k so that w i = v for the first n rows, w i = − v for the next n rows and w i = v for the last n rows. It easy to seethat w ∈ L ( A ( k ) ). We will show that w is in the Graver basis of A ( k ) . Assumeby contradiction that w / ∈ G ( A ( k ) ). Then w = z + c u for some nonzero vectors z , u ∈ L ( A ( k ) ). Since w i ∈ { v , − v , v } and belongs to G ( A ) it follows that z i or u i must equal w i for 1 ≤ i ≤ k . Therefore by summing up the rows of z we obtainthe relation(2) t v − t v + t v = or some nonnegative integers t , t , t such that t + t + t < n + n + n . However,Equation 2 implies immediately that t n = t n , t n = t n and t n = t n andsince n , n , n are pairwise relatively prime by assumption, it follows that n | t , n | t and n | t . Hence n + n + n ≤ t + t + t a contradiction. (cid:3) The following example shows that in general the inequality from Theorem 4.2 canbe strict.
Examples 4.3. (a) Let A = { , , } . Computations with 4ti2 show that themaximum 1-norm of the elements of G ( G ( A )) is 12 and thus g ( A ) equals the thelower bound of Theorem 4.2.(b) Let A = { , , } . Computations with 4ti2 show that the maximum 1-normof the elements of G ( G ( A )) is 30 and thus g ( A ) = 30, while the lower bound ofTheorem 4.2 is 22.For the rest of this section we want to briefly comment about the Markov com-plexity of generalized Lawrence liftings of monomial curves in A . Given A , r ≥ B ∈ M d × ( N ), the r -th generalized Lawrence lifting of A with B , Λ( A , B, r ),differs from A ( r ) in the last row block: B replaces I n , see [9, Definition 3.3]. Avector v belongs to L (Λ( A , B, r )) if and only if its row vectors v i ∈ L ( A ) for all i = 1 , . . . , r , and P ri =1 v i ∈ Ker( B ). Therefore L ( A ( r ) ) ⊂ L (Λ( A , B, r )) for all r ≥ m ( A , B ) as the largest type of any vector in the universal Markov basisof Λ( A , B, r ) as r varies. Example 4.4.
Let A = { , , } . It follows from Theorem 2.1 that I A is nota complete intersection, and M ( A ) consists of the following three vectors u =( − , , u = (1 , − ,
1) and u = (2 , , − m ( A , I ) = m ( A ) = 3.Let B = (1 3 0). One can show that m ( A , B ) = 2. We briefly indicate how toprove this equality, without details, since the proof is similar to the ones given inSection 2. More precisely, with the same techniques from Lemma 2.4 and Lemma 2.5,and applying Proposition 1.4, one can prove that for any r ≥ M (Λ( A , B, r )) consists of the vectors of type 1 with the nonzero row u andthe vectors of type 2 such that its two nonzero rows are u , u or u , − u , with u ∈ G ( A ) \ { u } . We also note that in this case m ( A , B ) equals the lower boundgiven by Ho¸sten and Sullivant in [9, Theorem 3.11].However if B = (1 n n ≥ m ( A , B ) ≥ (3 n + 1) / n ≡ m ( A , B ) ≥ n + 1 otherwise. Indeed,note that B · S ( A ) = ( n − − n n + 2) and in the Graver basis of this matrixthere is the circuit (0 , n +25 , n − ) if n ≡ , n + 2 , n −
1) otherwise.The previous example shows that while the Markov complexity m ( A ) is boundedabove by three for any monomial curve in A , the Markov complexity m ( A , B ) ofthe generalized Lawrence liftings of monomial curves in A is not bounded. Basedon computations with 4ti2 [1], we are tempted to raise the following question. Question 4.5.
Let A be a monomial curve in A so that I A is not a completeintersection. Is m ( A , B ) equal to the maximum 1-norm of the elements in theGraver basis of B · S ( A ), i.e. the lower bound given in [9, Theorem 3.11]? omputing Markov complexity is an extremely challenging problem, and a for-mula for it seems hard to find in general. We pose a final question based on extensivecomputational evidence. Question 4.6.
Let A ∈ M m × n ( N ). Is m ( A ) equal to the smallest integer r suchthat a minimal Markov basis of A ( r +1) does not contain elements of type r + 1? References
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Hara Charalambous, Department of Mathematics, Aristotle University of Thes-saloniki, Thessaloniki 54124, Greece
E-mail address : [email protected] Apostolos Thoma, Department of Mathematics, University of Ioannina, Ioannina45110, Greece
E-mail address : [email protected] Marius Vladoiu, Faculty of Mathematics and Computer Science, University ofBucharest, Str. Academiei 14, Bucharest, RO-010014, Romania, andSimion Stoilow Institute of Mathematics of Romanian Academy, Research groupof the project ID–PCE–2011–3–1023, P.O.Box 1–764, Bucharest 014700, Romania
E-mail address : [email protected]@gta.math.unibuc.ro