MT5821 Advanced Combinatorics

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1 MT5821 Advaced Combiatorics 9 Set partitios ad permutatios It could be said that the mai objects of iterest i combiatorics are subsets, partitios ad permutatios of a fiite set. We have spet some time coutig subsets; ow we tur to partitios ad permutatios. 9.1 Partitios For partitios, we are uable to write dow a simple formula for the coutig umbers, ad have to rely o recurrece relatios or other techiques. The Bell umber B() is the umber of partitios of a set of cardiality. There is o simple formula for B(), ad eve estimatig its size is quite hard. But there is oe thig we ca say. Every permutatio of {1,...,} has a cycle decompositio, a expressio as a product of disjoit cycles. The supports of the cycles partitio {1,...,}. So we have a map from permutatios to partitios, which is clearly surjective, sice ay subset supports a cycle. So B()! for all. We will refie this estimate shortly. Here is a recurrece relatio. Propositio 9.1 B(0) = 1 ad B() = ( ) 1 B( k) for > 0. k 1 Proof The iitial coditio is empty set theory. For the recurrece, look at the part of the partitio( which) cotais the elemet 1. Suppose that this part has size k. The there are ways to choose the k 1 remaiig k poits, ad the B( k) ways to partitio the poits that are left over. Multiplyig ad summig over k gives the result. This recurrece allows us to work out the expoetial geeratig fuctio for the Bell umbers, which has a ice form: 1

2 Propositio 9.2 B()x = exp(exp(x) 1). 0! Proof Let F(x) be the fuctio o the left of the equatio. The d dx F(x) = B()x 1 1 ( 1)! 0 = ( x = l l 0 l! = exp(x)f(x). x k 1 B( k)x k (k 1)! ( k)! )( ) B(m)x m m 0 (I the third lie we replaced the variables ad k by l = k 1 ad m = k. The give rage of values of ad k correspods to l ad m takig all o-egative iteger values.) Now we have a first-order liear differetial equatio with iitial coditio, which is easily solved. We fid that m! d (exp( exp(x))f(x)) = 0, dx so F(x) = Aexp(exp(x)); ad the iitial value F(0) = 1 gives A = exp( 1). We lear somethig about the magitude of Bell umbers from this theorem, which stregthes greatly our result that B()!: Propositio 9.3 The ratio B()/! teds to zero faster tha ay expoetial fuctio c for c > 0. Proof The radius of covergece of a power series a z is the umber r such that the series coverges for z < r ad diverges for z > r. (We do ot specify the behviour for z = r.) This icludes the cases r = 0 (the series diverges for all z 0) ad r = (the series coverges for all z). We use two facts from aalysis: 2

3 a coverget power series defies a aalytic fuctio iside the disc of covergece, ad the radius of covergece is the distace from the origi to the earest sigularity of the fuctio; the radius of covergece of a z is the reciprocal of limsupa 1/. Now the sum of B()z /! is aalytic for all z, so the radius of covergece 0 is ifiite; so lim (B()/!) 1/ = 0. This is equivalet to the statemet of the propositio. Here is aother applicatio. For the partitio umbers p(), we saw that 0 p()z = (1 z k ) 1. k 1 The right-had side is aalytic iside the uit circle, but has sigularities at every root of uity. So the radius of covergece is 1, whece limsup p() 1/ = 1. This meas that p() grows slower tha ay expoetial fuctio c with c > 1. O the other had, it ca be show that it grows faster tha ay polyomial. We refie the Bell umbers i the same way that the biomial coefficiets do for subset coutig. The Stirlig umber of the secod kid, S(,k), is the umber of partitios of a -set ito k parts. Thus, S(0,0) = 1 ad S(0,k) = 0 for k > 0; ad if > 0, the S(,0) = 0, S(,1) = S(,) = 1, ad S(,k) = 0 for k >. Clearly we have S(,k) = B() for > 0. The recurrece relatio replacig Pascal s is: Propositio 9.4 S(,k) = S( 1,k 1) + ks( 1,k) for 1 k. Proof We divide the partitios up ito two classes: Those i which is a sigleto part. This part is simply added to a partitio of {1,..., 1} with k 1 parts, so there are S( 1,k 1) of this type. 3

4 Those i which is i a part of size greater tha 1. Removig from this part, we have a partitio of {1,..., 1} with k parts. Give such a partitio, we ca choose arbitrarily which of the k parts we add to. So there are ks( 1,k) of this type. Addig gives the result. The Stirlig umbers ca be set out i a triagle like Pascal s. We will leftalig this triagle (that is, take the colum k = 1 to be vertical). The recurrece relatio says that each etry is obtaied by addig the etry to its orth-west ad the etry above multiplied by k: You may spot some patters here, especially if you cotiue the triagle a couple more steps. See the exercises for some of these. It turs out that we ca tur the recurrece ito a statemet about a geeratig fuctio, but with a twist. Let (x) k = x(x 1) (x k + 1) (k factors). The we have Theorem 9.5 x = S(,k)(x) k for > 0. Proof I will give two proofs of this result; the first oe ivolves maipulatios with the geeratig fuctios, the secod oe is more obviously combiatorial. 4

5 First proof S(,k)(x) k = Assume the result for 1. Them = x 1 x = x, S( 1,k 1)(x) k 1 (x k + 1) + k=2 S( 1,k 1)(x) k 1 (k 1) + ks( 1,k)(x) k 1 l=1 (l 1)S( 1,l 1)x l 1 sice the last two terms cacel. (We chaged the variable to l = k + 1 i the third term, ad omitted terms ivolvig S( 1,0) ad S( 1,) whose value is zero.) Secod proof Take a auxiliary set X of size x, where x. Now x is the total umber of -tuples of elemets from X. We cout these -tuples aother way. Each -tuple (x 1,...,x ) defies a partitio of {1,...,}, where we put i ad j ito the same part of the partitio if x i = x j. Havig chose a partitio, how do we recover the -tuple? Suppose that the partitio has k parts P 1,...,P k, ordered by the smallest elemet (so P 1 is the part cotaiig 1, P 2 the part cotaiig the smallest elemet ot i P 1, ad so o). Now we have to choose a value i X for the elemets x i for i P 1, a differet value for the elemets x i with i P 2, ad so o; i all, the umber of choices of k distict elemets of X i order, which is x(x 1) (x k + 1) = (x) k. Now there are S(,k) partitios with k parts to cosider. Thus we have S(,k)(x) k = x. This proves the theorem whe x is a positive iteger. But both sides are polyomials of degree ; if they take the same value for every positive iteger x, the they must coicide as polyomials. So the result is proved. It is possible to fid the ordiary geeratig fuctio for the idex : S(,k)y y k = k (1 y)(1 2y) (1 ky). Also, the expoetial geeratig fuctio for the idex is S(,k)x = k! 5 (exp(x) 1)k. k!

6 Summig over k gives the e.g.f. for the Bell umbers: B()x = exp(exp(x) 1). 0! However, I will ot give the proofs of these results here. 9.2 Permutatios The umber of permutatios of a -set (bijective fuctios from the set to itself) is the factorial fuctio! = ( 1) 1 for 0. The expoetial geeratig fuctio for this sequece is 1/(1 x), while the ordiary geeratig fuctio has o aalytic expressio (it is diverget for all x 0). The recurrece relatio for the factorials is 0! = 1,! = ( 1)! for 1. Ay permutatio ca be decomposed uiquely ito disjoit cycles. So we refie the cout by lettig u(,k) be the umber of permutatios of a -set which have exactly k cycles (icludig cycles of legth 1). Thus, u(,k) =! for > 0. The umbers u(,k) are the usiged Stirlig umbers of the first kid. The reaso for the ame is that it is commo to use a differet cout, where a permutatio is couted with weight equal to its sig (as defied i elemetary algebra, for example the theory of determiats). The sig of a permutatio has several defiitios, all equivalet: The sig of a permutatio is the determiat of the correspodig permutatio matrix. But this defiitio has the flaw that determiats are usually defied usig the cocept of sig! The sig of a permutatio g is ( 1) p, where g ca be writte as the product of p traspositios. This defiitio has the drawback that we eed to show that the parity of the umber of traspositios whose product is g is the same, for ay expressio. If g is a permutatio of {1,...,} with k cycles i its cycle decompositio (icludig cycles of legth 1), the the sig of g is ( 1) k. I regard this as the best defiitio. 6

7 I will sketch the proof that the secod ad third defiitios agree this also shows that the secod defiitio works. Let g be a arbitrary permutatio, decomposed ito cycles, ad let t be the traspositio (i, j). It is easy to show [do it!] that, if i ad j are i differet cycles of g, the these two cycles are cut at these poits ad stitched ito a sigle cycle i gt; ad coversely, if i ad j are i the same cycle of g, the this cycle is divided ito two cycles of gt. So the parity of k is chaged (up or dow) by 1 whe we multiply by a traspositio, ad hece the sig (accordig to the third defiitio) is reversed. Now the idetity has cycles, ad so has sig +1. Ay permutatio is a product of traspositios; if it is a product of p traspositios, the its sig is ( 1) p (after p reversals). Now the siged Stirlig umber of the first kid is defied to be s(,k) = ( 1) k u(,k). Note that ( 1) k is the sig of every permutatio couted by u(,k); so we are just coutig permutatios accordig to their sig. We have s(,k) = 0 for > 1. This is related to the algebraic fact that, for > 1, the permutatios with sig + form a subgroup of the symmetric group of idex 2 (that is, cotaiig half of all the permutatios), called the alteratig group. (The secod defiitio shows that the sig map is a homomorphism from the symmetric group to the multiplicative group {+1, 1}; for > 1, a traspositio has sig 1, so the map is oto. Its kerel is the alteratig group.) We will maily cosider siged Stirlig umbers below, though it is sometimes coveiet to prove a result first for the usiged umbers. As usual we take s(,0) = 0 for > 0 ad s(,k) = 0 for k >. Here is the recurrece relatio. Propositio 9.6 We have s(,) = 1, s(,1) = ( 1) 1 ( 1)!, ad s(,k) = s( 1,k 1) ( 1)s( 1,k) for 1 k. Proof s(, ) couts just the idetity permutatio, which has sig 1. O the other had, s(, ) couts the cyclic permutatios. Sice we may choose the startig poit of a cycle arbitrarily, we ca assume that the first etry i the cycle is 1; the other etries are the a arbitrary permutatio of {2,...,}, ad the sig of a cycle is ( 1) 1. 7

8 Now divide the permutatios with k cycles ito two classes, accordig to their actio o the poit. I the first class, is fixed, so the fixed poit is adjoied to a permutatio of {1,..., 1} with k 1 classes; there are u( 1,k 1) such permutatios. I the secod class, occurs i some cycle (..., j,,i,...) of legth greater tha 1. (We may have i = j here.) By simply bypassig, replacig this cycle by (..., j,i,...), we obtai a permutatio o {1,..., 1} also with k cycles. Coversely, give such a permutatio, we ca choose to isert aywhere i ay of the cycles; there are thus 1 ways this ca be doe, so the umber of permutatios is ( 1)u( 1,k). Thus u(,k) = u( 1,k 1) + ( 1)u( 1,k), ad puttig i the sigs gives the result. (Permutatios couted by u(,k) ad u( 1,k 1) have the same sig, while those couted by u( 1,k) have the opposite sig.) We ca produce a triagle of Stirlig umbers of the first kid similarly to what we did for those of the secod kid. This time, the multiplier o the southerly arrow is the row umber of the source row, rather tha the colum umber as before, ad with the sig chaged: From this, we fid a geeratig fuctio: Theorem 9.7 s(,k)x k = (x). Puttig x = 1 i this equatio shows that ideed the sum of the siged Stirlig umbers is zero for > 1. 8

9 Proof The proof of this ca be doe by maipulatig summatios i the same way that we did for the other kid of Stirlig umbers. If you have met the Orbit-Coutig Lemma for permutatio groups, a icer proof is possible. I will outlie this here. Suppose that a fiite group G acts o a fiite set T. We deote by fix(g) the umber of poits of the set T which are fixed by the elemet g G. We say that two elemets x,y of T are i the same orbit if there is a elemet of G carryig x to y. This is a equivalece relatio o X. (The idetity, iverse ad closure laws for G show respectively the reflexive, symmetric ad trasitive laws for the relatio.) The equivalece classes are called, aturally eough, orbits. Now the Orbit-coutig Lemma asserts that the umber of orbits is 1 G fix(g). g G I wo t prove this here, but will ow apply it. We take the group to be the symmetric group S, ad its actio is o the set of all -tuples of elemets from a auxiliary set X with X = x. Which -tuples are fixed by a permutatio g? Such a tuple must be costat o each cycle of g; so the umber is x c(g), where c(g) is the umber of cycles of g. Thus, the umber of orbits of G is 1! g Gx c(g) = 1! u(,k)x k. O the other had, two -tuples lie i the same orbit of the symmetric group if ad oly if they cotai each elemet of X the same umber of times. Thus the orbits correspod to uordered selectios of thigs from a set of size x. We saw that the umber of such selectios is ( ) + x 1 = x(x + 1) (x + 1).! Equatig the two expressios ad multiplyig by! gives x(x + 1) (x + 1) = u(,k)x k. Now replacig x by x ad multiplyig by ( 1) gives x(x 1) (x + 1) = 9 s(,k)x k.

10 We have proved the idetity whe x is a positive iteger (sice we eed a set of cardiality x i the argumet). But both sides are polyomials; so if the two expressios are equal for every positive iteger x, the they are equal as polyomials, as required. Note that this is the iverse of the relatio we foud for the Stirlig umbers of the secod kid. So the matrices formed by the Stirlig umbers of the first ad secod kid are iverses of each other. 9.3 The Möbius fuctio of the partitio lattice I this sectio we will use our kowledge of the Stirlig umbers of the first kid to calculate the Möbius fuctio of the poset of partitios. Let A be the set of all partitios of {1,...,}, so that A = B(), the Bell umber. For a,b A, we put a b if a refies b, that is, a splits the parts of b, but each part of a is cotaied i a sigle part of b. For example, a = {{1},{2},{3},{4,5,6}} refies b = {{1,2},{3,4,5,6}}, but does ot refie {{1,2},{3,4},{5,6}}. This poset is called the partitio lattice, ad deoted by P(). Theorem 9.8 Suppose that b has r parts p 1,..., p r, ad that a splits p i ito m i smaller parts for i = 1,...,r. Let m be the sum of the a i. The µ(a,b) = ( 1) m r (m 1 1)!(m 2 1)! (m r 1)!. Proof We begi with a couple of reductios. First reductio The partitios which refie b form the cartesia product of partitio lattices P( 1 ) P( r ). So it is eough to compute the Möbius fuctio for the case whe b has a sigle part, ad multiply the results together. Secod reductio The partitios i the iterval [a,b] do ot split the parts of a, which behave like atoms. So, if b has a sigle part ad a has m parts, the this iterval is isomorphic to P(m). We coclude that it is eough to compute µ(a,b), where b is the partitio of a m-set with a sigle part, ad a is the partitio of this set ito parts of size 1. If this umber is f (m), the the aswer to our origial problem is f (m 1 ) f (m r ). 10

11 We are goig to show that f (m) = ( 1) m 1 (m 1)!. The theorem will follow from this. Note that our proposed value of f (m) is the Stirlig umber of the first kid s(m,1). This is the clue that Stirlig umbers are ivolved! We review a couple of facts about Stirlig umbers: s(m,k) is ( 2) m k times the umber of permutatios of a m-set with k cycles. s(m,1) = ( 1) m 1 (m 1)!. m s(m,k) = 0 for m 2. Let a ad b be the partitio ito sigletos ad the partitio with a sigle part of a set of cardiality m. The µ(a,b) is the egative of the sum of the values of µ(a,c), for a c < b. Cosider a partitio c of {1,...,m} with k parts, havig sizes a 1,...,a k. For k > 1, the µ(a,c) is the product of ( 1) a i 1 (a i 1)! over all i. But this umber is the umber of permutatios havig the part of size a i as a sigle cycle, multiplied by the appropriate sig. So whe we take the product over i, we obtai the umber of permutatios with the give parts as cycles, multiplied by the sig of such permutatios. Summig over all partitios of size k the gives us s(m,k). So the sum over all such c is m s(m,k) = s(m,1), k=2 whece µ(a, b) = s(m, 1), as required. The proof is complete. Exercises 9.1. Show that S(,2) = 2 1 1, S(, 1) = ( 1)/2, ad S(, 2) = ( 1)( 2)(3 5)/ Show that s(, 1) = ( 1)/2. What is s(, 2)? 9.3. Show that cojugacy is a equivalece relatio i ay group. The ext exercise describes a deep coectio betwee partitios, Stirlig umbers of the secod kid, ad the expoetial fuctio o oe had, ad betwee permutatios, Stirlig umbers of the first kid, ad the logarithm fuctio 11

12 o the other had. The iverse relatio betwee the expoetial ad logarithm fuctios reflects the iverse relatio betwee the two kids of Stirlig umbers that we saw at the ed of sectio (a) Prove that the followig are equivalet for sequeces (a 1.a 2,...) ad (b 1,b 2,...), with expoetial geeratig fuctios A(x) ad B(x) respectively: (ii) b = S(,k)a k for 1; (i) B(x) = A(exp(x) 1). (b) Prove that the followig are equivalet for sequeces (a 1.a 2,...) ad (b 1,b 2,...), with expoetial geeratig fuctios A(x) ad B(x) respectively: (i) b = s(,k)a k for 1; (ii) B(x) = A(log(1 + x)). 12