Infinite generation of non-cocompact lattices on right-angled buildings

0.09 MB·English

by  Anne Thomas

1
Infinite generation of non-cocompact lattices
1
0 on right-angled buildings
2

n ANNETHOMAS
a
J KEVINWORTMAN

5
2
LetΓbeanon-cocompactlatticeonalocallyfiniteregularright-angledbuildingX.

] WeprovethatifΓhasastrictfundamentaldomainthenΓisnotfinitelygenerated.
R
Weusetheseparationpropertiesofsubcomplexesof X calledtree-walls.
G
.
h 20F05;20E42,51E24,57M07
t
a
m Treelatticeshavebeenwell-studied(see[BL]). Lessunderstoodarelatticesonhigher-
[ dimensional CAT(0) complexes. In this paper, we consider lattices on X a locally

finite, regular right-angled building (see Davis [D] and Section1 below). Examples
2
v of such X include products of locally finite regular or biregular trees, or Bourdon’s
5
building I [B], which has apartments hyperbolic planes tesselated by right-angled
3 p,q
2 p–gonsandallvertexlinksthecompletebipartite graph K .
q,q
4
. LetG beaclosed,cocompactgroupoftype-preservingautomorphismsofX,equipped
9
0 withthecompact-opentopology, andlet Γ bealatticein G. Thatis, Γ isdiscreteand
0 theseries P|StabΓ(φ)|−1 converges,wherethesumisoverthesetofchambers φ ofa
1
fundamentaldomainfor Γ. Thelattice Γ iscocompactin G ifandonlyifthequotient
:
v Γ\X iscompact.
i
X
If there is a subcomplex Y ⊂ X containing exactly one point from each Γ–orbit on
r
a X, then Y is called a strict fundamental domain for Γ. Equivalently, Γ has a strict
fundamental domainif Γ\X maybeembedded in X.
Anycocompactlatticein G isfinitelygenerated. Weprove:
Theorem 1 LetΓ beanon-cocompactlatticein G. IfΓ hasastrictfundamental
domain,thenΓ isnotfinitelygenerated.
WenotethatTheorem 1contrastswiththefinitegenerationoflatticesonmanybuildings
whosechambersaresimplices. Resultsof,forexample,Ballmann–S´wiatkowski[BS´],
֒
Dymara–Januszkiewicz [DJ], and Zuk [Zu], establish that all lattices on many such
buildingshaveKazhdan’sProperty(T).Hencebyawell-knownresultduetoKazhdan
[K],theselatticesarefinitelygenerated.
2 AThomasandKWortman
Our proof of Theorem1, in Section3 below, uses the separation properties of sub-
complexes of X which wecall tree-walls. These generalize thetree-walls (in French,
arbre-murs) of I , which were introduced by Bourdon in [B]. We define tree-walls
p,q
andestablish theirproperties inSection2below.
Thefollowingexamplesofnon-cocompactlatticesonright-angledbuildingsareknown
tous.
(1) For i = 1,2, let G be a rank one Lie group over a nonarchimedean locally
i
compactfieldwhoseBruhat–Titsbuildingisthelocallyfiniteregularorbiregular
tree T . Then any irreducible lattice in G = G × G is finitely generated
i 1 2
(Raghunathan [Ra]). HencebyTheorem1above,suchlattices on X = T ×T
1 2
cannot havestrictfundamental domain.
(2) Let Λ be a minimal Kac–Moody group over a finite field F with right-angled
q
Weyl group W. Then Λ has locally finite, regular right-angled twin buildings
X+ ∼= X−,and Λ actsdiagonallyontheproduct X+×X−. For q largeenough:
(a) ByTheorem0.2ofCarbone–Garland[CG]orTheorem1(i)ofRe´my[Re´],
thestabilizerin Λ ofapointin X− isanon-cocompactlatticein Aut(X+).
Any such lattice is contained in a negative maximal spherical parabolic
subgroup of Λ, which has strict fundamental domain a sector in X , and
+
soanysuchlatticehasstrictfundamental domain.
(b) By Theorem 1(ii) of Re´my [Re´], the group Λ is itself a non-cocompact
lattice in Aut(X+)×Aut(X−). Since Λ is finitely generated, Theorem1
above implies that Λ does not have strict fundamental domain in X =
X+×X−.
(c) By Section 7.3 of Gramlich–Horn–Mu¨hlherr [GHM], the fixed set G of
θ
certain involutions θ of Λ is a lattice in Aut(X ), which is sometimes
+
cocompact andsometimesnon-cocompact. Moreover,by[GHM,Remark
7.13],thereexists θ suchthat G isnotfinitelygenerated.
θ
(3) In[T],thefirstauthorconstructedafunctorfromgraphsofgroupstocomplexes
of groups, which extends the corresponding tree lattice to a lattice in Aut(X)
where X is a regular right-angled building. The resulting lattice in Aut(X)
has strict fundamental domain if and only if the original tree lattice has strict
fundamental domain.
Infinitegenerationforright-angledbuildings 3
Acknowledgements
The first author was supported in part by NSF Grant No. DMS-0805206 and in
part by EPSRC Grant No. EP/D073626/2, and is currently supported by ARC Grant
No. DP110100440. The second author is supported in part by NSF Grant No.
DMS-0905891. We thank Martin Bridson and Pierre-Emmanuel Caprace for helpful
conversations.
1 Right-angled buildings
In this section we recall the basic definitions and some examples for right-angled
buildings. We mostly follow Davis [D], in particular Section 12.2 and Example
18.1.10. Seealso[KT,Sections1.2–1.4].
Let (W,S) bearight-angled Coxetersystem. Thatis,
W = hS | (st)mst = 1i
where m = 1 for all s ∈ S, and m ∈ {2,∞} for all s,t ∈ S with s 6= t. We will
ss st
discussthefollowingexamples:
• W1 = hs,t |s2 = t2 = 1i ∼= D∞,theinfinitedihedral group;
• W2 = hr,s,t | r2 = s2 = t2 = (rs)2 = 1i ∼= (C2 ×C2)∗C2, where C2 is the
cyclicgroupoforder 2;and
• The Coxeter group W3 generated by the set of reflections S in the sides of a
right-angled hyperbolic p–gon, p ≥ 5. Thatis,
W3 = hs1,...,sp | s2i = (sisi+1)2 = 1i
withcyclicindexing.
Fix (qs)s∈S a family of integers with qs ≥ 2. Given any family of groups (Hs)s∈S
with |Hs| = qs,let H bethequotient ofthefreeproduct ofthe (Hs)s∈S bythenormal
subgroup generated bythecommutators {[h ,h] :h ∈ H ,h ∈H,m = 2}.
s t s s t t st
Now let X be the piecewise Euclidean CAT(0) geometric realization of the chamber
system Φ = Φ(cid:0)H,{1},(Hs)s∈S(cid:1). Then X is a locally finite, regular right-angled
building, with chamber set Ch(X) in bijection with the elements of the group H. Let
δ : Ch(X)×Ch(X)→ W bethe W–valueddistancefunctionandlet l : W → N be
W S
wordlengthwithrespecttothegeneratingset S. Denoteby d : Ch(X)×Ch(X)→ N
W
4 AThomasandKWortman
thegallerydistance l ◦δ . Thatis,fortwochambers φ and φ′ of X, d (φ,φ′) isthe
S W W
lengthofaminimalgalleryfrom φ to φ′.
Supposethatφandφ′ares–adjacentchambers,forsomes ∈ S. Thatis,δ (φ,φ′)= s.
W
The intersection φ∩φ′ is called an s–panel. By definition, since X is regular, each
s–panel is contained in q distinct chambers. For distinct s,t ∈ S, the s–panel and
s
t–panel of any chamber φ of X have nonempty intersection if and only if m = 2.
st
Each s–panelof X isreduced toavertexifandonlyif m = ∞ forall t ∈ S−{s}.
st
Fortheexamples W , W ,and W above,respectively:
1 2 3
• The building X1 is a tree with each chamber an edge, each s–panel a vertex
of valence q , and each t–panel a vertex of valence q . That is, X is the
s t 1
(q ,q)–biregular tree. Theapartmentsof X arebi-infinite raysinthistree.
s t 1
• Thebuilding X2 haschambersandapartmentsasshowninFigure1below. The
r–and s–panels are 1–dimensional andthe t–panels arevertices.
Figure1: Achamber(ontheleft)andpartofanapartment(ontheright)forthebuilding X .
2
• Thebuilding X3 haschambers p–gonsand s–panelstheedgesofthese p–gons.
If q = q ≥ 2 for all s ∈ S, then each s–panel is contained in q chambers,
s
and X , equipped with the obvious piecewise hyperbolic metric, is Bourdon’s
3
building I .
p,q
Infinitegenerationforright-angledbuildings 5
2 Tree-walls
We now generalize the notion of tree-wall due to Bourdon [B]. We will use basic
facts about buildings, found in, for example, Davis [D]. Our main results concerning
tree-wallsareCorollary3below,whichdescribesthreepossibilitiesfortree-walls,and
Proposition 6below,whichgeneralizes theseparation property 2.4.A(ii)of[B].
Let X be as in Section1 above and let s ∈ S. As in [B, Section 2.4.A], we define
two s–panels of X to be equivalent if they are contained in a common wall of type s
insome apartment of X. A tree-wall of type s isthen anequivalence class under this
relation. Wenotethatinorderforwallsandthustree-wallstohaveawell-definedtype,
it is necessary only that all finite m , for s 6= t, be even. Tree-walls could thus be
st
definedforbuildingsoftypeanyevenCoxetersystem,andtheywouldhaveproperties
similartothosebelow. Wewillhoweveronlyexplicitlyconsidertheright-angledcase.
Let T beatree-wall of X,oftype s. Wedefineachamber φ of X tobeepicormic at
T if the s–panel of φ is contained in T , and we say that a gallery α = (φ ,...,φ )
0 n
crosses T if,forsome 0≤ i < n,thechambers φi and φi+1 areepicormicat T .
By the definition of tree-wall, if φ ∈ Ch(X) is epicormic at T and φ′ ∈ Ch(X) is
t–adjacent to φ with t 6= s, then φ′ is epicormic at T if and only if m = 2. Let
st
s⊥ := {t ∈ S | m = 2} and denote by hs⊥i the subgroup of W generated by the
st
elements of s⊥. If s⊥ is empty then by convention, hs⊥i is trivial. Forthe examples
inSection1above:
• in W1,both hs⊥i and ht⊥i aretrivial;
• in W2, hr⊥i = hsi ∼= C2 and hs⊥i= hri ∼= C2,while ht⊥i istrivial;and
• in W3, hs⊥i i = hsi−1,si+1i∼= D∞ foreach 1≤ i ≤ p.
Lemma2 LetT beatree-wallofXoftypes.Letφbeachamberwhichisepicormic
atT andletA beanyapartmentcontainingφ.
(1) TheintersectionT ∩A isawallofA,henceseparatesA.
(2) Thereisabijectionbetweentheelementsofthegroup hs⊥i andthesetof
chambersofAwhichareepicormicatT andinthesamecomponentofA−T ∩A
asφ.
Proof Part(1)isimmediatefromthedefinitionoftree-wall. ForPart(2),let w ∈ hs⊥i
and let ψ = ψ be the unique chamber of A such that δ (φ,ψ) = w. We claim that
w W
ψ isepicormicat T andinthesamecomponent of A−T ∩A as φ.
6 AThomasandKWortman
For this, let s ···s be a reduced expression for w and let α = (φ ,...,φ ) be the
1 n 0 n
minimalgallery from φ = φ to ψ = φ oftype (s ,...,s ). Since w isin hs⊥i,we
0 n 1 n
have m = 2 for 1 ≤ i ≤ n. Hence by induction each φ is epicormic at T , and so
sis i
ψ = φ is epicormic at T . Moreover, since none of the s are equal to s, the gallery
n i
α doesnotcross T . Thus ψ = ψ isinthesamecomponent of A−T ∩A as φ.
w
It follows that w 7→ ψ is a well-defined, injective map from hs⊥i to the set of
w
chambers of A which are epicormic at T and in the same component of A−T ∩A
as φ. To complete the proof, we will show that this map is surjective. So let ψ be a
chamberof A whichisepicormicat T andinthesamecomponentof A−T ∩A as φ,
andlet w = δ (φ,ψ).
W
If hs⊥i is trivial then ψ = φ and w = 1, and we are done. Next suppose that the
chambers φ and ψ are t–adjacent, forsome t ∈ S. Sinceboth φ and ψ areepicormic
at T , either t = s or m = 2. But ψ is in the same component of A − T ∩ A
st
as φ, so t 6= s, hence w = t is in hs⊥i as required. If hs⊥i is finite, then finitely
many applications of this argument will finish the proof. If hs⊥i is infinite, we have
established thebasecaseofaninduction on n = l (w).
S
For the inductive step, let s ···s be a reduced expression for w and let α =
1 n
(φ ,...,φ ) be the minimal gallery from φ = φ to ψ = φ of type (s ,...,s ).
0 n 0 n 1 n
Since φ and ψ areinthesamecomponentof A−T ∩A and α isminimal,thegallery
α does not cross T . We claim that s is in s⊥. First note that s 6= s since α does
n n
not cross T and ψ = φ is epicormic at T . Now denote by T the tree-wall of X
n n
containing the sn–panel φn−1∩φn. Since α isminimalandcrosses Tn,thechambers
φ = φ and ψ = φ areseparated bythewall T ∩A. Thusthe s–panel of φ andthe
0 n n
s–panel of ψ are separated by T ∩A. As the s–panels of both φ and ψ are in the
n
wall T ∩A,itfollowsthatthewalls T ∩A and T ∩A intersect. Hence m = 2,as
n sns
claimed.
Now let w′ = wsn = s1···sn−1 and let ψ′ be the unique chamber of A such that
δ (φ,ψ′) = w′. Since s is in s⊥ and ψ′ is s –adjacent to ψ, the chamber ψ′ is
W n n
epicormic at T andinthesamecomponent of A−T ∩A as φ. Moreover s1···sn−1
isareducedexpressionfor w′,so l (w′)= n−1. Hencebytheinductiveassumption,
S
w′ isin hs⊥i. Therefore w =w′s isin hs⊥i,whichcompletestheproof.
n
Corollary 3 Thefollowingpossibilitiesfortree-wallsinX mayoccur.
(1) Everytree-walloftypes isreducedtoavertexifandonlyifhs⊥i istrivial.
(2) Everytree-walloftypes isfinitebutnotreducedtoavertexifandonlyifhs⊥i
isfinitebutnontrivial.
Infinitegenerationforright-angledbuildings 7
(3) Everytree-walloftypes isinfiniteifandonlyifhs⊥i isinfinite.
Proof Let T , φ, and A be as in Lemma2 above. The set of s–panels in the wall
T ∩A is in bijection with the set of chambers of A which are epicormic at T and in
thesamecomponent of A−T ∩A as φ.
FortheexamplesinSection1above:
• in X1,everytree-walloftype s andoftype t isavertex;
• in X2, the tree-walls of types both r and s are finite and 1–dimensional, while
everytree-walloftype t isavertex;and
• in X3,alltree-wallsareinfinite,andare 1–dimensional.
Corollary 4 LetT ,φ,andA beasinLemma2aboveandlet
ρ= ρ : X → A
φ,A
betheretractionontoA centeredatφ.Thenρ−1(T ∩A)= T .
Proof Let ψ be any chamber of A which is epicormic at T and is in the same
component of A−T ∩A as φ. ThenbytheproofofLemma2above, w := δ (φ,ψ)
W
is in hs⊥i. Let ψ′ be a chamber in the preimage ρ−1(ψ) and let A′ be an apartment
containing both φ and ψ′. Since the retraction ρ preserves W–distances from φ, we
have that δ (φ,ψ′) = w is in hs⊥i. Again by the proof of Lemma2, it follows that
W
thechamber ψ′ isepicormic at T . Buttheimageunder ρ ofthe s–panel of ψ′ isthe
s–panelof ψ. Thus ρ−1(T ∩A)= T ,asrequired.
Lemma5 LetT beatree-wallandletφ andφ′ betwochambersofX.Letα bea
minimalgalleryfromφ toφ′ andletβ beanygalleryfromφ toφ′. Ifα crossesT
thenβ crossesT .
Proof Suppose that α crosses T . Since α is minimal, there is anapartment A of X
whichcontains α,andhencethewall T ∩A separates φ from φ′. Chooseachamber
φ ofA whichisepicormicat T andconsidertheretraction ρ= ρ onto A centered
0 φ0,A
at φ . Since φ and φ′ arein A, ρ fixes φ and φ′. Hence ρ(β) isagalleryin A from φ
0
to φ′,andso ρ(β) crosses T ∩A. ByCorollary4above, ρ−1(T ∩A)= T . Therefore
β crosses T .
Proposition 6 LetT beatree-walloftypes. ThenT separatesX intoq gallery-
s
connectedcomponents.
8 AThomasandKWortman
Proof Fix an s–panel in T and let φ ,...,φ be the q chambers containing this
1 qs s
panel. Thenfor all 1 ≤ i < j ≤ q , theminimal gallery from φ to φ isjust (φ,φ),
s i j i j
andhence crosses T . ThusbyLemma5above, anygallery from φ to φ crosses T .
i j
Sothe q chambers φ ,...,φ liein q distinct components of X−T .
s 1 qs s
Tocomplete the proof, weshow that T separates X into at most q components. Let
s
φ be any chamber of X. Then among the chambers φ ,...,φ , there is a unique
1 qs
chamber, say φ , at minimal gallery distance from φ. It suffices to show that φ and
1
φ areinthesamecomponent of X−T .
1
Let α be a minimal gallery from φ to φ and let A be an apartment containing α.
1
Then there is a unique chamber of A which is s–adjacent to φ . Hence A contains
1
φ for some i > 1, and the wall T ∩A separates φ from φ . Since α is minimal
i 1 i
and d (φ,φ ) < d (φ,φ), the Exchange Condition (see [D, page 35]) implies that
W 1 W i
aminimal gallery from φ to φ maybe obtained byconcatenating α withthe gallery
i
(φ ,φ). Since a minimal gallery can cross T ∩ A at most once, α does not cross
1 i
T ∩A. Thus φ and φ areinthesamecomponent of X−T ,asrequired.
1
3 Proof of Theorem
Let G be as in the introduction and let Γ be a non-cocompact lattice in G with strict
fundamental domain. Fixachamber φ of X. Foreachinteger n ≥ 0 define
0
D(n) := {φ ∈ Ch(X)| d (φ,Γφ )≤ n}.
W 0
Then D(0) = Γφ ,andforevery n > 0 everyconnected component of D(n) contains
0
achamberin Γφ . ToproveTheorem1,wewillshowthatthereisno n> 0 suchthat
0
D(n) isconnected.
Let Y be a strict fundamental domain for Γ which contains φ . Foreach chamber φ
0
of X,denoteby φ therepresentative of φ in Y.
Y
Lemma7 Letφandφ′bet–adjacentchambersinX,fort ∈ S.Theneitherφ = φ′ ,
Y Y
orφ andφ′ aret–adjacent.
Y Y
Proof It suffices to show that the t–panel of φ is the t–panel of φ′ . Since Y
Y Y
is a subcomplex of X, the t–panel of φ is contained in Y. By definition of a strict
Y
fundamentaldomain,thereisexactlyonerepresentativein Y ofthet–panelofφ. Hence
theuniquerepresentative in Y ofthe t–panel of φ isthe t–panelof φ . Similarly,the
Y
uniquerepresentative in Y ofthe t–panelof φ′ isthe t–panelof φ′ . But φ and φ′ are
Y
Infinitegenerationforright-angledbuildings 9
t–adjacent, hence have the same t–panel, and so it follows that φ and φ′ have the
Y Y
same t–panel.
Corollary 8 ThefundamentaldomainY isgallery-connected.
Lemma 9 Forall n > 0,thefundamentaldomain Y containsapairofadjacent
chambersφ andφ′ suchthat,ifT denotesthetree-wallseparatingφ fromφ′:
n n n n n
(1) thechambersφ andφ areinthesamegallery-connectedcomponentofY −
0 n
T ∩Y;
n
(2) min{d (φ ,φ) |φ ∈Ch(X)isepicormicatT }> n;and
W 0 n
(3) thereisaγ ∈ StabΓ(φ′n) whichdoesnotfixφn.
Proof Fix n > 0. Since Γ is not cocompact, Y is not compact. Thus there exists a
tree-wall T with T ∩Y nonemptysuchthatforevery φ ∈ Ch(X) whichisepicormic
n n
at T , d (φ ,φ) > n. Let s be the type of the tree-wall T . Then by Corollary8
n W 0 n n
above, there is a chamber φ of Y which is epicormic at T and in the same gallery-
n n
connected component of Y −T ∩Y as φ , such that for some chamber φ′ which is
n 0 n
s –adjacent to φ , φ′ is also in Y. Now, as Γ is a non-cocompact lattice, the orders
n n n
ofthe Γ–stabilizers ofthechambers in Y areunbounded. Hencethetree-wall T and
n
chambers φn and φ′n maybechosensothat |StabΓ(φn)| < |StabΓ(φ′n)|.
Let φ , φ′ , T , and γ be as in Lemma9 above and let s = s be the type of the
n n n n
tree-wall T . Let α beagallery in Y −T ∩Y from φ to φ . Thechambers φ and
n n 0 n n
γ ·φ are in two distinct components of X −T , since they both contain the s–panel
n n
φ ∩ φ′ ⊆ T , which is fixed by γ. Hence the galleries α and γ · α are in two
n n n
distinct components of X−T ,andsothechambers φ and γ ·φ areintwodistinct
n 0 0
components of X −T . Denote by X the component of X −T which contains φ ,
n 0 n 0
andput Y =Y ∩X .
0 0
Lemma10 Letφ beachamberinX thatisepicormicatT .Thenφ isinY and
0 n Y 0
isepicormicatT ∩Y.
n
Proof We consider three cases, corresponding to the possibilities for tree-walls in
Corollary3above.
(1) If T isreducedtoavertex,thereisonlyonechamberin X whichisepicormic
n 0
at T ,namely φ . Thus φ = φ = φ andwearedone.
n n n Y
10 AThomasandKWortman
(2) If T is finite but not reduced to a vertex, the result follows by finitely many
n
applications ofLemma7above.
(3) If T isinfinite,theresultfollowsbyinduction, usingLemma7above,on
n
k := min{d (φ,ψ) |ψ isachamberof Y epicormicat T ∩Y}.
W 0 n
Lemma11 Foralln > 0,thecomplexD(n) isnotconnected.
Proof Fix n > 0, and let α be a gallery in X between a chamber in X ∩Γφ and
0 0
somechamber φ in X thatisepicormicat T . Let m bethelengthof α.
0 n
ByLemma7andLemma10above,thegallery α projectstoagallery β in Y between
φ and a chamber φ that is epicormic at T ∩Y. The gallery β in Y has length at
0 Y n
most m.
It follows from (2)of Lemma9 above that the gallery β in Y has length greater than
n. Therefore m > n. Hence the gallery-connected component of D(n) that contains
φ iscontained in X . Asthechamber γ·φ isnotin X ,itfollowsthatthecomplex
0 0 0 0
D(n) isnotconnected.
Thiscompletestheproof,as Γ isfinitelygeneratedifandonlyif D(n) isconnectedfor
some n.
References
[BS´] W.BallmannandJ.S´wiatkowski,On L2–cohomologyandproperty(T)forautomor-
֒
phismgroupsofpolyhedralcellcomplexes,Geom.Funct.Anal.7(1997),615–645.
[BL] H.BassandA.Lubotzky,TreeLattices.Prog.inMath.,176,Birkha¨user.
[B] M. Bourdon, Immeubles hyperboliques, dimension conforme et rigidite´ de Mostow,
Geom.Funct.Anal.7(1997),245–268.
[CG] L. Carbone and H. Garland, Existence of lattices in Kac–Moody groups over finite
fields,Commun.Contemp.Math.5(2003),813–867.
[D] M.Davis,TheGeometryandTopologyofCoxeterGroups,LMSMonographs,Princeton
UniversityPress,2008.
[DJ] J. Dymara and T. Januszkiewicz, Cohomology of buildings and their automorphism
groups,Invent.Math.150(2002),579–627.
[GHM] R.Gramlich,M.HornandB.Mu¨hlherr,Abstractinvolutionsofalgebraicgroupsand
ofKac-Moodygroups,toappearinJ.GroupTheory.