Math 301: The Erdős-Stone-Simonovitz Theorem and Extremal Numbers for Bipartite Graphs

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1 Math 30: The Edős-Stone-Simonovitz Theoem and Extemal Numbes fo Bipatite Gaphs May Radcliffe The Edős-Stone-Simonovitz Theoem Recall, in class we poved Tuán s Gaph Theoem, namely Theoem Tuán s Theoem Let k 3 Then exn, K k That is to say, if G is a gaph having at least On the othe hand, thee exist gaphs having k n k k n k k n k + edges, then G contains K k as a subgaph edges that do not contain K k as a subgaph Moeove, we showed that the extemal gaphs that is, the gaphs achieving the bound ae the complete balanced k -patite gaphs shown in Figue As these will come up again, we shall denote this gaph by T k n, called a Tuán gaph Ou goal fo the next few theoems is to extend this esult as best we can to othe classes of gaphs Fist, let s think about what was special about the Tuán gaphs These gaphs wok fo ou situation because we can goup the vetices up into k independent sets, and the Pigeonhole Pinciple allows us to say that if a K k wee pesent, it would have to have two vetices in one of these independent sets which is, of couse, impossible So pehaps, then, what is special about Tuán gaphs is the independent sets That is to say, it is cetainly clea that any gaph whose vetex set can be witten as a union of k independent sets can contain no copy of K k Pehaps this is the key to unlocking extensions of this extemal numbe So when can a gaph be witten as a union of k independent sets? Lemma Let G be a gaph on n vetices The vetex set of G can be patitioned into k independent sets if and only if G can be popely coloed with at most k colos The poof of this lemma is essentially tivial Note that if we popely colo the vetices of G, then any two vetices having the same colo must be independent Taking the patition as exactly the colo classes will yield the esult, and vice vesa: given a patition as above, we may colo each independent set with one colo to obtain a k -coloing Fundamentally, this is the poblem with finding a K k inside of T k n: the Tuán gaph is k - coloable, so any subgaph of T k n is also k -coloable But K k is not This leads to the following immediate genealization: Theoem Edős-Stone-Simonovitz Theoem Let H be a gaph with chomatic numbe χh k 3 Then exn, H k n k of edges in T k n + o n ; ie, the extemal numbe fo H is asymptotically equal to the numbe Futhemoe, it should be clea that the extemal gaphs will be the Tuán gaphs If that is not clea, pove it as an execise Use induction

2 V k- V V V 3 Figue : The Tuán Gaph Hee, we assume that V i and V j ae sets of vetices, whose sizes diffe by at most fo any i, j, and the solid lines between sets of vetices indicate that all edges ae pesent This gaph is a complete k -patite gaph, in that we have all edges between patition sets, and is balanced in the sense that the sizes of the vetex patition ae balanced The poof of this theoem is messy, so befoe we get into the details, let s sketch it out Since we ae looking at an asymptotic esult, we actually need to show that fo any ɛ > 0, thee exists some N sufficiently lage that if n > N, then exn, H k n k + ɛn Since ɛ is abitaily small, we essentially have witten that exn, H k n k + o n, just as we wanted So hee will be ou stategy: Choose some small ɛ Look at a gaph having n vetices and at least k n k + ɛn edges 3 Show that if n is lage enough, we can find, as a subgaph hee, the complete k-patite gaph having patite sets of size at least t fo any t We will specify late Actually, we will show it contains a copy of T k kt as a subgaph which is the same thing 4 Notice that since H is k-coloable, as long as t is lage then the size of each colo class, we can embed h inside T k kt, by embedding each colo class in one of the patite sets, and then using whicheve edges we need 5 Realize that these steps ae enough to pove the statement: we have shown that in any gaph G having at least k n k + ɛn with n lage enough, we can find a copy of H by lazily picking t V H 6 WIN THE GAME!!!! Cetainly, the difficulty of this stategy lies entiely in Step 3 This will be the bulk of the wok in poving the theoem Since this pat of the poof is haiy, we shall split it up into some lemmas Lemma Fix ɛ > 0, and let G be a gaph on n vetices having at least k n k + ɛn edges Then G contains a subgaph D having at least m ɛk n vetices, such that the degee of evey vetex in D k is at least k + ɛ m + Poof We shall poduce D using the following algoithm:

3 Initialize: D 0 G, n 0 n Given D i, n i : let v V D i be a vetex in D i having degee k k + ɛ n i 3 Define D i+ D i \{v}; that is, D i+ is obtained fom D i by emoving the vetex v Define n i+ V D i+ n i n i + 4 Repeat steps and 3 until no vetex v satisfies 5 Output D i Clealy, this algoithm will teminate eventually, since eventually we will eithe have a gaph whose vetices all have a lage degee, o a gaph with no vetices We d eally like it to be the fist thing, so let s conside how long the algoithm will un fo To do so, let s look caefully at the i th step Notice that in the i + st k step, we emove at most k + ɛ n i edges fom D i Hence, the subgaph D i is obtained by emoving at most i k k + ɛ n j j0 i k k + ɛ n j j0 k i k + ɛ n j j0 k k + ɛ ii ni k nn k + ɛ ni n n i n n i k n k + ɛ n i + n n i since i n n i edges fom G Also, the total numbe of edges in D i is at most n i n i Theefoe, the total numbe of edges in G is at most n i k n + k + ɛ n i + n n i On the othe hand, the hypothesis was that G contains at least k n k + ɛn edges Theefoe, it must be that k n k + ɛn n i k n + k + ɛ n i + n n i Theefoe, this pocess must stop wheneve the above inequality fails to hold That is, the pocess teminates when k n + ɛn > n i k n k + k + ɛ n i + n n i k n k + ɛ k k ɛ n k k + ɛ > n i k k ɛ n i k k + ɛ k k n ɛ n k + ɛ > n i k ɛ n i k + ɛ Notice, the left hand side of this inequality is just a constant when G is fixed, which it is, and the left hand side is a paabola that deceases when n i deceases If you use the quadatic fomula, you will find that this inequality is tue so long as n i < ɛk n That is to say, afte at most n ɛk steps, the algoithm must teminate, and thus the esult holds 3

4 A k- A A P Figue : The set P defined in the poof of Lemma 3 Note that if we conside a vetex in P, it has a lot of edges namely, t edges into each set A i Howeve, the edges need not always be to the same vetices Lemma 3 Let ɛ > 0, k, and let G be a gaph on n such that evey vetex of G has degee at least k k + ɛ n Fix t > 0 Then if n is sufficiently lage, T k kt is a subgaph of G Poof We shall wok by induction on k Note that the case of k is tivial, since T t is the empty gaph on t vetices Notice that k k k Suppose the esult is known fo k fo any choice of t Let s t ɛ, and hence G has sufficiently many edges to apply the induction hypothesis k Choose n lage enough that we can find a copy of T k k s in G Label the patite sets as A, A,, A k, and let A A A A k Let W be the set of vetices in V G that do not appea in A Define P {v W Nv A i t fo evey i}; that is, P is the set of vetices that have at least t edges into each A i This is illustated in Figue Notice that ou goal now is to constuct a set B k P, such that fo each i, thee is a subset B i A i with evey possible edge between B k and B i, and B i t This will be exactly a T k kt We have esticted to P, since these ae the only possible vetices to be included in such a set B k See Figue 3 Most of the est of the poof comes down to counting Fist, let us count the numbe of nonedges between W and A Fist, if a vetex v W is not in P, then thee exists some i such that the numbe of edges fom v to A i is at most t Hence, the numbe of nonedges between W and A is at least W P s t + W P s t k On the othe hand, the degee of evey vetex in G is at least k + ɛ n, and hence given any vetex k v, the numbe of nonedges involving v is at most n k + ɛ n n k ɛ Summing ove all vetices in A, we thus obtain that the numbe of nonedges between W and A is at most sk n k ɛ Taking these two bounds togethe yields sk n k ɛ W P s t 4

5 B A B A A k- B k- B k P Figue 3: The goal of the poof: to find a size t subset B i in each A i, and a set B k in P, with all the edges between B k and each B i Notice that we aleady have all the edges between B i and B j when i j and both i, j < k, since they ae subsets of the A i sets Recalling that s t ɛ and W n sk, this yields tk n k ɛ n sk P t ɛt Solving this inequality fo P and using lots of algeba, we obtain ɛk P n sk ɛ The impotant thing hee is that P is monotonically inceasing with n, even as A stays the same Hence, we may choose P as lage as we like Moeove, note that by the pigeonhole pinciple, as long as P is lage enough, we will be able to find a set B k in P satisfying the desied popeties Hence, by taking n sufficiently lage, we can poduce such B i, and the esult is demonstated Note that these two lemmas immediately pove the Edős-Stone-Simonovitz theoem, as we can fist find a good subgaph D in G by Lemma, and then find a T k kt inside D and hence inside G using Lemma 3 Once we have a T k kt as a subgaph of G, applying the obsevation made in the poof sketch then yields the esult Extemal numbes fo bipatite gaphs Note that the Edős-Stone-Simonovits Theoem woks fo any gaph H whose chomatic numbe is at least 3 This leaves out the entie class of gaphs having chomatic numbe exactly two Recall the following: Lemma 4 A gaph G has chomatic numbe if and only if G is bipatite Hence, we ae left asking, how do we compute extemal numbes fo bipatite gaphs? 5

6 The answe, unfotunately, is a lot of we don t know We do have the following bound fo any bipatite gaphs: Theoem 3 Let H be a bipatite gaph having v H vetices Then thee exists a constant c depending on H such that exn, H cn /v H To pove this bound, we shall make use of the following binomial identity While we will not pove this hee, you should be able to pove this youself if you know how to use Janson s inequality you can pove that n is convex in n Lemma 5 Let a, a,, a N be positive integes, with aveage A N ai Then a i N A That is to say, eplacing each a i with a copy of A only inceases the sum Poof Fo simplicity let s wite v H Notice that if K, is a subgaph of G, then so is H, since we may embed H just as we did in the Tuá n gaphs fo the poof of the Edős-Stone-Simonovits Theoem Now, suppose that fo each v V G, we take S v to be a set of neighbos of v Note that if thee exists a set of vetices, say A, such that fo all u, v A, S u S v, then we can fom a copy of K, inside of v having one patite set as A, and the othe patite set as S v fo any v A Hence, it must be that this constuction is impossible Theefoe, v V G dv n, since no set in [n] can be epesented in the sum o moe times Applying the bound fom the pevious Lemma, and setting eg equal to the numbe of edges in G, we thus obtain dv v V G n n dv eg/n n n n n Note that n n, and eg/n eg n use Stiling s fomula, fo example Theefoe, we have eg n n n Solving fo eg, we obtain eg / n / Note that the initial faction is independent of n, but does depend upon H; it is this faction that we take to be c in the statement of the theoem At this point, you might think Awesome, we did it! We got a bound! DONE And you e ight to celebate, this is useful, sot of But imagine if H has only a few vetices, say 3 The exponent on n in this case is n /3 n 5/3 But the only bipatite gaph on 3 vetices is the path of length, and we know that the extemal numbe fo this path is n, which is a fa, fa cy fom n 5/3 In fact, thee is NO bipatite gaph containing a cycle fo which exn, H is known We know a few of them asymptotically, but we don t know the constants Fo an example, let s conside the fist nontivial case: H C 4 By applying ou theoem caefully actually, by edoing the analysis in this somewhat special case, we obtain exn, C 4 n3/ + n 6

7 This uppe bound is asymptotically tight in SOME cases Specifically, it is known analysis complicated that exn, C 4 n3/ + on 3/, in the case that n p + p + fo some pime p I know, this n looks idiculous It comes fom analyzing cetain gaphs built out of finite fields, which ae always of ode p k fo a pime p This constuction is due to Füedi, and I d be happy to show it to you in office hous And in fact, in a constuction due to Klein, we have that exn, C 4 Θn 3/, that is, we ae asymptotically on the ight ode hee And while it is conjectued that the constant / is coect, we cannot pove it fo any n othe than some fancy foms involving pimes like the above 7