3 Fractional Ramsey Numbers

Transcription

1 27 3 Fractional Ramsey Numbers Since the definition of Ramsey numbers makes use of the clique number of graphs, we may define fractional Ramsey numbers simply by substituting fractional clique number into this definition Recall that the Ramsey arrow notation means that For any red/blue-coloring of the edges of, there must a either a red -clique or a blue - clique More precisely, whenever, then we must have or This edge-decomposition of may be thought of as an edge 2-coloring The Ramsey number! " # is the least positive integer for which this statement is true Very few exact values for! " # are known For instance,! "$% &$' is only known to be somewhere between 43 and 55 And while the growth rate of! &( is known to be an exponential function of, the best known lower and upper bounds on this growth are (roughly) ) *, and, respectively Now, we may define the fractional Ramsey arrow notation 12 #3( to mean that, whenever , we must have 7 :; or <=3 Then the fractional Ramsey number >2 #3? is the least positive integer for which 12 #3( Note that we may meaningfully take and 3 to be any real numbers greater than 1 All graphs in this chapter are implicitly assumed to be finite When we speak BC as an edge 2-coloring, it is implicitly assumed that we are invoking the condition on that D 7 FE D HG However, a little thought reveals that the above Ramsey definitions with or without this condition are equivalent We only note this because, in forthcoming constructions, it is sometimes convenient to take JI s which are not edgedisjoint However, if we find such KI s with 5I MLN I, removing edges from some KI s

2 28 to make them edge-disjoint does not cause this condition to be violated So we adhere to the convention of D 8E D 6G only as is convenient 31 The Value of Unlike the ordinary Ramsey numbers, the exact value of! >2 3( is known for any 2 #37 * Theorem 31 Let 2 #3 with 2 #3 Express and 3 as and 3, where and L%! " Let # %$'&)(*-, /, 21 Then! >2 #3?? 3 # Proof We first establish two basic facts about 4 5 & and # as given above (i) Either # 6, 7 8 or # 9, 'A%', so that either #;: or #2: % (ii) #, 7 L <% and #, L =%, so that # /, : L and # /, : L Now,! >2 3( is clearly symmetric in and 3, and! >2 3( >,? is easily checked for 3@A *B, so we restrict our attention to the case of and 3 both greater than 2 Otherwise, take 2 #3 & 5 and # as given above, and set % C # We establish that! >2 #3? in two steps: first, show that / >2 #3? ; then, construct a decomposition ED BC with MLC and MLC3 (which shows that / / >2 #3? is false) To show / 12 #3(, let K 7M be any edge 2-coloring of (so that D 2E D 4 G, 7 GF and vice versa) If :6HI, then :!J4, and we re done Since 5 KJ similarly implies 3, we may suppose that F K = and F 7 = Then by Lemma 13 and (i)

3 29 above, either #;:, in which case or #;: 7%, which gives F F?? # #? 3 #? # That one of these holds is exactly the statement / >2 3( To achieve 4ED with and ML and H = By Lemma 14 and (ii) above, this immediately gives L93, we take D /? 3 # / # / L;H and 4 / /, : We note that /, : H D L6!, and also that > /, : since # is at least 1 So / :, and applying (ii) again gives 7 /? # / # / L; 3 These are exactly the properties we required of 7 and, so / / >2 3( is false, and we re done Corollary 32 If = * are integers, then! " % / Note that, while! " &? grows exponentially in,! 12 only grows quadratically in Two plots of! >2 vs are shown in Figure 5

4 Figure 5: Two graphs of the function 3! 12 Although this is a step function, its growth roughly conforms to a parabola Large jumps occur at integer values of, while size 1 jumps occur at even intervals between integers 32 Multicolor Fractional Ramsey Numbers We may extend the definition of Ramsey numbers by using more than two colors on the edges of K We " & mean that, whenever 67(#, we must have 5I I for some * 1 The Ramsey number! " & is the least positive integer for which this holds Similarly, we take / 1 to mean that, whenever J '6, we must have 5I I for some * 1, and the fractional Ramsey number! > is the least positive integer for which this is true We may derive a recursive upper bound on! as follows: Theorem 33 Let * Then! >,! > D /, Proof Let,! > D /,, and let K M be any -

5 31 coloring of D Let? D If! > D, then contains a complete subgraph large enough to guarantee that 5I H I for some * / ;1, and we re done So we may suppose that! > D / Since F ; we have M F as required,! 1 D /,! > D / 9 In fact, there are no known instances where this upper bound does not provide the correct value of! For example, in the case of *, the given bound reduces to! >2 #3? $ & ( *,,? /, 3,, 3 /?21 which is easily shown to be the value derived for! 12 #3( in Theorem 31 Of course,! is symmetric in its arguments, so in order for the expression in Theorem 33 to yield the best possible bound, we must at each recursive step choose the best I value to play the role of in this expression All of this implies the following conjecture Conjecture 34 Let * Then for some * 1,! > 6,! 1 I D I /, I In the case that the arguments are integers, the correct choice of I is simply the largest argument, so for integers 6 D * we have! " I I / I( / / D /

6 32 On the other hand, it is easily shown 13 that! " & I " I /, so our upper bound is at least on the right order of magnitude There are several other instances where our conjecture is known to be true, the most significant being the case where the arguments are all the same integer Theorem 35 For *,! & / D / / We postpone the rather long proof of this theorem to mention a few other instances where Conjecture 34 is true In the following three cases, the value of! is that given by Conjecture 34, and we give D 6 6, where each AI is the indicated circulant graph, and I one of its independent sets of required size We take D I to be *! / * 1, and note that I! /, : I (by Lemma 13)! - * * * * * 1 * $ * - : $ 12 #3 *; LC2 #3@% L <1 *?!? "% $% #<1 #" :%$, *,-!<1 * % ; #" :%$,? &$% "<1 M *!? 1 #"! 12 #3 * % & * -!<1 6*%$2:, 13 See [6] LC% L93 - * * $% $ *,$% $ *; *'* 1,

7 33 * *? &$1 M * $? -,$% * ;1 4 6*%$2:%$, * $? #1 * '!<1 * $2:%$! 12 #3 $ LC% L93@ - L $ * -,$%!? $ 1 * #1 7 - : -, *,- $% "? 1 M * *? -,$% * ' * * * <1, - : ", * *?!% $? 2 * #<1 M * &$%,$% *E * $% * "<1, - :%$ There is a final general case where Conjecture 34 is true Let & * be integers Then if each of is sufficiently small, we have! & That this is the value indicated by Conjecture 34 is easily checked For the lower bound, consider B Every edge from this graph comes either from (between two copies of ) or (within a copy of I I ) If we let contain all such edges, and 4 all such edges, then 7 B and 47 B By Lemma 15, and More generally, if we take successive lexicographic products of the s, we get a complete graph on vertices And if we let I consist of all edges which come from 5I, then B)B 8 9 and

8 34 5I I L9 I I This shows that / & is false Finally note that, while choosing the largest I gives the best bound in the case that all arguments are integers, this is not in general the case To calculate! note that! ) ) and! - " # Then )' - ", first!! - ",! - ",! )', /, - " - $% but )' - " /, ) % The second bound is clearly better, even though the largest I was not used as the recursion point We now return to Proof of Theorem 35 We restrict our attention to the case of, as the 6* case is trivially true And since we ve already proved Theorem 33, all that remains to be shown is that! " &( = / D / / / As in the 2-color proof, this is accomplished by constructing 7D 78,; (an edge -coloring of the complete graph) such that I :L for all And as before, this is accomplished via the use of circulant graphs We use the more general form, and will generally suppress the subscript, as its value will be clear in context In the following, we will hold fixed and let vary Let / D / / /, the order of the complete graph whose edges we are -coloring Notice that < % /, and since /, we set & for consistency Since the 4I s will be chosen to be circulant graphs, they will be vertex transitive, and thus I by Lemma 13 Since we want 5I 4LH, we let F be the desired value of F I in the -color case

9 1 35 Specifically, F $ & (K*F F $ & (K*F F D L=1 D / D / with F Let (Henceforth, D and F are used interchangeably), so that * 1 Then in order to have K '6 where each 5I is a circulant graph, we require that the union of the connection distance sets of all the 4I s be exactly * ' 1 Note that in the following construction, the I s will not be edge-disjoint (the connection distance sets will not be disjoint) Let us superscript each 4I with its corresponding, and then let distances defining I, ie, I I! *! 1 Similarly! / *,! / 1 We begin our construction by defining I * I D # I / I D 1 I be the set of connection For a set of integers and an integer!, we define I /, I / I / I I / * I and then I * $ I I 1 I I I I * I Notice that, since, for we have that % D / 4 * D Then : * D, and we have / D *

10 * 36 D Therefore * D 1 * D And since / D D I I 1 * D # / D 1, and * D D, it follows that D 1 D Finally, note that Example To help illustrate the proof, we consider the case Here we present the items we have defined thus far for $ * - ( $ * - &% & & $ * -/ 1)2#3 1" %42!#('3 1) )# ",('3 1),3 1)2#3 1" %'3 1)2'95:5;!#77('3 1" %42!#('3 1) ,3 ',( $ - "!# ()% ) 1)2'95:5;!#77(2%#2,4'',, ' 92%,:%!'%(4)%'442!;'<!" (,()3 1"765656<(%'765676< :4#!#567656<<!,#3 1) )# ()%#3 We return to the example of = several times to further clarify our constructions

11 I I 37 We now use induction to verify that this construction actually does cover all connection distances ', (ie, that * 1, and therefore 4 6 ) We have * 1 as a base case, so we now suppose that * 1 Since * 1, we need only show that * # / ;1 Take L Since I I / * I so $ > / $, we compute that / D / D / / / " I / * I " D I D I -C I / I)D / / I D I D %,-N I I I Next, if I, then / / I, and I > / I $ I B > / I / I $ I $ I / * I D # I / I D 1 $ * I D # I / I)D 1 We also know that / L, so / I Since cover * 1 (our induction hypothesis), it then follows that cover * / # / E1 as well as * 1 (recall that I I ) But since, we have that * 1 * / # / ;1 * # / ;1, all of which is covered by, as desired Thus all connection distances are covered, and it follows that K 6 9 I I/B

12 38 We must now show that F I M F, so that (i) defining a set I * # / E1, (ii) showing that I is an independent set in (iii) showing that I F (i) Constructing I We first define the sets I * F2I / ;1 I I, and <L; We accomplish this by I * F2I / * 1 and I I I I We then define our independent set I to be I I I I I 6" / I! I I B I I 6" / * I B I! I I I B where! I I$* F8I / %, I L9 1 That is to say,! I is the largest possible value so that no element of I exceeds Specifically,! I D I D D I D To see this, add I (which is much smaller than the span of a I, but larger than the span of I ) to the leading in the last I in I We get! I I, I N I D I D I, " /, I /! I %,-N I D I /, I N I /! I I / D / / D I / D I D CHI,

13 so the given value of! I implies I is a subset of * / ;1 In the case of /, this gives! I, and we just have D D D D Finally, we set "* F / ;1 This completes our construction 39 Example Here, we list independent sets and their components in the case of and 2' A better illustration of these patterns for & may be found in Figure 6 1":'3 4 1:'3 1":5'3 1":53 1":555"'52#2,#3 1": <5:'3 1": <93 ) (4 1": <5: %,(4(#!# )%#3 * * * ( ) ",( 1":'3 1":(3 1":5'3 1":(554592; 2!#'%' :'3 1":555"'52#2;',; 3 See Figure 6 1":5 2%'42!'(9'5:#3 1":(554592; 2!#'%' : (%%4'%9' 4, 84)!'2!'2!%'(:('(('9,9,49,975:, 57:)!#5#77% :5",5( '75, 85 '!)5"%,' "%,%'754:574

14 4 74('3 1":555"'52#2; ) )%#2%; 2%%# 4, 84%442!%'<!;4<!!)(4'()!) 949)!'9(5:)!'57:('5:,97,7(579 (5",957:77955 :55,7%;:57%, 74:54574)#3 See Figure 6 1":567656<77: <%( %'7( (3 1":567656< )#3 (ii) I is an independent set of I Given a subset of the vertices of a circulant graph, is an independent set of if, for any two 2 3, the distance between them along their shorter around the circle (ie their connection distance ) is not in the set of connection distances of the graph We check that this condition is satisfied in each of three cases, dependent on the value of

15 D 2 D 2 D 2 h 2 D 2 2h 2 D 2 3h 2 D 2 4h 2 B 2 5h 2 41 p = 2, α 2 = 3, n 2 =11 I 2 2 = B p = 3, α 3 =11, n 3 = 43 3 r i,p =, I 2 = D 2 ( B 2 h 2 ) B 2 B 2 n 2 C 2 2n B 2 h 2 p = 4, α 4 = 43, n 4 = r i,p = k = 4, I 2 = D 2 ( D 2 h 2 ) ( D 2 2h 2 ) ( D 2 3h 2 ) ( D 2 4h 2 ) ( B 2 5h 2 ) I for * and *? - Note that, Figure 6: The desired independent sets I of while the vertices are drawn here in rows for clarity, they should be imagined to be laid out around the perimeter of a circle

16 (iia) is an independent set of Recall that *F 42 1 and * F / ;1 (where F / 5L= : * ) The elements of are therefore consecutive along the circle, and none are more than F / steps apart Since the smallest connection distance is F, no pair of vertices in can be adjacent (iib) D is an independent set of Recall that D D D *F D F / F D 1, and that copy of D To show that D is an independent set of D contains exactly one D, we consider all possible connection distances within D, and observe that none of them is found in D Note that D is just the union of translated copies of D and D Any pair of vertices in the same copy of or are at most F D / steps apart Since the smallest connection distance (element of D ) is F D, these vertices cannot be adjacent, so we turn our attention to vertices in different copies of and The smallest possible distance between such a pair of vertices occurs between the last vertex in one block ( or ) and the first vertex of the next block We now prove that, in all possible cases, this smallest possible distance is larger than F / F D (the largest value in D ) is last vertex in D D and is the first vertex in D =I, D with L= / In this case we have F D /, D and I, D, so the distance between them is / D / F D /, JF / F D I, as desired is the last vertex in D " / D and is the first vertex in D " / * D As above, the distance between and is D / F D /, JF / F D I

17 43 is the last vertex in D " / * D and is the first vertex in D D Here, F D / * " / * D and D " /, D / The distance between them is F / F D I, as desired is the last vertex in D D and is the first vertex in D In this case F / F D I, as desired D F D / and J Their distance is / D F D /, D ; any two vertices in different blocks of D have connection distance greater than F / F D, and so greater than any Recapping any two vertices in the same or block of D have connection distance less that F D, and so less than any value in D value in D (Keep these facts in mind for the following section) Thus D is an independent set of D (iic) I is an independent set of I for L / Recall that, like I I D, I is a collection of translated I s and I s So as before, the distance between any two vertices in distinct blocks of I is at least F2I / F8I All such I s and I s (except for the final copy of I ) are contained in translated copies of I, which are spaced at intervals of I If we consider some I, and we add or subtract I to it, we are essentially moving it to the same position within the next or the previous copy of I (so long as this movement does not push, or scroll, across the / to gap) Now, let 2 #38 I and let denote the distance between them along the shorter arc of the circle Without loss of generality, assume that 3CLH Note that / 3 or / > / 3?, depending on whether or not the shorter arc covers the / to

18 gap We must show that : I, and since I $ $ I is not in I I 44 I, it is enough to show that We first consider the case where / 3 (so the shorter arc between 3 and does not cover the / to gap) We may subtract multiples of I from to give I with the property that 3 L 3 I The distance between and 3 is $ I Now either and 3 are in the same copy of I or in consecutive copies of I In either case, they are either in the same I or I block (so, as before, L F8I D ) or in different I or I blocks (so F2I / F2I ) Therefore : adjacent in I I, and we conclude that and 3 are not We now turn our attention to the more difficult case of / 1 / 3(, where the shorter arc between and 3 covers the / to gap Here, we may subtract a multiple of I from 3 and add a multiple of I to to yield vertices and 3 where 3 is in the first copy of I in I and is either in the last copy of I (ie, I! I I ) or in the terminal I (ie, I! I G, I ) Setting / 1 / 3, we consider three possible cases Case (a): L I In this case, $ I Now, and 3 are at least as far apart as the last and first elements in I, which, as in the similar case from (iib), are at a distance of F8I / F2I I This is too large to be in I, so : I and : I Case (b): I! I I In this case we add I to I to get I This extra distance is subtracted from, but in moving across zero, it crosses an extra s relative position within its I This distance

19 is exactly I, which is the space between I s within I Therefore 45 s position in the first I is exactly one I block back 14 from the position of in the last I We still have behind 3 of 3 / since was at least I, but now both are contained in the first I at a distance $ have seen before that : Case (c): I 6! I %, I Write! I 8, I where I Adding I to this mod gives I Now is a distance between two elements of I, so that : I I I, ie, is I, and we s relative position in the terminal! I I, I I / I / I! I I, I / I I N I / B /,- / * I So after being moved from to > I $, we have " / * I * / F2I / * 1 B Notice that this set contains the I in the first (untranslated) I, and that 3 must be in this I (otherwise and 3 would have been closer than I, putting us in Case (a)) So and 3 are both in this set " / * I * / ' F8I / * 1 B, and the distance between any two elements in it is at most F I / L for all I I, so as before, Thus in all cases, I, ie, no connection distance between points in I is found in I, so I must be an independent set of I 6 I (iii) I IF 14 We do not allow to be in the first block of the last ; If it were, a shift forward by * would move it into the terminal in This situation is covered separately in Case (c) below

20 46 I! I %, I I! I %, " / * I I B F8I! I %, " / * F2I F2I / B F8I! I %, " F2I /, / F8I B F8I! I %, F8I /! I F2I! I %, I / I D / /, /! I I D / I D / /,! I %, I / *! I %, I D / I D / /! I I / *! I I, B I)D / I D / / D I C( / * D I D * I B I D / I D / / D I / D I D / / / B I D / I D / / D / D / / I D / I)D / / F so I is of the desired size, even for / (! ) Thus we have shown that has an independent set of size F, which is exactly what we needed in order to verify that I <L9, and so we are done I 33 A Relaxation of Fractional Ramsey Numbers When we take to mean that 4 78; implies or 4, we could equally well write this as IfH = and, then 7 or M We can now fractionalize the entirety of this statement, and take / 12 #3(

21 to mean If and, then A or 4 3 We then >2 3( to be the infimum of all for which this statement holds We may also define define! the multicolor version of this, where / 1 means that, if 6 and, then 5I I for some Likewise,! > is the infimum of all for which / 1 is true To achieve our result, we need the following lemma 47 Lemma 36 If B9, then Proof We prove this statement in the form positive integers such that 5I I : I2 = 7 Let ( be 5I : I for * (the proof of Lemma 11 guarantees that such integers exist) Let I be a proper I -fold coloring of I using a set of I colors, where I is the set of I colors assigned to, and if D KI then &I BE I 6G We may now construct a -fold coloring of the vertex the set of colors, using a set of colors: assign to & Now, if D, then D 7 or D, which in turn implies that either ' E G or E & G This guarantees that, & and -fold coloring of This shows that?, and so & are disjoint, and so we have a proper %& (? Theorem 37 For real numbers *, we have! 1

22 48 Proof For any with and any decomposition =, if we suppose that I AL I for all, then Lemma 36 (applied repeatedly) implies that L This is a contradiction, and so we must have as desired Therefore We prove the lower bound by inducting on 5I I for some, / 1, and! 1 ; BASE : That! > is immediate INDUCTION HYPOTHESIS Suppose that! > D D : Take any 4LC We must find a graph and a decomposition J# A where A and 5I AL I for all To start, we choose rational #' and # such that * L #, LC D * L # L and # # By our induction hypothesis, there is a graph with decomposition D 6 D such that D #, but I L I for each / Recalling that #, we take D CB (wherein each vertex of is replaced with a copy of D ), so we have # C# by Lemma 15 Let have vertices, and D have vertices We may partition the edges of into the set of edges within copies of D, and the set of edges between copies of D The former yields a graph of the form D B 9 D B B D B while the later is 4EB So if we set 5I: I B for /, and

23 49 KB, then 7B 9 Further, 5I 9 I = I LC I for / This construction shows that ; 4 ; %# L / > is false, and so! 1 9 Since this relation is implied by our induction hypothesis, we are done 34 -Ramsey Numbers We may define the -Ramsey number of a graph by replacing in the Ramsey definition with instead of Recall from Section 11 that A -clique of is a function max st * * The weight, or value, of this -clique is minimum 15 value of taken over all -cliques of * 1 1 such that & for all &, and is the We let / >2 3( stand for the statement F6, then A or 3 Then the -Ramsey number! >2 #3? is the least positive integer for which this statement is true Because M < for any positive integer, it immediately follows that! 12 #3(! >2 #3?! 12 #3( In general, because! >2 #3? involves a discrete optimization invariant similar to!, we expect computation of these values to be difficult (as opposed to the relative ease of calculating! 12 #3? ) We do, however, have two principle results 15 Since we are not discussing infinite graphs, we do mean minimum and not infimum

24 * * 5 regarding the limiting behavior of! The first of these tells us that! &(, like! &(, grows exponentially in, although the bound achieved on this growth rate is considerably smaller Lemma 38 The edge set of K contains at least / / - edge-disjoint triangles Proof It is a long-known result 16 that there is a Steiner triple system on elements iff or $! a collection In other words, for sets of size or $!, we may find of size 3 subsets such that every pair of elements appears in exactly one member of This is equivalent to saying that partitions into triangles Such a partition necessarily has F /, :! the edges of a KD subgraph into / > / - :! triangles Even if $!, we may still partition triangles, giving the desired result Note that, while the above value is not always best possible, it still approaches :! asymptotically, which clearly is the best possible limit Theorem 39 If positive integers and satisfy $ D then! " ( C D L Proof We start by defining max st * 1 which is identical to, except that we are only allowed to assign weights or 1/2, and not 1, to vertices We refer to feasible solutions to this program as half-cliques Note 16 First shown by Kirkman around 185; see [1]

25 that, in any such half-clique, the set of vertices receiving weight 1/2 must be triangle-free, and in fact, is 1/2 times the number of vertices in the largest triangle-free induced subgraph of Now, the set of positively weighted vertices in any 2-clique must also be triangle free, but here we are using weights 1/2 and 1 Multiplying 51 by 2 is equivalent to assigning weight 1 to every vertex in the largest triangle-free subgraph, so we must have * We now employ a probabilistic technique similar to those used to calculate lower bounds for ordinary Ramsey numbers (see, for instance, [13]) Fix, and let us red/blue color the edges of K, giving any edge red or blue with probability 1/2, independent of the coloring of any other edges For any size subset of 7, define the events * D has no blue triangle 1 * D has no red triangle 1 * K contains a monochromatic weight :'* half-clique 1 Now, the probability that any given triangle is not all blue is 7/8, and any size set contains at least / " / - :! $;: D D Pr* 1 Pr So we have edge-disjoint triangles by Lemma 38, so Pr* * Pr* 1AL * $ 1 Pr* If we choose small enough so as to make this quantity less than 1, then Pr* D D 1AL 1, so there must exist some edge 2-coloring of with no monochromatic half-cliques of

26 L * weight : * That is, there is some 7 ; with 5I L : * for * Since 5I * 5I, we know that / " &( is false Let us perform a rough asymptotic analysis of versus to find a lower bound on D D $2:, then! &( If we take and $2: $ D LG becomes * So we have the approx- Then * :%$, which gives, approximately, L : $ imate bound of!, " &? :%$ * * $ $ LG 52 While this is not nearly as good as the ) * lower bound for! &(, we do know that! " &(! " ( in general, and this does at least establish the exponential growth of!' &( Since as 5 for any graph (Theorem A2), we wonder if the same holds for - and fractional Ramsey numbers, ie does! 12 #3(! 12 #3(? Note that, since! and! are integer valued, convergence occurs iff there exists a for which implies! =! Theorem 31! >2 #3?! >2 #3? as K if and only if >2 3( is not a discontinuity point of the function! Proof Recall 2 #3 & 5 and # 6$ & (*-, /, 21 of Theorem 31 Notice that if we hold and fixed, and may range freely between multiples of : and :, respectively, without changing the value of # Thus! >2 3( is constant over these rectangular regions of the plane These rectangles are closed along their upper and right edges, and open on their lower and left edges All discontinuity points of! >2 #3? lie on these edges, though not all

27 53 points on these edges are discontinuity points To be more precise, >2 #3? is a discontinuity point of! if! > 4 #3 4! 12 #3? for all 4 Lemma 311 If >2 #3? is a discontinuity point of! with! then there exists a decomposition 7 H such that 12 #3(, % and ;3 Proof Suppose to the contrary Then for every such decomposition, either or I 3 There are only a finite number of such decompositions, so let 4>$ & (* / 2 / 3K1, taken over all such positive values for all such decompositions Thus for any we know that 4 or 3 4 But this tells us that! > 4 3 4, contradicting the fact that 12 #3? is a discontinuity point of! Our supposition to the contrary is therefore false For discontinuity point 12 #3(, take the indicated by Lemma 311 For *, let I I : I, where I : I is a lowest terms fraction Let be any positive integer such that neither nor divide Then there is no integer I such that I : I 5I &I7: &I7: I Thus I Since must always be a multiple of :, we cannot have 7 L 9 and ML =3 and so! &12 #3(! >2 #3? Since we may choose such to be arbitrarily large, we have!! at this >2 #3? Lemma 312 For any graph and, / :

28 54 Proof Start with a maximum fractional clique, and round all weights on vertices down to the nearest multiple of : less than : has been removed The result is a valid -clique, and total weight Let! >2 3( be constant over all > B #3@>3 #3 B Choose any specific > and 3 >3 #3, so that >2 #3? is not a discontinuity point of! Let %$'&)(* / 2 #3 / 31, and choose such that : L Now, for any decomposition K 2C, either M9 or 4 <;3 Then for any, either 7 / : 9 / 9 / > / or / : ;3 / ;3 / 3 / 3? 3 so / 12 #3? Thus! >2 #3?! 12 #3( for all, and so!! at >2 3( Clearly, there is still much work that could be done in the area of -Ramsey numbers Even though! >2 (presumably) grows exponentially in, for almost any fixed we have! >2 M! >2, a function which only grows quadratically in There seems to be some interesting ground to cover in comparing the growth rates of! &12 #3( in and 3 versus 35 Lovász- Ramsey Numbers The fractional clique number of a graph is a relaxation of the ordinary clique number; the Lovász- number of a graph, denoted, represents a weaker relaxation of clique

29 55 number That is, for any finite graph was introduced by Lovász in 1977 (see [1]), and has been well studied since then (see [7] for an overview of history and results) We will define Lovász- Ramsey numbers by replacing clique number with the Lovász- number 17 To define, we first need to define orthogonal labelings An orthogonal labeling of a graph D is an assignment of a unit 18 vector to each such that whenever D That is, adjacent vertices are assigned perpendicular vectors, and for all These vectors may be of any fixed dimension The cost of a vector in such a labeling is defined to be, where? is the first entry of The cost of the labeling, denoted, is just the vector of costs (whose -th entry is ) Recall the integer (ID) and linear (DP) duals defining and from Section 11 Let us refer to the feasible regions of these programs as and, respectively We now define the region * 3@ 3 for all orthogonal labelings of, 37 1 Similarly to and, we define It is easy to show 19 that max st 3@, from which 17 Note: Our definitions of and orthogonal labelings of are more commonly taken to be those of We switch the roles of these two quantities to maintain consistency with our previous work As Ramsey numbers are symmetric in the usage of and, our choice of definitions will not affect our Ramsey results 18 Requiring unit vectors is not standard in the definition of orthogonal labelings, but is done without loss of generality for our purposes; see [7] 19 See Lemma 2 from [7]

30 ! 56 immediately follows We now take / >2 #3? to mean that, whenever 7, we must have M9 or <=3 Then the -Ramsey number! >2 3( is the least positive integer for which / 12 #3? While we cannot compute! exactly as we did with!, we can show that it s value is nearly the same as! Our result follows easily from the following lemma 2 Lemma 313 For a graph on vertices, ; if is vertex transitive, then equality holds Theorem 314! 12 #3?! >2 #3?, 3 3 *! 3 /!! Proof Fix 2 Let >2 3(, and let K be any edge 2-coloring of K Then or, and so >2 #3? Thus it follows that >2 3( 12 #3(, 3, 3!! Now take 3, and any decomposition 7 =4 (with 7 4 ) Either A or AL implies that A by Lemma 313 Thus >2 3( Since 12 #3( is very nearly as large as, we have fairly tight bounds on >2 #3? It is interesting to note the closeness of the values of! to!, even though lies somewhere between and Note that we did not use the second part of Lemma 313 If we could establish values of for a large class of vertex transitive graphs (as we did with : for ), we could likely achieve even more accurate bounds on! 2 See Lemma 23 and Theorem 25 from [7]