Contents lists available atScienceDirect
Journal
of
Algebra
www.elsevier.com/locate/jalgebra
Extensions
of
simple
modules
over
Leavitt
path
algebras
✩Gene Abramsa,∗, Francesca Manteseb, Alberto Tonoloc
a DepartmentofMathematics,UniversityofColorado,ColoradoSprings,
CO 80918,USA
b
DipartimentodiInformatica,UniversitàdegliStudidiVerona,I-37134 Verona, Italy
cDipartimentoMatematica,UniversitàdegliStudidiPadova, I-35121,Padova,
Italy
a r t i c l e i n f o a bs t r a c t
Article history:
Received31August2014 Availableonline13March2015 CommunicatedbyLuchezarL. Avramov
DedicatedtoAlbertoFacchinion theoccasionofhissixtiethbirthday
MSC:
primary16S99
Keywords:
Leavittpathalgebra
Let E be a directed graph, K any field, and let LK(E) denote the Leavitt path algebra of E with coefficients in K. Foreach rational infinite path c∞ of E weexplicitly construct aprojectiveresolutionof thecorrespondingChen simple left LK(E)-module V[c∞]. Further, when E is
row-finite, for each irrational infinitepath p of E weexplicitly construct aprojectiveresolutionof thecorrespondingChen simple left LK(E)-module V[p]. For Chen simple modules
S,T we describe Ext1
LK(E)(S,T ) by presenting an explicit
K-basis. ForanygraphE containingatleastonecycle,this
✩ ThefirstauthorispartiallysupportedbyaSimonsFoundationCollaborationGrantsforMathematicians
Award #208941. The second and third authors are supported by Progetto di Eccellenza Fondazione Cariparo “Algebraicstructuresandtheirapplications: Abelianandderived categories,algebraic entropy andrepresentationofalgebras”andProgettodiAteneo“CategorieDifferenzialiGraduate”CPDA105885. PartofthisworkwascarriedoutduringavisitofthefirstauthortotheUniversitàdegliStudidiPadova. Thefirstauthorispleasedtotakethisopportunitytoagainexpresshisthankstothehostinstitution,and itsfaculty,foritswarmhospitalityandsupport bytheVisitingScientist2012-13grant oftheUniversità degliStudidiPadova.
* Correspondingauthor.
E-mailaddresses:abrams@math.uccs.edu(G. Abrams),francesca.mantese@univr.it(F. Mantese),
tonolo@math.unipd.it(A. Tonolo).
http://dx.doi.org/10.1016/j.jalgebra.2015.01.034
Chensimplemodule description guarantees the existence of indecomposable left LK(E)-modulesofanyprescribedfinitelength.
© 2015ElsevierInc.All rights reserved.
1. Introduction
Given any directed graph E and fieldK, one mayconstruct the Leavitt path alge-bra ofE with coefficients in K (denotedLK(E)),as firstdescribed in[2]and[3]. Since their introduction, various structural propertiesof thealgebras LK(E) have been dis-covered,with asignificant numberofthe resultsinthesubjecttaking onthe following form:LK(E) hassomespecified algebraicproperty if and onlyifE has somespecified graph-theoretic property. (The structure of the field K often plays no role in results of this type.) A few (of many) examples of such resultsinclude a description of those Leavittpathalgebraswhicharesimple;purelyinfinitesimple;finitedimensional;prime; primitive;exchange; etc.
Although there are graphs for which the structure of corresponding Leavitt path algebra is relatively pedestrian (e.g., is a direct sum of matrix rings either over K,
or over the Laurent polynomial algebra K[x,x−1], or some combination thereof), the less-mundaneexamplesof Leavittpathalgebras exhibit somewhatexoticbehavior.For instance,theprototypical LeavittpathalgebraA= LK(Rn) (n≥ 2),whicharisesfrom the graph Rn having one vertex and n loops, has the property that A ∼= An as left (or right) A-modules. Analogous “super decomposability” propertiesare also found in other important classes of Leavitt path algebras. These types of structural properties lead to a dearth (if not outright absence) of indecomposable one-sided LK(E)-ideals, whichsubsequentlymakes thesearchforsimple (and,moregenerally,indecomposable) modulesoverLeavittpathalgebrassomewhatofachallenge.
For a graph E, an infinite path in E is a sequence of edges e1e2e3· · ·, for which
s(ei+1)= r(ei) foralli∈ N.In[6],Chenproduces,foreachinfinitepathp inE,asimple leftLK(E)-moduleV[p].Further,Chen describes,foreach sinkvertex w of E,asimple
leftLK(E)-moduleNw.
InSection2we produce explicitprojective resolutions forChen simplemodules. As aresult,we will seeinTheorem 2.8 thatV[c∞] isfinitely presented forany closed path
c.Further,in Theorem 2.20we givenecessary and sufficient conditionson arow-finite graphE whichensurethatV[p]isnotfinitelypresentedforanirrationalinfinitepathp.In
Section3wedescribetheextensiongroupsExt1(S,T ) correspondingtoanypairofChen simplemodulesS,T .Usingsomegeneralresultsaboutuniserialmodulesoverhereditary rings,weconcludebyshowing (Corollary 3.25)how ourdescriptionofExt1(S,S)
guar-anteestheexistenceofindecomposableLK(E)-modulesofany prescribedfinitelength. Wesetsomenotation.A(directed)graphE = (E0,E1,s,r) consistsofavertexset E0,
anedge set E1, andsource and range functionss,r : E1 → E0. Forv ∈ E0,the set of
arefinitesets.E iscalledrow-finite incases−1(v) isfiniteforeveryv∈ E0.Apath α in E is asequence e1e2· · · en ofedgesinE forwhichr(ei)= s(ei+1) forall1≤ i≤ n− 1. Wesaythatsuchα haslength n,andwewrites(α)= s(e1) andr(α)= r(en).Weview
eachvertexv∈ E0asapathoflength0,anddenotev = s(v)= r(v).Wedenotetheset
of paths inE by Path(E).A path σ = e1e2· · · en inE is closed incaser(en)= s(e1).
Following [6] (but not standard in the literature), a closed path σ is called simple in
case σ= βm forany closed pathβ andintegerm≥ 2. Asink in E isavertex w∈ E0
for whichthesets−1(w) isempty,while aninfiniteemitter in E isavertexu∈ E0for whichtheset s−1(u) isinfinite.
For any field K and graph E the Leavitt path algebra LK(E) has been the focus of sustained investigation since 2004. We give here a basic description of LK(E); for additional information,see [2] or [1]. Let K be afield,and let E = (E0,E1,s,r) be a
directedgraphwith vertex setE0 and edgeset E1. TheLeavitt path K-algebra LK(E)
of E with coefficients in K is theK-algebra generated bya set {v | v ∈ E0},together
with aset ofsymbols{e,e∗| e∈ E1},whichsatisfythefollowingrelations:
(V) vu= δv,uv forallv,u∈ E0, (E1) s(e)e= er(e)= e foralle∈ E1,
(E2) r(e)e∗ = e∗s(e)= e∗ foralle∈ E1, (CK1) e∗e= δe,er(e) foralle,e ∈ E1,and
(CK2) v ={e∈E1|s(e)=v}ee∗ foreveryv∈ E0 forwhich0<|s−1(v)|<∞.
An alternate description of LK(E) may be given as follows. For any graph E let
E denote the “double graph” of E, gotten by adding to E an edge e∗ in a reversed directionforeachedgee∈ E1.Then LK(E) istheusualpathalgebraK E,modulothe idealgenerated bytherelations(CK1)and(CK2).
It is easy to show that LK(E) is unital if and only if |E0| is finite; in this case,
1LK(E)=
v∈E0v.Every elementof LK(E) maybewrittenas n
i=1kiαiβi∗, where ki is a nonzero element of K, and each of the αi and βi are paths inE. If α ∈ Path(E) then we mayviewα∈ LK(E),and will oftenreferto such α asareal path in LK(E); analogously, forβ = e1e2· · · en ∈ Path(E) weoftenrefertotheelementβ∗= e∗n· · · e∗2e∗1
of LK(E) as a ghost path in LK(E). The map KE → LK(E) given by the K-linear extension of α → α (for α ∈ Path(E)) is an injection of K-algebras by [1, Corollary 1.5.12].
Theideaspresentedinthefollowingfewparagraphscomefrom [6]; however,someof thenotationweuseherediffersfrom thatusedin[6],inordertomakeourpresentation morenotationallyconsistentwiththegeneralbodyof literatureregardingLeavittpath algebras.
Let p beaninfinite path in E; thatis, p isasequence e1e2e3· · ·,where ei ∈ E1 for all i ∈ N, and for which s(ei+1) = r(ei) for all i ∈ N. We emphasize that while the phrase infinite path in E mightseemto suggestotherwise,aninfinite pathinE is not anelementofPath(E),normayitbeinterpretedasanelementofthepathalgebraKE
norof theLeavittpath algebraLK(E).(Such apathis sometimescalled aleft-infinite
pathintheliterature.)Wedenotetheset ofinfinitepathsinE byE∞.
Forp= e1e2e3· · · ∈ E∞andn∈ N wedenote byτ≤n(p),oroftenmoreefficientlyby
pn,the(finite)pathe1e2· · · en,whilewedenotebyτ>n(p) theinfinitepathen+1en+2· · ·. Wenote thatτ≤n(p) isanelement ofPath(E) (andthusmaybe viewedas anelement ofLK(E)),andthatp istheconcatenationp= τ≤n(p)· τ>n(p).
Letc beaclosed pathinE.Thenthepathccc· · · isaninfinitepathinE, whichwe denote by c∞. We call an infinite path of the form c∞ a cyclic infinite path. For c a
closedpathinE letd bethesimpleclosedpathinE forwhichc= dn.Then c∞= d∞ aselements ofE∞.
If p and q are infinite paths in E, we say thatp and q are tail equivalent (written p∼ q) incase there exist integers m,n for which τ>m(p) = τ>n(q); intuitively,p ∼ q
incase p andq eventually become thesame infinitepath. Clearly ∼ is an equivalence relationonE∞, andwelet[p] denote the∼ equivalenceclass oftheinfinitepathp.
The infinite path p is called rational in case p ∼ c∞ for some closed path c. By a previousobservation,we mayassumewithoutloss of generalitythatsuch c isasimple closedpath. Inparticular, forany closedpathc we havethatc∞is rational.Ifp∈ E∞
isnotrationalwe sayp isirrational.
Example1.1.LetR2 denote the“rosewith twopetals”graph
•v
e f .
Thenq = ef ef f ef f f ef f f f e· · · isanirrationalinfinitepathinR∞2 .Indeed,itiseasyto showthatthereareuncountablymanydistinctirrationalinfinitepathsinR∞2 .Wenote additionallythatthereareinfinitelymanysimpleclosedpathsinPath(R2),forinstance,
anypathoftheform efifori∈ Z+.
LetM bealeftLK(E)-module.Foreachm∈ M wedefinetheLK(E)-homomorphism ˆ
ρm: LK(E)→ M,givenbyρˆm(r)= rm.Therestrictionoftheright-multiplicationmap ˆ
ρmmayalsobeviewedasanLK(E)-homomorphismfromanyleftidealI ofLK(E) into
M .WhenI = LK(E)v for somevertex v ofE,wewilldenote ρˆm simplybyρm. Following [6], for any infinitepath p in E weconstructasimple leftLK(E)-module
V[p],as follows.
Definition1.2.Letp beaninfinitepathinthegraphE,andletK beanyfield.LetV[p]
denote the K-vector space having basis [p], that is, having basis consisting of distinct elements of E∞ which are tail-equivalent to p. For any v ∈ E0, e ∈ E1, and q =
f1f2f3· · · ∈ [p],define v· q = q if v = s(f1) 0 otherwise, e· q = eq if r(e) = s(f1) 0 otherwise, and
e∗· q =
τ>1(q) if e = f1
0 otherwise.
Then the K-linearextension toall ofV[p] of this actionendows V[p] with thestructure
of aleftLK(E)-module.
Theorem 1.3. (See[6,Theorem 3.3].)LetE beany directed graph andK anyfield. Let p∈ E∞.ThentheleftLK(E)-moduleV[p]describedinDefinition 1.2issimple.Moreover,
if p,q∈ E∞,then V[p]∼= V[q] asleft LK(E)-modulesif andonly ifp∼ q,whichhappens
precisely whenV[p]= V[q].
We will refer to a module of the form V[p] as presented in Theorem 1.3 as a Chen
simple module.
Forany sinkw inagraphE,Chenin[6]presentsaconstruction,similarinflavorto theonegiveninDefinition 1.2,ofasimpleleftLK(E)-moduleNw.Hethen showsthat
NwisisomorphicasaleftLK(E)-moduleto theleftidealLK(E)w ofLK(E) generated by w. Observe that, for any sink w, the ideal LK(E)w is spanned by the paths in E
ending inw. Moreoverfor any i ∈ Z+, we get that wi = w andthus we canconsider
w = w∞ as an element in E∞. For these reasons, for any sink w of E, we refer to
Nw= LK(E)w asaChensimplemoduleand,forconsistency,wedenoteNwbyV[w∞].
Remark 1.4. By invoking apowerful result of Bergman, it wasestablished in [3, The-orem 3.5] that,when E is row-finite, then LK(E) is hereditary, i.e., every left idealof
LK(E) isprojective.Thispresumablycouldmakethesearchforprojectiveresolutionsof variousLK(E)-modulessomewhateasier,inthattheprojectivityofleftidealsisalready a given.However, muchof thestrength ofour resultsliesinour explicitdescriptionof the kernels of germane maps; for instance, it is these explicit descriptions which will allowusto analyzetheExt1 groupsoftheChensimplemodules.
A significant majorityof thestructural propertiesof aLeavittpathalgebras LK(E) donotrelyonthespecificstructureof thefieldK.Theresultscontainedinthis article are no exceptions. So while each of the statements of the results made herein should also contain the explicit hypothesis “Let K be any field”, we suppress this statement throughout forefficiency.ForafieldK,K× denotesthenonzeroelements of K. 2. ProjectiveresolutionsofChensimplemodulesoverLK(E)
Thegoalofthissectionisto presentanexplicitdescriptionofaprojectiveresolution of S, where S isaChen simplemodule overthe Leavittpath algebraLK(E). Such an explicit description will provide astrengthening of somepreviously established results (see [4, Proposition 4.1]), as well as provide the necessary foundation for subsequent results.Asweshallsee,thedescriptionofprojectiveresolutions,aswellasthedescription oftheExt1groups,ofChensimplemoduleswillproceedbasedonwhichofthefollowing three typesdescribesthemodule:
(1) V[w∞]∼= LK(E)w where w isasink,
(2) V[c∞] where c isasimpleclosed path;
(3) V[q] whereq isanirrationalinfinitepath.
Letv beanyvertexinE.Sincev isanidempotent,theleftidealLK(E)v isaprojective left LK(E)-module.Therefore projective resolutions ofChen simple modulesof type1 areeasy:
Proposition 2.1 (Type (1)). Let w be a sink in E. Then the Chen simple left module V[w∞] is projective.
Proof. WehaveV[w∞]∼= LK(E)w asleftLK(E)-modulesby[6]. 2
WenowbegintheprocessofdescribingprojectiveresolutionsofChensimplemodules ofthesecondtype,namely,oftheform V[c∞] forc asimpleclosedpath.
Notations. Letc= e1e2· · · etbe asimpleclosedpathinE,withv = s(e1)= r(et). (1) For0≤ i≤ t we defineci:= e1e2· · · ei anddi:= ei+1ei+2· · · et (wherec0= v = dt
andct= c= d0).Then clearlyc= cidi foreach0≤ i≤ t. (2) Forn≥ 0 weletc−n denote(c∗)n,and letc0denote v = s(c).
(3) Anelement μ ofLK(E) issaidto beastandardformmonomial in casethereexist
α,β∈ Path(E) forwhichμ= αβ∗.Wedenotethesetofstandardformmonomialsin
LK(E) byS.Forμ= αβ∗astandardformmonomialwedefiner(μ):= r(β∗)= s(β); thatis,r(μ) istheuniqueelementv∈ E0 forwhichμv = μ.Define
S1(c) :={μ ∈ S | μ · cN= 0 in LK(E) for some N ∈ N}, and S2(c) :=S \ S1(c).
Although S1(c) and S2(c) depend on c, we will often simply write S1 and S2 for
thesesets.
Byanalyzing the form of monomials inLK(E), we get the following description of theelementsofS1 andS2.
Lemma 2.2. Let0= μ∈ S. If c is asink then μ∈ S1 if and only if r(μ)= c.If c is a
closed pathe1e2· · · et,then μ∈ S1 if andonlyif μ is ofone ofthefollowingtwoforms:
(1) r(μ)= s(c) (i.e., μ· s(c)= 0 inLK(E)),or
(2) μ = μf∗c∗i(c∗)n for some n,i ∈ Z+, some μ ∈ S, and some f ∈ E1 for which
s(f )= s(ei+1) but f = ei+1.
Consequently, 0= μ∈ S2 ifand only if μ= αci∗(c∗)n forsome path α in E,and some
Lemma 2.3.Letc be aclosedpath inthegraph E,andletv = s(c).
(1) For any z∈ Z wehave cz− v ∈ L
K(E)(c− v).
(2) Supposeμ∈ S1(c).Then μ∈ (u∈E0\{v}LK(E)u)LK(E)(c− v).
Proof. (1) If z = 0 we have s(c)− v = 0 = 0(c− v). For z > 0 we have cz − v = (cz−1+ cz−2+· · · + c+ v)(c− v).Forz < 0 wehavecz− v = −cz(c−z− v),whichisin
LK(E)(c− v) bythepreviouscase.
(2) Supposeμ∈ S1(c).Ifr(μ)= v thenμ∈u∈E0\{v}LK(E)u.On theother hand, supposer(μ)= v,andthatμ·cN = 0 forsomeN ∈ N.Butr(μ)= v givesμv = μ,sothat withthehypothesisμ=−μ(cN−v),whichgivesthatμ∈ LK(E)(cN−v)⊆ LK(E)(c−v) bythepreviousparagraph. 2
Remark 2.4. Let p= e1e2· · · be an infinite path in E. If p = τ>r(p) for some r > 0, then p isarationalpathoftheformp= c∞,where c istheclosedpathe1e2· · · er.This follows fromtheobservationthatp= τ>r(p) impliesp= τ>ir(p) foralli∈ N.
Lemma 2.5. Letc be asimple closed pathe1e2· · · et in E withs(c)= r(c)= v. Suppose
α and β arepathsinE forwhich0= α · c∞= β· c∞ inV[c∞].ThenthereexistsN ∈ Z+
forwhich α = βcN orβ = αcN.
Consequently, α· c∞= β· c∞ in V[c∞] impliesα− β ∈ LK(E)(c− v).
Proof. Assume α = f1f2· · · fand β = g1g2· · · gm,where thefi and gj areedges inE. If= m,from α· c∞= β· c∞ wegetα = β.Soassumem> ;wehave
g1· · · gg+1· · · gm= f1· · · fcne1e2· · · ek with m− = k + n× t,k≤ t. Ifk < t,fromα· c∞= β· c∞ weget
c∞= ck+1· · · ct· c∞= τ>k(c∞);
then byRemark 2.4wewouldhavec∞= (e1· · · ek)∞,acontradiction sincec issimple. Therefore k = t andβ = αcn+1.Thecasem< isidentical.
Forthesecondstatement,notethat0= α · c∞= β· c∞givesr(α)= r(β);denotethis common vertex byv. Soifα = βcn then α− β = β(cn− v),which isinLK(E)(c− v) byLemma 2.3(1).Thecaseβ = αcn isidentical. 2
Proposition2.6.LetE beany graph.Letc beasimpleclosedpathinE,andletv denote s(c) = r(c). Let ρc∞ : LK(E)v → V[c∞] and ρˆc∞ : LK(E) → V[c∞] denote the right
multiplication by c∞.Then
Ker(ρc∞) = LK(E)(c− v) and Ker(ˆρc∞) = ⎛ ⎝ u∈E0\{v} LK(E)u ⎞ ⎠ ⊕ LK(E)(c− v).
Proof. Since(c− v)· c∞ = c∞− c∞ = 0 inV[c∞], we get LK(E)(c− v) ⊆ Ker(ρc∞). Wenowproceed toshowthatKer(ρc∞)⊆ LK(E)(c− v).Fornotationalconveniencewe denotetheleftidealLK(E)(c− v) ofLK(E) byJ .
Soletλ∈ Ker(ρc∞),andwrite
λ =
μ∈M
kμμ
where M ⊆ S is some finite set of distinct standard form monomials in LK(E), and
kμ∈ K×.ByLemma 2.3wemayassumethatM⊆ S2;thatis, byLemma 2.2,wemay
assumethat,foreachμ∈ M,μ= αμc∗iμ(c∗)nμ forsomepathαμ,some0≤ i
μ ≤ t,and somenμ ≥ 0. Sowehaveλ=μ∈Mkμαμc∗iμ(c ∗)nμ.Byhypothesisλ· c∞= 0 inV [c∞],so that μ∈M kμαμc∗iμ(c ∗)nμ· c∞= 0 in V [c∞]. But(c∗)n· c∞= c∞ inV [c∞] forany n∈ Z.So μ∈M kμαμc∗iμ· c ∞= 0 in V [c∞].
Also,c∗i · c∞= di· c∞ inV[c∞] forany 0≤ i≤ t. So
μ∈M kμαμdiμ · c ∞= 0 in V [c∞]. Nowdefine λ= μ∈M kμαμdiμ.
Thenthepreviousequationgivesthatλ∈ Ker(ρc∞).
We claim that λ ∈ J if and only if λ ∈ J. To show this, we show thatλ = λ as elementsofLK(E)/J .Wenote firstthatc∗i = di inLK(E)/J ; thisfollowsimmediately fromtheobservationthatdi− c∗i = ci∗(c− v)∈ J.ButtheninLK(E)/J wehave
λ = μ∈M kμαμc∗iμ(c ∗)nμ = μ∈M kμαμc∗iμ(c∗) nμ = μ∈M kμαμc∗iμv byLemma 2.3(1)
= μ∈M
kμαμc∗iμ
=
μ∈M
kμαμdiμ by the above note
=
μ∈M
kμαμdiμ
= λ.
Thus in order to show that λ ∈ J, it suffices to show that λ ∈ J, i.e., that λ = 0 in LK(E)/J . Butλ ∈ Ker(ρc∞), i.e., μ∈Mkμαμdiμ · c∞ = 0 in V[c∞]. Now partition
M=
t=1Mtinsuchawaythatμ∼ μ ∈ Mt (forsomet) ifandonlyifαμdiμ· c∞ =
αμdiμ· c
∞ inV
[c∞].
ByLemma 2.5,ifμ∼ μ then αμdiμ = αμdiμ inLK(E)/J ;wedenote thiscommon
element ofLK(E)/J byxt. Now μ∈Mkμαμdiμ· c∞= 0 gives t=1 μ∈Mt kμαμdiμ· c∞= 0,
whichbythelinearindependenceofsetsofdistinctelements oftheform α· c∞inV[c∞]
givesμ∈Mtkμ= 0 foreach1≤ t≤ .Butthen
λ= μ∈M kμαμdiμ = t=1 μ∈Mt kμαμdiμ = t=1 μ∈Mt kμαμdiμ = t=1 μ∈Mt kμxt = t=1 ( μ∈Mt kμ)xt= t=1 (0)xt= 0,
whichestablishesthatKer(ρc∞)⊆ LK(E)(c−v),asdesired.Theclaimaboutρˆc∞follows easily fromLK(E)=
u∈E0\{v}LK(E)u⊕ LK(E)v. 2
Lemma 2.7. Let E beany graph. Letc be a simple closed path in E based atthe vertex v, andlet r∈ LK(E)v. Thenr(c− v)= 0 inLK(E) ifand onlyif r = 0.Inparticular,
themap
ρc−v : LK(E)v→ LK(E)(c− v)
Furthermore,if E isafinite graph,then themap
ˆ
ρc−1: LK(E)→ LK(E)(c− 1)
isanisomorphism ofleft LK(E)-modules.
Proof. Letr∈ LK(E)v.Ifr(c−v)= 0 thenrc= rv = r,whichrecursivelygivesrcj = r foranyj≥ 1.Nowwriter =ni=1kiαiβi∗,wheretheαi andβiareinPath(E).Wenote that,for any m ∈ N, ifβ ∈ Path(E) has length at most m, then β∗ has theproperty thatβ∗cm iseither0 oranelementof Path(E) inLK(E).Now letN be themaximum length of the paths in the set {β1,β2,. . . ,βn}. Then the above discussion shows that
rcN isanelement ofL
K(E)v oftheform n
i=1kiγi,where γi ∈ Path(E) for1≤ i≤ n; thatis, rcN ∈ KE. ButrcN = r, so thatr∈ KE. However,theequation rc= r (i.e.,
r(c− v)= 0) hasonlythezerosolutioninKE byadegreeargument.Sor = 0.
Thesecondstatementisestablishedinanalmostidentical manner. 2
We now haveall the tools to describe aprojective resolution for themodules V[c∞]
where c isasimple closedpath, thus completingthestudy of thesecond typeof Chen simplemodule.
Theorem 2.8 (Type(2)).Let E be any graph. Let c be a simple closed path in E, with v = s(c).Then the Chensimple moduleV[c∞] isfinitely presented. Indeed, a projective
resolutionof V[c∞] isgivenby 0 LK(E)v ρc−v LK(E)v ρc∞ V[c∞] 0 .
If E isafinite graph,an alternateprojectiveresolutionof V[c∞] isgivenby
0 LK(E) ˆ ρc−1 LK(E) ˆ ρc∞ V[c∞] 0 .
Proof. V[c∞]isasimpleleftLK(E)-modulebyTheorem 1.3,andc∞= vc∞isanonzero elementinV[c∞].Sothemapρc∞: LK(E)v→ V[c∞]issurjective.ByProposition 2.6we
haveKer(ρc∞)= LK(E)(c−v).WegetthefirstshortexactsequencesincebyLemma 2.7 themap
ρc−v: LK(E)v→ LK(E)(c− v)
isanisomorphismofleftLK(E)-modules.Moreoversincev isidempotent,LK(E)v isa projectiveleftLK(E)-module.
Assume now that E is a finite graph. Let us see that Ker( ˆρc∞) = LK(E)(c− 1). SinceKer( ˆρc∞) clearly containsu foranyu= v ∈ E0,we havec− 1= c−u∈E0u=
u=−u(c−1)∈ LK(E)(c−1).Sincec−v = v(c−1)∈ LK(E)(c−1),usingProposition 2.6 we have shown that each of the generators of Ker( ˆρc∞) is in LK(E)(c− 1). But by Lemma 2.7, ρˆc−1 : LK(E)→ LK(E)(c− 1) isan isomorphismof left LK(E)-modules, thus establishingtheresult. 2
Corollary 2.9. Let E be any graph. Let c be a simple closed path in E, with v = s(c). Then theChensimplemoduleV[c∞] has projective dimension 1.
Proof. FromTheorem 2.8wegettheexactsequence
0 LK(E)v
ρc−v
LK(E)v ρc∞
V[c∞] 0 .
Since v is an idempotent in LK(E), the left module LK(E)v is projective and hence
V[c∞]hasprojectivedimension≤ 1.TheleftmoduleV[c∞]isnotprojective,otherwisethe
abovesequencesplitsandLK(E)v wouldcontainadirectsummandisomorphictoV[c∞];
inparticularLK(E)v wouldcontainanonzeroelementα (theelementcorrespondingto
c∞)suchthatcα = α andhencecnα = α foreachn∈ N.Thisisimpossiblebyadegree argument. 2
BeforewepresentaprojectiveresolutionofthethirdtypeofChensimplemodule,we study,inthesituationwhereE isrow-finite,rightmultiplicationbyanyofthemonomial generatorsofV[c∞]forc asimpleclosedpathorasink.Wefirstintroducesomenotation
whichwill beusefulthroughout theremainderofthesection.
Definition 2.10. Let E be any graph. Let β = e1e2· · · en be a path in E. For each 1≤ i≤ n letβi denote e1e2· · · ei.Foreach0≤ i≤ n− 1 let
Xi(β) ={f ∈ E1 | s(f) = s(ei+1), and f = ei+1}.
Theelements ofXi(β) arecalled theexits ofβ ats(ei+1).Notethat,foragiveni,itis possible thatXi(β)=∅.Foreachi≥ 0 letJi(β) bethe leftidealof LK(E) definedby setting
Ji(β) = f∈Xi(β)
LK(E)f∗βi∗.
(So possibly Ji(β) ={0}, precisely when Xi(β) =∅.) When the path β is clear from context,wemaydenoteXi(β) (resp.,Ji(β))byXi (resp.,Ji).
Now let p= e1e2e3· · · ∈ E∞ be aninfinite path inE. Letp0 denote s(e1), andfor
eachi≥ 0 letpi+1 denote τ≤i+1(p)= e1e2· · · ei+1.Foreachi≥ 0 we define
Definition 2.11. Let E be any graph. Let β = e1e2· · · en be a path inE for which no vertexofβ is aninfiniteemitter.For0≤ i≤ n− 1 let
Fi(β) = f∈Xi(β)
f f∗∈ LK(E).
Note that this sum is finite by the hypothesis on β. (We interpret Fi(β) as0 in case
Xi(β)=∅.)Inparticular, bythe(CK2)relationwehave
s(ei+1)− Fi(β) = ei+1e∗i+1 for0≤ i≤ n− 1,and by(CK1)thatFi(β)ei+1 = 0.
Lemma2.12.LetE beanygraph. Letα = e1e2· · · en beapath inE forwhichno vertex
of α isaninfinite emitter.Letαi denotee1e2· · · ei foreach 1≤ i≤ n (so inparticular
α = αn).Supposeq,x∈ LK(E) satisfy theequation qα = x inLK(E). Then
q = xα∗+ qαn−1Fn−1(α)α∗n−1+· · · + qα1F1(α)α∗1+ qF0(α).
Proof. Multiplybothsidesoftheequationqα = x by e∗n,toget
xe∗n= qαe∗n= qe1· · · en−1ene∗n = qe1· · · en−1(s(en)− Fn−1(α)). Multiplyingthefinal termandswitchingsides,thisgives
qe1· · · en−1= xe∗n+ qe1· · · en−1Fn−1(α).
Multiplyingnowbothsidesofthisdisplayedequationontherightbye∗n−1,and proceed-inginthesameway,weeasilyget
qe1· · · en−2 = xe∗ne∗n−1+ qe1· · · en−1Fn−1(α)e∗n−1+ qe1· · · en−2Fn−2(α). Continuinginthisway,after n stepswereach
q = xe∗ne∗n−1· · · e∗1 + qe1· · · en−1Fn−1(α)e∗n−1· · · e∗1 + · · · + qe1F1(α)e∗1 + qF0(α)
asdesired. 2
Ofcourse,ifc isasimpleclosedpathorasink,anynonzeroelementoftheChensimple moduleV[c∞] generatesV[c∞];thisisinparticulartrueofany“monomial”elementαc∞,
whereα isapathinE forwhichr(α)= s(c).Wedescribeheretheprojectiveresolution correspondingto suchelements,incaseE isrow-finite.
Theorem2.13(Types(1)& (2)). LetE beany graph.Letc beasimple closedpath ora sink in E,with v = s(c). Letα = e1e2· · · en be any path inE for which novertex of α
is aninfinite emitter,andforwhichr(α)= r(en)= v.Letu denotes(α)= s(e1). Then
thefollowingis aprojectiveresolutionof theChensimple LK(E)-moduleV[c∞]:
0 LK(E)(αcα∗− u) LK(E)u ραc∞
V[c∞] 0 .
Proof. Sinceuα = α,wehavethatραc∞(u)= αc∞isanonzeroelementoftheChen sim-plemoduleV[c∞],sothatραc∞ issurjective.SoweneedonlyestablishthatKer(ραc∞)=
LK(E)(αcα∗−u).Sinceραc∞(αcα∗−u)= (αcα∗−u)αc∞= αcc∞−αc∞= 0,itremains
onlyto showthatKer(ραc∞)⊆ LK(E)(αcα∗− u).
So let q ∈ Ker(ραc∞); specifically, qαc∞ = 0. But then qα ∈ Ker(ρc∞), which, by Theorem 2.8,isprecisely LK(E)(c− v).So
qα = r(c− v)
forsomer∈ LK(E).ByLemma 2.12,wehave
q = r(c− v)α∗+ qαn−1Fn−1(α)α∗n−1+· · · + qα1F1(α)α∗1+ qF0(α).
Using this representation of q, it suffices to show that each of the summands on the right hand side is an element of LK(E)(αcα∗ − u). Since easily we get (c− v)α∗ =
α∗(αcα∗−u),wehavethatr(c−v)α∗∈ LK(E)(αcα∗−u).Butforeach0≤ i≤ n−1 we have Fi(α)α∗iα = Fi(α)ei+1· · · en = 0 (using theobservation madeinDefinition 2.11). Using this, we see that qαiFi(α)α∗i = −qαiFi(α)α∗i(αcα∗− u), so that qαiFi(α)α∗i ∈
LK(E)(αcα∗− u) foreach0≤ i≤ n− 1,thuscompletingtheproof. 2
We now describe a projective resolution of the third type of Chen simple module, namely, onecorresponding to anirrationalinfinitepath.Whereas aChensimple corre-spondingtoarationalpathisalwaysfinitelypresented,wewillseethatthedetermination of thefinite-presentednessofaChensimplecorrespondingto anirrationalinfinitepath will dependonthestructureofthegraphitself.
Lemma2.14.LetE beanygraph.Letp beanirrationalinfinite pathinE withs(p)= v,
and let ρp : LK(E)v → V[p] be the map r → rp. Let x ∈ Ker(ρp). Then there exists
nx ∈ N such that xτ≤nx(p) = 0 in LK(E). In other words, if xp = 0 in V[p], then
xpnx = 0 inLK(E) forsome finiteinitial segmentpnx of p.
Proof. Let x = mi=1kiαiβi∗ ∈ Ker(ρp), where αi,βi ∈ Path(E). Denote by N the maximumlengthoftheβi,i= 1,. . . ,m.Wehave
ρp(x) = m i=1 kiαiρp(β∗i) = m i=1 kiαiρτ>N(p)(β ∗ iτ≤N(p)).
Sincethelengthofeachβi islessthanor equalto N ,ti:= βi∗τ≤N(p) is eitherzeroora realpath.Therefore
0 = ρp(x) = m i=1 kiαiρτ>N(p)(ti) = ρτ>N(p)( m i=1 kiαiti) = ρτ>N(p)( m =1 hγ) = m =1 hγτ>N(p),
wheretheγ(1≤ ≤ m)aredistinctelementsoftheformαitiinPath(E),andh∈ K. Since p is irrationaland γ (1≤ ≤ m) are distinct paths, weclaim thatthe infinite paths γτ>N(p) (1 ≤ ≤ m) are distinct elements of V[p], as follows. Assume to the
contrary thatγiτ>N(p)= γjτ>N(p) forsome i= j;necessarily γi and γj havedistinct lengthssi andsj.Assumesi− sj = s> 0;then
γiτ>N(p) = γjκiτ>N(p) = γjτ>N(p),
andhenceκiτ>N(p)= τ>N(p),whereκiisasuitableelementofPath(E) havinglengths. Thereforeτ>N(p)= τ>s(τ>N(p))= τ>s+N(p). Butthis propertyimpliesbyRemark 2.4 thatp isrational,contrary to hypothesis.Thusthe γτ>N(p) (1≤ ≤ m)aredistinct infinitepaths.
Consequently, the set {γτ>N(p) | 1 ≤ ≤ m} is linearly independent over K, so the previously displayed equation 0 = m=1 hγτ>N(p) yields that h = 0 for each 1≤ ≤ m.Therefore xτ≤N(p) = m i=1 kiαiβi∗τ≤N(p) = m i=1 kiαiti = m =1 hγ= 0, asdesired. 2
Lemma 2.15. Let E be any graph. Suppose β is a path of length n in E for which no vertex of β is an infinite emitter, and for whichs(β) = v. Foreach 0≤ i ≤ n− 1 let
Ji(β) be theleft ideal of LK(E) given in Definition 2.10. If x∈ LK(E)v has xβ = 0, thenx∈n−1i=0 Ji(β).
Proof. Write β = e1e2· · · en. For each 1≤ i ≤ n let βi = e1e2· · · ei. So β = βn, and
thus by hypothesis we are assuming that xβn = 0. Then using the (CK2) relation at theverticess(e1),s(e2),· · · ,s(en) inorder(thisispossiblebythehypothesisonβ),and
interpretingemptysumsas0,weget
x = xv = x( f∈X0(β) f f∗+ β1β∗1) = x( f∈X0(β) f f∗) + xβ1r(β1)β1∗
= x( f∈X0(β) f f∗) + xβ1( f∈X1(β) f f∗+ e2e∗2)β1∗ = x( f∈X0(β) f f∗) + xβ1( f∈X1(β) f f∗)β1∗+ xβ2r(β2)β2∗ =· · · = x( f∈X0(β) f f∗) + xβ1( f∈X1(β) f f∗)β1∗+· · · + xβn−1( f∈Xn−1(β) f f∗)βn∗−1 + xβn−1ene∗nβn−1∗ = x( f∈X0(β) f f∗) + xβ1( f∈X1(β) f f∗)β1∗+· · · + xβn−1( f∈Xn−1(β) f f∗)βn∗−1+ xβnβ∗n = x( f∈X0(β) f f∗) + xβ1( f∈X1(β) f f∗)β1∗+· · · + xβn−1( f∈Xn−1(β) f f∗)βn∗−1+ 0,
with thefinal statementfollowingfromthehypothesisthatxβ = xβn = 0.Thus
x = f∈X0(β) (xf )f∗+ f∈X1(β) (xβ1f )f∗β∗1+· · · + f∈Xn−1(β) (xβn−1f )f∗βn−1∗ ∈ n−1 i=0 Ji(β). 2
Lemma 2.16.LetE beafinite graph,andletp= e1e2· · · ∈ E∞ beanirrational infinite
path in E. Then Xi(p) is nonempty for infinitely many i ∈ Z+. Consequently, in this
case, Ji(p) isnonzeroforinfinitely manyi∈ Z+.
Proof. SupposetothecontrarythatthereexistsN ∈ N forwhichXi(p)=∅ foralli≥ N.
Since E0 isfinite,there exist t,t ≥ N, t< t, forwhich s(et)= s(et).But Xt(p)=∅
thengiveset= et,andinasimilarmanneryieldset+= et+forall∈ Z+.Ifd denotes
theclosed pathetet+1· · · et−1, thenwegetp∼ d∞,thedesiredcontradiction. 2 WenotethatLemma 2.16isnotnecessarilytruewithoutthefinitenesshypothesison thegraph.Forinstance,letMNbethegraph
• e1
• e2
• e3
· · ·
and letp∈ MN∞ be theirrationalinfinitepathe1e2· · ·.ThenXi(p)=∅ for alli≥ 0. Corollary 2.17. Let E be any graph. Let p ∈ E∞ be an irrational infinite path in E for which no vertex of p is an infinite emitter, and for which s(p) = v. Let
ρp: LK(E)v→ V[p] be the map r → rp. For each i ≥ 0 let Ji(p) be the left ideal of
LK(E) giveninDefinition 2.10.Then
Ker(ρp) = ∞ i=0Ji(p).
Proof. Clearly, for every i ≥ 0,each element of Ji(p) is inKer(ρp). Now suppose x ∈ LK(E) has xp = 0 in V[p]. By Lemma 2.14, xβ = 0 where β = pn = τ≤n(p) for some
n∈ N.Then Lemma 2.15, togetherwith the definitionofJi(p) forp∈ E∞, givesthat Ker(ρp)=
∞ i=0Ji(p).
Nowsupposeni=0ri= 0 inLK(E),whereri∈ Ji(p) for0≤ i≤ n.Byconstruction,
ripn = 0 foralli< n.Ontheotherhand,foranyf ∈ Xn(p),f∗p∗npnpn∗ = f∗p∗n,sothat
rnpnp∗n = rn for all rn ∈ Jn(p). Thus multiplying both sidesof the proposed equation n
i=0ri = 0 ontheright bypnpn∗ givesrn = 0.Using thissameideaiteratively,we get
ri= 0 forall0≤ i≤ n,sothatthesumisindeeddirect. 2
WenotethatCorollary 2.17isnotnecessarilytruewithoutthefiniteemitter hypoth-esisontheverticesof p.Forinstance,letF be thegraph
•w ∞ •v e1
• e2
• e3
· · ·
wherethere areinfinitelymanyedges {fi |i∈ Z+} fromv tow.Letp betheirrational infinitepath e1e2· · ·,andletρp : LK(E)v→ V[p] as usual. Then easilyx= v− e1e∗1 ∈
Ker(ρp).However,x∈/∞i=0Ji(p),sinceotherwisethiswouldyieldthatv isafinitesum of e1e∗1 plusterms ofthe form rififi∗ forri ∈ LK(E), whichcannot happen as v isan infiniteemitter.
Remark2.18.Corollary 2.17shows thatifE isrow-finite andp isan irrationalinfinite path, thenKer(ρp) is generated by thoseghost paths ofLK(E) which annihilatesome (finite)initialpathofp.Effectively,thisisthemaindifferencebetweentherationaland irrationalcases; inthe rationalcase, where p= d∞ and s(p)= v, there are additional elements inKer(ρp) whichare notof this form, namely,elements of the form r(d− v) wherer∈ LK(E).
Lemma2.19.Letp beanirrationalinfinite pathinanarbitrarygraph E.Forf ∈ E1 let
vf denotethevertex r(f ).Then, foreach i≥ 0,
Ji(p) = f∈Xi(p) LK(E)f∗p∗i ∼= f∈Xi(p) LK(E)vf,
asleft LK(E)-modules. In particular,each Ji(p) is aprojectiveleft LK(E)-module. Proof. BydefinitionJi(p)=f∈Xi(p)LK(E)f∗p∗i.Weclaimthesumis direct.So sup-pose0=f∈Xi(p)rff∗pi∗, withrf ∈ LK(E) for eachf ∈ Xi(p). Without loss wemay
assumethateachexpressionrff∗isnonzero,sothatwemayfurtherassumewithoutloss thatrfvf = rf foreachf ∈ Xi(p).Takeg∈ Xi(p);bymultiplying0=f∈Xi(p)rff
∗p∗ i ontherightbypig,andusingthe(CK1)relation,weget0= rgg∗g = rg· vg= rg.Thus the sumis direct,so thatJi(p)=⊕f∈Xi(p)LK(E)f∗p∗i. Butfor g ∈ Xi(p) it iseasy to show thatLK(E)g∗p∗i ∼= LK(E)vg,bythemapx → xpig. 2
Theorem2.20(Type(3)).LetE beanygraph. Letp∈ E∞ beanirrational infinitepath inE forwhichnovertexofp isaninfiniteemitter.ThentheChensimpleLK(E)-module
V[p] isfinitely presented if andonlyif Xi(p) is nonemptyonly forfinitelymanyi∈ Z+.
In particular,if E isafinite graph,then V[p] isnotfinitely presented.
Proof. Let v denotes(p).Weconsider theexactsequence
0 Ker(ρp) LK(E)v ρp V[p] 0 .
By Corollary 2.17 we have that Ker(ρp) =⊕∞i=0Ji(p). Furthermore, each Ji(p) is pro-jective by Lemma 2.19, so the given exact sequence is a projective resolution of V[p].
Therefore V[p] isfinitely presented ifandonly ifJi(p) isnonzeroonlyfor finitely many
i∈ Z+,i.e.Xi(p) isnonemptyonlyforfinitelymanyi∈ Z+.
For the particular case, when E is finite then by Lemma 2.16 Ji(p) is nonzero for infinitelymanyi. 2
Corollary 2.21. If E is a finite graph, and p∈ E∞ is an irrational infinite path in E, then theChensimple LK(E)-moduleV[p] has projectivedimension1.
Proof. FromCorollary 2.17wegettheexactsequence 0 ∞i=0Ji(p) LK(E)v
ρp
V[p] 0 .
Sincev isanidempotentinLK(E),theleftmoduleLK(E)v isprojective;byLemma 2.19 also ∞i=0Ji(p) isprojectiveandhenceV[p]hasprojectivedimension≤ 1.SinceE isfinite,
Ji(p) is not zero for infinitely many i and hence ∞i=0Ji(p) is not finitely generated. ThentheleftmoduleV[p]isnotprojective, otherwise
∞
i=0Ji(p) wouldbeanotfinitely generated directsummand ofacyclic module:contradiction. 2
Remark2.22.LetMN bethegraph
• e1
• e2
• e3
· · ·
considered previously,and letp∈ MN∞ be the irrationalinfinite path e1e2e3· · ·.Then
Xi(p)=∅ foralli≥ 0.SobyCorollary 2.17,theChensimplemoduleV[p] isisomorphic
Example2.23.WereconsiderthegraphR2 andirrationalinfinitepathq = ef ef f ef f f e
· · · ∈ R∞
2 described inExample 1.1.Then,as R2 isfinite,Theorem 2.20yieldsthatthe
ChensimplemoduleV[q] isnotfinitelypresented.
Remark2.24. Wenote thatTheorems 2.8 and 2.20strengthen and sharpen [4, Propo-sition 4.1], most notablybecause we havebeen ableto explicitly describe a projective resolutionofeachoftheChensimplemodules.
In[5,Theorem 4.12]itisshownthatforanygraphE,foranyvertexv∈ E0,L(E)v is
asimpleleftidealifandonlyifv isalinepoint,i.e.inthefullsubgraphofE generated
by{u∈ E0 | there is a path from v to u} thereare nocycles,and thereareno vertices
which emit morethan one edge. Our results allowus to recover [5, Theorem 4.12], as follows.
Corollary2.25. LetE be any graph.Let u∈ E0. ThenL
K(E)u issimple if and onlyif
u isalinepoint.
Proof. Therearethreepossibilities:
(1) thereisapathα∈ Path(E) with s(α)= u andforwhichr(α)= w isasinkinE;
(2) thereis apath α∈ Path(E) withs(α)= u and forwhich r(α)= v isthesource of asimpleclosedpathc;
(3) thereisaninfiniteirrationalpathq forwhichs(q)= u.
Ifinα (cases1and2)or inq (case3)there isaninfiniteemitterx,then LK(E)x is notasimplesubmoduleofLK(E)v (see[5,Lemma 4.3]).Thereforewecanassumethat
α (cases1and 2)andq (case3)havenoinfiniteemitter.
Cases1and2.ByTheorem 2.13LK(E)u isasimplemoduleifandonlyifαcα∗= u, wherec iseitherasimpleclosedpathorasink.Ifc isasimpleclosedpath,byadegree argument αcα∗ is not a vertex. If c is a sink and α = e1· · · e then αcα∗ = αα∗ =
e1· · · ee∗· · · e∗1 isequaltou ifandonlyifeie∗i = s(ei) fori= 1,. . . ,,i.e.ifandonlyif
s(ei) isthesourceofonlyoneedge,i.e.u isalinepoint. Case3. ByCorollary 2.17, LK(E)u issimple ifand only if
∞
i=0Ji(p) = 0 and the latterisequivalenttop havingnoexits,i.e.u isalinepoint. 2
3. ExtensionsofChensimplemodules
In this section we use the results of Section 2to describe Ext1LK(E)(S,T ), where S
andT areChen simplemodulesover theLeavittpathalgebraLK(E) correspondingto afinitegraphE.Asaconsequence,thiswillallowusto(amongotherthings)construct classesofindecomposablenon-simpleLK(E)-modules.
WegivehereashortreviewofExt1;seee.g.[8]formoreinformation.LetR bearing, andletM,N beleftR-modules.Suppose
0 Q μ P f M 0
is ashort exactsequencewithP projective. Thenthere isanexactsequenceofabelian groups HomR(P, N ) μ∗ HomR(Q, N ) Δf Ext1R(M, N ) 0,
where μ∗(ϕ) = ϕ◦ μ for ϕ ∈ HomR(P,N ), and Δf is the “connecting morphism”. If μ is viewed as an inclusion of submodules, then μ∗(ϕ) = ϕ|Q, the restriction of ϕ
to Q. Exactness yields that Ext1R(M,N ) = 0 if and only μ∗ is surjective. Moreover, Ext1R(M,N )= 0 ifandonlyifthereexists anon-splittingshort exactsequence
0→ N → L → M → 0,
i.e., L is a non-trivial extension of N by M . For instance if M,N are simple left
R-modules, then Ext1R(M,N ) = 0 if and only if there exist indecomposable left
R-modules of length 2which are extensions of N by M . Finally, observe thatif R is
aK-algebraoverafieldK,thentheabeliangroupExt1R(M,N ) hasanaturalstructure of K-vectorspace foranyleftR-module M andN .
Weoutlineourapproach.There arethreetypesofChensimplemodules:thoseofthe formV[w∞] forasinkw;oftheformV[c∞] forasimpleclosedpathc;andoftheformV[p]
foranirrationalinfinitepathp.LetT denoteanyChensimplemodule.InLemma 3.1we makethe(trivial)observationthatExt1LK(E)(V[w∞],T )= 0;inTheorem 3.13wedescribe
Ext1LK(E)(V[c∞],T );and inTheorem 3.21 wedescribe Ext1LK(E)(V[p],T ).Werecall that
we areassumingw = w∞∈ E∞ foranysinkw.
Lemma 3.1 (Type (1)). Let E be any graph. Let w be a sink in E, and let T denote any left LK(E)-module. ThenExt1LK(E)(V[w∞],T )= 0,i.e.any extensionof V[w∞] by T
splits.
Proof. This follows immediately from the fact that V[w] ∼= LK(E)w is a projective
LK(E)-module(seeProposition 2.1). 2
Definition 3.2.LetT beaChensimplemodule.DenotebyU (T ) theset
U (T ) :={v ∈ E0 | vT = {0}} = {v ∈ E0 | there exists t ∈ T with vt = 0}.
Remark 3.3. Let T be a Chen simple module and let q = e1e2· · · ∈ E∞ such that
T = V[q]. ThenU (T ) consistsofthose verticesv for whichthereis apath α∈ Path(E)
having s(α) = v and r(α) = s(ei) for some i ≥ 1. Equivalently, a vertex v ∈ U(T ) if andonlyifthereisaninfinitepathtail-equivalenttoq startingfromv.HenceU (T ) isa featureoftheChensimplemoduleT thatcanbereaddirectlyfrom thegraphE.
Definitions3.4.LetE beany graphand letd beasimpleclosedpathinE.
Foranyp∈ E∞ wesaythatp isdivisibleby d ifp= dp forsomep∈ E∞. Foranyq∈ E∞,wedefine theset
L(d,q):={p ∈ E∞ | p ∼ q, s(p) = s(d), and p is not divisible by d} ⊆ V[q],
whereV[q]istheChensimpleLK(E)-modulegenerated byq.
Aninfinitepathp isdivisiblebyasimpleclosedpathd∈ E ifandonlyifd= t≤(p), where isthelengthof d.The set L(d,q) consists ofthose infinitepaths whichstartat
s(d),and whicheventuallyequalsome tailofq, butdo notstartoutby traversingthe closed pathd. Observethatthesubset L(d,q) of V[q] does notdependon q butonly on
theequivalenceclass[q].LetT = V[q];ifq isnottailequivalentto d∞,thenthereexists
q∼ q suchthatd |q andhenceT hasageneratornotdivisiblebyd.
Remark3.5. Letd beasimpleclosedpathinE andq∈ E∞.
(1) Supposeq isnottailequivalenttod∞ andconsidertheChensimplemoduleV[q];
we canassume withoutloss ofgenerality that q is notdivisible byd. The set L(d,q) is
notemptyifandonlyifs(d) belongstoU (V[q]);insuchacaseany0= t∈ V[q]forwhich
s(d)t = t is a linear combination of infinite paths tail equivalent to q whose sources coincidewith s(d).Inparticular, takinginaccount thedivisibilitybyd ofthese infinite paths,t canbewritteninauniquewayas
t = t0+ dt1+ d2t2+· · · + dsts,
where theti areK-linearcombinations of elements inL(d,q) and ts= 0. We call s≥ 0 the d-degree oft andwe denoteitbydegd(t).
(2)Suppose q = d∞.Then L(d,d∞)= ∅ ifand onlyif thereexists acycle c= d with
s(c)= s(d).Any0= t∈ V[d∞] forwhichs(d)t= t canbe writteninauniquewayas
t = kd∞+ t0+ dt1+ d2t2+· · · + dsts,
where the ti ∈ V[d∞] are K-linearcombinations of elements inL(d,d∞) and ts= 0. We calls≥ 0 the d-degree of t andwe denoteitbydegd(t).
Inparticular,any 0= t∈ L(d,q) hasd-degreeequalto 0.Weemphasizethat,incase
q = d∞,thed-degreeoftheelementd∞ofV[d∞] iszerotoo:degd(d∞)= 0.Thed-degree isnotdefinedon0.
Example3.6.WerevisitthegraphR2 givenby
•v
e f .
Consider the simple closed path e and the rational infinite path f∞. Then L(e,f∞) =
{p∈ R∞
paths {fiejf∞ | i≥ 1,j ≥ 0}.(There are additionalelements of L(e,f∞), for instance,
f ef ef∞.) Moreover,consider anelement of theform ejfief∞ ∈ V[f∞], withi ≥ 1 and
j ≥ 0.Thendege(ejfief∞)= j.
Ontheotherhand,L(f,f∞)={p∈ R∞2 |p∼ f∞and pis not divisible by f} contains
the infinitepaths {eif∞ | i≥ 1}. Note thattheelement f∞ of V
f∞ is definedto have degf(f∞)= 0.
Recall thatLK(E) isaringwitunityifandonlyifE isfinite.
Lemma 3.7. Let E be a finite graph. Let d be a simple closed path and q ∈ E∞. Let t∈ V[q],and considertheequation inthevariableX
(d− 1)X = t.
The equation admits asolutionin V[q]if oneof thefollowingholds:
(1) s(d)t= 0;
(2) t= dnp− p for some p∈ L
(d,q) andn≥ 0.
Proof. (1)iseasy,sinceifs(d)t= 0 thendt= 0,andhenceX =−t isasolution.(2)is nearly aseasy,sincewehave(d− 1)ni=0−1dip= dnp− p= t, andhenceX =ni=0−1dip
is asolution. 2
Lemma 3.8. Let E be a finite graph. Let d be a simple closed path and let q ∈ E∞. Assumeeither q = d∞ orq isageneratorofV[q] notdivisiblebyd. Let0= t∈ V[q],and
consider theequation
(d− 1)X = t.
Assume t = dnt for some n ≥ 0 and some 0 = t ∈ V
[q] for which s(d)t = t and
degd(t)= 0.Thentheequation has nosolution inV[q].In particular:
(1) the equation (d− 1)X = t hasno solution inV[q]whenever t∈ L(d,q),and
(2) the equation (d− 1)X = d∞ hasno solution inV[d∞].
Proof. Let v = s(d). Since vd= d and t= dnt, weget vt = t.Soif x is asolutionof (d− 1)X = t,then we would havev(d− 1)x= t, so that (d− 1)vx = t;thus we may assumewithoutloss thatvx= x.Hencetheequationyields
x = dx− t,
x = kd∞+ x0+ dx1+· · · + dsxs,
where the xis are linearcombination of elements in L(d,q), and k = 0 in caseq = d∞.
Then,usingd· d∞= d∞,weget
t = dnt = (d− 1)x = dx − x
= kd∞− kd∞− x0+ d(x0− x1) +· · · + ds(xs−1− xs) + ds+1xs =−x0+ d(x0− x1) +· · · + ds(xs−1− xs) + ds+1xs.
Weclaimthatthis isimpossible.Setx−1= 0= xs+1. Bytheuniquenessofthe decom-position inRemark 3.5,since degd(t)= 0, onegetst = xn−1− xn and xi−1− xi = 0 for any i = n, −1 ≤ n ≤ s+ 1. Then we have 0 = x0 = x1 = · · · = xn−1 and
t =−xn=· · · − xs+1= 0, contradiction. 2
AssumeE isafinite graphandd asimpleclosedpathinE.Inordertocomputethe groupsExt1LK(E)(V[d∞],T ) foranyChensimplemoduleT ,wecanconsidertheprojective
resolutionofV[d∞] 0→ LK(E) ˆ ρ(d−1) −→ LK(E) ˆ ρd∞ −→ V[d∞]→ 0 ensuredbyTheorem 2.8.
Lemma 3.9. LetE be a finite graph. Letd be a simple closed path in E and letT be a Chensimple module.Consider theexact sequence
HomLK(E)(LK(E), T )
ˆ
ρ(d−1)∗
HomLK(E)(LK(E), T )
π
Ext1LK(E)(V[d∞], T ) 0
whereρˆ(d−1)∗(φ)= φ◦ ˆρd−1,andπ istheconnectinghomomorphism. Then
π( ˆρt) = 0 if and only if the equation (d− 1)X = t has a solution in T.
Consequently, Ext1LK(E)(V[d∞],T )= 0 if and only if (d− 1)X = t has a solution in T
foreveryt∈ T .
Proof. Byexactnessitfollowsthatπ( ˆρt)= 0 ifand onlyifthereexists x∈ T suchthat ˆ
ρt= ˆρ(d−1)∗( ˆρx) = ˆρx◦ ˆρ(d−1)= ˆρ(d−1)x
i.e.ifandonlyiftheequation (d− 1)X = t hasasolutioninT .
Theorem 3.10 (Type (2)). Let E be a finite graph. Let d be a simple closed path in E and letT beaChen simplemodule. Thenthefollowingare equivalent:
(1) Ext1LK(E)(V[d∞],T )= 0.
(2) s(d)∈ U(T ).
Proof. (1)⇒ (2)Ifs(d)T = 0,thenforanyt∈ T wehaves(d)t= 0,sobyLemma 3.7(1) theequation (d− 1)X = t admitsasolutionforanyt∈ T .ApplyingLemma 3.9(2),we getthatExt1LK(E)(V[d∞],T )= 0.
(2) ⇒ (1)FirstassumeT = V[d∞].AsobservedinRemark 3.5,T admitsagenerator
q notdivisiblebyd andL(d,q)isnotempty.Letp∈ L(d,q).ByLemma 3.8,theequation
(d− 1)X = p has nosolutioninV[q]andso,byLemma 3.9,π( ˆρp)= 0.
On theotherhand,supposeT = V[d∞].ByLemma 3.8,theequation(d− 1)X = d∞
hasnosolutioninV[d∞],andso,againinvokingLemma 3.9, π( ˆρd∞)= 0.
In either case we have established the existence of a nonzero element in Ext1LK(E)(V[d∞],T ). 2
Corollary 3.11. Let E be a finite graph. For any simple closed path d,
Ext1LK(E)(V[d∞],V[d∞])= 0.
Example 3.12.WeagainrevisitthegraphR2:
•v
e f .
Let q beany element ofR∞2 . Letd beany (oftheinfinitely many)simpleclosedpaths in R2. Since clearly Condition (2) of Theorem 3.10 is satisfied for V[q], we get that
Ext1LK(R2)(V[d∞],V[q])= 0. 2
Havingnowestablishedconditionswhichensurethatthereexistnontrivialextensions of theChen simplemoduleT bythesimplemodule V[d∞], wenow giveamoreexplicit
descriptionofthenumberofsuchextensions.
Proposition 3.13. LetE be any finite graph. Letd be asimple closed path inE and let T be aChensimple module. Assumeq∈ E∞ such thatT = V[q].
(1) SupposeT = V[d∞].ThendimKExt1LK(E)(V[d∞],T )=|L(d,q)|.
(2) Ontheotherhand, dimKExt1LK(E)(V[d∞],V[d∞])=|L(d,d∞)|+ 1.
Proof. We considertheexactsequence
HomLK(E)(LK(E), V[q])
ˆ
ρ(d−1)∗
HomLK(E)(LK(E), V[q])
π
