On Galois coverings and tilting modules

0.39 MB·English

by  Patrick Le Meur

On Galois
overings and tilting modules
∗†
Patri
k Le Meur
7
February 2, 2008
0
0
2

n
Abstra
t
a A k T
J Let be a basi

onne
ted (cid:28)nite dimensional algebra over an algebrai
ally
losed (cid:28)eld . Let be
A G

2 a basi
tilting -modulewith arbitrary (cid:28)niteprojAe
tive dimensioGn. For a (cid:28)xed group we
omparethe
1 set of iso
lasses of
onne
ted Galois
overings of with group and th→−e set of iso
lasses of
onne
ted
End (T) G K
Galois
overings of A with group . Using the Hasse diagram A (see [17℄ and [23℄) of basi
A T
] tilting -modules, we give su(cid:30)
ient
onditions on under whi
h there is a bije
tion between these two
T A
sets (these
onditions are always veri(cid:28)ed when is of (cid:28)nite representation type). Then we apply these
R A EndA(T)
results to studywhenthesimple
onne
tednessof impliestheone of (see [5℄). Finally, using
.
h an argument due to W. Crawley-Boevey, we prove that the type of any simply
onne
ted tilted algebra
t is a tree and that its (cid:28)rst Ho
hs
hild
ohomology group vanishes
a
m
Introdu
tion
[

k A k
4 Let be an algebrai
ally
losed (cid:28)eld and let be a (cid:28)nite dimensional -algebra. In order to study the
mod(A) A A
v
ategory of (cid:28)nite dimensional (left) -modules we may assume that is basi
and
onne
ted.
mod(A) T
7 In the study of , tilting theory has proved to be a powerful tool. Indeed, if is a basi
tilting
A B = End (T) A B
4 -module and if we set A , then and have many
ommon properties: Brenner-Butler
mod(A) mod(B)
6
Theorem establishes an equivalen
e between
ertain sub
ategories of and (see [10℄, [16℄
9 A B
and[21℄), and haveequivalentderived
ategories (see[14℄) and(inparti
ular) theyhaveisomorphi
0
Grothendie
k groups and isomorphi
Ho
hs
hild
ohomologies. In this text we will study the following
6 A B
problem relating and :
0
A B (P )
/ is it possible to
ompare the Galois
overings of and those of ? 1
h
A=kQ Q T
at Asanexample,if x Qwith a(cid:28)niteBqu=ivkeQrw′ ithoutQor′iented
y
leandifQ isanAPR-tiltingmodule
m asso
iatedtxoasink of (sQee[6℄)Qth′en where isobtainedfrom byreverAsingallthearrows
endings at . In parti
ular and have the same underlying graph and therefore has a
onne
ted
G B
:
v Galois
overing with group if and only if the sameholAds for . A k
i Re
all that in order to
onsider Galois
overings of we always
onsider as a -
ategory. When
X C →A mod(A) C
isaGalois
overing,itispossibletodes
ribepartof intermsof -modules(seeforexample
mod(C) mod(A) C
r [9℄and[12℄). Thisdes
riptionisusefulbe
ause iseasiertostudythan ,espe
iallywhen
a A
issimply
onne
ted(thislastsituationmayo

urwhen isof(cid:28)niterepresentationtype,see[12℄). Noti
e
that simple
onne
tedness and tilting theory have already been studied together through the following

onje
ture formulated in [5℄:
A =⇒ B (P )
2
is simply
onne
ted is simply
onne
ted
A T
More pre
isely, the above impli
ation is true if: is of (cid:28)nite representation type and is of proje
tive
A=kQ Q B
dimensionat mostone(see [2℄),or if: (with a quiver)and istame(see [5℄,see also [3℄for a
(P ) (P ) A
1 2
generalisationtothe
aseofquasi-tiltedalgebras). Thetwoproblems and arerelatedbe
ause
C →A C
issimply
onne
ted if and onlyif thereisnoproper Galois
overing with
onne
tedandlo
ally
bounded (see [19℄).
(P ) T
1
In order to study the question we will exhibit su(cid:30)
ient
onditions for to be of the (cid:28)rst kind
F
C −→ A T F
w.r.t. a (cid:28)xed Galois
overing . Indeed, if is of the (cid:28)rst kind w.r.t. , then it is possible to
B T

onstru
t a Galois
overing of . Under additional hypotheses on ,theequivalen
e
lass of this Galois
∗
adress: DépartementdeMathématiques,E
olenormalesupérieuredeCa
han,61avenueduprésidentWilson,94235Ca
han,
Fra†n
e
e-mail: plemeurdptmaths.ens-
a
han.fr
1
F

Fo:veCri→ngAis uniqFu′e:lyC′d→eteArmined by the equivalen
e
lass of . Here we say that two Galokis
overings
and areequivalentif andonlyifthereexistsa
ommutativesquareof -
ategories
k
and -linear fun
tors: C ∼ // C′
F F′
(cid:15)(cid:15) (cid:15)(cid:15)
A ∼ // A
wherehorizontalarrowsareisomorphismsandwherethebottomhorizontalarrowrestri
tstotheidentity
A A B
map on the set of obje
ts of . For simpli
ity, let us say that and have the same
onne
ted Galois
G Gal (G) Gal (G) Gal (G)
A B A

overingswithgroup ifthereexistsabije
tionbetweenthesets and where
Gal (G) C →A C →B
B
(resp. ) stands for the set of equivalen
e
lasses of Galois
overings (resp. ) with
G C
group andwith
onne
tedandlo
allybounded. Withthisde(cid:28)nition,weprovethefollowingtheorem
(P )
1
whi
h is themainresult of this textand whi
h partially answers :
T A B =End (T) G
A
Theorem 1. Let be a basi
tilting -module, let and let be group.
→− →−
T′ ∈ K K T End (T) End (T′)
A A A A
1. If lies in the
onne
ted
omponent of
ontaining , then and
G
have the same
onne
ted Galois
overings with group .
→−
T K A DA A B
A
2. If lies in the
onne
ted
omponent of
ontaining or , then and have the same
G

onne
ted Galois
overings with group .
→−
K A A B
A
In parti
ular, if is
onne
ted (whi
h happens when is of (cid:28)nite representation type) then and
G G
have the same
onne
ted Galois
overings with group , for any group .
→−
K T A
A A
Here is the Hasse diagram asso
iated with the poset of basi
tilting -modules (see [17℄ and
A A
[23℄). Re
all(see[12℄)thatwhen isof(cid:28)niterepresentationtype, admitsa
onne
tedGalois
overing
G G π (A)
1
with group if and only if is a fa
tor groupof thefundamental group of theAuslander-Reiten
A A B
quiver of with its mesh relations. Theorem 1 allows us to get the following
orollary when and
are of (cid:28)niterepresentation type. We thankIbrahimAssemfor having pointedoutthis
orollary.
T A B = End (T) A B
A
Corollary 1. Let be a basi
tilting -module and let . If both and are of (cid:28)nite
A B
representation type, then and have isomorphi
fundamental groups.
(P )
2
Theorem 1 also allows us to prove thefollowing
orollary related to .
T A B=End (T)
A
Corollary 2. (see [2℄ and [3℄) Let be a basi
tilting -module and let .
→− →−
T′ ∈ K K T End (T)
A A A
1. If lieEsnindth(eT
′o)nne
ted
omponentof
ontaining ,then: issimply
onne
ted
A
if and only if is simply
onne
ted.
→−
T K A DA A
A
2. If liesin the
onne
ted
omponent of
ontaining or then: issimply
onne
ted ifand
B
only if is simply
onne
ted.
→−
K A A
A
In parti
ular, f is
onne
ted (e.g. is of (cid:28)nite representation type, see [17℄), then: is simply
B

onne
ted if and only if is simply
onne
ted.
→−
K A
A
Theorem1usestheHassediagram →−inordertoprove(inparti
ular)thatabasi
tilting -module
K
A
lyinginaspe
i(cid:28)

onne
ted
omponentof isofthe(cid:28)rstkindw.r.t. agiven
onne
tedGalois
overing
A A
of . OMn the other hand, aEnxta1rg(Mum,eMnt)d=ue0to W. Crawley-Boevey (see [13℄) proves that any rAigid -
module (i.e. su
h that A ) is of the (cid:28)rst kind w.r.t. any Galois
overing of with
Z car(k)=0 Z/pZ car(k)=p p
group if (resp. if , prime). Using this argumentwe are able to prove the
(P )
2
following proposition. Itgivesapartialanswer to andtoA.Skwro«ski'squestionin[24, Problem1℄
wetheritistruethatatriangularalgebraissimply
onne
tedifandonlyifits(cid:28)rstHo
hs
hild
ohomology
group vanishes.
A = kQ Q T
Proposition 1. (see [3℄, [5℄) Assume that with a quiver without oriented
y
le. Let be a
A B=End (T) B Q
A
basi
tilting -moduleandlet (hen
e, istiltedoftype ). Thenthefollowingimpli
ation
holds:
B =⇒ Q HH1(B)=0
is simply
onne
ted is a tree and
B B
Re
all thatthe impli
ation of Proposition 1 has beenprovedin [5℄ for tametilted and in [3℄ for
tamequasi-tilted.
The text is organised as follows. In Se
tion 1 we will give the de(cid:28)nition of all the notions mentioned
aboveandwhi
hwillbeusedfortheproofofFT′heoBrem1. InSe
tion2wewilldetailtFhe:
Con→strAu
tioAnand
give some properties of the Galois
overing of starting from a Galois
overing of and
2
A T A
aTba1si
Ttilting -module . In thisFstudy, we will introdFu′
e the fol2lowingChypotheseFs.oTn the -module
T: ) is of the (cid:28)rst kind w.r.t. (this ensures thatF′ exists), ) theF-module ob3taψin.eNd ≃froNm
by restri
ting the s
alaNrs is bTasi
(this ensures that is
ψo:nAne
−∼→tedAif is
onne
ted) )
for any dire
t summand of and for any automorphismF′ whi
h restri
ts to the identity
map on obje
ts (this ensures that the equivalen
e
lass of does depend only on the equivalen
e
lass
F
of ). These three hypotheses la
k of simpli
ity, therefore, Se
tion 3 is devoted to (cid:28)nd simple su(cid:30)
ient
A T

onditionsforthebasi
tilting -modu→−le toverifythese. Inparti
ular,wewillprovethatthe
ondition
T K A
A
(cid:16) lies in the
onne
ted
omponent of
ontaining (cid:17) (cid:28)ts ourrequirements. Sin
eourmainobje
tive
A
is to establish a
orresponden
e between the equivalen
e
lasses of the
onne
ted Galois
overin→−gs of
B T K
A
→a−nd those of , we will need to (cid:28)nd
onditions for to lie in both
onne
ted
omponents of and
K A B T B
B

ontaining and respe
tively (re
all that →−is also a→−basi
tilting -module). This will be done
K K
in Se
tion4 where we→−
omparetheHasse diagrams A and B. Inparti
ular, we will prove thatth→−ere
K A T K
A B
is an oriented path in starting at and ending at if and only if there is an oriented path in
B T
starting at and ending at . This equivalen
e will be used in Se
tion 5 in order to prove Theorem 1,
Corollary 1 and Corollary 2. Finally, in Se
tion 6, we prove Proposition 1.
Iwouldliketoa
knowledgeEduardoN.Mar
osforhisstimulatingremarks
on
erningtheimpli
ation
(P )
2
during theCIMPA s
hool Homologi
al methods and representations of non-
ommutative algebras in
Mar del Plata, Argentina (February2006).
1 Basi
de(cid:28)nitions and preparatory lemmata
k k C
Reminder on -
ategories (see [9℄ for more details). A -
ategory is small
ategory su
h that for
x,y∈Ob(C) C x y k
y x
any theset ofmorphismsfrom to isa -ve
torspa
e andsu
hthatthe
omposition
C k k C
of morphisms in is -bilinear. A -
ategory is
alled
onne
ted if and only if there is no non trivial
Ob(C)=E⊔F C = C =0 x∈E,y∈F
y x x y
partition sku
h that for anyk F. : E → B F′: E′ → B
All fun
tors bketween -
ategorieFs are Fsu′pposed to be -linear. If and are
fun
torsbetween -
ategories, then and are
alledequivalentifthereexistsa
ommutativediagram:
E ∼ //E′
F F′
(cid:15)(cid:15) (cid:15)(cid:15)
B ∼ //B
wherehorizontalarrowsareisomorphismsandwherethebottomhorizontalarrowrestri
tstotheidentity
Ob(B) k k C
mapon . A lo
ally bounded -
ategory is a -
ategory verifyingthe following
onditions:
C
. distin
tobje
ts in are not isomorphi
,
x∈Ob(C) k C C
. for any ,the -ve
torspa
es y∈Ob(C) y x and y∈Ob(C) x y are (cid:28)nite dimensional,
L L
x∈Ob(C) k C
x x
. for any ,the -algebra is lo
al.
A k A
For example, let be a basi
(cid:28)nite dimensional -algebra (basi
means that is the dire
t sum of
A {e ,...,e }
1 n
pairwise non-isomorphi
inde
omposable proje
tive -modules)and let bea
ompleteset of
A k
pairwise orthogonal primitive idempotents. Then
an be viewed as a lo
ally bounded -
ategory as
e ,...,e A e e e Ae i,j
1 n i j j i
follows: aretheobje
tsof ,thespa
eofmorphismsfrom to isequalto forany
A
and the
omposition of morphismsis indu
ed by the produ
t in . Noti
e that di(cid:27)erent
hoi
es for the
e ,...,e k
1 n
primitiveidempotents giverisetoisomorphi
-
ategories. Inthistextweshallalways
onsider
A k
su
h an algebra as a lo
ally bounded -
ategory.
k C k C k M: C →
Modules over -
ategories. If is a -
ategory, a (left) -module is a -linear fun
tor
MOD(k) MOD(k) k C M → N
where is the
ategory of -ve
tor spa
es. A morphism of -modules is a
k C MOD(C)
-linear natural transformation of fun
tors. The
ategory of -modulesis denotedby .
C M M(x)
A -module is
alled lo
ally (cid:28)nite dimensional (resp. (cid:28)nite dimensional) if and only if is
x∈Ob(C) M(x)
(cid:28)nitedimensionalforany (resp. x∈Ob(C) is(cid:28)nitedimensional). The
ategoryoflo
ally
LC Mod(C) mod(C)
(cid:28)nite dimensional (resp. (cid:28)nite dimensional) -modules is denoted by (resp. ). Noti
e
C =A Mod(C)=mod(C)
thatif as above, then .
IND(C) Ind(C) ind(C) MOD(C)
We shall write (resp. , resp. ) for the full sub
ategory of (resp. of
Mod(C) mod(C) C M =N ... N N ∈ind(C)
1 t i
,resp. of )ofinde
omposable -modules. Finally,if with
i M N1,...,Nt L L
for any , then is
alled basi
if and only if are pairwise non isomorphi
.
3
A k A
Tilting modules. Let be a basi
(cid:28)nite dimensional -algebra. A tilting -module (see [10℄, [16℄
T ∈mod(A)
and [21℄) is a module verifying thefollowing
onditions:
(T1) T pd (T)<∞
A
has (cid:28)niteproje
tive dimension (i.e. ),
(T2) Exti (T,T)=0 i>0 T
A for any (i.e. is selforthogonal),
(T3) mod(A) 0→A→T →...→T →0 T ,...,T ∈add(T)
1 r 1 r
thereisanexa
tsequen
e in : with (this
T ,...,T T
1 r
last propertymeansthat are dire
tsums of dire
tsummandsof ).
(T1) (T2)
A module whi
h satis(cid:28)es
onditions and above is
alled an ex
eptional module. Assume that
T A T End (T) f.t = f(t)
A
is a tilting -module. Then, is also a tilting -module for the following a
tion:
f ∈End (T) t∈T T A
A
for and . Assume moreover that is basi
as an -module and (cid:28)x a de
omposition
T =T ⊕...⊕T T ,...,T ∈ind(A) End (T)
1 n 1 n A
with . This de(cid:28)nesa de
omposition oftheunitof into a
B :=End (T) k
A
sumofprimitivepairwise orthogonalidempotentssothat isalo
allybounded -
ategory
{T ,...,T } i,j B
as follows: the set of obje
ts is 1 n and for any the spa
e of morphisms Tj Ti is equal to
Hom (T ,T ) x∈Ob(A) T(x) B
A i j
. For any , is aninde
omposable -module:
B → MOD(k)
T ∈Ob(B) 7→ T (x)
i i
u∈ B 7→ T (x)−u−→x T (x)
Tj Ti i j
T = T(x) k
and x∈Ob(A) . Finally, thefollowing fun
tor is an isomorphismof -
ategories:
L
ρ : A −→ End (T)
A B
x∈Ob(A) 7−→ T(x)∈Ob(End (T))
B
T(u)
u∈ A 7−→ T(x)−−−→T(y)
y x
End (T)
A
For more details on the above properties and for a more general study of , we refer the reader
to [10℄, [14℄, [16℄ and [21℄.
T A T
A A
Let bethesetofbasi
tilting -modulesuTp∈toTisomorphism. Then isendowedwTit⊥hapTartialorder
A
introdu
edin [23℄ and de(cid:28)nedas follows. If , theright perpendi
ular
ategory of is de(cid:28)ned
by(see [7℄): T⊥ ={X ∈mod(A) | (∀i>1) Exti (T,X)=0}
A
′
T′ ∈T T 6 T′ T⊥ ⊆ T ⊥
If A is another basi
tilting module, we write provided that . In parti
ul→−ar, we
T 6A T ∈T K
A A
have forany . In→−[17℄,D.HappelandL.UngerhaveprovedthattheHassed→−iagram of
T K T T →T′ K
A A A A
Tis=asXfollowTs. TheXve∈rtii
neds(iAn) T′a=reYtheeTlementsYin∈inda(nAd)thereisanarrow in ifandonly
if: u vwith , with andthereexistsanonsplitexa
tsequen
e
0 → X −→LM −→ Y → 0 mod(A) LM ∈ add(T) u v
in with . In su
h a→−situation, (resp. ) is the left (resp.
add(T) X Y K
A
right) -approximationof (resp. ). Formoredetailson ,wereferthereaderto[17℄and[18℄.
k G G k E
Galois
overingsof -
ategories. Let beagroup. Afree -
ategoryisa -
ategory endowed
G→Aut(E) G Ob(E)
with amorphismofgroups su
hthattheindu
eda
tion of on isfree. Inthis
ase,
E →E/G E G k
there exists a (unique)quotient of by in the
ategory of -
ategories. With this property,
B G F: E →B
aGalois
overingof withgroup isbyde(cid:28)nitionafun
tor endowedwithagroupmorphism
G→Aut(F)={g∈Aut(E)|F ◦g=F}
and verifying thefollowing fa
ts:
G→Aut(F)֒→Aut(E) E G
. thegroup morphism endows with a stru
ture of free -
ategory,
F
E/G−→B F
. thefun
tor indu
edby is an isomorphism.
G→Aut(F)
This de(cid:28)nition implies that thEe group morphism isxon∈e-Otob(-oBn)e (a
tualFly−o1n(xe)
an show that
this is an isomorphismFwhenx is
onne
tFed−)1.(FM(oxr)e)o=verGf.oxr any x∈Ob(Eth)e set is non empty
and
alled the(cid:28)ber of at . It veri(cid:28)es for any .
Were
allthatGalois
overingsareparti
ular
asesof
overingfun
tors(see[9℄). A
overingfun
tor
k F: E → B x,y ∈ E F
0
is a -linear fun
tor su
h that for any , the following mappings indu
ed by are
bije
tive:
y′Ex→ F(y)BF(x) yEx′ → F(y)BF(x)
and
y′∈FM−1(F(y)) x′∈FM−1(F(x))
Remarkthata
overingfun
toris notsupposedto restri
t to a surje
tive mappingonobje
ts. However,
k
a
overing fun
tor is an isomorphism of -
ategories if and only if it restri
ts to a bije
tive mapping on
obje
ts. Using basi
linear algebra argumentsit is easy to prove thefollowing useful lemma:
4
p,q k q◦p p,q q◦p
Lemma 1.1. Let be -linear fun
tors su
h that the
omposition is de(cid:28)ned. Then and
are
overing fun
tors as soon as two of them are so.
F: E →B G B E
If isaGalois
overingwithgroup andwith
onne
tedthen neednotbe
onne
ted. In
E = E E E i
i i
su
ha
ase, if wherethe 'sarethe
onne
ted
omponentsof ,thenforea
h ,thefollowing
i‘∈I
fun
tor:
F :E ֒→E →B
i i
is a Galois
overing with group:
G :={g∈G | g(Ob(E ))∩Ob(E )6=∅}={g∈G| g(Ob(E ))=Ob(E )}
i i i i i
i,j ∈ I G G G
i j
Moreover, if then the groups and are
onjugated in and there exists a
ommutative
diagram:
Ei? ∼ //Ej
?? (cid:127)(cid:127)
? (cid:127)
? (cid:127)
Fi ??(cid:31)(cid:31) (cid:127)(cid:127)(cid:127)(cid:127)(cid:127) Fj
B
G {E |i∈I}
i
wherethehorizontalarrowisanisomorphism. Thisimpliesthat a
tstransitivelyontheset
E B
of the
onne
ted
omponentsof . Noti
e thatall these fa
ts maybe false if is not
onne
ted.
B
Two Galois
overingsof are
alledequivalentifandonlyif theyareisomorphi
as fun
torsbetween
k
-
ategories (see above, this implies thatthe groups of the Galois
overings are isomorphi
). The equiv-
F [F]
alen
e
lass of a Galois
overing will be denoted by . Finally, we shall say for short that a Galois
E →B E B

overing is
onne
ted if and only if is
onne
ted and lo
ally bounded(this impliesthat is

onne
ted and lo
ally bounded,see [12, 1.2℄).
k B k
Simply
onne
ted lo
ally bounded -
ategories. Let be a lo
ally bounded -
ategory. Then
B B
is
alled simply
onne
ted if and only if there is no proper
onne
ted Galois
overing of (proper
meanswithnontrivialgroup). Thisde(cid:28)nitionisequivalenttotheoriginalone(see[20℄forthetriangular
B

ase and [19, Prop. 4.1℄ for the non-triangular
ase) whi
h was introdu
ed in [1℄: is simply
onne
ted
π (Q ,I)=1 kQ /I ≃B B
1 B B
ifandonlyif foranyadmissiblepresentation of (see[20℄forthede(cid:28)nitionof
π (Q ,I)
1 B
).
F:E →B
Basi
notions on
overing te
hniques(see [9℄ and[22℄). Let bea Galois
overingwith
G G E G MOD(E) M ∈MOD(E) g∈G
ggMrou:p= F. ◦Tgh−e1 ∈-aM
tiOonDo(nE) gives risetoFana
tion of on : ifF : MOD(E) →anMdOD(B), then
λ
. Moreover, de(cid:28)nes two additive fun
tors (the
F.: MOD(B) → MOD(E)
push-down fun
tor) and (the pull-up fun
tor) with the following properties
(for more details we refer thereader to [9℄):
F.M =M◦F M ∈MOD(B)
. for any ,
. if M ∈ MOD(E), then (FλM)(x) = x′∈F−1(x)M(x′) for any x ∈ Ob(B). If u ∈ yEx, then the
(F M)(F(u)) M(g.x)L g.x∈F−1(F(x))=G.x M(g.u): M(g.x)→
λ
restri
tion of to (for )isequalto
M(g.y)
,
F F.
λ
. and are exa
tand sendproje
tive modules to proje
tive modules,
F E ≃ B F.B ≃ E E B E x 7→ E
. λ g∈G and , where (resp. ) is the -module y∈Ob(E) y x (resp. the
B-moduLle x7→ y∈Ob(B) yBx), L
F.F = LgId
. λ g∈G MOD(E)
X ∈MLOD(B) X ≃F Y Y ∈MOD(E) F F.X ≃ X
. if veri(cid:28)es λ for some , then λ g∈G ,
F (mod(E))⊆mod(B) F (Mod(E))⊆Mod(B) F.(Mod(B))⊆Mod(E) L
λ λ
. , , ,
D◦F.=F.◦D D◦F ≃F ◦D D=Hom (?,k)
λ|mod(E) λ |mod(E) k
. and where is theusual duality,
F F.
λ
. is left adjoint to ,
D◦F ◦D F. Hom (F.M,N)≃
λ E
. isrightadjointto (inparti
ular, thereisa fun
torialisomorphism
Hom (M,F N) M ∈MOD(B) N ∈mod(E)
B λ
for any and any ).
M,N ∈MOD(E) F
λ
. for any , thefollowing mappingsindu
edby are bije
tive:
Hom (gM,N)→Hom (F M,F N) Hom (M, gN)→Hom (F M,F N)
E B λ λ E B λ λ
and
gM∈G gM∈G
5
These properties give the following result whi
h will be usedmanytimes in this text:
M ∈MOD(E) M ∈MOD(B) F M
λ
Lemma1.2. If (resp. ) has(cid:28)niteproje
tive dimension,thensodoes
F.(M)
(resp. ).
M ∈MOD(E) N ∈MOD(B) j >1
Let , and . There is an isomorphism of ve
tor spa
es:
Extj(F M,N)≃Extj(M,F.N)
B λ E
M ∈mod(E)
Moreover, if then there is an isomorphism of ve
tor spa
es:
Extj(F.N,M)≃Extj(N,F N)
E B λ
F. F
λ
Proof: The (cid:28)rst assertion is due to the fa
t that and are exa
t and send proje
tive modules
F. F F.: D(MOD(B)) → D(MOD(E))
λ
to proje
tive modules. For the same reasons, and indu
e and
F : D(MOD(E)) → D(MOD(B)) (F ,F.) (F.,F )
λ λ λ
respe
tively and the adjun
tions and Exatjt(tXh,eYle)v=el
of module
ategories give rise to adjun
tions at the level of derived
ategories. Sin
e E
Hom (Y,X[j]) (cid:4)
D(MOD(E))
we get theannoun
ed isomorphisms.
k F
Remark that an isomorphism of -
ategories is a parti
ular
ase of Galois
overing. When is an
F. F
λ
isomorphism, and haveadditional properties as shows thefollowing lemmawhose proof isa dire
t

onsequen
e of thede(cid:28)nition of thepush-down and pull-upfun
tors.
F: E → B k F.F = Id
λ MOD(E)
Lemma 1.3. Assume that is an isomorphism of -
ategories. Then and
F F.=Id
λ MOD(B)
.
F: E → B G B M
Modules of the (cid:28)rst kind. Let be a Galois
overing with group . A -module is
F N M

alledofthe(cid:28)rstkindw.r.t. ifandonlyifforanyinde
omposabledire
tsummand of thereexists
N ∈MOD(E) N ≃F N ind (B) mod (B)
λ 1 1
su
h that . We will denote by (resp. ) the full sub
ategory of
ind(B) mod(B) F ind (B)
b (resp. of )ofmodulbesofthe(cid:28)rstkindw.r.t. . Noti
ethefollowingpropertiesof 1 :
M ∈ind (B) N ∈MOD(E) M ≃F N N ∈ind(E)
1 λ
. if and verify , then ,
M ∈ind (B) N,N′ ∈MOD(E) M ≃F N ≃F N′ g∈G
1 λ λ
. Nif′ ≃ gN and verify , thenthere exists su
h that
.
B E = E E E
If is
onne
ted and if i∈I i, where the i's are the
onne
ted
omponents of , then an
B M ‘ F
inde
omposable -module is of the (cid:28)rst kind w.r.t. if and only if it is of the (cid:28)rst kind w.r.t.
F : E ֒→ E → B i ∈ I
i i
for any . More pre
isely, we have the following well know lemma where we keep
theestablished notations.
M ∈ind(B) M ∈ind(E) F M ≃M i∈I
λ
Lemma 1.4. Let . If is su
h that , then there is a unique su
h
M ∈ind(E ) M ≃(F ) M j ∈I g∈G
that i . In su
h a
ase, wce have i λ . Morecover, if then there exists su
h
g(E )=E g gM ∈ind(E ) (F ) gM ≃M
that c i j, and for any su
h we have: c j and j λ .
c c
A k n
Throughout this text will denote a basi
and
onne
ted (cid:28)nite dimensional -algebra and will
K (A)
0
denote therankof its Grothendie
k group .
2 Galois
overings asso
iated with modules of the (cid:28)rst kind
Throughout this se
tion we will use thefollowing data:
F: C →A G
- a Galois
overing with group
T = T ... T ∈ mod(A) T ∈ ind(A) A
1 n i
- (with ) a basi
tilting -module of the (cid:28)rst kind w.r.t.
F L L
,
λ : F (T )→T T ∈ind(C) i∈{1,...,n}
i λ i i i
- an isomorphism with ,for every .
B=EndbA(T) b
Let . With these data, we wish to:
F G B
1.
onstru
ta Galois
overing Ti,λi with group of ,
b
F T λ F
2. studythedependen
eoftheequivalen
e
lass of Ti,λi onthedata i, i andonthe
hoi
eof in
[F] b
its equivalen
e
lass , b
T B
3. repeatthe
onstru
tion madeat the(cid:28)rststepstartingfrom (viewedas a basi
tilting -module)
F End (T)
andtheGalois
overing Ti,λi. ThiswillgiveaGalois
overingof B whi
hwillbe
ompared
F b ρ : A−∼→End (T)
A B
with using theisomorphism .
6
F
b
2.1 Constru
tion of the Galois
overing Ti,λi
End ( gT ) k
Let C g,i i be the following -
ategory:
L b { gT | g∈G, i∈{1,...,n}} gT g′T
i i j
. th(ei,sget)6=of(ojb,jge′
)ts is ( and are
onsidered as di(cid:27)erent obje
ts
if ), b b b
gT hT Hom (gT , hT )
i j C i j
. thespa
e of morphismsfrom to is equal to ,
b b b mobd(C)
. the
omposition is indu
edbythe
omposition of morphismsin .
C gT F.T
Remark 2.1. 1. The -modules g,i i and are isomorphi
.
L
G C b k End ( gT )
2. If is a (cid:28)nite group, then is a (cid:28)nite dimensional -algebra. In parti
ular, C g,i i and
EndC(F.T) are isomorphi
k-algebras. L b
G mod(C) End ( gT ) G
3. The -a
tion on naturally endows C g,i i with a stru
ture of free -
ategory.
L
End ( gT ) G = 1b i
4. C i,g i is lo
ally bounded if and only if Ti for any . This is equivalent to say that
F.T L C b
is a basi
b -module.
λ ,...,λ
1 n
The isomorphisms de(cid:28)nethefollowing fun
tor:
F : End ( gT ) −→ B
Ti,λi C g,i i
b LgT 7−→ T
i b i
gT −→ub hT 7−→ T −λ−j−−F−λ−u−−λ−−i→1 T
i j i j
b b
F : End ( gT )→B G
Lemma 2.2. The fun
tor Ti,λi C i,g i is a Galois
overing with group .
b L
C′ Enbd ( gT ) F′: C′ → B F
Proof: For simpli
ity, we shall write for C i,g i and for Ti,λi. Re
all (see
G C′ L F′◦g=F′ Fb′ F′
Remark2.1)that a
tsfreelyon . Moreover,wehave b by
onstru
tionof . So, de(cid:28)nes
k k
a
ommutativediagram of -
ategories and -linear fun
tors:
C′ ⋆
CC ( )
CCCF′
C
C
(cid:15)(cid:15) C!!
C′/G //B
F′
C′ → C′/G F
λ
WherFe′ is the quotient funC
′t→or.CF′/rGom the properties veri(cid:28)ed by (see Se
tion 1) wFe′infer
that is a
overing fun
tor. Sin
e is a also
overing fun
tor we dedu
e that so is (see
F′ Ob(C′)/G={gT | g∈G, i∈{1,...,n}}/G→
i
OLebm(Bm)a=1.{1T).,F..in.a,Tlly,} rFes′tri
ts to a bije
tive mappinFg′ G (cid:4)
1 n so is an isomorphism. Thus, is a Galois
ovebring with group .
F End ( gT )
Sin
e Ti,λi isaGalois
overing,itisnaturaltoaskwhether C i,g i is
onne
tedornot. The
b L
following lemma partially answers this question. b
C End ( gT )
Lemma 2.3. If is not
onne
ted, then C i,g i is not
onne
ted.
L
C′ End ( bgT ) C
Proof: For simpli
ity let us write for C i,g i . Assume that is not
onne
ted and let
C = x∈ICx wheretheCx'sarethe
onne
ted
omLponentbsof C. Fori∈{1,...,n}, we haveTi∈ind(C),
‘ x ∈I T ∈ind(C )
so there exists a unique i su
h that i xi . Let us set: b
G b={g∈G| g(C )=C }
x1 x1 x1
i∈{1,...,n} G {C | x∈I} g ∈G g (C )=C
Let , sin
e a
ts transitively on x , there exists i su
h that i x1 xi
g ∈G
(in parti
ular 1 x1). Therefore:
(∀i∈{1,...,n}) gi−1Ti ∈mod(Cx1)
O C′ b
Let us set to be thefollowing set of obje
ts of :
O:={gT | i∈{1,...,n} gg ∈G }⊆Ob(C′)
i and i x1
O b
Remarkthat satis(cid:28)es thefollowing:
7
O6=∅ T ∈O
1
. be
ause .
. Sin
e C is not
bonne
ted and sin
e G a
ts transitively on {Cx | x ∈ I} we have Gx1 ( G. Let
g∈G\G gg 6∈G gT 6∈O O(Ob(C)
x1, then 1 x1 and 1 . Hen
e .
. For anygTi ∈Ob(C′), wCe′ havegTib∈O if and oOnly if gTi ∈ind(Cx1O).bA(Cs′)a\O
onsequen
e, thereis no
non zero mborphismin betweenban obje
t in and abn obje
t in .
C′ (cid:4)
As a
onsequen
e, is not
onne
ted.
b
F F,T ,λ
2.2 Independen
e of the equivalen
e
lass of Tbi,λi on the data i i
[F ] F
Inthetwofollowing lemmas,weexaminethedependen
eoftheequivalen
e
lass Ti,λi of Ti,λi onthe
T ,...,T ,λ ,...,λ F [F] b b
1 n 1 n

hoi
e of and on the
hoi
e of in its equivalen
e
lass .
b b i ∈ {1,...,n} µi: FλTi → Ti Ti ∈ ind(C)
Lemma 2.4. For ea
h , let be an isomorphism with . Then
F F
Ti,λi and Ti,µi are equivalent.
b
Proof: We need to exhibit a
ommutativesquare:
EndC( i,g gTi) ϕ // EndC( i,g gTi) (⋆)
L L
F F b
Ti,µi Ti,λi
B(cid:15)(cid:15) ψ //B(cid:15)(cid:15) b
ϕ,ψ ψ(x)=x x∈Ob(B)={T ,...,T } i∈{1,...,n}
1 n
where FλaTrei ≃isoTmio≃rpFhλisTmis and where for any θi: Ti −∼→ giTi gi ∈.GLet ϕ.
We have , so there exists an isomorphism with . Let us de(cid:28)ne
by: ϕb: End ( gT ) −→ End ( gTb)
C i,g i C i,g i
LgTi 7−→ gLgiTi b
gTi −→u hTj 7−→ ggiTi−−h−θj−−u−−bg−θi−−→1 hgjTj
ϕ k θ b b F θ : F T → F T
i λ i λ i λ i
Then is an isomorphism of -
ategories. Noti
e that de(cid:28)nes an isomorphism .
ψ
So we
an de(cid:28)ne by: b
ψ: B −→ B
T 7−→ T
i i
u
T −→T 7−→ ψ(u)
i j
ψ(u)
where is the
omposition:
T −λ−−i−→1 F T −F−λ−θ−i−→1 F T −µ→i T −→u T −µ−−j−→1 F T −F−λ−θ→j F T −λ→j T
i λ i λ i i j λ j λ j j
ψ kb Obb(B) ϕ ψ
So isanisomorphismof -
ategories whi
h restri
ts to theidentitymapon . Moreover and
(⋆) (cid:4)
make
ommutative.
T
In the following lemma, we show that, under additional hypotheses on , the equivalen
e
lass of
F F [F]
Ti,λi does notdependonthe
hoi
e of in .
b F′: C′ →A G F
Lemma 2.5. Assume that is a Galois
overing (with group ) equivalent to and assume
T
that veri(cid:28)es the following
ondition:
H ψ.T ≃T i ψ: A−∼→A
A,T i i
( ):(cid:17) for any and for any isomorphism whi
h restri
ts to the identity map on
Ob(A)
.(cid:17)
T F′ i∈{1,...,n} µ : F′T →T
TThe∈nindi(sC′o)f the (cid:28)rsFt′kind w.r.tF. . For ea
h let i λ i i be an isomorphism with
i . Then Ti,µi and Ti,λi are equivalent.
b F F′
Proof: Let us (cid:28)xan isomorphism between and :
C′ ϕ //C
F′ F
A(cid:15)(cid:15) ψ //A(cid:15)(cid:15)
8
ν: Aut(C′)→Aut(C) Aut(C′)=G Aut(C)=
Letus set to betheisomorphismof groups(re
all that and
G
): ν: Aut(C′) → Aut(C)
g 7→ ϕ◦g◦ϕ−1
g ∈Aut(C)=G g ∈Aut(C′)=G g MOD(C)
ReM
alOl Dth(aCt′)any (resp. )MdeO(cid:28)nDe(sCa′)n→auMtomOoDr(pAh)ism of (resp.
of ). Therefore we have an equalityof fun
tors :
(∀g∈Aut(C′)) ϕ ◦g=ν(g)◦ϕ
λ λ
θ : ψ.T →T i T =ϕ.T ϕ T =T
i i i i i λ i i
Letus(cid:28)xanisomorpψhi.sψm=Id ,forea
h ,andψlFet′u=sFseϕt . Inparti
ular:
(see Lemma1.3). Sin
e λ MOD(A) (lo
.
it.) and ,we infebr that: b
F′T =ψ.ψ F′T =ψF ϕ T =ψ.F T
λ i λ λ i . λ λ i λ i
i µ : F′T →T b
Therefore, we get for ea
h anisomorphism i λ i i equal to the
omposition:
µ : F′T =ψF T −ψ−.−λ→i ψT −θ→i T
i λ i . λ i . i i
T F′ b
This proves that is of the (cid:28)rst kind w.r.t. . A

ording to the pre
eding subse
tion, this de(cid:28)nes the
G
Galois
overing with group :
FT′i,µi: EndC′( gTi)→B
Mg,i
F′ F
Thanks to Lemma2.4 we only need to prove that Ti,µi and Ti,λi are equivalent.
ϕλ b
First, we have thefollowing fun
tor indu
edby :
ϕ: EndC′( i,g gTi) −→ EndC( i,g gTi)
gLT 7−→ ν(g)T L=ϕ gT
i i λ bi
gT −→u hT 7−→ ν(g)T −ϕ−λ−→u ν(h)T
i j ib j
ν: G→G bϕ ϕ.=Id b ϕ.ϕ =Id
λ MOD(C) λ MOD(C)
Sin
e isanisomorphismandbe
auseoftheequalities and
ϕ
(see Lemma1.3), thefun
tor is an isomorphism.
ψ
λ
Se
ondly,we have the following fun
tor indu
edby :
ψ: B −→ B
T 7−→ T
i i
T −→u T 7−→ T −ψ−λ−(−θ−j−−1u−θ−i→) T
i j i j
ψ ψ.=ψ.ψ =Id ψ
λ λ MOD(A)
Sin
e ,thefun
tor isawellde(cid:28)nedisomorphismandrestri
tstotheidentity
Ob(B)
map on . Therefore, we have a diagram whose horizontal arrows are isomorphisms and whose
bottomhorizontal arrow restri
ts to theidentitymapon theset of obje
ts:
EndC′( i,g gTi) ϕ // EndC( i,g gTi)
L L
F′ F b
Ti,µi Ti,λi
B(cid:15)(cid:15) ψ //B(cid:15)(cid:15) b
gT −→u hT
i j
This diagram is
ommutative,indeed, for any we have:
ψF′ (u) = ψ(µ F′(u)µ−1)=ψ (θ−1µ F′(u)µ−1θ )
Ti,µi j λ i λ j j λ i i
= ψ (θ−1θ ψ.(λ )F′(u)ψ.(λ )−1θ−1θ )
λ j j j λ i i i
= λ (ψ F′)(u) λ−1 ψ ψ.=Id
= λj (Fλϕλ)(u) λi−1 be
ause Fλϕ=ψF′MOD(C)
j λ λ i be
ause
= F (ϕ (u))=F ϕ(u)
Ti,λi λ Ti,λi
b b
F′ F (cid:4)
This proves that Ti,µi and Ti,λi are equivalent.
b
T A
→− Later, we shall prove that if is a basi
tilting -module lying in the
onne
ted
omponent of
K A H
A A,T

ontaining , then the hypothesis ( ) in the pre
eding lemma is automati
ally veri(cid:28)ed. As a
A

onsequen
e,forthesetilting -modules,theproperty(cid:16)tobeofthe(cid:28)rstkindw.r.t. anequivalen
e
lassof
A H A
A,T
Galois
overingsof (cid:17) doesmakesense. However,thehypothesis( )isnotveri(cid:28)edforany -module
T
as the following exampleshows.
9
A
Example 2.6. Let be the path algebra of the following quiver:
2
(cid:0)@@ >>
(cid:0) >
b (cid:0) >c
(cid:0) >
(cid:0) >
(cid:0) >
(cid:0) (cid:30)(cid:30)
1 //3
a
n = 3 k ψ: A −∼→ A ψ(x) = x
Here and we have an isomorphism of -
ategories: su
h that for any
x∈Ob(A) ψ(a)=a+cb ψ(b)=b ψ(c)=c i T A
i
, , and . For any integer , let be the following -module:
k
(cid:0)@@ >>
(cid:0) >
1(cid:0)(cid:0) >>1
(cid:0) >
(cid:0) >
(cid:0) (cid:30)(cid:30)
k // k
i
Then:
T T i
i i+1
- and are not isomorphi
, for any ,
car(k)6=2 T ,T ,T
1 2 3
- if , then are pairwise non isomorphi
,
ψ.T =T i
i i+1
- for any .
car(k)6=2 H T =T ⊕T ⊕T T
A,T 1 2 3
In parti
ular, if , thien hypothesEisxt(1(T ,T) i)s≃nokt satis(cid:28)edτfo(rT )≃T . Remark that
is not tilting. Indeed, for any , we have A i i be
ause A i i.
Remark2.1,Lemma2.4 and Lemma2.5 justify the following de(cid:28)nition:
H
A,T
De(cid:28)nition 2.7. Assume that the hypothesis ( ) is satis(cid:28)ed (see Lemma 2.5). The equivalen
e
lass
[F] F A T =T ... T [F] T ∈ind(A)
1 n i
of and the basi
tilting -module of the (cid:28)rst kind w.r.t. (with )
L L B G F
uniquely de(cid:28)ne an equivalen
e
lass of Galois
overing of with group and whi
h admits Ti,λi as a
[F]T: EndC(F.T)→B b
representative. This equivalen
e
lass will be denoted by .
[F] ([F] )
2.3 Comparison of and T T
C′ End ( gT ) F′ F
For short, let us write for C g,i i and for Ti,λi . In this subse
tion we shall not assume
that the hypothesis (HA,T)λof:LFemTLm−a∼→2T.5bisi∈sa{ti1s,(cid:28).e.d.,,nex}
ebpt for the last proposition. Starting froFm′ F
i λ i i
aBndfromtheisomorphisms , ,wehavFe′
onstru
tedtheGalois
overing Fo′f′
.EOnndem(Ta)y≃tryAto perform thesambe
onsFtr′u′
tionFstarting from in order to get a Galois
oTvering
B
of Fa′nd eventually
ompaLre: Ob(Aw)it→h O.bI(nC)this purpose, we need to prFov:eOtbh(aCt)→isOobf(Ath)e
(cid:28)rstkindw.r.t. . Letus(cid:28)xalifting ofthesurje
tive mapping .
x∈Ob(A) T[(x) C′
For , let be the -modulesu
h that:
T[(x)(gT )=T (g−1L(x)) gT ∈Ob(C′)
i i i
- for any ,
T[(x) gbT −→u bhT T (g−b1L(x))−u−L−(−x→) T (h−1L(x)) u∈ C′
- “ i j” is equal to i j for any hTj gTi.
b b
b i∈b{1,...,n} b b
Therefore, for any , we have:
F′T[(x) (T )= T[(x)(gT )= T (g−1L(x))= F T (x)
λ i i i λ i
“ ” gM∈G gM∈G “ ”
b b b
(µ ) : F′T[(x) (T )→(T(x))(T ) (F T )(x)−(−λi−)→x T (x)
So we mayset x Ti “ λ ” i i to be equalto λ i i .
Lemma 2.8. The linear isomorphisms (µx)Ti (i∈{1,...,n}) de(cid:28)ne anbisomorphism of B-modules:
µ : F′T[(x)−∼→T(x)
x λ
µ B u ∈ C′ F′(u) ∈
Proof: We only need to prove that x is a morphism of -modules. Let tTj sTi so that
TjBTi, and let us prove that thefollowing diagram
ommutes: b b
F′T[(x) (T ) (µx)Ti=(λi)x //T (x) ⋆
“ λ ” i i ( )
“Fλ′T[(x)”(F′(u)) (T(x))(F′(u))=F′(u)x
F′T[(x(cid:15)(cid:15)) (T ) (µx)Tj=(λj)x // T ((cid:15)(cid:15)x)
λ j j
“ ”
10