arXiv:1507.06587v2 [math.CO] 11 Aug 2015ChromaticfunctorsofgraphsMasahikoYoshinagaAugust12,2015AbstractFinitegraphsthathaveacommonchromaticpolynomialhavethesamenumber of regular n-colorings. Anatural questionis whetherthereexistsanatural bijectionbetweenregular n-colorings. Wead-dressthisquestionusingafunctorial formulation. LetGbeasimplegraph. Thenfor eachset Xwe canassociate aset of X-colorings.Thisdenesafunctor,chromaticfunctorfromthecategoryofsetswithinjectionstoitself. Therstmainresultveriesthattwonitegraphs determineisomorphicchromatic functors if andonlyif theyhavethesamechromaticpolynomial.Chromaticfunctorscanbedenedforarbitrary, possiblyinnite,graphs. Thisfactenablesustoinvestigatefunctorial chromaticthe-oryforinnitegraphs. WeprovethatchromaticfunctorssatisfytheCantor-Bernstein-Schr oder property. We alsoprove that countableconnectedtreesdetermineisomorphicchromaticfunctors. Finally,wepresentapairofinnitegraphsthatdeterminenon-isomorphicchro-maticfunctors.1 IntroductionAsimple graphG=(V, E) consists of aset of vertices V andaset ofunorientededgesE2Vthatconnect twovertices. Aregularn-coloringof Gisamapc: V [n] ={1, 2, . . . , n}suchthat{v, v}Eimpliesc(v) = c(v).It isknownthat, for anitegraph G = (V, E),thereexistsapolynomial(G, t) Z[t] suchthat thenumber of regular n-coloringsequals (G, n)([2, 5]), where (G, t) is called the chromatic polynomial of G. The minimumDepartment of Mathematics, Hokkaido University, North 10, West 8, Kita-ku, Sapporo060-0810,JAPANE-mail: yoshinaga@math.sci.hokudai.ac.jp1positive integer n satisfying(G, n) = 0 is called the chromatic number of Ganddenotedby(G).TwonitegraphsG1andG2aresaidtobechromaticallyequivalentif(G1, t) = (G2, t) ([2]). Non-isomorphic graphs may be chromatically equiv-alent, forexample, if G=(V, E)isaniteconnectedtree, then(G, t)=t(t 1)#V 1. Thus, connectedtreesthathavetheprescribed#V arechro-matically equivalent. Therefore, it is a natural question to ask whether thereexistsacertainbijectionbetweenthesetsofregularn-coloringsofchromat-icallyequivalentgraphs.To address the above question, we introduce a functorial formulation. LetXbe a set. We considerXto be a set of colors, and dene the set of regularX-coloringby(G, X) := {c : V X| {v1, v2} E= c(v1) = c(v2)}. (1)LetG1andG2begraphs. If#(G1, X)=#(G2, X), thenthereexistsabijectionXX: (G1, X) (G2, X).However, if weconsider(G1, X)asafunctorwithrespecttoX, wemayimposefunctoriality. Thenatural bijectionshouldbeformulatedasaniso-morphism : (G1, ) (G2, ) (2)of functors(chromaticfunctors). If thechromaticfunctorsareisomorphicinthe sense of (2), thenbysubstitutingX=[n], we canconclude thatthegraphsarechromaticallyequivalent. Themainresultassertsthattheconversealsoholds.Theorem1.1. (Theorem2.1)Let G1and G2be nite graphs. Then they are chromatically equivalent if andonlyifassociatedchromaticfunctorsareisomorphicasfunctors(2).The advantage of chromatic functors over polynomials is that we candene chromatic functors also for graphs with innite vertices. We say graphsG1andG2, whicharepossiblyinnitegraphs, arechromaticallyequivalentifassociatedchromaticfunctorsareisomorphicinthesenseof(2). Wewillshow several results on chromatic functors for innite graphs in 3. The rstresultisthefollowingCantor-Bernstein-Schr oderproperty.Theorem1.2. (Theorem3.3)LetG1andG2begraphs. SupposethereexistsurjectivegraphhomomorphismsG1 G2andG2 G1. ThenG1andG2arechromaticallyequivalent.2Asacorollarytotheabovetheorem, wecanshowthatconnectedun-bounded countable trees determine isomorphic chromatic functors, which canbeconsideredasageneralizationoftheresultonthechromaticpolynomialofconnectednitetrees.In3.3, wewill characterizeisomorphismclasses of chromaticfunctorsfor innite graphs withnitechromatic numbers. Roughlyspeaking, theisomorphismclassof thefunctor(G, )ischaracterizedbythechromaticnumber (G) and cardinality #(G, [(G)]) if G is a countable innite graphwith(G) < (Theorem3.10).If(G) = ,thechromaticfunctorsdetectmoresubtleinformation. In3.4,weexhibittwographsG1andG2whichsatisfy(G1) = (G2),#(G1, [(G1)]) = #(G2, [(G2)]),however,G1andG2arenotchromaticallyequivalent.2 Chromaticfunctors2.1 DenitionsandthemainresultLetG1=(V1, E1)andG2=(V2, E2)besimplegraphs. Ahomomorphism : G1G2is a map : V1V2suchthat {v1, v2} E1={(v1), (v2)}E2. Ahomomorphism: G1G2issaidtobesur-jectiveif themapof vertices : V1V2is surjective. Wedenotethecategoryof graphsandhomomorphisms, orsurjectivehomomorphisms, byGraph,orGraphsurj.Let G = (V, E) be a graph and Xa set. Dene the set (G, X) XVby(G, X) := {c : V X| {v1, v2} E= c(v1) = c(v2)}. (3)Thisdenesabi-functor : GraphSetinjSet, (G, X) (G, X), (4)whereSet(resp. Setinj)isthecategoryofsetsandmaps(resp. injections).Notethat (G, X) is covariant withrespect toXandcontravariant withrespect toG. If werestrict thefunctor toGraphsurj, thenwehaveabi-functor : GraphsurjSetinj Setinj.3These functors canalsobeformulatedinterms of graphhomomorphisms([4]). LetusdenotethecompletegraphwiththesetofverticesXbyKX.Then,thereisanaturalisomorphism(G, X) HomGraph(G, KX).LetGbeasimplegraph. ThenGdeterminesafunctor(G, ) : SetinjSetinj,whichwecallthechromaticfunctorassociatedtoG. Ifwerestrictthe chro-maticfunctortothecategorySetbijofsetsandbijections,weobtain(G, ) : Setbij Setbij.Hence, thecardinality#(G, X)dependsonlyonthecardinality#X. Asmentioned in 1, for a nite graph G, there exists a polynomial (G, t) Z[t]suchthat#(G, [n]) = (G, n),for n>0, whichis calledthechromaticpolynomial of G. If (G1, ) (G2, ) as functors SetinjSetinj, then obviously, (G1, [n]) = (G2, [n]).In particular, they have the same chromatic polynomial. The following resultassertsthattheconverseisalsotrue.Theorem2.1.Let G1and G2be nite graphs. Then the following are equiv-alent.(i) (G1, t) = (G2, t).(ii) (G1, ) (G2, )asfunctorsSetinj Setinj.Theorem2.1willbeprovedlaterin2.2.Denition2.2. Let G1and G2be (possibly innite) graphs. G1and G2aresaidtobechromaticallyequivalent if(G1, ) (G2, ),asfunctorsSetinjSetinj.Notethatfornitegraphs,byTheorem2.1,thisdenitionisequivalenttotheclassicaldenitionofchromaticequivalence[2].Remark2.3. Thefunctorialformulationinthispaperisrelatedtotheno-tion of combinatorial species[3, 1], which considers a combinatorial objectas afunctor SetbijSetbij. However, we consider thecategorySetinjinsteadof Setbij. This requires stronger functorialitythancombinatorialspecies.42.2 StablepartitionsofchromaticfunctorsLetG = (V, E)beasimplegraph.Denition2.4. = {| }iscalledastablepartitionofGif :=

= V ,and{v1, v2} E, v11, v22=1=2. (Inother words,v1, v2 = {v1, v2}/ E.)WedenotethesetofallstablepartitionsbySt(G) = { | isastablepartitionofG}.Example 2.5. Let Xbe aset. There are twoextremal partitions: thetrivial partition {X} and the nest partition X:= {{x} | x X} into singleelements. The stable partition St(KX) of the complete graph KXconsists ofasinglepartitionSt(KX) = {

X}.Proposition2.6. LetG = (V, E)bea(possiblyinnite)graph. Then(G, )

St(G)(K, ). (5)Proof. LetXbeasetand c : V XaregularX-coloring. Thecoloringcdeterminesastablepartitionc:= {c1(x) | x X, c1(x) = },ofG. Thispartitiondeterminesasurjectivegraphhomomorphismc: G Kc, (6)which sends v Vto c1(c(v)) c. By construction, the coloring c : G KXfactorsthrough(6). Therefore,wehaveamap(G, X)

St(G)(K, X).Conversely, foranypartitionSt(G), thereisasurjectivegraphhomo-morphism G K. Because is contravariant with respect to the graphs,wehave(G, X) (K, X). Thisinduces(G, X)

St(G)(K, X).Itisapparentthatthesemapsareinversetoeachother.5Now,letusprove(i)=(ii)ofTheorem2.1. LetG = (V, E)beanitegraph. Thenthereareonlynitelymanystablepartitions. WedecomposeSt(G)asSt(G) =

k1Stk(G),where,Stk(G) = { St(G) | # = k}.UsingProposition2.6,wehave(G, )

k1(Kk, )#Stk(G). (7)Thisimplies(G, t) =

k1#Stk(G) (Kk, t). (8)Recall that (Kk, t) =t(t1) . . . (tk+1). Hence, the polynomials(Kk, t), (k1)arelinearlyindependent. Therefore, from(8), thenum-ber#Stk(G)isuniquelydeterminedbythechromaticpolynomial(G, t).NowassumethattwographsG1andG2satisfy(G1, t)=(G2, t). Itfollows that #Stk(G1) =#Stk(G2) for anyk 1. Thenusing(7), wemayconcludethattheassociatedchromaticfunctorsareisomorphic. ThiscompletestheproofofTheorem2.1.2.3 UniquenessofstablepartitionsWeprovetheuniquenessofthestablepartition. ThefollowingisessentiallytheYonedalemma.Lemma2.7. LetXandY besets. Supposethat : (KX, ) (KY, )isanisomorphism of functors. Thenisinducedfromabijection: YX,thatis, = .Proof. We denote the image of idX (KX, X) by the mapX: (KX, X)(KY, X)by:=X(idX). Similarly, set:=1Y(idY )(KX, Y ). Letc(KX, Z) be an arbitrary coloring. By denition, c : X Zis an injection.Bythecommutativityofthefollowingdiagram,(KX, X) X(KY, X)c

c(KX, Z) Z(KY, Z),6wehave Z(c) =Z(c(idX)) = c( X(idX)) = c() = c = (c).Similarly, we have1= . Using the same diagram with Z= Yand c = ,we have = idY . Similarly, = idX. Thus,and are bijective.LetX={x| xX}andY ={y| yY }bepartitionsof

X:=

xX xand

Y:=

yYyrespectively. ThenXdeterminesafunctorFX:=

xX(Kx, ) : Setinj Setinj.Supposethereexistsabijection :

X

Y thatpreservespartitions,thatis,therestriction|xinducesabijection|x: x yforsomey Y .Thenclearlyinducesanisomorphismoffunctors : FX FY.Thenextresultassertsthattheconversealsoholds.Proposition2.8. Let : FX FYbeanisomorphismoffunctors. Thenthereexistsabijection :

X

Y thatpreservespartitions.Proof. Letx0 X. Considertheidentitymapidx0: x0 x0asacoloringidx0(Kx0, x0). Thenthereexistsauniquey0Y suchthat(idx0)(Ky0, x0).Letc(Kx0, Z)beanarbitrarycoloringofKx0. Sincec:x0Zisaninjection,wemayconsiderthefollowingcommutativediagram.(Kx0, x0) x0 yY(Ky, x0)c

c(Kx0, Z) Z yY(Ky, Z)By the commutativityof the diagram, wehaveZ(c) = c((idx0)), whichiscontained in (Ky0, Z). Hence, induces a homomorphism : (Kx0, ) (Ky0, ). Itisanisomorphism, hencebyLemma2.7, inducedfromabi-jectionx0y0. ThisenablesustoconcludethattheisomorphismisinducedfromtheisomorphismofpartitionsX Y .73 Chromatictheoryforinnitegraphs3.1 RelativeCantor-Bernstein-SchrodertheoremIn this section, we prove the following relative version of the Cantor-Bernstein-Schr oder(CBS)theorem.Theorem3.1. Let f: X1X2andg: Y1Y2bemapsof sets. Leti1, i2, j1,andj2beinjectionsthatmakethefollowingdiagramcommutative.X1i1Y1f

gX2i2Y2X1j1Y1f

gX2j2Y2(9)AssumethatforanysubsetZ X2andW Y2,itholdsthatg1(i2(Z)) = i1(f1(Z)), f1(j2(W)) = j1(g1(W)). (10)Thenthereexistbijectionsr: X Y( = 1, 2)suchthatthefollowingdiagramiscommutative.X1r1Y1f

gX2r2Y2(11)Remark3.2. Theconditions(10)aresatisedifthediagrams(9)areberproducts.Proof. DenethesubsetCn( = 1, 2andn 0)byC0:= X \ j(Y),Cn+1:= j(i(Cn)),andsetC:=

n=0Cn. Letusdenethemapr: X Y( = 1, 2)asfollows.r(x) =i(x), ifx C,j1(x), ifx/ C.Itiswellknownthatr: X Yisbijective. Itremainsforustoprovecommutativityr2 f= g r1. (12)8First,letusprovef1(Cn2) = Cn1byinductiononn. Inthecasen = 0,sinceX2= C02 j2(Y2),wehaveX1= f1(X2) = f1(C02) f1(j2(Y2))= f1(C02) j1(g1(Y2))= f1(C02) j1(Y1).Hence,f1(C02) = C01. Furthermore,f1(Cn+12) = f1(j2(i2(Cn2)))= j1(g1(i2(Cn2)))= j1(i1(f1(Cn2)))= j1(i1(Cn1))= Cn+11.Thus, weobtainf1(C2)=C1. Itfollowsthecommutativityrelation(12)fromthatof(9).3.2 CBSpropertyforchromaticfunctorsInthissection,weprovethatchromaticfunctorssatisfytheCBSproperty.Theorem3.3. Let G1andG2be(possiblyinnite)graphs. Supposethatthere exist surjective graph homomorphisms : G1G2, and: G2G1.ThenG1andG2arechromaticallyequivalent.Proof. Letf:XY beaninjectionofsets. Thenwehavethefollowingcommutativediagrams.(G1, X)(G2, X)f

f(G1, Y )(G2, Y )(G1, X)(G2, X)f

f(G1, Y )(G2, Y )(13)Note that all maps in (13) are injective. It is easyto see that both diagramsareberproducts. ApplyingTheorem3.1(seealsoRemark 3.2),thereexistbijectionsrXandrYsuchthatthefollowingdiagramiscommutative.(G1, X)rX(G2, X)f

f(G1, Y )rY(G2, Y )(14)9Thesystemofbijections{rX}determinesanisomorphismoffunctorsr: (G1, ) (G2, ).Thiscompletestheproof.Remark3.4. Wecan alsoprovethepreviousresultusingstablepartitions.Namely, agraphsurjectivehomomorphismG1G2induces aninjectionSt(G1) St(G2)foranycardinality. Similarly, G1 G2inducesaninjectionSt(G1) St(G2). Then the CBS Theorem concludesthat thereexists a bijection St(G1) St(G2) for any cardinality . Then it followsfromProposition2.6thatG1andG2arechromaticallyequivalent.Denition3.5. Thenatural treeTN= (V, E)(Figure1)isdenedbyV=N = {0, 1, 2, . . . , n, . . . },andE= {{n, n + 1} | n N}.s s s s s s q q q0 1 2 3 4 5Figure1: ThenaturaltreeTNProposition3.6. LetG=(V, E)beaconnectedcountablegraph. Thenthereexistsasurjectivegraphhomomorphism : TNG.Proof. Since V is countable andGis connected, there exists asequence{vi0, vi1, . . . , }of verticessuchthat{vis, vis+1}Eand{vis}sNcovers V .ThenN s vis Vinduces a surjective graph homomorphism : TNG.LetG = (V, E)beaconnectedgraph. ThenthesetVofverticesadmitstheadjacencymetric,denotedbydG(, ). ThediameterofGisdenedbysupx,yVdG(x, y).ThegraphGissaidtobeunboundedifthediameteris.Theorem3.7. Let G=(V, E) beaconnectedcountableunboundedgraphthat has nocycle of oddlength(inother words, (G) =2). ThenGischromaticallyequivalenttothenatural treeTN.10Proof. Let p V . Thenbythe assumptionthat there are nocycles ofoddlength, themapdG(p, ) : V Ninduces agraphhomomorphismdG(p, ): GTN. Theunboundednessof Gimpliesthatthehomomor-phism dG(p, ) : G TNis surjective. Proposition 3.6 and CBS-type result(Theorem3.3)concludethatGandTNarechromaticallyequivalent.Corollary 3.8.Connected countable unbounded trees are chromatically equiv-alenttoTN.3.3 InnitegraphswithnitechromaticnumbersDenition3.9. Thenatural wheel WN= (V, E)(Figure2)isdenedbyV=N = {0, 1, 2, . . . , n, . . . },andE= {{0, n} | n > 0}.s s s s s s q q q0 1 2 3 4 5Figure2: ThenaturalwheelWNThegraphWNisacountableconnectedtreewithdiameter2. SinceWNhasanitediameter,wecannotuseCorollary3.8. However,inthissection,wewillprovethatWNisalsochromaticallyequivalenttoTN.Let G = (V, E) be a countable innite graph (we do not make any assump-tionsaboutdiameterandconnectivity). Wealsoassumethatthechromaticnumber(G)isnite. Forinstance,thegraphWnsatisestheseconditions((Wn)=2). Thenweprovethattheisomorphismclassof thechromaticfunctor(G, )isdeterminedbythecardinality#(G, [(G)]).Theorem3.10. LetG1=(V1, E1)andG2=(V2, E2)becountableinnitegraphswiththesamenitechromaticnumber(G1) = (G2) = n. Supposethat#(G1, [n]) = #(G2, [n]),(possiblyinnitecardinality). ThenG1andG2arechromaticallyequivalent.11Proof. We shall use stable partitions of chromatic functors (Proposition 2.6).Itissucienttoshowthat#Stk(G1) = #Stk(G2), (15)forallk 0. Since#(G, [k]) =k

i=1

ki

i! #Sti(G),#Stk(G1)=#Stk(G2)holdsforkn. Letc(G1, [n])beanelement.Since#V1=0, thereexists i0[n] suchthat #c1(i0) =0. ForanysubsetX c1(i0),letusdenethemap c : V [n + 1]byc(v) =c(v) ifc(v) = i0c(i0) ifc(v) = i0andv Xn + 1 ifc(v) = i0andv/ X.Then c(G1, [n + 1]). Themap cchangesaccordingtothechoiceofthesubsetX c1(i0). Hence,wehave#Stn+1(G1) = 20. Similarly,wehave#Stk(G1) = #Stk(G2) = 20forn < k 0. Thiscompletestheproof.Remark3.11. WecanalsoproveTheorem3.7usingTheorem3.10.Example 3.12. Let G1, G2, andG3be graphs of Figure 3. Thentheyhavethesamechromaticnumber, (G1)=(G2)=(G3)=3. Notethat#(G1, [3])=20and#(G2, [3])=#(G3, [3])=6. Hence, byTheorem3.10,G2andG3arechromaticallyequivalent,however,G1isnot.3.4 ExamplesofnonchromaticallyequivalentgraphsAs presentedinthe previous section, the isomorphismclass of the func-tor (G, ) is determinedbythe chromatic number (G) andcardinality#(G, [(G)]), whenever (G) < . However, this is not the case for graphswithinnitechromaticnumber(G),twoofwhicharepresentedhere.Example3.13. LetXbeaninniteset. Recall thatKX=(V, E)isthecompletegraphwithverticesX. Namely, V=XandE= {e2X|#e=2}. Lete0={x0, x1}Ebeanedge, anddeneE:=E \ {e0}. Considerthe graph KX= (V, E) obtained from KXby deleting an edge e0. Then KXandKXsatisfythefollowing.12G1q q q s s s s s s s s s s ss s s s s s s s s s

G2q q q s s s s s s s s s s ss s s s s s s s s s

q q q q q qG3s s s s s s s s s s ss s s s s s s s s s ss s s s s s s s s s ss s s s s s s s s s s

Figure3: G1, G2andG3(i) (KX) = (KX) = #X.(ii) #(KX, X) = #(KX, X) = 2#X.(iii) KXandKXarenotchromaticallyequivalent.Itiseasilyseenthat(i)and(ii)hold. Letusprove(iii). First,considerthestable partition of KX. St(KX) consists of two partitions, 1:= X:= {{x} |x X}and2:= {{x} | x X, x = x0, x1} {{x0, x1}}.Both have the same cardinality #1= #2= #X. Therefore, the chromaticfunctorisisomorphicto(KX, ) (KX, ) (KX, ).However, bytheuniquenessofstablepartitions(Proposition2.8),itcannotbeisomorphicto(KX, ).Acknowledgements. The main part of this work was conducted during theauthors stay at Hiroshima University in June 2015. The author thanks Pro-fessorsIchiroShimada, Shun-ichi Kimura, MakotoMatsumoto, NobuyoshiTakahashi, AkiraIshii, andYuyaKodafor manyinspiringcomments. Inparticular, the ideas of considering Setinjinstead of Setbijand the distinctionbetweenbounded and unbounded treeswererealized duringour discussions.Theauthorisalsograteful toProfessorTakahiroHasebeformanysugges-tions(2.3,Remarks3.4and3.11). ThisworkwaspartiallysupportedbyaGrant-in-AidforScienticResearch(C)25400060, JSPS.13References[1] F. Bergeron, G. Labelle, P. Leroux, Combinatorial species andtree-likestructures. Encyclopediaof MathematicsanditsApplications, 67.CambridgeUniversityPress,Cambridge,1998.xx+457pp.[2] Birkho, G. D.; Lewis, D. C. Chromatic polynomials. Trans. Amer.Math.Soc.60,(1946). 355-451.[3] A. Joyal, Unetheoriecombinatoiredesseriesformelles. Adv. inMath.42(1981),no.1,1-82.[4] D. Kozlov, Combinatorial algebraictopology. AlgorithmsandCompu-tationinMathematics,21.Springer,Berlin,2008.xx+389pp.[5] R. C. Read, An introduction to chromatic polynomials. J. CombinatorialTheory4(1968)52-71.[6] R. P. Stanley, Asymmetric functiongeneralizationof the chromaticpolynomialofagraph.Adv.Math.111(1995),no.1,166-194.14