Math 216A Notes, Week 3

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1 Math 26A Notes Week 3 Scrie: Parker Williams Disclaimer: These otes are ot early as polishe (a quite possily ot early as correct as a pulishe paper. Please use them at your ow risk.. Posets a Möius iversio Previously we cosiere partially orere sets with the ati symmetric relatio which satisfies x x x y y z = x z x y y x = y = x We the efie a fuctio ζ(x y i the followig way { if x y ζ(x y = otherwise This ζ the forms a upper triagular matrix shoul we orer rows a colums properly. This will prove valuale as may prolems we wish to cosier such as iclusio exclusio ca e thought of as simply solvig ζx = y which the simply gives us x = ζ y The catch here is that we kow ζ has a iverse we still ee to actually compute it. For example if our poset was the susets of {... } the ζ has imesios 2 y 2 a while upper triagular matrices are easy to ivert the size of the matrix makes eve a ice matrix ifficult to ivert. If our poset is structure however we may e ale to use this structure to compute µ without ivertig the full matrix... Summig over Itervals. Note that if µ(x y is the x y etry of ζ the all it really meas to e the iverse is that µ shoul satisfy { if x = y ζ(x zµ(z y = otherwise z Last time we sketche the proof that µ(x y = uless x y. This allows us to the restrict the sum for x z z y We the have provie x y x z y µ(z y = a via the same argumet we may otai agai provie x z x z y µ(x z = This ca e thought of as sayig that the sum of the Möius fuctio over ay iterval is.2. Example: susets of the two elemet set. We costruct the followig Hasse iagram { 2} {} {2} We have µ(

2 a it folows that µ( {} + µ( = so µ( {} = followig this logic we otai further that µ( {2} = Applyig the iterval argumet oe more time with y = { 2} we ca further otai µ( { 2} = Cotiuig owars µ has the form {} {2} { 2} {} {2} { 2} Oe way of speeig up the calculatio is to oserve that the Möius fuctio oly cares µ(x y oly cares aout what happes i the part of the Poset etwee x a y (sice the recursive calculatio oly ivolves that part of the Poset. For example sice the two chais. { 2} {} {} are isomorphic we have µ( {} = µ({} { 2}. This suggests that posets such as the suset lattice which cotai eormous amouts of symmetry may have particularly simply Möius fuctios..3. Example: The Itegers from to. Take... with staar. {}. {2} {} We have µ( = (recall that µ(x x = always sice µ is the iverse of ζ a ζ is upper triagular with o the iagoal. Sice ay iterval must a up to we have µ( 2 = a µ( y = for all y > 2. Oservig the isomorphism y (y x = x i geeral we have { } if x = y µ(x y = if x = y otherwise 2

3 What the iverse formula the says is that x g(x = f(y the f(x =. y= x µ(y xg(y = g(x g(x.4. Prouct posets a the Möius fuctio of the suset lattice. y= Defiitio. Give two posets S a T with relatios S a T relatio (s t (s 2 t 2 s S s 2 a t T t 2 so the prouct poset S T with the ca e thought of as ( ( ( ( This ca also e thought of as a represetatio of the susets of { 2} where the first cooriate asks if is i the set a the seco cooriate asks if 2 is i the set. More geerally we ca thik of the suset lattice of {... } as... times Ie is i my suset is 2 i my suset all the way up to. The key thig aout this is that it turs out that if we kow the Möius fuctio for each part of the prouct we ca write the Möius fuctio for the whole prouct. Theorem. I the prouct poset S T µ((s t (s 2 t 2 = µ S (s t µ T (s 2 t 2 Proof. It suffices to show that the µ ζ matrix is the ietity for µ as efie i this theorem. Cosier the ((s t (s 2 t 2 etry of the this matrix which is give y (s 3t 3 µ((s t (s 3 t 3 ζ((s 3 t 3 (s 2 t 2 The summa is oly ozero i the rage (s t (s 3 t 3 (s 2 t 2. Oce we restrict to that rage ζ is always so we may remove it from the summa a otai (s t (s 3t 3 (s 2t 2 3 µ((s t (s 3 t 3

4 A y our assumptio our µ factors as it is a prouct as well as our relatio factors so the whole sum will factor. µ S (s t µ T (s 2 t 2 t t 3 t 2 s s 3 s 2 µ S (s t µ T (s 2 t 2 s s 3 s 2 t t 3 t 2 which is simply the prouct ( ( if s = s 2 if t = t 2 = if s if s s 2 if t t t = s 2 t 2 2 The cetral oservatio to this proceure is that we ca uil igger posets from smaller posets a if we uersta how the Möius fuctio ehaves o the smaller posets we uersta how it ehaves o the igger poset a with this uerstaig we ca otai more complex iversio formulas Returig to our example we may ow cosier the Möius fuctio o the suset lattice. We kow that µ(s T = uless S T. If S T we have µ(s T = ( T \S sice whe we multiply the Möius fuctios we get a ( for each time T iffers from S. The iversio formula the ecomes M(S = T S I particular oserve that if S = the f(t = f(s = T S T \S M(T ( f( = T M(T ( T Which is simply iclusio exclusio..5. Example: Divisor lattice. We ow cosier a ew lattice that will agai e the prouct of lattices we uersta. Defiitio. The ivisor lattice of is the poset whose elemets are all the positive itegers which ivie havig the orer relatio x y if a oly if x is a ivisor of y. For example here is the ivisor lattice of We ca thik of this as eig isomorphic to

5 I geeral if = p a pa pa k the the ivisor lattice is isomorphic to k a a k..... k copies Usig the previously compute Möius fuctio for the itegers uer we immeiately otai if x y µ(x y = ( k if x y is ot square free is the prouct of k istict primes if x y We will sometimes write µ( as simply µ( correspoig to the classical Möius fuctio from umer theory which it equals. Note also that µ( as µ (. The Möius iversio formula for this lattice the ecomes that f( = g( = g( = ( f(µ We will ow look at some familiar umer theoretic ietities that ca e otaie via this formula. Let φ eote Euler s fuctio. We first ote Claim. φ( = m Proof. Cosier a map θ efie o {... } efie y θ(x = gc(x. If the the elemets i the preimage of are of the form y where gc(y =. I particular the preimage of has size φ(. Everythig must e mappe somewhere a as there are thigs total we have = ( φ Applyig Möius iversio to the aove claim we otai φ( = µ ( = µ( which ca e writte elegatly as φ( = µ(.6. Coutig circular sequeces of s a s. Our goal is to cout how may circular sequeces of s a s ca e otaie if we cosier two sequeces that are rotatios of each other to e the same. For example for = 4 we have the followig sequeces If it were t for the rotatio equivalece we woul simply have 2 sequeces. For ay sequece S havig perio there are precisely sequeces (icluig S equivalet to. Lets efie f( equal the umer of rotatio-iequivalet sequeces of perio. By the aove oservatios we kow that f( = 2. 5

6 Applyig Möius iversio to this relatio we otai The expressio we really care aout is f( = f( µ(2 Applyig the aove relatio to each summa we otai f( = µ(j2 j j This is a sum over chais j. We ca rewrite it as a sum parameterize y j a l := j = jl. Doig so (a reorerig the sum so l is o the outsie we otai f( = jl µ(j2l = l l Applyig our claim from the previous sectio we see that ( 2 l φ. l j l l 2 l l j l µ(j. j (so that If is prime this has the particularly ice form (2 + 2(. Note that whe we applie our Möius iversio after a uch of cacelatio we were left with a formula where all the summas are positive! This suggests that maye there is a alterate approach where each term of the sum has some sort of comiatorial meaig which we ll see i the ext sectio. 2. Coutig orits of a group G actig o a set X We ow are goig to thik of these prolems i more algeraic terms. Suppose that we have a group G a it is actig o a set X y permutig it. Theorem. (Bursie s Lemma For each g G let ψ(g e the umer of fixe poits of G actig o X. The the umer of orits of G s actio o X satisfies # orits = G ψ(g I other wors the umer of orits is just the average umer of fixe poits. Proof. We cout the umer of pairs (x g X G where g(x = x two ifferet ways. Metho g G has ψ(g solutios so the total is simply g ψ(g Metho 2 take ay x X if x is i the orit of size O x the x is fixe y G permutatios so the total umer of pairs is thus x G = G O x g G orits x orit O x O x = ψ(g = G #umer of orits g G Note that i Metho 2 we eee that G was a group i orer to guaratee that x was sprea evely over its orit y G s actio. 6

7 2.. Example: Z/Z actig o colorigs. Here G is Z a X is the collectio of all 2 fuctios. G acts y rotatio. If G rotates our set y k the the fixe poits are precisely those fuctios with f(x = f(x + k for all x. It follows from elemetary umer theory that this happes precisely whe f is perioic with perio a ivisor of gc(k so y Bursie s lemma the umer of orits is k= 2 gc(k Sice for each there are φ ( choices of k with gc(k = we ca rewrite this as ( 2 φ a simpler way of gettig at the formula from the previous sectio. 3. Weighte Colorigs 3.. The Cycle Iex. Our goal here is to geeralize Bursie s lemma to allow for coutig of weighte colorigs. Defiitio. The cycle iex of a group G actig o a set X is give y Z g (x... x = x z(g x z2(g 2... x z(g G where z j (g eotes the umer of cycles of legth j i the permutatio of X y g. g G Despite the otatio the cycle iex epes oth o the group a o the actio itself. For example cosier the permutatio of S 4 actig o { 2 3 4} i the staar way. The cojugacy classes here are Cojugacy Class Numer of Perm. Cycle Types (234 6 oe 4-cycle (23(4 8 oe 3-cycle oe -cycle (2(34 3 two 2-cycles so (2(3(4 6 oe 2-cycle two -cycles ((2(3(4 four -cycles Z S4 (x x 2 x 3 x 4 = 24 (x4 + 6x 2 x 2 + 3x x x 3 + 8x x 3 + 6x New framework; Colorigs. Let G e ay group A e ay set a let B e a set fiite set of colors. For each B associate a weight w( To each fuctio f : A B associate a weight W (f = a A w(f(a We say two fuctios are equivalet if f (a = f 2 (g(a for some g G a all a A. Our previous example correspoe to the case where all weights were equal to. We ow wat to compute W (f Note that that f f 2 class. Theorem. (Weighte Bursie s classes = W (f = W (f 2 so it makes sese to talk aout the weight of a equivalet classes W (f = G g G g(f=f where Φ W (g eotes the total weight of the fixe poits of g. Φ W (g Agai this reuces ow to the previous versio of Bursie s lemma i the case where W is ietically. Proof. (sketche Cout the total weight of pairs (g f where f is a fuctio fixe y g two ifferet ways i the exact same way as i the proof of the uweighte versio. 7

8 Note that the fixe poits of g are precisely those colorigs which are costat o every cycle of g s actio o A. I the simplest case where g has oe cycle of legth k the the fixe poits are just the costat colorigs a the total weight of such colorigs is Φ W (g = w( k. If istea g has cycles of legths k k 2... k l the we ca write l Φ W (g = w( i ki = (... l i= ( l w( ki i= ( z(g ( z2(g ( z(g = w( w( 2... w( where z j (g is the umer of cycles of legth j i the actio of g. Pluggig this ito Bursie s Lemma we see Theorem. (Pólya classes W (f = Z G ( B w( B w( 2... B w( 3.3. Returig to circular sequece of s a s. Now let us cosier the circular colorig prolem from efore ow with w( = a w( = x. We start y computig the cycle iex of Z/Z. We retur to the familiar fact that φ( elemets have gc( k =. It follows that the weighte sum of classes is Z G ( + x + x x = φ(( + x. Applyig the iomial theorem to this we otai φ( r= ( x r r As with geeratig fuctios we ca otai aitioal iformatio y extractig iiviual coefficiets from this. For the x k coefficiet is the force to ivie k. Sice our sum was alreay restricte to terms which also ivie ecessarily ivies gc( k. Thus our x k coefficiet is. gc(k Comiatorially this is coutig circular sequeces zeroes a oes cotaiig exactly k oes. φ( ( k 8