GRUPOS NANTEL BERGERON

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1 Drft of Septemer 8, 2017 GRUPOS NANTEL BERGERON Astrt. 1. Quik Introution In this mini ourse we will see how to ount severl ttriute relte to symmetries of n ojet. For exmple, how mny ifferent ies with four olors n we onstrut?......??? There re mny possiilities n it is hr to nswer without the tools we will uil. Another typil exmple is how mny ifferent nekles n we uil using 3 kin of es?...?? To hieve this we nee to lern out groups of symmetries, tion of groups, osets, et We will see muh more interesting enumertions n pplitions. 2. Simetris y groupos We re intereste in unerstning ll the symmetries of n ojet we stuy n the lgeri strutures on those symmetries Conjuntos e Simetris. Suppose you hve soli squre tthe in the mile: We n turn the squre n get k the sme thing. This is symmetry. Reserh of Bergeron supporte in prt y NSERC n York Reserh Chir. 1

2 2 BERGERON We wnt to list ll the possiilities: Is there more? When re two symmetries onsiere the sme?? To keep trk of wht hppen, We n nme the verties of the squre to see wht hppen to them. This is seret mrking, it is not prt of the ojet, it is only there to help us keep trk of wht hppen: We n then enoe wht hppen to the squre y only listing wht hppen to,,, n : σ 0 =, σ 1 =, σ 2 =, σ 3 =. This is mthemtil moel of the symmetries of the squre. Now we sy tht two symmetries re equl if strting with the sme mrking, they en up with the sme mrking fter

3 GRUPOS 3 the symmetry. = This moel is useful to unerstn the strutures of those symmetries n fous on wht interests us t the moment (the verties t the orner of the squre). This esrie in some wy ll the symmetries of the squre. I think of these s mps. This is not unique, we will onsier soon other esriptions of the sme symmetries n see tht they n hve ifferent presenttions opertions on symmetries. We now wnt to esrie opertion we n o on symmetries. First, omposing symmetries: Wht hppen if I o rottion, n then nother one from wht we hve one? Now using our mthemtil moel: I think of these s omposition of mps: σ 1 : {,,, } {,,, } Wht we hve seen ove is tht = σ 1 σ 1 = σ 2. = σ 2 : {,,, } {,,, }

4 4 BERGERON it is the sme mps. Another exmple = In ll se omposing rottions gives us nother rottion. Now using our mthemtil moel we n look t ll ses The ientity symmetry: You remrke tht there is speil symmetry tht oes nothing : enoe y I = This symmetry is speil in the sense tht if we ompose y I on the right or on the left we o not hnge nything: I σ = σ I = σ Inverting symmetries: Finlly, you see tht ny symmetry n e unone: = I

5 GRUPOS 5 ere we hve tht = We sy tht it is the inverse. If σ is symmetry, we enote its inverse y σ 1. Using nottion s ove, we hve σ 1 0 = σ 0 σ 1 1 = σ 3 σ 1 2 = σ 2 σ 1 3 = σ Astrt groups. When we tlk out the symmetries of n ojet we hve seen four importnt fetures (1) We hve set of symmetries (2) We n ompose symmetries n it gives k symmetry in our set (3) There is speil symmetry lle the ientity. (4) Every symmetry hs n inverse In mthemtil terms, we hve group. More formlly, group is: (1) A set G (2) An ssoitive opertion m: G G G [we often write m(, ) = or m(, ) = + ] (3) A unique element 1 G: for ll g G (4) Every g G hs n inverse g 1 : ssoitive: () = () 1g = g1 = g gg 1 = g 1 g = 1 Exmple 1. C 4 = {1,, 2, 3 } where we ssume 4 = 1 so m It is ssoitive. We hve the speil element 1 n every element hs n inverse 1 = 3, 2 = 2, 3 =.

6 6 BERGERON Exmple 2. C 2 C 2 = {1,,, } where we ssume 2 = 1, 2 = 1 n = so m It is ssoitive. We hve the speil element 1 n every element hs n inverse 1 =, 1 =, () 1 =. Exmple 3. S 3 = {1, s, t, st, ts, sts} where we ssume s 2 = 1, t 2 = 1 n sts = tst (is it ll?), so m 1 s t st ts sts 1 1 s t st ts sts s s 1 st t sts ts t t ts 1 tst s st st st sts s ts 1 t ts ts t sts 1 st s sts sts st ts s t 1 It is ssoitive. We hve the speil element 1 n every element hs n inverse 1 =, 1 =, () 1 = Symmetri group. We hve seen ove tht symmetries of n ojet n e enoe y ijetive mps on finite set. This rises two questions Is it possile to relize ny (strt) group with ijetive group (equivlently, is it true tht strt groups re the symmetries of something?) Is the set of ll ijetions of finite set group? Let us first nswer the seon question. Let [n] = {1, 2,..., n} n onsier the set S n = {σ σ : [n] [n] ijetion} We sy tht the element of S n re permuttions. For exmple S 3 = {123, 132, 213, 231, 312, 321} where we enoe the permuttions y its list of vlues. Tht is using the nturl orer 1 < 2 <... < n, we list the vlues σ(1), σ(2),..., σ(n). For exmple the permuttion 231 is the mp {1, 2, 3} {1, 2, 3} We hve the ientity permuttion I = 123 n S n n given two permuttions σ, π S n we n ompose the two mps n we get permuttion σ π S n. Moreover, for every permuttion σ S n we n fin σ 1 suh tht σ σ 1 = σ 1 σ = I.

7 GRUPOS 7 This gives us nie group n we ll it the symmetri group Representtion y permuttion. Let us now onsier the question of seeing n strt group s group of symmetry. Look t Exmple 1. You n hek tht the mp ; 2341; ; is reliztion of the group C 4. Now ompre this with our strting exmple of rotting the squre n see tht up to hnging the nmes, we hve the sme group of symmetries. The strt group C 4 is relize using permuttion ( suset of S 4 ). {1234, 2341, 3412, 4123} S 4. We nee to mke some oservtions. When we hve {I} S n n is group y itself we sy tht is permuttion sugroup of S n. For group G, if we hve mp ϕ: G S n suh tht ϕ(1) = I n ϕ() = ϕ() ϕ(), then we sy tht ϕ is homomorphism of G. You n hek tht in this se {I} ϕ(g) S n is then permuttion sugroup of S n n we sy tht ϕ(g) is permuttion representtion of G. If moreover G = ϕ(g) then we sy tht ϕ(g) is permuttion reliztion of G. Exerises. Ex.2.1 For C 4 in Exmple 1, fin homomorphism ϕ: C 4 S 5. Fin 5 points on the squre tht re mps to themselves fter rottions n visulize your onstrution of ϕ. Ex.2.2 Put numers on the verties n the sies of the squre use this to efine homomorphism φ: C 4 S 8. Now you see tht the sme group my hve mny ifferent permuttion reliztions. Ex.2.3 Drw the two igonls of the squre n ll them 1 n 2: 1 2 use this to efine homomorphism φ: C 4 S 2. Is this permuttion representtion ut is it permuttion reliztions? Why? Ex.2.4 Consier the group C 2 C 2 in Exmple 2. Fin wy to see it s the symmetry of n ojet. Use tht to give homomorphism ϕ: C 2 C 2 S 4 n give permuttion reliztions of this group. Ex.2.5 Show tht S 3 in Exmple 3 is the sme s S 3 in Setion 2.4. Tht is fin homomorphism ϕ: S 3 S 3 Wht re the permuttions ϕ(s) n ϕ(t)? Cn you fin geometri ojet suh tht S 3 esrie the symmetries of tht ojet? Ex.2.6 Some enumertion. Wht is the rinlity of S n. Tht is how mny permuttions of [n] is there?

8 8 BERGERON Ex.2.7 Tke geometril ojet in R 3 (prism, ue, tetrher,...). Desrie its symmetry. Then, put lel on your ojet to get permuttions reliztion of tht group of symmetries. 3. Group tions n enumertion In the exerises of Setion 2, we hve seen tht to get permuttion reliztion of group we onsier n ojet to visulize the symmetries n then nme some portions of the ojet to get the permuttion. Now I will isuss how to o this systemtilly Ation. An tion of group G on finite set X is mp G X X suh tht - 1.x=x - (gh).x=g.(h.x) Remrk tht the nottion here put emphsis on the ft tht G o something to element of X. Look t exerise 1.3 where we use C 4 = {1,, 2, 3 } s the symmetries of the squre This gives us n tion of C 2 on X = {1, 2}..1 = 2.2 = = = = = 1 We see tht for fix g G, the mp X X given y x g.x is permuttion of X. It is esy to hek tht this mp is invertile sine g 1.(g.x) = (g 1 g).x = 1.x = x. So ny time we hve n tion, we will hve permuttion representtion. When o we know if it is reliztion or not? In the exmple ove we see tht {1, 2 } gives the sme permuttion (the ientity) on X. As well, {, 3 } lso gives the sme permuttion on X. We n prtition e group C 4 into lsses oring to the permuttion they give { } {1, 2 }, {, 3 } This gives us lot of lue on wht hppen. Before we relly o this let me first give some generl propositions. Proposition 4. Given finite group G n sugroup of G. Let (1) g = {gh h }. {g } g G is prtition of G (2) g = for ll g G (3) G is ivisile y n { g g G } = G

9 GRUPOS 9 A sugroup here, s efore, is suset {1} G suh tht is itself group insie G. You see tht (3) follows from (1) n (2). For ny g G we see tht one to one orresponene g h gh g 1 gh = h gh This gives us tht (2) is lwys true. Now (1) nee to e well unerstoo. Wht re wee relly sying? Lets look t C 4 ove n remrk tht = {1, 2 } is inee sugroup. Now we for the set of sets 1 = {1, 2 }; = {, 3 }; 2 = {1, 2 }; 3 = {, 3 }. So when we write {g } g C 4 = { } {1, 2 }, {, 3 } we men o not repet element tht re the sme. So wht (1) is sying is tht g 1 g 2 = g 1 = g 2. Proposition 5. Given finite group G ting on finite set X. Let = {g G g.x = x for ll x X}. (1) is sugroup of G. (2) The tion gives permuttion reliztion if n only if = 1 (3) We n tke X = G n the tion is left multiplition, in this se we get permuttion reliztion. I will leve s n exerise to show (1). If you hve never one tht it is goo to o it. If you hve seen tht efore it is very esy. The item (2) is sutle. When we hve n tion G X X you hve seen in exerise tht we n uil homoporphism ϕ: G S X. For fix g, the permuttion ϕ(g) is given y how g permutes X uner the mp x g.x. Now we hve ϕ(g 1 ) = ϕ(g 2 ) g 1 = g 2. inee g 1.x = g 2.x g2 1 g 1.x = g2 1 g 2.x = x This is true for ll x if n only if g2 1 g 1. Tht is g2 1 g 1 = (sine is sugroup) n thus g 1 = g 2. You see tht we hve s mny permuttions in ϕ(g) s we hve element in { g g G } = G ϕ(g) = so { g g G } = G = G = 1. Now (3) is muh esier. Remrk tht the tion here is G G G the usul multiplition ut we think of G s eing the set we t on. The ft tht it is n tion is ler. If we hve gh = h then it is ler tht g = ghh 1 = hh 1 = 1. so = {1} only.

10 10 BERGERON Remrk 6. The Proposition 5 (3) tell us tht ny strt group n e relize (in t lest one wy) s permuttion group. This is known s Cyley s theorem. Lets o some exmples. Tke C 2 C 2 = {1,,, } (see Exmple 2) n X = {1, 2, 3, 4}. We n efine ϕ(): X X 1.1 = = = = 4 ϕ(): X X 1.1 = = = = 3 ϕ(): X X 1.1 = = = = 3 You see tht ( ϕ() ) 2 ( ) 2 = I, ϕ() = I n ϕ() = ϕ()ϕ() = ϕ()ϕ(). ene this is well efine tion n we o hve homomorphism. Moreover only ϕ(1) = I n we hve permuttion reliztion of C 2 C 2. If inste we tke φ(): X X 1.1 = = = = 3 φ(): X X 1.1 = = = = 3 φ(): X X 1.1 = = = = 4 we see gin tht ( φ() ) 2 ( ) 2 = I, φ() = I n φ() = φ()φ() = φ()φ() n this is ifferent tion of the sme group on X. But this time φ(1) = φ() = I n here we o not hve permuttion reliztion sine = {1, } The ue. We woul like to hve muh igger exmple to ply with. Let us onsier the group B 3 of symmetry of ue pir of opposite verties pir of opposite eges pir of opposite fes (4 suh pirs) (6 suh pirs) (3 suh pirs) This group hs 1 ientity mp, 4 2 rottion of 120 n 240 for eh pir of opposite verties, 6 rottion of 180 for eh pir of opposite eges n 3 3 rottion of 90, 180 n 270 for eh pir of opposite fes. Tht is 24 = This group is strting to e omplite n it woul tke some time to write own the full tle of multiplition. We will soon nme ll its element.

11 GRUPOS 11 To help us, let us first nme ll the omponents of the ue The group B 3 ts on the set 5 6 i l e h C 1 2 k 8 E 7 A B F g j 4 3 f D 8 verties 12 eges 6 fes X = {1, 2, 3, 4, 5, 6, 7, 8,,,,, e, f, g, h, i, j, k, l, A, B, C, D, E, F } For exmple the permuttion orresponing to the rottion of 120 roun the line through the verties 3 n 5 is e f g h i j k l A B C D E F h g k l j e f i B D F A C E It is sometime etter to write tht into isjoint yle nottion. For this we simply follow wht hppen to the element s we iterte the sme symmetry: n we hve yle. We simply write (1 6 8) n we ssume the yle lose up. So the permuttion ove is (1 6 8)(2 7 4)(3)(5)( h j)( g)( k f)(e l i)(a B D)(C F E) The tion we esrie gives us permuttion representtion ϕ: B 3 S 26. This is permuttion reliztion of B 3 (you n see why?). In this permuttion representtion the element ove hs 8 yles of length 3 n 2 yle of length 1. We ll tht the yle struture of the element. Cyle struture will ply very importnt role in Poly Theory. Of ourse if we hnge the permuttion representtion, we get ifferent yle strutures. Exerises. Ex.3.1 Show tht X in Proposition 5 is sugroup. Ex.3.2 Let G = S 3. Fin ll ifferent tions of S 3 on X = {,, }. In eh se esrie the orresponing homomorphism ϕ: S 3 S 3. Desrie the yle struture eh elements of S 3, for eh permuttion representtions you onstrute. Ex.3.3 Look gin t Ex.2.2. Desrie the yle struture of eh element of C 4 for this permuttion representtion. Ex.3.4 D 3, D 4, D 5, D 6 : Stuy the iherl groups. These re the group of symmetries D 3 = S 3 D 4 D 5 D 6

12 12 BERGERON where here we n rotte the figure n we n lso tke it off the tle, flip it, n put it k (refletion). Ex.3.5 A 4 : stuy the group of rottion symmetries of tetrher orits n ounting points. When we stuie the tions of B 3 on the set X of verties, eges, fes of the ue you noties tht symmetry sens vertex to nother vertex, n ege to nother ege n fe to fe. A symmetry preserve the types. This le us to efine the notion of orit of point x X. Before we strt let us onsier the smller exmple. Lets look t the symmetries of We see tht ny symmetry nee to fix the point 1. One the other hn, we n lwys fin symmetry tht will sen ny point of {2, 3, 4, 5} to nother point of {2, 3, 4, 5}. We sy tht the group of symmetry ting on the points {1, 2, 3, 4, 5} hs two orits: {1} n {2, 3, 4, 5}. We see tht the orits enoe the nture of the points on our ojets. So the orits gives us interesting informtion out the symmetries of the ojets. We will now efine orits n see how to ount points within n orit. Given group G ting on set X, we sy tht the orit G.x of point x X is the set G.x = {g.x g G} X. This is the set of points of X we n reh from x vi the tion of G. We sw ove exmple of orits. ow to ount the numer of points insie the orits of x? Tht is n we fin nie formul for G.x. We hve lrey enountere this prinipl. Let St(x) = {g G g.x = x} G As you see G.x X n St(x) G. The set St(x) is in ft sugroup of G. This gives us nie wy to ompute G.x: Proposition 7 (Lgrnge s theorem). For x X, we hve G.x = 4 G St(x) The ie is gin to onstrut n expliit ijetion etween the two sets G.x {gs g G}

13 GRUPOS 13 where S = Stg x (G). For ny y G.x, there re mny possile g G suh tht y = g.x. α: G.x {gs g G} y gs where g is ny suh tht g.x = y Of ourse we hve to mke sure tht it is well efine. Tht is if y = g.x = h.x h 1 g.x = x gs = hs The mp in the other iretion is given y β : {gs g G} G.x gs g.x The two mps onstrut orresponene etween the two sets. If we go k to our smll exmple. The group of symmetries of is C 4 (you see this?). We notie 2 orits:{1} n {2, 3, 4, 5}. For 1, we see tht every symmetry fix 1, ene C 4 St(1) = 4 4 = 1 For ny of the points 2, 3, 4 or 5, only the ientity symmetry fix suh point. We get If we look t Setion 3.2 the group B 3 C 4 St(2) = 4 1 = pir of opposite verties pir of opposite eges pir of opposite fes (4 suh pirs) (6 suh pirs) (3 suh pirs)

14 14 BERGERON ts on the set of verties, eges n fes of the ue: 5 6 i l e h C 1 2 k 8 E 7 A B F g j 4 3 f D 8 verties 12 eges 6 fes We see tht extly three symmetries fix the vertex 1. ene B 3.1 = B 3 = = 8. Similrly only two symmetries fix the ege, hene B 3. = B 3 2 Finlly four symmetries fix fe hene = 24 2 = 12. B 3.A = B 3 = = 6. This is useful to ount verties, eges, fes, of ojet with lot of symmetries. It is even more useful for things we nnot see (like in higher imensions). Exerises. Ex.3.6 Show tht St(x) is sugroup. Ex.3.7 In Proposition 7, show tht α β = I n β α = I. Ex.3.8 Count the numer of points, eges n fes of the following ojets using Proposition 7. Ex.3.9 The ojet re not lwys fully symmetri n points, eges n fes re roken into smller orits (ifferent kin of points, eges, fes). Unerstn the orits to ount the numer of points, eges n fes of the following ojets using Proposition 7 n group of symmetries.

15 GRUPOS 15 Ex.3.10 Count the numer of verties, eges, 2-fes, 3-fes,..., of herperue of imension n. [hint. fin group of symmetry of the hyperue tht ontins extly one orit for eh type of fes] 3.4. ounting numer of orits. At this point I hope you re onvine tht using group theory is powerful to ount things relte to symmetry. Our next step is to ount the numer of orits. As we hve seen ove, we n see this s ounting the numer of type of points we hve. Lets look gin t C 4 ting on the squre pyrmi n onsier its verties, eges n fes. g f C 5 h E 4 e A B 2 3 D The group C 4 is ting on the set Y = {1, 2, 3, 4, 5,,,,, e, f, g, h, A, B, C, D, E}. This set eompose into isjoint orits. We enote this s { } Y/C 4 = {1}, {2, 3, 4, 5}, {,,, }, {e, f, g, h}, {A, B, C, D}, {E} 1 We see tht the numer of orits ount the ifferent types of things. As for the numer of element in n orits, there is very eutiful formul to ount the numer of orits of n tion. Let G ts on set X n efine. F ix(g) = {x X g.x = x} Proposition 8 (Burnsie s lemm). For G ting on set X, we hve X/G = 1 F ix(g) G This is not hr to see fter few mnipultions of the onepts we hve seen F ix(g) = g G g G x X = x X g G δ x,g.x = x X g G δ x,g.x St(x) G = G.x x X The lst equlity is nie trik. = G x X 1 G.x = G X/G

16 16 BERGERON So if we wnt to ount the numer of orit of n tion, we simply nee to fin the rinlity of the set F ix(g) for eh g G. Lets look gin t C 4 = {1,, 2, 3 } ting on Y ove. We fin tht F ix(1) = Y F ix() = {1, D} F ix( 2 ) = {1, D} F ix( 3 ) = {1, D} ene Y/C 4 = 1 24 ( ) = 4 4 = 6 I Alwy mrvel t the ft tht this works so well. Exerises. Ex.3.10 Look t the tions you efine in Ex.3.8 n Ex.3.9 n use Proposition 8 to ount the numer of orits Ex.3.11 Consier the tions of C 4 = {1,, 2, 3 } on the set Z = {1, 2, 3, 4, 5,, } s the five verties {1, 2, 3, 4, 5} n two igonl segment {, } piture ellow. Use Proposition 8 to ount the numer of orits Coloring n Poly theory Suppose your frien Lur wnts to rete gme with piees tht re tiles like The gme will e to tile 2 3 or so tht olor mthes where the tile touh. A goo gme will en with tiling like this:... But she woul like to know how mny kin of tile we nee, n how mny ifferent goo gme will e possile. Lets try to help oloring of n tion. When we hve finite set X, oloring of X with ertin olors is funtion w : X {olors}. Inee if you hve set like X = {1, 2, 3} n your hve two olors re n lue, you n o 8 ifferent olorings. Nmely 123; 123; 123; 123; 123; 123; 123; 123;

17 GRUPOS 17 This is the sme s funtions {1, 2, 3} {re, lue}. If C is set of olor, then we enote y C X the set of ll oloring of X y C. Tht is the set of ll funtions C X = {w : X C funtions} For our exmple ove with X = {1, 2, 3} n C = {re, lue} we get C X = {123, 123, 123, 123, 123, 123, 123, 123} Now, when group G ts on X, then G lso ts on the lrger set of ll olorings C X in nturl wy. Inee, if w C X, let (g.w)(x) = w(g 1.x) n it works. ere g.w is new funtion uile from w. we nee to use the inverse of g in this efinition so tht the tehnility of the efinition of tion works for the tion on C X. ((hg).w)(x) = w((hg) 1.x) = w((g 1 h 1 ).x) = w(g 1.(h 1.x)) = (g.w)(h 1.x)) = (h.(g.w))(x) I put this here to show you the tehnility in showing tht (hg).w = h.(g.w). Try to unerstn eh equlity? This tehnility is importnt for things to work ok, ut will not ply ig role in the en. Lets work with our exmple n the group of permuttion S 3 ting on X = {1, 2, 3} (it permutes the numers). Rell tht S 3 = {123, 132, 213, 231, 312, 321} n eh σ S 3 is permuttion σ : X X n we hve n tion. Let C = {re, lue} n onsier w C X. For exmple 123 is the funtion w(1) = re, w(2) = lue n w(3) = re. Let σ = 231 n hek tht σ 1 = 312. So the new funtion σ.w is efine s follow: (σ.w)(1) = w(σ 1 (1)) = w(3) = re (σ.w)(2) = w(σ 1 (2)) = w(1) = re (σ.w)(3) = w(σ 1 (3)) = w(2) = lue So σ.w = 123. If we look t the orits of X uner the tion of S 3 we see tht there is only one orit (X is single orit). But if we look t the orits of the tion on C X very ifferent piture emerge: C X /S 3 = { {123}, {123, 123, 123}, {123, 123, 123}, {123} } We n represent this on tringle S 3 = D 3 is the symmetry of the tringle llowing flips n rottion) The set C X /S 3 n e piture s { { } { C X /S 3 =,, X = } { } { } },,,,, As you n see, eh orit represent one possile type of nekle with three es n two ifferent olor. So we n use Proposition 8 to ount the numer of possiilities. First we

18 18 BERGERON nee to figure out wht is F ix(g) for eh symmetry on the set C X. Let us first look t our exmple. For I = 123 S 3 it is esy s 123 lwys fix ll oloring F ix(i) = C X = F ix(123) = C X = 2 3 = 8 Then we look t the other permuttions: F ix(132) = {{123, 123, 123, 123} = F ix(132) = 4 = 2 2 F ix(213) = {{123, 123, 123, 123} = F ix(213) = 4 = 2 2 F ix(231) = {{123, 123} = F ix(231) = 2 = 2 1 F ix(312) = {{123, 123} = F ix(312) = 2 = 2 1 F ix(321) = {{123, 123, 123, 123} = F ix(321) = 4 = 2 2 Using Proposition 8 we get the right nswer: C X /S = ( ) = 6 6 = 4 You notie tht F ix(g) is lwys power of C = 2, this is not n ient. Proposition 9. Let G t on X. As ove we get n tion of G on C X. For g G, we hve tht w F ix(g) w(x) = w(g.x) = w(g 2.x) = w(g 3.x) =... for ll x X Tht is, w is onstnt over the yles of G for the tion on X Be reful here, we hve two tions to onsier. The tion of G on X n the tion of G on C X. We re intereste for the set F ix(g) for the tion of G on C X, ut to ompute it, we use the tion of G on X. We see here tht we re intereste in the isjoint yles of g on X. Rememer tht the yle of g G re otine y looking wht hppen when we pply suessive power of g to the element of X. x g.x g 2.x until we get t x. If w is fixe y g, tht is (g.w)(x) = w(x), then pplying g 1 oth sie we get w(x) = (g 1.w)(x) = w(g.x) Applying g 1 gin to the equlity ove we get w(g.x) = (g 1.w)(x) = (g 2.w)(x) = w(g 2.x) n we n ontinue like tht to show one of the implition of the Proposition 9. In yle nottion, the permuttion 213 is (1 2)(3). Look gin t F ix(213) n inee ll oloring in the set hve the sme olor for the yle (1 2) n for the yle (3). On the other hn, 231 is single yle (1 2 3) n only 2 oloring re onstnt on this yle. I leve it to you to see the onverse of the theorem ove. Wht we hve seen so fr is tht when G ts on C X we hve F ix(g) = C y X (g) where y X (g) is the yle eomposition of g for the tion on X. Theorem 10. Given G ting on X. We hve tht G ts on C X. Let = C, C X /G 1 = y X(G) G You unerstn tht this follow from Proposition 8 n Proposition 9. g G

19 GRUPOS 19 Exerises. Our frien Lur tht just lerne out this theorem is pretty sure we n use tht to solve her question. Lets o tht together...first some wrm up. C Ex.4.1 Let B 3 t on the ue A B F n let X = {A, B, C, D, E, F } e the set E D of fes. Consier C = {re, lk, lue, green} n B 3 ting on the oloring C X. Now nswer the first question we h: how mny ifferent ies with four olors n we onstrut. Ex.4.2 Now nswer the seon questions we hve seen: how mny nekles of 8 es of 3 ifferent kin n we mke? [hint: use D 8 ting on n otogone] Ex.4.3 Solve the two questions of Lur. Keep in min tht the tiles re olore on one sie only Poly Theory. Toy, Lur si tht it is ie to hve the sme olor on single tile more thn twie. It woul mke the gme more interesting if we o not llow the sme olor more thn twie on eh tile. But now ll our ounting nee to e reone? She is not sure we n solve her prolem nymore. But she her tht hemist nme Poly h similr prolem n foun solution. The moleule C 6 4 Br 2 is 6 toms of ron (C) rrnge on n hexgon n the toms of hyrogen () n Bromine (Br) re tthe in irulr wy roun it. But in nture there is more thn one possiility n the moleule my hve ifferent ehvior epening of the onfigurtion Br Br Br Br e wnte s well to unerstn the possile onfigurtion of more omplex moleule n for this he neee to refine the ounting priniple n e le to ount oloring with speifie type of olors (how mny of eh olor). For this we nee to refine our ounting reoring whih olors ws use. Lets follow n exmple. Rell tht D 6 = {1, r, r 2, r 3, r 4, r 5, s 1, s 2, s 3, s 4, s 5, s 6 } t on the hexgon s epite ellow n this efine n tion on the verties X = {1, 2, 3, 4, 5, 6}: Br 4 5 Br s 6 r 2 r s 5 r s 4 r 4 r5 s 3 s 2 s 1 2 1

20 20 BERGERON We hve seen in Teorem 10 tht the numer of yles for eh element of the group ply n importnt role in ounting olorings. If we wnt to refine our ounting, we now hve to stuy extly the yle type of eh element of D 6 using the permuttion representtion with respet to X. For this we will use monomil to enoe the yle type of eh element. For exmple, if n element hs 2 yles of length 1, 3 yles of length 2 n 1 yle of length 4, then we will enoe this with monomil x 2 1x 3 2x 4 In generl the exponent of x i is the numer of yles of length i in our permuttion representtion of the element. For D 6 ting on X = {1, 2, 3, 4, 5, 6} ove we hve elements of D 6 Cyle type yle monomil 1 (1)(2)(3)(4)(5)(6) x 6 1 r ( ) x 6 r 2 (1 3 5)(2 4 6) x 2 3 r 3 (1 4)(2 5)(3 6) x 3 2 r 4 (1 5 3)(2 6 4) x 2 3 r 5 ( ) x 6 s 1 (1)(4)(2 6)(3 5) x 2 1x 2 2 s 2 (1 2)(3 6)(4 5) x 3 2 s 3 (2)(5)(1 3)(4 6) x 2 1x 2 2 s 4 (2 3)(1 4)(5 6) x 3 2 s 5 (3)(6)(2 4)(1 5) x 2 1x 2 2 s 6 (3 4)(2 5)(1 6) x 3 2 We now put ll this informtion together in polynomil (the sum of ll the monomil we proue P D6,X(x 1, x 2, x 3, x 4, x 5, x 6 ) = 12( 1 ) x x 6 + 2x x x 2 1x 2 2 We ll this the yle inex of the permuttion representtion of D 6. The ftor 1 is there 12 s it will ply role in the finl result. This polynomil oes not use the vrile x 4 n x 5 in this se. In generl, onsier group G ting on finite set X where n = X. The yle inex of the permuttion representtion of G on X is P G,X (x 1, x 2,..., x n ) = 1 x y X,1(g) 1 x y X,2(g) 2 x y X,n(g) n G g G where y X,i (g) is the numer of yle of length i in the permuttion representtion of g with X. We remrk right wy tht if we olor the tion with = C olors, Theorem 10 n e written s follow C X /G = PG,X (,,..., ). You see this?

21 GRUPOS 21 For D 6 P D6,X(,,,,, ) = 1 ( ) = 1 ( ) Now the exponent of in the right hn sie, ount extly the numer of yle s in Theorem 10. If we wnt to refine our ounting we n use the yle inex polynomil tht knows out the size of eh yles. Rememer tht oloring must respet the yles of n element in orer to e fixe y this element. The vrile x i in the yle inex polynomil ontriute yle of length i. Inste of sustituting x i we o n try to keep trk of whih olor ws use n how mny time. The wy to enoe tht with generting funtion is to o x i i 1 + i i We unerstn the right hn sie s hoosing (the + ) olor from { 1, 2,..., } n using it i-times. Lets look wht hppen when we o this with P D6,X hving two olors {, }: P D6,X( +, 2 + 2, 3 + 3, 4 + 4, 5 + 5, ) = 1 ( ( + ) 6 + 2( ) + 2( ) 2 + 4( ) 3 + 3( + ) 2 ( ) 2) 12 Lets expn eh monomil: ( + ) 6 = = ( ) 2 = ( ) 3 = ( + ) 2 ( ) 2 = ( )( ) = Eh monomil expnsion ount the numer of oloring fixe y n element of D 6. For exmple, s 5 D 6 hs yles (3)(6)(2 4)(1 5). We know tht oloring will e fixe one we hoose olor for 3, 6, {2, 4} (sme olor) n {1, 5} (sme olor). The yle monomil is x 2 1x 2 2. When we sustitute x 1 + n x we proue ll the hoies of hoosing olor (+)(+)( )( ) for 3, 6, {2, 4} n {1, 5} respetively. So the expnsion Gives us refinement of ounting the olorings tht re fixe y s 5 tking into ount how mny times olor is use. For exmple the oeffiient 3 in front of 4 2 mens there re 3 oloring tht will e fixe y s 5 with four s n two s. When we expn ( + )( + )( )( ) we n otin 4 2 in three ifferent wys: ( + )( + )( )( ) ( + )( + )( + 2 )( ) ( + )( + )( )( + 2 ) Eh wys of getting the monomil 4 2 orrespon to extly one hoie of oloring tht is fixe y s 5 seleting whih olor we put on eh yles 3, 6, {2, 4} n {1, 5} respetively. They re (liste in the sme wy we got the monomils ove)

22 22 BERGERON where here I put re for s n lue for the s. So if we ontinue our expnsion of P D6,X P D6,X( +, 2 + 2, 3 + 3, 4 + 4, 5 + 5, ) = 1 12 ( ) 2 = When we o this, in front of eh monomil we use Proposition 8 for eh istriution of olor seprtely. ene P D6,X = gives us the istriution of possile oloring epening on how mny times eh olor is use. Inee Poly s question is nswere s there is extly three possile oloring of n exgon using 2 Br s n 4 s. It is the oeffiient of 2 4 in P D6,X fter the sustitution x i i + i. We sum up ll our generl unerstning s follow Theorem 11 (Poly). Given group G ting on set X. Let = C n onsier the tion of G on C X. () Compute P G,X n P G,X (,,..., ) = C X /G () Compute the sustitution x i i 1+ i 2+ + i in P G,X. The oeffiient of in the resulting polynomil is the numer of oloring of X with olor i use i times. Exerises. Now we will try to solve Lur s question. Agin, lets wrm up little it efore Ex.4.4 For eh permuttion representtion we hve enounter this week. () Give the yle inex polynomils () Count the numer of oloring using 3 olors () Count the numer of oloring using 3 olors, ut not repeting ny olor in oloring () Count the numer of oloring using 3 olors, ut hving one olor repete one. Ex.4.5 Count the numer of wy to onstrut the Lur s tile with no olors use more thn twie. Ex.4.6 Cn you ount the numer of Lur s goo 2 3 tiling using only the tiles in Ex.4.6? Ex.4.6 Cn you o oloring in higher imension? (oloring of polytopes)? 1 2 1

23 GRUPOS 23 Nntel Bergeron, Deprtment of Mthemtis n Sttistis, York University, Toronto, Ontrio M3J 1P3, Cn E-mil ress: ergeron@mthstt.yorku. URL: