Fractional Local Dimension

FRACTIONAL LOCAL DIMENSION arXiv:1906.05839v2 [math.CO] 16 Oct 2020 HEATHER C. SMITH AND WILLIAM T. TROTTER Abstract. The original notion of dimension for posets was introduced by Dushnik and Miller in 1941 and has been studied extensively in the literature. In 1992, Brightwell and Scheinerman developed the notion of fractional dimension as the natural linear programming relaxation of the Dushnik-Miller concept. In 2016, Ueckerdt introduced the concept of local dimension, and in just three years, several research papers studying this new parameter have been published. In this paper, we introduce and study fractional local dimension. As suggested by the terminology, our parameter is a common generalization of fractional dimension and local dimension. For a pair (n, d) with 2 ≤ d < n, we consider the poset P (1, d; n) consisting of all 1-element and d-element subsets of {1, . . . , n} partially ordered by inclusion. This poset has fractional dimension d + 1, but for fixed d ≥ 2, its local dimension goes to infinity with n. On the other hand, we show that as n tends to infinity, the fractional local dimension of P (1, d; n) tends to a value FLD(d) which we will be able to determine exactly. For all d ≥ 2, FLD(d) is strictly less than d + 1, and for large d, FLD(d) ∼ d/(log d − log log d − o(1)). As an immediate corollary, we show that if P is a poset, and d is the maximum degree of a vertex in the comparability graph of P , then the fractional local dimension of P , is at most 2 + FLD(d). Our arguments use both discrete and continuous methods. 1. Introduction A non-empty family R of linear extensions of a poset P is called a realizer of P when x ≤ y in P if and only if x ≤ y in L for each L ∈ R. In their seminal paper, Dushnik and Miller [8] defined the dimension of P , denoted dim(P ), as the least positive integer d for which P has a realizer R with |R| = d. For an integer n ≥ 2, the standard example Sn is a height 2 poset with minimal elements A = {a1 , a2 , . . . , an } and maximal elements B = {b1 , b2 , . . . , bn } such that ai < bj in Sn if and only if i 6= j. As noted in [8], dim(Sn ) = n for all n ≥ 2. Dimension is clearly a monotone parameter, i.e., if Q is a subposet of P , then dim(Q) ≤ dim(P ). Accordingly, a poset which contains a large standard example as a subposet has large dimension. On the other hand, posets which do not contain the standard example S2 are called interval orders, and it is well known that interval orders can have arbitrarily large dimension. In this paper, we introduce and study a new parameter for posets which we call fractional local dimension. To discuss our new parameter, we need some additional notation and terminology. A partial linear extension, abbreviated ple, of a poset P is a linear extension of a subposet of P . We do not require the subposet to be proper, so a linear extension is also a ple. We let ple(P ) denote the family of all ple’s of P . When P is a poset, a function w assigning to each M ∈ ple(P ) a real-valued weight w(M ), with 0 ≤ w(M ) ≤ 1, will be called a local weight function for P . For each u ∈ P , we then define the local measure of u for w, denoted µ(u, w) by setting: X µ(u, w) = {w(M ) : M ∈ ple(P ), u ∈ M }. In turn, we define the local measure of P for w, denoted µ(P, w) by setting: Date: October 10, 2020. 2010 Mathematics Subject Classification. 06A07, 05C35, 05D10. Key words and phrases. Dimension, local dimension, fractional dimension, fractional local dimension. 1 2 SMITH AND TROTTER µ(P, w) = max{µ(u, w) : u ∈ P }. A local weight function w for P will be called a fractional local realizer of P provided for each ordered pair (u, v) of elements (not necessarily distinct) of P , if u ≤ v in P , we have X {w(M ) : M ∈ ple(P ), u ≤ v in M } ≥ 1, and if u is incomparable to v in P , we have X {w(M ) : M ∈ ple(P ), u > v in M } ≥ 1. The fractional local dimension of P , denoted fldim(P ), is then defined by: fldim(P ) = min{µ(P, w) : w is a fractional local realizer of P }. Since fldim(P ) is the solution to a linear programming problem posed with coordinates which are integers, there is an optimal solution in which all weights are rational numbers. Readers may note that our parameter becomes fractional dimension if we only allow positive weights on linear extensions. This concept was introduced in 1992 by Brightwell and Scheinerman [5] and has been studied by several groups of researchers in the intervening years (see [11], [27] and [2], for example). On the other hand, if we only allow integer values of 0 or 1 for weights, then we obtain local dimension. This is a relatively new concept introduced by Ueckerdt [28] in 2016, but in the past two years, this concept is already the subject of several research papers (see [1], [25] and [4], for example). This research on local dimension has also sparked renewed interest in a related parameter, known as Boolean dimension introduced in 1989 by Nešetřil and Pudlák [21] (see also [13], [10], [20], [1] and [25], for example). In the discussion to follow, we will denote the fractional dimension, local dimension and Boolean dimension of a poset P by fdim(P ), ldim(P ) and Bdim(P ), respectively. We list below some basic properties of fractional local dimension. The first three are immediate consequences of the definition. The remaining items in the list are non-trivial. However, in each case, a proof can be obtained via a relatively straightforward modification of an argument, or arguments, given elsewhere in the literature. Accordingly, for each of these statements, we give citations to the relevant papers. Our focus in this paper is on results which do not have analogues in the literature and seem to require proof techniques which are completely novel in the combinatorics of posets. (1) For any poset P , fldim(P ) ≤ fdim(P ) and fldim(P ) ≤ ldim(P ). (2) If P is a poset and Q is a subposet of P , then fldim(Q) ≤ fldim(P ). (3) If P is a poset and P ∗ is the dual of P , then fldim(P ∗ ) = fldim(P ). (4) If P is a poset, then the following statements are equivalent: (a) dim(P ) = 2; (b) fdim(P ) = 2; (c) ldim(P ) = 2; (d) fldim(P ) = 2 and (e) Bdim(P ) = 2. The proof for fractional local dimension is an easy extension of ideas from the proof for fractional dimension given in [5] and the proof for local dimension is given in [1]. (5) If P is a poset with at least two points, then for every x ∈ P , fldim(P ) ≤ 1+ fldim(P − {x}). Again, the proof is a blending of ideas from the proof of the analogous result for fractional dimension (see either [5] or [2]) and the proof for local dimension given in [1]. The question as to whether the analogous result holds for Boolean dimension is still open. (6) For all n ≥ 3, ldim(Sn ) = 3, as noted by Ueckerdt [28]. This observation was part of his motivation in developing the concept of local dimension. On the other hand, as noted in [5], fdim(Sn ) = n for all n ≥ 2. It is a nice exercise to show that fldim(Sn ) < 3 for all n ≥ 3. (7) If P is an interval order, then it is shown in [5] that fdim(P ) < 4. In [27], it is shown that this inequality is asymptotically tight. Both statements also hold for fractional local dimension, and the argument for the tightness of the inequality is a technical, but relatively FRACTIONAL LOCAL DIMENSION 3 straightforward, extension of the proof in [27]. On the other hand, it is shown in [1] that both local dimension and Boolean dimension are unbounded on the class of interval orders. (8) The following conjecture, known as the Removable Pair Conjecture, has been open for nearly 70 years: If P is a poset with at least three points, then there is a distinct pair {x, y} of points of P such that dim(P ) ≤ 1 + dim(P − {x, y}). In [2], Biró, Hamburger and Pór gave a very clever proof of the analogous result for fractional dimension, and it is straightforward to adapt their argument to obtain the parallel result for fractional local dimension. However, the Removable Pair Conjecture is open for both local dimension and Boolean dimension. 2. Forcing Large Fractional Local Dimension We have already noted that the fractional local dimension of a standard example is less than 3 and the fractional local dimension of an interval order is less than 4. However, there does not appear to be an elementary argument to answer whether or not fractional local dimension is bounded on the class of all posets. Accordingly, our primary goal for the remainder of this paper is to analyze the fractional local dimension of posets in a well-studied family. As a by-product of this work, we will learn that there are indeed posets which have arbitrarily large fractional local dimension. For a pair (d, n) of integers with 2 ≤ d < n, let P (1, d; n) denote the height 2 poset consisting of all 1-element and d-element subsets of {1, . . . , n} ordered by inclusion. It is customary to ignore braces in discussing the minimal elements of P (1, d; n) so that the minimal elements are just the integers in {1, . . . , n}. Accordingly, u < S in P (1, d; n) when u ∈ S. We will abbreviate dim(P (1, d; n)) as dim(1, d; n), with analogous abbreviations for other parameters. Here are some of the highlights of results for posets in this family: √ (1) Dushnik [7] calculated dim(1, d; n) exactly when d ≥ 2 n. (2) Combining the upper bound of Spencer [23] with the lower bound of Kierstead [17], we have for fixed d:   dim(1, d; n) = Ω 2d log log n and dim(1, d; n) = O(d2d log log n). (3) For d = 2, the value of dim(1, 2; n) can be computed exactly for almost all values of n (see the comments in [3]). (4) It is shown in [1] that for each fixed d ≥ 2, ldim(1, d; n) → ∞ with n. (5) In [5], it was noted that for each pair (d, n) with n > d ≥ 2, fdim(1, d; n) = d + 1. For fixed d ≥ 2, fldim(1, d; n) is non-decreasing in n and bounded from above by d + 1, so the following limit exists: FLD(d) = lim fldim(1, d; n). n→∞ Although we know 2 < FLD(d) ≤ d+1 for all d ≥ 2, it is not immediately clear that FLD(d) < d+1. While this fact will emerge from our results, we know of no simple proof. Also, it is not clear whether FLD(d) grows with d. Again, we will show that it does, but there does not appear to be an elementary proof of this fact either. For the remainder of this section, fix an arbitrary integer d ≥ 2. To introduce our main results, we require some additional notation and terminology. Define a real-valued function f (x) on the closed interval [0, 1] by setting f (x) = (1 − x)d . Then there is a unique real number β with 0 < β < 1 so that f (β) = β. For example, when d = 2, √ β = (3 − 5)/2 ∼ 0.381966. 4 SMITH AND TROTTER We then consider real-valued functions r(x) and g(x), both having the closed interval [β, 1] as their domain, defined by: i d+1 1 h (1) and r(x) = (d + 1)x − dx d − (1 − x)d+1 d+1 (d + 1)r(x) − x g(x) = (2) . (d + 1)r(x) − x2 Let xbst and xbal be the (uniquely determined) numbers in [β, 1] such that (1) the value of g(x) is maximized when x = xbst , and (2) r(x) = x2 /2 when x = xbal . Simple calculations show for all d ≥ 2, β < xbst < xbal < 1. In turn, we define quantities cbal and cbst by: x evaluated when x = xbal and cbal = r(x) (d − 1)(x − x2 ) evaluated when x = xbst . cbst = 2 + (d + 1)r(x) − x2 With these definitions in hand, we can now state our main theorem. Theorem 2.1. For every d ≥ 2, FLD(d) = min{cbal , cbst }. Readers may note that our proof for Theorem 2.1 shows that FLD(d) is the minimum of the two quantities cbst and cbal but does not tell us which of the two is the correct answer. However, with some supplementary analysis, it is straightforward to determine which of the two values is actually the minimum: Theorem 2.2. When 2 ≤ d ≤ 3, FLD(d) = cbal , and when d ≥ 4, FLD(d) = cbst . For a representative sampling of choices for d, the following table gives the values of xbst , xbal , cbal and cbst , with all values truncated to six digits after the decimal point. d xbst xbal cbal cbst 2 0.790715 0.800466 2.498543 2.499190 3 0.667279 0.675604 2.960314 2.960495 4 0.560972 0.589913 3.390330 3.389027 5 0.478918 0.526764 3.796767 3.790859 10 0.289481 0.356142 5.615731 5.560625 100 0.048626 0.071263 28.065087 26.812765 1000 0.007055 0.010988 182.020489 172.548703 10000 0.000932 0.001500 1333.454911 1269.860071 100000 0.000116 0.000191 10458.564830 10018.329933 1000000 0.000014 0.000023 85721.802797 82526.406188 Many extremal problems in combinatorics are relatively straightforward if there is a single extremal example. In our problem, there are several natural candidates, and as suggested by the statement of our main theorem, the optimum solution is always to be found among two of these examples. The following elementary proposition, proved with standard techniques of calculus, shows the asymptotic behavior of FLD(d). Proposition 2.3. If d ≥ 4, then FLD(d) = d log d−log log d−o(1) . Arguments for the supporting lemmas for our main theorems require us to prove formally some highly intuitive statements. Some of these statements involve properties of posets and ple’s while others involve properties of real-valued functions, their derivatives, and intervals where their concavity is upwards and downwards. In cases where elementary combinatorial arguments, algebraic manipulations and techniques from undergraduate analysis suffice, we will sometimes give only the FRACTIONAL LOCAL DIMENSION 5 statement of lemmas and propositions. In modestly more complex cases, we will simply sketch the details of the proof. The remainder of the paper is organized as follows. In Section 3, we give the proof of the lower bound on FLD(d), i.e., we show that FLD(d) is at least the minimum of the quantities cbst and cbal . We consider this lower bound our main result. In Section 4, we give a construction showing that the lower bound is tight, a construction which capitalizes on the details of our proof for the lower bound. In particular, since the lower bound is the minimum of two quantities, we develop constructions which show that each of them is an upper bound on FLD(d). In Section 5, we explain how our main theorem implies that the fractional local dimension of a poset P is at most 2 + FLD(d) when P is a poset in which d is the maximum degree of its comparability graph. Finally, in Section 6, we close with some brief comments on open problems for fractional local dimension. 3. Proof of the Lower Bound Fix an integer d ≥ 2. In the first part of the proof, we also fix an integer n which is extremely large in comparison to d. With the values of d and n fixed, we abbreviate P (1, d; n) to P . Throughout the proof, we will denote members of ple(P ) with the symbols M and L, sometimes with subscripts or primes. When we use M , the ple may or may not be a linear extension, but when we use L, the ple is required to be a linear extension. To provide readers with an overview1 of where the argument is headed, we will identify three ple’s which will be called Mbal , Mbst and Msat . Then we find that an optimal fractional local realizer will assign positive weight only to ple’s which are isomorphic to one of these three. In fact, either (1) positive weight will be placed only on copies of Mbal ; or (2) positive weight will be placed only on copies of Mbst and Msat . 3.1. The Relaxed Linear Programming Problem LP. Recall that finding the value of fldim(1, d; n) amounts to solving a linear programming problem. Taking advantage of the natural symmetry of the poset P , we formulate a relaxed version of this problem using the following notation: For each M ∈ ple(P ), (1) (2) (3) (4) a(M ) counts the number of elements of Min(P ) which are in M . b(M ) counts the number of elements of Max(P ) which are in M . r(M ) counts the number of pairs (u, S) ∈ Min(P ) × Max(P ) with u > S in M . q(M ) counts the number of pairs (u, S) ∈ Min(P ) × Max(P ) with u < S in M . We note that the count q(M ) is for all pairs (u, S) with u < S in M , regardless of whether or not u ∈ S. Therefore, we have the following basic identity which holds for an arbitrary ple M : r(M ) + q(M ) = a(M ) · b(M ). Here is the reduced problem which we will refer to as LP: 1As is to be expected, this overview is a slight distortion of the truth, but the errors disappear as n → ∞. 6 SMITH AND TROTTER LP. Minimize the quantity c for which there is a local weight function w : ple(P ) → [0, 1] satisfying the following constraints: X A(w) = a(M ) · w(M ) ≤ cn. M ∈ple(P )   n B(w) = b(M ) · w(M ) ≤ c . d M ∈ple(P )   X n−1 R(w) = r(M ) · w(M ) ≥ n . d M ∈ple(P )   X n Q(w) = q(M ) · w(M ) ≥ n . d X M ∈ple(P ) We refer to these four inequalities as constraints A, B, R and Q, respectively2. The following observation relates the optimal solution for LP and the value of fldim(1, d; n). Observation 3.1. If w is a fractional local realizer of P and c = µ(P, w), then the pair (c, w) is a solution for LP. Therefore if c0 is the minimum cost for LP, then fldim(1, d; n) ≥ c0 . P the fact that |A(w)| = n, Proof. Observe that A(w) = a∈A µ(a, w). By the definition of c, and P P n
we see A(w) ≤ cn. Likewise, B(w) = b∈B µ(b, w) ≤ b∈B c = c d . In P (1, d; n), each a ∈ A is
P elements of B. By the definition of a fractional local realizer, {w(M ) : incomparable with n−1 d
M ∈ ple(P ), u > v in M } ≥ 1 for each incomparable pair (u, v). Therefore R(w) ≥ n n−1 d . Similarly, the definition of a fractional local realizer requires that every ordered pair (u, v) of P distinct elements with u ≤ v or (v, u) ∈ Inc P satisfies {w(M ) : M ∈ ple(P ),
u ≤ v in M } ≥ 1.
Since there are n nd of these ordered pairs in Min P × Max P alone, Q(w) ≥ n nd . Therefore (c, w) is a solutions for LP.  Accordingly, to obtain a lower bound for fldim(1, d; n), we focus on finding a good lower bound on c0 . Although, the value of c0 depends both on d and n, we will show that with d fixed and n → ∞, c0 tends to the minimum of cbst and cbal . 3.2. Properties of Optimum Solutions. Let W0 denote the family of all local weight functions w such that (c0 , w) is an optimum solution for LP. In this part of the argument, we develop four tie-breaking rules: TB(1), TB(2), TB(3), and TB(4), which will be used to select a local weight function w ∈ W0 satisfying several useful properties. Since the rules will not be presented at the same time, we use the “computer science” tradition for notation so that W0 consists of those optimum local weight functions which remain tied after all conditions developed to this point in the argument have been applied. The following elementary proposition is stated for emphasis. Proposition 3.2. If w ∈ W0 , at least one of the two constraints A and B is tight, and at least one of the two constraints R and Q is tight. Here is our first tie-breaking rule for the selection of w: TB (1). Maximize the quantity R(w) + Q(w) − A(w) − B(w). We note that TB(1) prefers optimum solutions where there is a “surplus” in the constraints R and Q and/or “slack” in the constraints A and B. The following elementary proposition follows from tie-breaker TB(1). 2Writing the constraints in this in-line form serves to facilitate several arguments for properties of solutions. However, later in the proof, we will consider them in the “scaled” form obtained by dividing both sides of each constraint by the terms involving n and/or d. FRACTIONAL LOCAL DIMENSION 7 Proposition 3.3. For M ∈ ple(P ), if w ∈ W0 and w(M ) > 0, then both a(M ) and b(M ) are positive, so that at least one of q(M ) and r(M ) are positive. Proof. For contradiction, suppose a(M ) = 0. (The case for b(M ) = 0 is similar.) Therefore b(M ) > 0 and q(M ) = r(M ) = 0. Create a new local weight function w′ which agrees with w on all ple’s except w′ (M ) = 0. Observe A(w′ ) = A(w), B(w′ ) = B(w) − b(M )w(M ), Q(w′ ) = Q(w), and R(w′ ) = R(w). Therefore w′ is a solution to LP with cost c′ ≤ c0 . Since w(M ) and b(M ) are positive, w′ is preferred to w by TB(1), a contradiction.  So for the remainder of the argument, we discuss only ple’s M ∈ ple(P ) for which both a(M ) and b(M ) are positive. This guarantees that at least one of q(M ) and r(M ) is positive. If M is a ple of this type, there is a uniquely determined integer s = s(M ), called the block-length, with 0 ≤ s ≤ n − d so that M has the following block-structure: (3) M = [A0 < B1 < A1 < B2 < A2 < · · · < Bs < As < Bs+1 ], where (1) {B1 , B2 , . . . , Bs+1 } is a family of pairwise disjoint subsets of Max(P ) with Bi 6= ∅ when 1 ≤ i ≤ s, and (2) {A0 , A1 , . . . , As } is a family of pairwise disjoint subsets of Min(P ) with Ai 6= ∅ when 1 ≤ i ≤ s. Note that we allow either or both of A0 and Bs+1 to be empty. Also note that q(M ) and r(M ) are determined entirely by the block-length s of M and the sizes of the blocks in the block-structure of M . Neither value is affected by the arrangement of elements in a block. For the balance of the argument, whenever we refer to a ple M of P , we will use the letter s = s(M ) to denote the block-length of M and we will assume that the block-structure of M is given by (3). Also, when we refer to a pair
(b, a), it will always be the case that a and b are positive integers with 1 ≤ a ≤ n and 1 ≤ b ≤ nd . The statement of the following proposition, which again follows immediately from TB(1), uses a convention for subscripts that will be used for the balance of this section. When we are discussing a finite family {M1 , M2 , . . . , Mt } of ple’s, for each i ∈ {1, . . . , t}, we will use the terms in (ai , bi , si , ri , qi , wi ) as abbreviations for the terms in (a(Mi ), b(Mi ), s(Mi ), r(Mi ), q(Mi ), w(Mi )). If we make changes to a local weight function w to form a local weight function w′ , the value of w′ (Mi ) is abbreviated as wi′ . Proposition 3.4. Let w ∈ W0 . Then there do not exist ple’s M1 and M2 such that: (1) w1 > 0, (2) r1 /a1 ≤ r2 /a2 , r1 /b1 ≤ r2 /b2 , q1 /a1 ≤ q2 /a2 , and q1 /b1 ≤ q2 /b2 , and (3) at least one of the four inequalities in the preceding requirement is strict. Proof. For contradiction, suppose there exist M1 and M2 with the listed properties. First assume both r2 and q2 are positive. By property (2), we can choose ε with max{r1 /r2 , q1 /q2 } ≤ ε ≤ min{a1 /a2 , b1 /b2 }. By (3), the inequality between ε and one of these four fractions is strict. Define a new weight function w′ which agrees with w on all ples except w1′ = 0 and w2′ = w2 +εw1 . Now the following inequalities verify that w′ is a solution to LP. a1 A(w′ ) = A(w) − w1 a1 + εw1 a2 ≤ A(w) − w1 a1 + w1 a2 = A(w). a2 b 1 B(w′ ) = B(w) − w1 b1 + εw1 b2 ≤ B(w) − w1 b1 + w1 b2 = B(w) b2 r1 R(w′ ) = R(w) − w1 r1 + εw1 r2 ≥ R(w) − w1 r1 + w1 r2 = R(w) r2 q1 Q(w′ ) = Q(w) − w1 q1 + εw1 q2 ≥ Q(w) − w1 q1 + w1 q2 = Q(w) q2 So w′ is a solution to LP and, by (3), one of these inequalities is strict. Therefore R(w′ ) + Q(w′ ) − A(w′ ) − B(w′ ) > R(w) + Q(w) − A(w) − B(w). 8 SMITH AND TROTTER Thus w′ is preferred to w by TB(1), a contradiction. Now assume r2 = 0. The argument for when q2 = 0 is almost identical. By property (2), r1 = 0 and therefore q1 and q2 are positive by Proposition 3.3. Since r1 = r2 = 0, property (3) states that q1 /a1 < q2 /a2 or q1 /b1 < q2 /b2 . Select ε with q1 /q2 ≤ ε ≤ min{a1 /a2 , b1 /b2 }, noting that the inequality between ε and one of these three fractions is strict. When we define w′ as above, we see R(w′ ) = R(w), A(w′ ) ≤ A(w), B(w′ ) ≤ B(w), and Q(w′ ) ≥ Q(w) where one of the three inequalities is strict. As a result, w′ is preferred to w by TB(1), a contradiction.  For a pair (b, a), we let ple(b, a) denote the set of all ple’s M with a(M ) = a and b(M ) = b. We say that a ple M ∈ ple(b, a) is maximal-preserving if there is no M ′ ∈ ple(b, a) with q(M ′ ) > q(M ). Determining the structure of maximal-preserving ple’s is trivial, since a ple M is maximal-preserving if and only if it has block-length 0 and block-structure M = [B0 < A1 ]. Also, if M ∈ ple(b, a) and M is maximal-preserving, then r(M ) = 0 and q(M ) = ab. Lemma 3.5. Let M1 ∈ ple(b, a). If M1 is not maximal-preserving, there exists a ple M2 ∈ ple(b, a) with q2 = q1 + 1. Proof. If s1 = 0, then M1 is maximal-preserving, so we know that s1 ≥ 1. Therefore, both B1 and A1 are non-empty and r1 > 0. Let S be any set in A1 , and let u be any element of B1 . Form a ple M2 by setting: M2 = [A0 < B1 − {S} < {u} < {S} < A1 − {u} < B2 < A2 < · · · < As1 < Bs1 +1 ]. Clearly, M2 satisfies the requirements of the lemma.  Lemma 3.6. If w ∈ W0 , then constraint R is tight. Proof. Let w ∈ W0 . We assume that constraint R is not tight for w and argue to a contradiction. By Proposition 3.2, constraint
Q must be tight. Thus there is a positive number δ so that R(w) = n−1
n d + δ while Q(w) = n nd . Let M1 be any ple with r1 and w1 both positive. Then M1 is not maximal-preserving. Let M2 be the ple provided by Lemma 3.5. Let ε = min{w1 , δ/2}. Form a local weight function w′ by making the following two changes to w: set w1′ = w1 − ε and w2′ = w2 + ε. Then A(w′ ) = A(w) and B(w′ ) = B(w). Furthermore, R(w′ ) = R(w) − ε and Q(w′ ) = Q(w) + ε. Since ε ≤ δ/2, (c0 , w′ ) is a feasible solution for LP with neither constraint R nor constraint Q being tight. The contradiction completes the proof of the lemma.  Analogous with our treatment for maximal-preserving ple’s, we will say that a ple M ∈ ple(b, a) is maximal-reversing if there is no ple M ′ ∈ ple(b, a) with r(M ′ ) > r(M ). We are now ready for the analogue of Lemma 3.5. Lemma 3.7. Let M ∈ ple(b, a) and let s be the block-length of M . Then M is maximal-reversing if and only if M satisfies the following five properties: Property 1. Property 2. If 0 ≤ i ≤ s, u ∈ Ai and S ∈ Bi+1 , then u ∈ S. If 1 ≤ i ≤ s and Bi+1 6= ∅, then |Ai | = 1. Property 3. If 1 ≤ i ≤ s and Bi+1 6= ∅, then Bi consists of all sets S ∈ Max(P ) such that (1) Ai−1 ⊂ S and (2) S contains no elements of Ai+1 ∪ Ai+2 ∪ · · · ∪ As .
Property 4. |Bs+1 | = max{0, b − n−1 d }. Property 5. |A0 | = max{0, a − (n − d)}. Furthermore, if M1 ∈ ple(b, a) is not maximal-reversing, then there is M2 ∈ ple(b, a) with r2 = r1 + 1. FRACTIONAL LOCAL DIMENSION 9 Proof. Before we present the proof, we pause to make two observations. First, it is easy to see that a ple satisfying Properties 1, 2 and 3, and one of Properties 4 and 5 satisfies all five properties. We have elected to state the hypothesis in this form so that the argument will be symmetric. Second, given a pair (b, a), if M1 and M2 belong to ple(b, a) and both M1 and M2 satisfy all five properties, then it is easy to see that M1 and M2 have the same block-length and the same block-structure. Therefore, r1 = r2 and q1 = q2 . Accordingly, to complete the proof of the lemma, we need only show that if M1 ∈ ple(b, a) violates one of the five properties in the hypothesis, then there is a ple M2 ∈ ple(b, a) with r2 = r1 + 1. Now suppose that M1 does not satisfy Property 1. Then there is some i with 0 ≤ i ≤ s1 for which there is an element u ∈ Ai and a set S ∈ Bi+1 such that u 6∈ S. In this case, we form a ple M2 ∈ ple(b, a) by setting: M2 = [A0 < B1 < A1 < B2 < · · · < Ai − {u} < {S} < {u} < Bi+1 − {S} < < Ai+1 < Bi+2 < · · · < As1 < Bs1 +1 ]. Then r2 = r1 + 1. Accordingly, we may assume that M1 satisfies Property 1. Now suppose that M1 violates Property 2. Then there is some i with 1 ≤ i ≤ s1 such that Bi+1 6= ∅ and |Ai | ≥ 2. Let u and u′ be distinct elements of Ai . Then u, u′ ∈ S for every set S ∈ Bi+1 by Property 1. Fix a set S ∈ Bi+1 . Since i ≥ 1, we know Bi 6= ∅. Let S ′ be any set in Bi , and let v be any element of S ′ − S. Then let T be the set obtained from S by removing u′ and replacing it with v. Clearly, the set T does not belong to M1 . Form M2 ∈ ple(b, a) by setting: M2 = [A0 < B1 < A1 < B2 < · · · < Ai − {u′ } < {T } < {u′ } < Bi+1 − {S} < < Ai+1 < Bi+2 < · · · < As1 < Bs1 +1 ]. Again, we note that r2 = r1 + 1. Accordingly, we may assume that M1 satisfies Properties 1 and 2. Now suppose that M1 violates Property 3. Let i be the least integer with 1 ≤ i ≤ s1 for which Bi+1 6= ∅ and there is a set T such that (1) Ai−1 ⊂ T ; (2) T contains no element of Ai ∪ Ai+1 ∪ · · · ∪ As1 ; and (3) T 6∈ Bi . Clearly, the set T is not in M1 . Let S be any set in Ai+1 . Form M2 ∈ ple(b, a) by setting: M2 = [A0 < B1 < A1 < B2 < · · · < Bi ∪ {T } < Ai < Bi+1 − {S} < < Ai+1 < Bi+2 < · · · < As1 < Bs1 +1 ]. It follows that M1 satisfies Properties 1, 2 and 3. In view of our observations at the start of the proof, we are left with
the case that M1 violates both Properties 4 and 5. So the inequalities |Bs1 +1 | ≥ max{0, b − n−1 d } and |A0 | ≥ max{0, a − (n − d)} must both be strict. Now let u ∈ A0 and S ∈ B1 . It follows that there is a set T ∈ Max(P ) such that T contains no element of {u} ∪ A1 ∪ · · · ∪ As1 . Then T is not in M1 . Form a ple M2 by setting: M2 = [A0 − {u} < {T } < {u} < B1 − {S} < A1 < B2 < A2 < · · · < Bs1 < As1 < Bs1 +1 ]. Since r2 = r1 + 1, the contradiction completes the proof of the lemma.  For the balance of the proof, given a pair (b, a), we let M (b, a) denote a maximal-reversing ple from ple(b, a), and we abbreviate r(M (b, a)) and q(M (b, a)) as r(b, a) and q(b, a), respectively. Now for the second tie-breaking rule: TB (2). Minimize the number t2 of ple’s in ple(P ) which are assigned positive weight by w and are neither maximal-preserving nor maximal-reversing. Lemma 3.8. The value of t2 is zero, i.e., there is a local weight function w ∈ W0 such that if w(M ) > 0, then either M is maximal-preserving or maximal-reversing. 10 SMITH AND TROTTER Proof. We argue by contradiction. Let w ∈ W0 . Then there is a ple M1 with w1 > 0 such that M1 is neither maximal-preserving nor maximal-reversing. Since M1 is not maximal-preserving, r1 > 0. If q1 = 0, the block-length of M1 is 1 and the block-structure of M1 is [B1 < A1 ] which implies M1 is maximal-reversing. It follows that q1 > 0. Applying Lemma 3.5 repeatedly to M1 , let i2 be the largest integer for which there is a ple M2 such that a2 = a1 , b2 = b1 , q2 = q1 + i2 (and r2 = r1 − i2 ). Note that i2 is positive and M2 is maximal-preserving. Applying Lemma 3.7 repeatedly to M1 , let i3 be the largest integer for which there is a ple M3 such that a3 = a1 , b3 = b1 , r3 = r1 + i3 (and q3 = q1 − i3 ). Again, we note that i3 > 0 and M3 is maximal-reversing. We determine a local weight function w′ by making the following three changes to w: Set w1′ = 0, w2′ = w2 + w1 i3 /(i2 + i3 ) and w3′ = w3 + w1 i2 /(i2 + i3 ). With these values A(w′ ) = A(w), B(w′ ) = B(w), Q(w′ ) = Q(w) and R(w′ ) = R(w), so that w′ ∈ W0 . However, the value of t2 has gone down by one. This observation completes the proof.  Here is an another easy consequence of TB(1) and TB(2). Lemma 3.9. Let w ∈ W0 and let M be a ple with w(M ) > 0. Then the following statements hold:
(1) If M is maximal-preserving, then a(M ) = n and b(M ) = nd so that M = L = [Min(P ) < Max(P )] is a linear extension of P .
(2) If M is maximal-reversing, s = s(M ) and k = |A1 | + |A2 | + · · · + |As |, then |B1 | = n−k d .
Proof. To prove the first statement, we assume that M1 satisfies w
1 > 0, r1 = 0 and a1 +b1 < n+ nd . We note that q1 = a1 b1 and r1 = 0. Therefore (1) q1 /a1 = b1 ≤ nd and (2) q1 /b1 = a1 ≤ n. Then let M2 be a maximal-preserving linear extension of P with block-structure M2 = [Min(P ) < Max(P )]. It follows that the pair (M1 , M2 ) violates Proposition 3.4. For the second statement, let M1 be a maximal-reversing ple with w1 > 0, and let k = |A1 | + |A2 | + · · · + |As1 |. If a1 ≥ n − d, then k = n − d and |B1 | = 1, as required. Now suppose that n−k
a1 < n − d. Then A0 = ∅. If s1 ≥ 2, then |B1 | = d follows from Property 3. Now suppose s1 = 1. Then M1 has block-structure [B1 < A1 ]. Therefore r1 = a1 b1 and q1 = 0. n−a1
n−k
Suppose that |B1 | < d = . Let M2 be a ple obtained from M1 simply by adding to B1 d a set S which does contain any of the minimal elements in A1 and does not belong to B1 . Then a2 = a1 , b2 = b1 + 1, r2 = r1 + b1 and q2 = q1 = 0. It follows that the pair (M1 , M2 ) violates Proposition 3.4.  We will say that a maximal-reversing ple M of block-length s is block-regular if the following two conditions are met: (1) Either A0 = ∅ or a = n.

s| (2) Either b = nd or b = n−|A . d Here is our third tie-breaking rule: TB (3). Minimize the number t3 of ple’s in ple(P ) which are assigned positive weight by w and are maximal-reversing but not block-regular. Lemma 3.10. The value of t3 is 0, i.e., there is an optimum local weight function w ∈ W0 so that if M ∈ ple(P ) and w(M ) > 0, then either M is a maximal-preserving linear extension of P , or M is block-regular. Proof. We argue by contradiction and assume that t3 > 0. Let w ∈ W0 and consider the non-empty family F of all ple’s assigned positive weight which are maximal-reversing but not block-regular. We show that we can devise a sequence of weight shifts to decrease the value of t3 . Describing these shifts involves three cases, and we give full details for one of them. First, suppose that M1 ∈ F, FRACTIONAL LOCAL DIMENSION 11
n−|As1 |
. Note that 1 ≤ a1 ≤ n − d and 1 ≤ b1 < n−1 d . In this case, we must have 0 < |Bs1 | < d−1 this requires s1 ≥ 2. Let M2 be a block-regular ple with a2 = a1 and s2 = s1 − 1. Also, let M3 be a block-regular ple with a3 = a1 and s3 = s1 . Note that b2 < b1 < b3 . Then there are positive numbers λ2 and λ3 with λ2 + λ3 = 1 so that b1 = λ2 b2 + λ3 b3 . Simple calculations then show that r1 = λ2 r2 + λ3 r3 and q1 = λ2 q2 + λ3 q3 . Form a local weight function w′ from w by making the following three changes: w1′ = 0, w2′ = w2 + λ2 w1 and w3′ = w3 + λ3 w1 . Then A(w′ ) = A(w), B(w′ ) = B(w), R(w′ ) = R(w) and Q(w′ ) = Q(w), so w′ ∈ W0 . However, the value of t3
has gone down by 1. The contradiction shows that if M1 ∈ F, then either n − d < a1 < n or
n−1 n < b1 < d . d

n−1
n
If n−1 < b1 < n−1 d d , then we shift weight from M1 to M2 = M ( d , a1 ) and M3 = M ( d , a1 ). If n − d < a1 < n, we shift weight from M1 to M2 = M (b1 , n − d) and M3 = M (b1 , n). After at most two applications of such shifts, the value of t3 has been decreased.  Considering the structure of LP, the following terminology is natural. Let M ∈ ple(b, a) We will say that:
(1) M is balanced if bn = a nd
. (2) M is A-heavy if bn < a nd
. (3) M is B-heavy if bn > a nd .
We note that linear extensions of P are balanced, and so is the block-regular ple M ( n−1 d , n − d). Due to the limitations of integer arithmetic, we may expect that many maximal-reversing ple’s are not balanced. On the other hand, the following lemma asserts that collectively, the family of ple’s assigned positive weight is balanced. Lemma 3.11. If w ∈ W0 , then both constraints A and B are tight. Proof. We argue by contradiction and show that constraint B must be tight. The argument for constraint A is symmetric. Since constraint B is not tight, constraint A must be tight by Proposi
tion 3.2. Accordingly, there is a positive real number δ so that A(w) = nc0 and B(w) = nd c0 − δ. In turn, this implies that w assigns positive weight to at least one ple which is block-regular and A-heavy.
Now suppose that w assigns positive weight to a block-regular ple M1 = M (b, a) with a1 < n−1 d . Let M2 be the maximal-reversing (but not block-regular) ple M (b1 + 1, a1 ). We note that r2 > r1 and q2 > q1 . Let ε1 = min{w1 , δ/2}. Then set ε2 = max{ε1 r1 /r2 , ε1 q1 /q2 }. Then make the following two changes to w: Set w1′ = w1 −ε1 and w2′ = w2 +ε2 . Note first that R(w′ ) = R(w)−ε1 r1 +ε2 r2 ≥ R(w). Similarly, Q(w′ ) ≥ Q(w). On the other hand, since ε2 < ε1 , we have: A(w′ ) = A(w) − ε1 a1 + ε2 a2 = A(w) − ε1 a1 + ε2 a1 < A(w) = nc0 . Also, we have: B(w′ ) = B(w) − ε1 b1 + ε2 b2 = B(w) − ε1 b1 + ε2 (b1 + 1) ≤ B(w) − ε1 b1 + ε1 (b1 + 1) = B(w) + ε1 ≤ B(w) + δ/2   n < c0 . d 12 SMITH AND TROTTER It follows that w′ ∈ W0 but neither
constraint A nor constraint B is tight, contradicting Propofor every block-regular ple M assigned positive weight by sition 3.2. This forces b(M ) ≥ n−1 d w.
It follows that w assigns positive weight to the block-regular ple M1 = M ( n−1 d , n). Let M2 =

n M ( n−1 d , n − d) and M3 =
M ( d , n). We note that r1 = r2 = r3 and q2 < q1 < q3 . Let ε1 = min{w1 , δ/(2 nd )}. Then set ε2 = ε1 (q3 − q1 )/(q3 − q2 ) and ε3 = ε1 (q1 − q2 )/(q3 − q2 ). Form w′ by making the following three changes. Set w1′ = w1 − ε1 , w2′ = w2 + ε2 and w3′ = w3 + ε3 . Since ε1 = ε2 + ε3 , we have R(w′ ) = R(w). An easy calculation shows that we also have Q(w′ ) = Q(w). For constraint A, we have: A(w′ ) = A(w) − ε1 n + ε2 (n − d) + ε3 n < A(w) − ε1 n + ε2 n + ε3 n = A(w) = nc0 . For constraint B, we have:       n−1 n−1 n + ε2 + ε3 d d d     n n−1 = B(w) + ε3 − d d   n < B(w) + ε1 d ≤ B(w) + δ/2   n < c0 . d B(w′ ) = B(w) − ε1 Again, these calculations show that w′ ∈ W0 but neither constraint A nor constraint B is tight. The contradiction completes the proof.  We close this subsection with a two useful observations about maximal preserving ple’s. The elementary arguments are left as exercises. Proposition 3.12. Fix an integer a with 1 ≤ a ≤ n. Then the ratio q(b, a)/r(b, a) is strictly increasing in a. Furthermore, over all pairs
(b, a), the maximum value of q(b, a)/r(b, a) is d and this occurs if and only if a = n and b = nd . 3.3. Transitioning to the Continuous Setting. In the next part of the discussion, we will keep the value of d fixed, but we will transition from the discrete to continuous settings by analyzing the behavior of solutions as n → ∞. Since our emphasis is on combinatorics and not analysis, we proceed in an informal manner, but it should be clear to the reader how our arguments can be recast in a completely formal setting.
Fix a block-regular ple M = M (b, a) with 1 ≤ a ≤ n − d and b ≤ n−1 d . With M fixed for the moment, we abbreviate the quantities in (a(M
), b(M ), r(M ), q(M ), s(M )) as (a, b, r, q, s), respectively. We also set m = |As |, so that b = n−m . d We then have the following exact formula for r = r(M ):         n−a n−a+1 n−m−1 n−m r= + + ··· + +m d d d d       n−m n−m n−a (4) . =m + − d d+1 d+1     n−m n−a i 1 h (n + md − d) − (n − a − d) = d d d+1

We define quantities x and y by setting x = b/ nd and y = a/n. Since b = n−m , it follows that d (in the limit) x = (1 − m/n)d . Since a ≥ m, it follows that we have the following restrictions on FRACTIONAL LOCAL DIMENSION 13 the pair (x, y): (5) 0 ≤ x, y ≤ 1 y ≥ 1 − x1/d . Previously, we analyzed the quantity r(b, a). Now we study the ratio r(x, y) which we define by setting: r(b, a)
. r(x, y) = n nd We then have the following formula for r(x, y): i d+1 1 h (6) r(x, y) = (d + 1)x − dx d − (1 − y)d+1 . d+1 It is not necessary to develop a parallel formula for q(x, y) since q(x, y) = xy − r(x, y). In the continuous setting, the linear programming problem LP becomes: Minimize the quantity c subject to the following constraints: A(w) = X y(M ) · w(M ) ≤ c. X x(M ) · w(M ) ≤ c. X r(M ) · w(M ) ≥ 1. X q(M ) · w(M ) ≥ 1. M ∈ple(P ) B(w) = M ∈ple(P ) R(w) = M ∈ple(P ) Q(w) = M ∈ple(P ) We continue to refer to these inequalities as constraints A, B, R and Q, respectively. 3.4. Ratio Curves. Let ρ be a non-negative number. We call the set of all pairs (x, y) such that q(x, y)/r(x, y) = ρ the ρ-ratio curve. One special case is ρ = 0 and we note that the 0-ratio curve is just the set of pairs (x, y) with y = 1 − x1/d . Lemma 3.13. For each ρ ∈ R0 , the pairs (x, y) on the ρ-ratio curve determine a decreasing function of x. Proof. The equation q(x, y)/r(x, y) = ρ is equivalent to xy/r(x, y) = 1 + ρ. Using implicit differendy at the tiation and abbreviating r(x, y) as r, we obtain the following formula for the derivative dx point (x, y): dy y[r − x(1 − x1/d )] =− . dx x[r − y(1 − y)d ] In this quotient, when 0 < x, y < 1, the numerator is positive since r > x(1 − x1/d ). Note that, in the discrete setting, this is just saying that r > bm. Similarly, the denominator is positive since r > y(1 − y)d . In the discrete setting, if b1 is the size of the bottom block of maximal elements, dy < 0 and the pairs on the ρ-ratio this inequality is simply the statement that r > ab1 . Therefore dx curve form a decreasing function in x.  Lemma 3.14. Let ρ be a fixed non-negative number and consider the set of all pairs (x, y) on the ρ-ratio curve with x, y > 0. Among these pairs, the quantity r(x, y) is maximized and the quantity (x + y)/r(x, y) is minimized when x = y. 14 SMITH AND TROTTER Proof. We outline the proof for the first statement. The argument for the second is similar. On dy the ρ-ratio curve, y is a function of x, and we have a formula for dx from the proof of the preceding lemma. It follows that on this ratio curve, r(x, y) is a function of x. Again, using implicit differentiation, we have: dy dr = 1 − x1/d + (1 − y)d . dx dx dr From this expression, it is straightforward to verify that dx is positive when x < y, 0 when x = y and negative when x > y. Therefore, on the ρ-ratio curve, r(x, y) is maximum when x = y.  Lemma 3.15. If M = M (x, y), w ∈ W0 and w(M ) > 0, then M is balanced. Proof. We argue by contradiction and suppose that w assigns positive weight to a ple which is not balanced. Since constraints A and B are tight, it follows that there are ple’s M1 = N (x1 , y1 ) and M2 = N (x1 , y2 ), both with positive weight, so that M1 is B-heavy and M2 is A-heavy. For i ∈ {1, 2}, let ρi = qi /ri . Let M3 = M (x3 , y3 ) be the point on the ρ1 -ratio curve where x3 = y3 . From the preceding lemma, we know that r3 > r1 . Also, we know that 2x3 /r3 < (x1 + y1 )/r1 . Also, let M4 = N (x4 , y4 ) be the point on the ρ2 -ratio curve where x4 = y4 . From the preceding lemma, we know that r4 > r2 . Also, we know that 2x4 /r4 < (x2 + y2 )/r2 . Let w′ be the local weight function obtained from w by making the following changes: w1′ = w2′ = 0, w3′ = w3 + ε1 r3 /r1 and w4′ = w4 + ε2 r4 /r2 . Then R(w′ ) = R(w), Q(w′ ) = Q(w), A(w′ ) < A(w) and B(w′ ) < B(w). This is a contradiction since neither constraint A nor constraint B is tight.  For the balance of the argument, since we will be concerned exclusively with ple’s on the balancing line (where x = y), we abbreviate r(x, x) to r(x). Similarly, q(x, x) will be written as q(x), and the ple M (x, x) will now be just M (x). Also, we will only be concerned with constraints R, Q and B. Now for our fourth tie-breaker. TB (4). Minimize the number t4 of ple’s in ple(P ) which are assigned positive weight by w. In view of the statement of our main theorem, and our outline for the proof, it is clear that ultimately we will show that t4 ≤ 2, The following proposition consists of some relatively straightforward calculus exercises. Proposition 3.16. (1) The function r(x)/x, defined on the interval (0, 1], is concave downwards on the interval (0, 1], has a vertical asymptote at x = 0 and achieves a local maximum value at x = δ with β < δ < xbst . (2) The function q(x)/x is monotonically increasing on [β, 1]. Since q(x)/x and r(x)/x are increasing on the closed interval [β, δ], the following result is a straightforward application of Proposition 3.4 in the continuous setting. Lemma 3.17. There is no local weight function w ∈ W0 which assigns positive weight to a ple M = M (x) with β ≤ x < δ. Next, we identify three balanced ple’s, to be denoted Mbst , Mbal and Msat . The subscripts in this notation are abbreviations, respectively, for best, balanced and saturated. Recall, from Section 2, that xbst is the unique point in the interval [β, 1] where the function g(x) = [(d + 1)r(x) − x]/[(d + 1)r(x) − x2 ] achieves its maximum value. Also, xbal is the unique point from [β, 1] where r(x) = q(x). Straightforward calculations show β < δ < xbst < xbal < 1. Set Mbst = M (xbst ) and Mbal = M (xbal ).
With d fixed and n → ∞, the block-regular ple M ( nd , n) converges to the linear extension which we now denote as M (1). We note that r(1) = 1/(d + 1) and q(1) = d/(d + 1). To be consistent with our notation for Mbst and Mbal , we set xsat = 1 and Msat = M (1). FRACTIONAL LOCAL DIMENSION 15 Lemma 3.18. There is no w ∈ W0 which assigns positive weight to a ple M = M (x) with xbal < x < 1. Proof. We argue by contradiction and assume some w ∈ W0 assigns positive weight w0 to a ple M0 = M (x0 ) with xbal < x0 < 1. Let M1 = Mbal and M2 = Msat . We will form a local weight function w′ from w by removing all weight from M0 and increasing the weight on M1 and M2 so that R(w′ ) = R(w), Q(w′ ) = Q(x) and B(w′ ) < B(w). To accomplish this task, we must find positive values v1 = w1′ − w1 and v2 = w2′ − w2 that satisfy the following three requirements: w0 r0 ≤ v1 r1 + v2 /(d + 1) w0 q0 ≤ v1 q1 + v2 d/(d + 1) w0 x0 > v1 x1 + v2 . We consider the first two requirements as equations. Noting that q1 = r1 , this results in the following solution: v1 = w0 (dr0 − q0 )/[(d − 1)r1 ] v2 = (d + 1)w0 (q0 − r0 )/(d − 1). From Proposition 3.12, we know q0 /r0 < d which shows that v1 > 0. Also, since xbal < x0 < 1, we know q0 > r0 which implies v2 > 0. Using the fact that v2 can be also be written as (d + 1)(w0 r0 − v1 r1 ), the third inequality is equivalent to: (d + 1)r1 − x1 (d + 1)r0 − x0 > . 2 (d + 1)r1 − x1 (d + 1)r0 − x20 However, we know the function g(x) is decreasing when x ≥ xbst , so this last inequality holds. The contradiction completes the proof of the lemma.  Lemma 3.19. Let w ∈ W0 . Then there is no maximal-preserving linear extension which is assigned positive weight by w. Proof. Suppose to the contrary that some w ∈ W0 assigns positive weight w0 to the maximalpreserving linear extension L0 = [Min(P ) < Max(P )]. Since constraint R is tight, there must be some ple M1 with w1 > 0 and r1 > q1 . Let M2 = Msat . We will then obtain w′ from w by setting w0′ = w0 − v0 , w1′ = w1 − v1 and w2′ = w2 + v2 . We require 0 < v0 ≤ w0 , 0 < v1 ≤ w1 and v0 /v1 = dr1 − q1 . Clearly, appropriate values of v0 and v1 can be chosen. We then set v2 = (d + 1)v1 r1 . With these choices, it follows that R(w′ ) = R(w) and Q(w′ ) = Q(w). However, B(w′ ) = B(w) − v0 − v1 x1 + v2 = B(w′ ) − v1 (dr1 − q1 ) − v1 x1 + (d + 1)v1 r1 = B(w′ ) − v1 [dr1 − (x21 − r1 ) − x1 + (d + 1)r1 ] = B(w) − v1 (x1 − x21 ) < B(w). Recall that all ple’s with positive weight are balanced, by Lemma 3.15, so we also have A(w′ ) < A(w), a contradiction which completes the proof of the lemma.  Lemma 3.20. Let w ∈ W0 . If w assigns positive weight to a ple M = M (x) with δ ≤ x < xbal , then x = xbst . Proof. Suppose to the contrary that w assigns positive weight to a ple M0 = M (x) with δ ≤ x < xbal and x 6= xbst . Since r0 > q0 , w must also assign positive weight to M2 = Msat . We form w′ by making three changes. First, we set w0′ = w0 − v0 , and we will require that 0 < v0 ≤ w0 . Also, 16 SMITH AND TROTTER there may be an additional restriction on the size of v0 , a detail that will soon be clear. Second, we will take M1 = Mbst and set w1′ = w1 + v1 with v1 > 0. Third, we will take M2 = Msat and set w2′ = w2 + v2 , however we place no restrictions on the sign of v2 , except that we need w2′ ≥ 0, so if v2 is negative, we need |v2 | ≤ w2 . This will be handled if necessary by restricting the size of v0 . Our choices will be made so that R(w′ ) = R(w), Q(w′ ) = Q(w) and B(w′ ) < B(w). This requires: (7) v0 r0 = v1 r1 + v2 /(d + 1), (8) v0 q0 = v1 q1 + v2 d/(d + 1), (9) v0 x 0 > v1 x 1 + v2 . Solving the first two equations, we obtain v1 = (dr0 − q0 )/(dr1 − q1 ) so that v1 > 0. Note that this equation can be rewritten as v1 = [(d + 1)r0 − x20 ]/[(d + 1)r1 − x21 ]. We also obtain: v2 = (d + 1)(v0 r0 − v1 r1 ). Accordingly, when v2 < 0, we can maintain |v2 | ≤ w2 by scaling the system 7 to keep v0 sufficiently small. After substituting the specified values, the third constraint v0 x0 > v1 x1 + v2 becomes g(x1 ) > g(x0 ), which is true since x0 6= x1 . Since w′ can be obtained with the desired properties, we have contradicted w ∈ W0 , completing the proof of the lemma.  With these observations in hand, our linear programming problem reduces to: Minimize the quantity c subject to the following constraints: B(w) = w(Mbst )xbst + w(Mbal )xbal + w(Msat ) ≤ c R(w) = w(Mbst )r(xbst ) + w(Mbal )r(xbal ) + w(Msat )/(d + 1) ≥ 1 Q(w) = w(Mbst )q(xbst ) + w(Mbal )q(xbal ) + w(Msat )d/(d + 1) ≥ 1 By increasing w(Mbal ), we can decrease w(Mbst ) and w(Msat ) or vice versa to see that there are optimal solutions with at most two non-zero values among the variables w(Mbst ), w(Mbal ) and w(Msat ). Furthermore, the form of the constraints tell us that the only possible solutions are (1) to put weight only on Mbal , or (2) to put positive weight only on Mbst and Msat . The values of c0 given in the statement of our main theorem reflect the calculations of c0 which these two options would yield. With these remarks, the proof of our lower bound is complete. 4. Proof of the Upper Bound In this section, we show that the lower bound obtained in the preceding section is also an upper bound. Our lower bound is the minimum of two quantities, so in this section we show that each of the two options provides an upper bound. As before, our
treatment is informal. Recall that M (x) is a block-regular ple with x = b/ nd = a/n. In particular, Mbal = M (xbal ), Mbst = M (xbst ), and Msat = M (1) where xbal and xbst were defined in Section 2. We start with the first option in which all weight is put on the ple M0 = Mbal so that x0 = xbal . We have r0 = r(x0 ) = x20 /2. Therefore w0 = w(M0 ) = 1/r0 = 2/x20 . Also, we have c0 = w0 x0 = x0 /r0 = 2/x0 . To construct a fractional local realizer for P , we then distribute the weight w0 evenly among all ple’s which are isomorphic to M0 . With this rule, we will have: (1) For a given pair (S, S ′ ) of distinct sets in Max(P ), the total weight assigned to ple’s M with S > S ′ in M is w0 x20 /2 = 1. (2) For a given pair (u, u′ ) of distinct sets in Min(P ), the total weight assigned to ple’s M with u > u′ in M is w0 x20 /2 = 1. FRACTIONAL LOCAL DIMENSION 17 (3) For a pair (u, S) ∈ Min(P ) × Max(P ) with u 6∈ S, the total weight assigned to ple’s M with u > S in M is w0 r0 = 1. (4) For a pair (u, S) ∈ Min(P ) × Max(P ) with u 6∈ S, the total weight assigned to ple’s M with u < S in M is w0 (x20 − r0 ) = w0 r0 = 1. We are left to consider pairs (u, S) ∈ Min(P ) × Max(P ) with u ∈ S. For the ple M0 , the fraction 1/d of Min(P ) that is in the top box is 1 − x0 , so the total weight assigned to ple’s M with u < S in 1/d M is w0 x0 [x0 − (1 − x0 )]. We need this quantity to be at least 1, but this is equivalent to requiring 1/d that r0 = x20 /2 ≥ x0 (1 − x0 ), and a straightforward computation shows that this inequality is valid. For the second option, we take M1 = Mbst , x1 = xbst , M2 = Msat and x2 = 1. Now we have: w1 r1 + w2 /(d + 1) = 1, (10) w1 q1 + w2 d/(d + 1) = 1, w1 x1 + w2 = c0 . We take all ple’s of P which are isomorphic to M1 and distribute the weight w1 evenly among them. Similarly, we take all ple’s that are isomorphic to M2 and distribute the weight w2 evenly among them. We then have: (1) For a given pair (S, S ′ ) of distinct sets in Max(P ), the total weight assigned to ple’s M with S > S ′ in M is w1 x21 /2 + w2 /2. However, if we add the first two of the equations in (10) and divide by 2, we obtain: w1 (r1 + q1 )/2 + w2 /2 = w1 x21 /2 + w2 /2 = 1. is: (2) For a given pair (u, u) of distinct sets in Min(P ), the total weight assigned to ple’s M with u > u′ in M is w1 x21 /2 + w2 /2. Again, this sum is 1. (3) For a pair (u, S) ∈ Min(P ) × Max(P ) with u 6∈ S, the total weight assigned to ple’s M with u > S in M is w1 r1 + w2 /(d + 1) = 1. (4) For a pair (u, S) ∈ Min(P ) × Max(P ) with u 6∈ S, the total weight assigned to ple’s M with u < S in M is w1 (x21 − r1 ) + w2 d/(d + 1). This reduces to the second equation of (10). Again, we are left to consider pairs (u, S) ∈ Min(P )×Max(P ) with u ∈ S. The desired inequality 1/d w1 x1 [x1 − (1 − x1 )] + w2 ≥ 1. To obtain this inequality, we first observe that the inequality d + 1 − dw1 r1 ≥ 1 is equivalent to 1 ≥ w1 r1 , but this inequality holds strictly in view of the first equation of (10). The first two equations in (10) imply w2 = 2 − w1 x21 . It follows that: 1/d 1/d w1 x1 [x1 − (1 − x1 )] + w2 = w1 x1 [x1 − (1 − x1 )] + 2 − w1 x21 1/d = −w1 x1 (1 − x1 ) + 2 ≥ −w1 r1 + 2 ≥ 1. With these observations, the proof of the upper bound is complete. 5. Posets of Bounded Degree Let d be a non-negative integer. One of the long-standing problems in dimension theory is to determine as accurately as possible the quantity MD(d) which we define to be the maximum dimension among all posets in which each element is comparable with at most d other elements. Trivially, MD(0) = MD(1) = 2. With a little effort, one can show that MD(2) = 3. On the other hand, it is not at all clear that MD(d) is well-defined when d ≥ 3. 18 SMITH AND TROTTER However, in 1983, Rödl and Trotter[24] proved that MD(d) is well-defined and satisfies MD(d) ≤ 2d2 + 2. Subsequently, Füredi and Kahn [12] proved in 1986 that MD(d) = O(d log2 d), and it is historically significant that their proof was an early application of the Lovász local lemma. From below, the standard examples show MD(d) ≥ d+1. However, in 1991, Erdős, Kierstead and Trotter [9] improved this to MD(d) = Ω(d log d). From 1991 to 2018, there were no improvements in either bound, but after a gap of nearly 30 years, Scott and Wood [22] have recently made the following substantive improvement in the upper bound. Theorem 5.1. If d ≥ 104 and d → ∞, then √ MD(d) ≤ (2e3 + o(1))(d log d) e2 log log d
log log d. As a consequence, we know MD(d) ≤ d log1+o(1) d. Of course, the analogous problem can be considered for any of the variants of dimension, and in some cases, the following phenomenon is then observed: There is a bound on the parameter for the class of height 2 posets in which each maximal element is comparable with at most d minimal elements independent of how many maximal elements are comparable with a single minimal element. As our results show, fractional local dimension is one such parameter. Fractional dimension and Boolean dimension are two others. Curiously, local dimension is not, since ldim(1, 2; n) goes to infinity with n. For general posets, one can ask for the maximum value MFLD(d) of the fractional local dimension of a poset in which each element is comparable with at most d others. Taking the split of such a poset, the fractional local dimension cannot increase more than 1, so we have FLD(d) + 2 as an upper bound. However, we have no feeling as to whether this is (essentially) best possible. 6. Closing Comments One natural open problem for this new parameter is to determine the maximum fractional local dimension among all posets on n points. The answer to the analogous question for fractional dimension is ⌊n/2⌋ when n ≥ 4, and the standard examples show this inequality is tight. On the other hand, it is shown in [18] that the maximum local dimension of a poset on n points is Θ(n/ log n), and we suspect that this is also the answer for fractional local dimension. Other open questions revolve around the interplay between fractional local dimension and width. For any poset with width w ≥ 3, the dimension is at most w [6]. This inequality is tight for fractional dimension precisely when P contains the standard example Sw [11]. However, we don’t know if this inequality is tight for local dimension as standard examples have local dimension 3 [1]. 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Mathematics & Computer Science Department, Davidson College, Davidson, North Carolina 28035 Email address: hcsmith@davidson.edu School of Mathematics, Georgia Institute of Technology, Atlanta, Georgia 30332 Email address: trotter@math.gatech.edu