Mathieu Dutour Sikiric, Ellis Graham and Achill Schurmann- The homology of PSL4(Z)

8/3/2019 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.

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Mathieu Dutour Sikiric, Ellis Graham and Achill Schurmann- The homology of PSL4(Z)