Mathieu Dutour Sikiric, Ellis Graham and Achill Schurmann- The homology of PSL4(Z)
Transcript of Mathieu Dutour Sikiric, Ellis Graham and Achill Schurmann- The homology of PSL4(Z)
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The homology of PSL4(Z )
Mathieu Dutour Sikiric
Institut Rudjer Boskovic,Croatia
Ellis Graham
NUI Galway
Achill Schurmann TU Delft, Netherland
March 30, 2010
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I. G -moduleand group resolutions
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G -modulesWe use the GAP notation for group action, on the right.A G -module M is a Z -module with an action
M G M (m , g ) m .g
The group ring Z G formed by all nite sums
g G
g g with g Z
is a G -module.If the orbit of a point v under a group G is {v 1, . . . , v m }, thenthe set of sums
m
i =1 i v i with i Z
is a G -module.We can dene the notion of generating set, free set, basis of aG -module. But not every nitely generated G -module admitsa basis.
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Resolutions
Take G a group.
A resolution of a group G is a sequence of G -modules(M i )i 0:
Z M 0 M 1 M 2 . . .
together with a collection of G -linear operators
d i : M i M i 1 such that Ker d i = Im d i 1What is useful to homology computations are free resolutionswith all M i being free G -modules.The homology is then obtained by killing off theG -action of a
free resolution, i.e replacing the G -modules (Z
G )k
byZ k
,replacing accordingly the d i by d i and getting
H i (G ) = Ker d i / Im d i 1
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Examples of resolutions
If C m = < x > is a cyclic group then a resolution (R n )n 0 withR n = Z C m is obtained by taking the differential d 2n+1 = 1 x and d 2n = 1 + x + + x m 1
For the dihedral groups D 2m of order 2m with m odd thereexist periodic resolutions.If a group G admits an action on a m-dimensional polytope P which is xed point free then it admits a periodic freeresolution of length m .The technique is to glue together the cell complexes, which
are free:
Z C 0(P ) C 1(P ) C m 1(P ) C 0(P ) . . .
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ProductsIf G and G are two groups of resolution R and R then thetensor product S = R R is a resolution for G G . That is
S n = i + j = n R i R j
with the differential at R i R j
D ij = d i + ( 1)i d j
If N is a normal subgroup of G and we have a free resolutionR for N and a free resolution R for the quotient G / N then itis possible to put together the twisted tensor productS = R R and get a resolution for G . That is
S n = i + j = n R i R j
with the differential at R i R j D ij = d 0 + d 1 + + d i
with d k : R i R j R i k R j + k 1
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The p -rank
A p -group G is called elementary Abelian if it is isomorphic to
(Z p )n
with n being called the rank of G .If G is a nite group then the p -rank of G is the largest rankof its elementary Abelian p -subgroups.For a nite group G of maximum p -rank m the dimension of its free resolutions grows at least likenm 1.For products C m 1 . . . C m p of cyclic groups the dimensions of resolutions is at least n+ p p .One example:
gap> GRP:=AlternatingGroup(4);;gap> R:=ResolutionFiniteGroup(GRP, 20);;gap> List([1..20], R!.dimension);[ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 11, 11, 12, 12, 11, 11,12, 12, 12, 13 ]
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II. The CTC Walllemma
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Using non-free resolutions
Suppose we have a G -resolution,
Z M 0 M 1 M 2 . . .
which is not freeThen we nd free resolutions for every moduleM i and sum it
to get a resolution of the formR 0, 2 R 1, 2 R 2, 2
R 0, 1 R 1, 1 R 2, 1
R 0, 0 R 1, 0 R 2, 0
The differentials involved ared k : R i , j R i k , j + k 1 and thesummands are
S n = i + j = n R i , j
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CTC Wall lemmaWe denote d 1 the operator of the R i , 0 R i 1, 0.We can nd free resolutions of the R i , 0 G -modules byG -modules
R i , 0 R i , 1 R i , 2 . . .
with the boundary operators being named d 0.Then we search for operators d k : R i , j R i k , j 1+ k such that
D =
i =0
d i
realize a free resolutions of G -modules i + j = k R i , j .It suffices to solve the equations
k
h=0
d h d k h = 0
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CTC Wall lemma
One way is to have the expression
d k = h0(k
h=1
d h d k h )
with h0 a contracting homotopy for the d 0 operator, i.e. anoperator h0 : R i , j R i , j +1 such that d 0(h0(x )) = x if x belongs to the image of d 0.This gives a recursive method for computing rst d 1 from therelations
d 1d 0 + d 0d 1 = 0Then d 2, d 3, . . .CTC Wall also gives a contracting homotopy for the obtainedresolution.
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CTC Wall lemma for polytopes: right cosets
Take R k , 0, which is sum of orbitsO i of faces of dimensionk .We compute resolutions R k , i , l of Stab f l with f l representativeof O l .
R k , i = r l =1 R k , i , l
The matrix of the operator d 0 : R k , i R k , i 1 is then a blockmatrix of the d 0 : R k , i , l R k , i 1, l For the contracting homotopy of a vector v R k , i :
Decompose v into components v l R k , i , l Z G Decompose v l into right cosets
v l =
s
v s , l g s
with g s G distinct right Stab f l -cosets and v s , l R k , i , l .Apply the contracting homotopy h0 of the resolution R k , i , l tov s , l and sum
h0(v l ) =s
h0(v s , l )g s
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CTC Wall lemma for polytopes: twistingIf F is a face of a polyhedral tesselation,H = Stab( F ) andf Stab( F ) then we need to consider whether or not f
inverts the orientation of F .This dene a subgroup Rot(F ) of Stab(F ) formed by allelements stabilizing the face F and its orientation.We dene a twisting accordingly:
(f ) = 1 if f Rot(F ) 1 otherwise
The twisting acts on Z G and it can twist accordingly agiven resolution of Stab(F ).
For the contracting homotopy, the solution to the twistingproblem is accordingly:
(d i )y = x d i (y ) = (x )
Twisting does not change the dimension of the resolutions,but it does change their homology.
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III.The PSL4(Z )case
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Resolutions forA5, A4 and S 4
The symmetry group of the Icosahedron is the group
W (H 3) =Z
2 A5. The subgroup A5 acts on it withstabilizersC 5 for vertices (1 orbit) nontwistedC 2 for edges (1 orbit) twistedC 3 for faces (1 orbit) nontwisted
Thus by applying the CTC Wall lemma we get a resolution(R n )n 0 with dim R n = n + 1. This is about the best possiblesince the 2-rank is 2.Similarly the symmetry group of the Tetrahedron is S 4 and A4acts on the Tetrahedron with stabilizers C 3, C 2 and C 3.The symmetry group of the cube is W (B 3) of size 48 and itcontains two conjugacy classes of subgroups isomorphic to S 4,one comes from the tetrahedron and the other has stabilizersC 4, C 2 and C 3.
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Perfect forms in dimension 4
Tessel1: Initially there are 2 orbits of perfect forms so fulldimensional cells are:
O 1: full dimensional cell with 64 facets and stabilizer of size288.O 2: full dimensional cell with 10 facets and stabilizer of size 60.
Tessel2: Now split both O 1 by adding a central ray. We thenget as orbits of full dimensional cells:
O 1, 1: full dimensional cell with 10 facets and stabilizer of size6.O 1, 2: full dimensional cell with 10 facets and stabilizer of size2.O 2, 1: full dimensional cell with 10 facets and stabilizer of size6.
Tessel3: Every cellO 1, 1 is adjacent to a unique cell O 2, 1. Jointhem:
O 1: full dimensional cell with 18 facets and stabilizer of size 6.O 2: full dimensional cell with 10 facets and stabilizer of size 2.
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The stabilizers inTessel1
For the rst tesselation Tessel1, we have the following orbits of faces and stabilizers:
0 A5, G 288 = ( A4 A4) : C 2.1 S 3, S 3 S 3.2 C 2 C 2, C 2 C 2.3 C 2 C 2, C 2 C 2, S 4, S 4.4 S 3, D 8, S 4, C 2 S 3 S 3.5 D 24, S 4, A5.6 G 96 = (( C 2 C 2 C 2 C 2) : C 3) : C 2.
Since we have the quotients
G 288 / A4 S 4 and G 96/ C 2 C 2 S 4
by combining everything we can nd resolutions of asymptoticallyoptimal dimensions for all stabilizers.
l
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For Tessel30 S 3, S 3 S 3.1 C 2 C 2, C 2, C 2, C 2.2 C 2, C 2, C 2, C 2 C 2, S 3, C 3, C 2 C 2, C 2 C 2, C 2, C 2, C 2,
S 3, C 4.3 S 3, C 2, C 2, 1, C 2 C 2, 1, C 2 C 2, C 2 C 2, S 4, A4, S 4, S 4, 1,
C 2, 1, D 12, S 3, D 12 , S 4.
4 1, S 4, D 10, 1, C 2, S 3, C 2, D 10 , C 2, D 8, C 2, C 2 S 3 S 3, C 4,S 3, C 2, C 2, C 2, C 2, C 2 C 2.
5 C 2, C 2 C 2, C 2, 1, S 4, A5, 1, S 3, D 24, A4, C 2, D 8, C 3, D 8,D 8, C 2.
6 C 2, C 2, ((C 2 C 2 C 2 C 2) : C 3) : C 2, C 2, C 3, S 3, C 2, S 3,S 4, (C 3 C 3) : C 2.
7 C 2, C 2 C 2, S 3, C 2 D 8.8 S 3, S 4.
9 A5, (A4 A4) : C 2.
L di i l l
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Low dimensional results
By using the tesselation Tessel3 we can compute the followinghomology groups:H 0(PSL4(Z ), Z ) = ZH 1(PSL4(Z ), Z ) = 0H 2(PSL4(Z ), Z ) = 3 Z 2
H 3(PSL4(Z
),Z
) = 2Z
4 +Z
+ 2Z
3 +Z
5H 4(PSL4(Z ), Z ) = 4 Z 2 + Z 5H 5(PSL4(Z ), Z ) = 13 Z 2
Further computations are difficult and constrained byChoosing tesselations inuence the occurring stabilizers.
Dimensions is one factor, another is the length of theboundaries.Memory is the main problem of such computations.
C h l i
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Cohomology rings
If G is a nite group then H (G , Z ) is a ring and also for anyprime p H (G , Z p ).This ring can be computed very efficiently by rst computingthe cohomology of the sylow p -subgroup and then bycomputing the stable classes.
But this is valid only for nontwisted cohomology. If u and u are cohomology classes inH n (G , Z ) and H n (G , Z ) thentheir cup products u u is a cohomology class inH n+ n (G , Z ).
So, one possible strategy if f
H (G
,Z
) is to use thefollowing identities:
f H (G , Z ) H (G , Z ) and f H (G , Z ) H (G , Z )
Th i i t t l
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The equivariant spectral sequence
It is also named the Leray spectral sequence, Normalizerspectral sequence, etc.Denote by Repr p (X ) the list of representative of orbits of p dimensional faces of the polyhedral complexX .In the E 1 page of the spectral sequence we have
E 1pq = F Repr p (X )H q (Stab( F ), Z )
The differential d i that allows to dene the higher spectralsequences are determined by the CTC Wall lemma.
5 t f th h l g I
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5-part of the homology I
We have
H n (PSL4(Z ), Z )(5) =Z 5 if n = 3 + 4 k Z 5 if n = 4 + 4 k
0 otherwise
First of all A5 is a subgroup of PSL4(Z ) and it appears in thetop dimensional cells.Thus we have mapping in homologyH (A5, Z ) H (PSL4(Z ), Z ) and another mapping incohomology H (PSL4(Z ), Z ) H (A5, Z ).
Since A5 is directly embedded in the resolution, we can readilycheck that the mapping H n (A5, Z ) H n (PSL4(Z ), Z ) areinjective if n 3.So we get that for all k , H 4k (PSL4(Z ), Z )(5) contains anon-zero class.
5 part of the homology II
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5-part of the homology IIFor the tesselation Tessel1 the only stabilizer groups withorder divisible by 5 areA5 and they have no twisting.
Thus we get the following picture for the E 1 page:0 Z 5 0 0 0 0 Z 5 70 0 0 0 0 0 0 60 0 0 0 0 0 0 5
0 0 0 0 0 0 0 40 Z 5 0 0 0 0 Z 5 30 0 0 0 0 0 0 20 0 0 0 0 0 0 1Z Z 3 Z 4 Z 4 Z 2 Z 2 Z 2 06 5 4 3 2 1 0
Either d 5 is non-zero and it kills twoZ 5 or it is zero and theystay.But we know that two H 8(PSL4, Z )(5) is nontrivial thus d 5has to be zero which proves the result.
Towards the 3 part
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Towards the 3-part
For the 3-part we have the relations
H 1(G 288 , Z )(3) = 0 H 6(G 288 , Z )(3) = 3 Z 3H 2(G 288 , Z )(3) = Z 3 H 7(G 288 , Z )(3) = 5 Z 3
H 3(G 288 ,Z
)(3) = 3Z
3 H 8(G 288 ,Z
)(3) = 0H 4(G 288 , Z )(3) = 0 H 9(G 288 , Z )(3) = 0H 5(G 288 , Z )(3) = 0 H 10(G 288 , Z )(3) = 5 Z 3
So, by doing the same computation as for A5 we can prove
that the mapping H n (G 288 , Z )(3) H n (PSL4(Z ), Z )(3) is asurjection for n 5.