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Euler characteristics of arithmetic groups

We develop a general method for computing the homological Euler characteristic of finite index subgroups G of GL_m(O_K) where O_K is the ring of integers in a number field K. With this method we find, that for large, explicitly computed dimensions m, the homological Euler characteristic of finite index subgroups of GL_m(O_K) vanishes. For other cases, some of them very important for spaces of multiple polylogarithms, we compute non-zero homological Euler characteristic. With the same method we find all the torsion elements in GL_3(Z) up to conjugation. Finally, our method allows us to obtain a formula for the Dedekind zeta function at -1 in terms of the ideal class set and the multiplicative group of quadratic extensions of the base ring.


Introduction
The homological Euler characteristic of a group Γ with coefficients in a representation V is defined by (see [S], [B1]). The main theoretical result of our thesis is a general method that allows us to calculate χ h (Γ, V ). Toward the end of the introduction we describe briefly this method. Before that we list the most important results of the thesis which demonstrate the scope of the method. Our first result is about vanishing of homological Euler characteristics. We also compute the homological Euler characteristic of the arithmetic subgroups Γ 1 (3, N ) and Γ 1 (4, N ) of GL 3 (Z) and GL 4 (Z), respectively, where Γ 1 (m, N ) is the subgroup of GL m (Z) that fixes the vector [0, . . . , 0, 1] mod N .
Theorem 0.3 The homological Euler characteristic of Γ 1 (3, N ) and of Γ 1 (4, N ) for N not divisible by 2 and 3 is given by χ h (Γ 1 (3, N ), Q) = − 1 12 ϕ 2 (N ) + 1 2 ϕ(N ), where ϕ(N ) is the Euler ϕ-function, and ϕ 2 (N ) is the multiplicative arithmetic function generated by ϕ 2 (p a ) = p 2a (1 − 1 p 2 ) for a ≥ 1. With the same technique we find all the torsion elements in GL 3 Z up to conjugation (see proposition 3.2). Using our method we, also, compute χ h (GL m (Z), S n V m ) for m = 3 and m = 4. The computation of χ h (GL 3 (Z), S n V 3 ) (see theorem 6.4) agrees with the computation of the H i (GL 3 (Z), S n V 3 ) in [G1], which was used for computation of dimensions of spaces of certain multiple polylogatithms. Also the Euler characteristic of Γ 1 (3, N ) with trivial coefficients when N is an odd prime, greater that 3, agrees with the computation of H i inf (Γ 1 (3, N ), Q) in the corrected version of [G1]. We compute the homological Euler characteristic of GL 2 (Z[i]) and GL 2 (Z[ξ 3 ]) with coefficients in the symmetric powers of the standard representations. Also we compute the homological Euler characteristic of Γ 1 (2, a) for an ideal a in Z[i] and Z[ξ 3 ], respectively. These computations can be used for finding dimensions of spaces of certain elliptic polylogarithms, as it was explained to me by professor A. Goncharov.
Theorem 0.4 (a) If 1 + i does not divide a the homological Euler characteristic of Γ 1 (2, a) ⊂ GL 2 (Z[i]) is given by where ϕ Z[i] (a) is the multiplicative function defined on the ideals of Z[i], generated by ).
In general the method works for any arithmetic subgroup of GL m (O K ), where O K is the ring of integers in a number field K.
Our approach is the following: we generalize a result of K. Brown [B2] that relates the torsion elements in the group up to conjugation to the homological Euler characteristic of the group. Namely, where the sum is over all torsion elements T of Γ up to conjugation and C(T ) is the centralizer of T in Γ. Let us recall the definition of orbifold Euler characteristic of Γ, denoted by χ(Γ), which we simply call Euler characteristic. If Γ is a torsion free group then χ(Γ) = χ h (Γ). If Γ has torsion consider a finite index torsion free subgroup Γ ′ . Then χ(Γ) is defined by Arithmetic groups do have a finite index torsion free subgroup. So for them the Euler characteristic is defined. The main properties of the Euler characteristic that we are going to use are: |G| for a finite group G, Given an exact sequence 0 → Γ 1 → Γ → Γ 2 → 0, we have χ(Γ) = χ(Γ 1 )χ(Γ 2 ), and χ(SL m (O K )) = ζ K (−1) . . . ζ K (1 − m).
The first two properties can be found in K. Brown's book [B1]. And the last one is a difficult result due to Harder [H].
In order to use the Brown's formula we need to know the torsion elements in the group. We develop a method for finding the torsion elements in GL m (O K ), and consider another form of the above formula that requires very few of the torsion elements ( in general they can be quite a large number). This method involves linear algebra over number rings. More precisely is describes a normal form of matrices over number ring. First we deal with matrices with irreducible characteristic polynomial which lead to a relation to ideal classes (proposition 2.1). Then we examine matrices with reducible characteristic polynomials which leads to resultants (corollary 3.5 and proposition 3.7). Then we examine the relation between the torsion elements in GL m (O K ) and the torsion elements in an arithmetic subgroup Γ of GL m (O K ). That gives us the homological Euler characteristic of Γ with coefficients in a representation.
We obtain another interesting application of our method for computation of the values of the Dedeking zeta functions at −1. Now we are going to explain how this is related to our method. First, the Dedekind zeta function at −1 vanishes for number fields which are not totally real. For totally real number fields K we consider the arithmetic group SL 2 (O K ). By a result of Harder [H] we have that the orbifold Euler characteristic of SL 2 (O K ) is the Dedekind zeta function at −1, ζ K (−1). Also the homological Euler characteristic is always an integer. Let us denote the homological Euler characteristic by an integer N , i.e N = χ h (SL 2 (O K )).
We also introduce the following notation. Let be the set of ideal classes in O K [ξ] which are free as O K -modules.
Theorem 0.5 Let K be a totally real number field. Let R be the ring of integers in K. Then the Dedekind zeta function at −1 can be expressed as The organization of the thesis is the following. In section 1 we deals with linear algebra over rings of algebraic integers. We give a method for classification of the matrices with integer coefficients for the ones that are diagonalizable over the complex numbers. In section 2 we give a method of computing the centralizer of a matrix which leads to explicit formulas for the homological Euler characteristics of GL m Z, GL m (Z[i]), GL m (Z[ξ 3 ]) and GL 2 (O Q( √ −d) ), see respectively theorems 2.10, 2.11, 2.12 and 2.13. In section 3 we find the torsion elements in GL 2 Z and GL 3 Z up to conjugation. In section 4 is very computational. We compute many resultants and centralizers needed for the homological Euler characteristics of various group. In section 5 we find the homological Euler characteristics of GL m (O K ) with coefficients the symmetric powers of the standard representation. In section 6 we find the homological Euler characteristics of arithmetic groups Γ 1 (m, N ) and Γ 1 (m, a). In section 7 we examine the groups SL 2 (O K ) for totally real number fields K which gives a relation to the Dedekind zeta function of the field. And in section 8 we prove the generalization of Brown's formula.
Acknowledgements I would like to thank most of all my advisor professor Alexander Goncharov for constant encouridgement for my work and for giving me a series of problems that lead to this thesis. I would like to thank professor Michael Rosen, Bruno Harris, and Steven Lichtenbaum for the discussions we had. I would like to thank professor Thomas Goodwillie for refering me to Brown's formula. And also I would like to thank Amir Jafary for the many useful conversations we had that contributed to this paper.

Conjugacy classes in GL m (O K )
In this section we describe the conjugacy classes of elements in GL m (O K ), where O K is the ring of integers in a number field K. We approach the description of conjugacy classes in the following way. First we deal with matrices in GL m K. Then we examine matrices in GL m (O K ) whose characteristic polynomial is irreducible over K. They are described by ideal classes of a larger ring. And, finally, we construct an algorithm for matrices with reducible characteristic polynomial that allows to consider instead matrices of smaller dimension. Using this inductive step we can describe completely conjugacy classes of certain type, knowing the conjugacy classes of matrices of smaller dimension.

Conjugacy classes in GL m K
First we recall a few statements about matrices with coefficients in an infinite field K.
Proposition 1.1 Let A and B be matrices with coefficients in an infinite field K. If A and B are conjugate to each other inside GL mK , whereK is the algebraic closure of K, then they are conjugate to each other inside GL m K. Since M is a solution of a linear system over fields, we have that We know that there is Q ∈ GL mK such that AM = M B. Equivalently, there is Q such that Q ∈ VK and det Q = 0. Let Then the set det 0 K is Zariski closed subset of V K that is not the entire space, because K has infinitely may elements. Let P ∈ V K − det 0 K . Then AP = P B, and P ∈ GL m K. Thus, A and B are conjugate to each other as elements of GL n K.
We are going to use a well known statement.
For 2 × 2-matrix can checked by direct computation The one can use induction on the size of the matrix.

Block-triangular form
This subsection deals with the case when the characteristic polynomial of a matrix is reducible over the rational numbers. The partition of a matrix into blocks will be done in the following way: Given an m × m matrix A, let m = m 1 + . . . + m k be a partition of n. Then A can be thought of as a k × k block-matrix whose A ij -entry, i, j = 1, . . . k is a block (and a matrix) of size m i × m j . This will be the type of block-matrices that we consider. Note that the blocks A ii are square matrices.
where A ij are matrices such that A ii has irreducible over K characteristic polynomial and A ij = 0 for i > j.
Proof. The plan for the proof is the following: First we consider the invariant subspaces over K, where K is the algebraic closure of K, in order to find the invariant spaces over K. Then we construct a good basis over R. Let A be matrix from GL m (O K ). Let λ be an eigenvalue of A. Let v 1,λ be one of the eigenvectors corresponding to λ. Let v i,λ , i = 2, · · · , k be adjoint vectors to v 1,λ so that We can choose the coefficients of v 1,λ to be from the field K(λ), since it is determined from the linear system Then similarly, all v i,λ 's have coefficients in the field K(λ). Let λ 1 , · · · , λ d be the Galois conjugates to λ. And let We can consider the Galois conjugates of v 1,λ . If σǫGal(L/K) then v σ 1,λ = v 1,λ σ is an eigenvector with eigenvalue λ σ and adjoint vectors v σ i,λ = v i,λ σ for i = 2, · · · , k. We can consider invariant subspaces with coefficients in L, namely the vector space over L spanned by the vectors v 1,λ , · · · , v i,λ which we are going to denote by Therefore the matrix A and the Galois group Gal(L/K) send the vector space (over L) V i to itself. Then we can take the Galois invariant subspace of V i , or equivalently the vectors with coordinates in K. Let Now we have a filtration of vector spaces over K, namely, is a a finitely genrated, torsion free module over a principal ideal domain. Choose a basis {e 1 , · · · , e i 1 } of V 1,O K . One can extend it to a basis {e 1 , · · · , e i 1 , e i 1 +1 , · · · , e i 2 } of V 2,O K . In the same way one can extend the basis successively so that at the end we have a basis {e 1 , · · · , e i k } of V k,O K whose restriction to V i,O K gives again a basis. Finally, take the basis {e 1 , · · · , e i k } and extend it to a basis {e 1 , · · · , e m } of O m K . Let P be a matrix whose column vectors are {e 1 , · · · , e m } in the same order. Then P −1 AP has the following properties: . When this process is over, we have the matrix A conjugated by a matrix Q so that the resulting matrix Q −1 AQ has the following property: each block on the diagonal has irreducible characteristic polynomial, and each block under the diagonal is zero.
If we relax the condition that O K is a principal ideal domain then instead of matrices A ij interpreted as homomorphism of free modules, we should have A ij to be a homomorphism of projective modules.
Definition 1.4 Let P be a finitely generated torsion free module over an integral Noetherian ring R of dimension 1. Let K be the field of fractions of R. Let A be an endomorphism of P , A ∈ End R (P ).
Then A can be extended to an endomorphism of a finite dimensional vector space P ⊗ R K. Then the characteristic polynomial of A is defined to be the characteristic polynomial of the induced map in End K (P ⊗ R K).
The examples of such rings that we consider are orders in a number rings. If the ring is also integrally closed, that is it is a Dedekind domain, we have the following generalization of the previous theorem.
Theorem 1.5 Let R be a Dedekind domain. And let P be a finitely generated projective module over R. Assume that the field of fractions K has characteristic 0. Let, also, A be an endomorphism of P . Then in a suitable basis we have where A ij ∈ Hom(P j , P i ), with P = P 1 ⊕ . . . ⊕ P d , so that A ij = 0 for i > j and A ii has irreducible over K characteristic polynomial.
Proof. Consider A as an endomorphism of P ⊗ R K. Let λ 1 be an eigen values of A. Let λ 1 , . . . , λ r be the Galois conjugates of λ 1 . Let v 1 be an eigenvector in P ⊗ R K corresponding to the eigenvalue λ 1 . Let v 1 , . . . , v r be the Galois conjugates of v 1 .
Note that the endomorphism A send V R to itself. We claim that V R and P/V R are projective R-modules and therefore direct sumand of P . Let π : P → P/V R . Suppose P/V R has torsion. Let T be the torsion submodule of P/V R . Then V R is of finite index in π −1 (T ). Therefore Then R/V R is torsion free. We are working over a Dedekind domain. Then the finitely generated torsion free modules are projective. Thus, P/V R is a direct summand of P . Set P d = V R . Note that A induces the zero homomorphism in Hom(P/P d , P d ). Proceed by induction util P is factorized into projective modules P = P 1 ⊕, . . . , ⊕P d so that A induces the trivial map in Hom(P j , P i ) for i > j.

Irreducible blocks
We have the following characterization of matrices with an irreducible characteristic polynomial. For principal ideal domains the proof is simpler. Proof. The idea of the proof is to consider calO K -endomorphism of ideals in O K [λ], where λ is a root of f (t) in an algebraic closure of K. Let I be an ideal in It is a free O K -module because O K is a principal ideal domain. And let be a basis of I. Then the multiplication by λ is an endomorphism of the ideal I.
where a ij ∈ O K . If A = (a ij ) and − → α = (α 1 , . . . , α m ) t then − → α is an eigenvector of A with eigenvalue λ. Also the matrix A = (a ij ) has characteristic polynomial f , because one of the eigenvalues of A is λ and because a ij ∈ O K .
If we choose an ideal from the same ideal class we have I ′ = α ′ I for some α ′ ∈ K[λ]. Form the basis be a different basis of I. Let also and let B = (β ij ).
Let β = (β 1 , · · · , β m ) and α = (α 1 , · · · , α m ) be vectors. Then Then the matrix corresponding to the basis is BAB −1 , which is conjugate to A. In this way we associate a matrix with integer coefficients to an element of the ideal class. Conversely, if A is a matrix with coefficients in O K and with irreducible over K characteristic polynomial f , we want to associate an element of the ideal class of O K [t]/(f (t))., which is a free O K -module. If we start with the ideal (1) with is the corresponding matrix, where λ is a root of the characteristic polynomial f (t) = t m + a m−1 t m−1 + · · · + a 0 .
Using proposition 1.1, we can find P ǫGL m K such that A = P A 0 P −1 . We can also assume that P has integer entries otherwise we can consider N · P where N is a suitable integer. Let and its automorphism with respect to the given basis is the matrix A. Before we prove that let − → α = (α 1 , · · · , α m ) t and − → λ = (1, λ, · · · , λ m−1 ) t be vectors. Then As a consequence the O K -module I is an ideal, since each of its basis elements transforms via where a ij ∈ O K . Also, A corresponds to the multiplication by λ. Thus A corresponds to the ideal class of I.
Proposition 1.7 Let f (t) be an irreducible monic polynomial. And let P be a projective O K -module. Consider the set of all endomorphisms {A i } of P which have characteristic polynomial f (t) and which are not conjugate to each other via an automorphism of P . This set is parametrized by the ideal classes in O K [t]/(f (t)) which are isomorphic to P as O K -modules.
Proof. The idea of the proof is to consider where λ is a root of f (t) in an algebraic closure of K. Let I be an ideal in which is isomorphic to P as O K -module. Let A be an endomorphism of I induces by multiplication by λ. Tensoring the endomorphism with the field K we obtain that λ is an eigenvalue. Thus, f (t) is the characteristic polynomial of A. Also, if B is an O K -automorphism of P , then is it also an O K -automorphism of I. Then A sends BI to BI. Therefore, the induced endomorphism on I is BAB −1 . Thus A is determined up to conjugation. If J is in the same ideal class as I, then J = αI, for α ∈ K(λ) × . That induces an isomorphism between the O K -modules I and J. Thus, an endomorphism on one lead to an endomorphism of the other. This is the one of the direction of the correspondance between ideal classes and endomorphisms. Given an endomorphism of P with characteristic polynomial f (t), we want ot associate an ideal class in O K [λ]. Note that the ring O K [A] is isomorphic to O K [λ], since f (A) = 0 and f (λ) = 0. The ring O K [A] acts on P , because P is an O Kmodule, and also A acts on P . Therefore P is an . It remains to show that torsion free O K [λ]-modules of rank 1 correspond to ideal classes. Consider the imbedding P → P ⊗ O K [λ] K(λ) followed by the isomorphism θ : P ⊗ O K [λ] K(λ) → K(λ). The image og the composition is a fractional ideal in K(λ). All the choises for the isomorphism are parametrized by K(λ) × . Therefore, the torsion free O K [λ]-modules of rank 1 correspond to the set of ideal classes module K(λ) × which is precisely the ideal classes in O K [λ].
is a integrally closed. Then the conjugacy classes of matrices in GL m (O K ) with characteristic polynomial f (t) are in one-to-one correspondence with the elements in Proof. It is known that for number rings O K we have where Cl(K) is the ideal class group of O K (see [M]). Then the ideal classes in

Reducible blocks
Using the theorem 1.5 we obtain the following: If A is an m × m matrix with coefficients in O K , or more generally, an endomorphism of projective O K -module of rank m, having an irreducible characteristic polynomial then A is conjugated by an element of GL m (O K ), or of Aut(P ), to a 2 × 2-block endomorphism Let A 11 and A 22 be automorphisms of projective modules P 1 and P 2 , with P 1 ⊕ P 2 = P.
Now we are going to describe a method that simplifies the block A 12 and leaves A 11 and A 22 unchanged. We can assume that Conjugate A with an automorphism of the some block form. that is B 11 ∈ Aut(P 1 ), B 22 ∈ Aut(P 2 ) and B 12 ∈ Hom(P 2 , P 1 ). We want the conjugation by P to preserve A 11 and A 22 that is We do have that A 12 is changed to A ′ 12 . Then . Denote by C(A ii ) the centralizer of A ii . Then B 11 ∈ C(A 11 ) and B 22 ∈ C(A 22 ). The block B 12 could be any map in Hom O K (P 2 , P 1 ). Let be a map from the space Hom O K (P 2 , P 1 ) to itself. It is a linear. Then the relation between A ′ 12 and A 12 can be written as Let also Q mod = Hom O K (P 2 , P 1 )/P mod . Then Hom O K (P 2 , P 1 ), P mod and Q mod are Proof. Let B 11 ∈ C(A 11 ) and B 22 ∈ C(A 22 ). Then Obviously, Hom O K (P 2 , P 1 ) is a C(A 11 ) × C(A 22 )-module. Thus, the quotient, as abelian group, Q mod = Hom O K (P 2 , P 1 )/P mod has the structure of a C(A 11 )×C(A 22 )module.
Proposition 1.10 Let A 11 ∈ Aut(P 1 ) and A 22 ∈ Aut(P 2 ). Suppose A 11 and A 22 have no common eigenvalues. Then the matrices are conjugate to each other if and only if the projection of A 12 and A ′ 12 onto the finite set C(A 11 ) × C(A 22 )\Q mod coincide.
In order to prove the above proposition, we need the following definition and lemma. Consider the linear space of n × m-matrices, or more generally of homomorphisms. Multiplication on the left and multiplication on the right are linear transformations of this space. So this transformations can be written as matrices, or as endomorphisms. Now we are going to set the notation for these matrices, after tensoring with K. Let A = (A ij ) and B = (b kl ) be two matrices. We shall write the tensor A⊗ B as a matrix in the following way: The matrix can be considered as a block-matrix with each block being the size of A, and he (k, l)-block being b kl A. This tensor product has the following properties: Let A be an m × m matrix acting by left multiplication on M ǫM at m,n , i.e on the m × n matrices. Consider M as a vector by arranging the column vectors of M one below the other. Then the left multiplication can be expressed as a matrix. And this matrix is precisely A ⊗ I n . If we multiply on the right by an n×n matrix B then such a linear transform can be expressed by I m ⊗B t . Another property that is easy to check is (A ⊗ B)(C ⊗ D) = (AC) ⊗ (BD).
Lemma 1.11 If i (t − α i ) and j (t − β j ) are the characteristic polynomials of A 11 ∈ GL m 1K and A 22 ∈ GL m 2K then the characteristic polynomial of In particular, If A 11 and A 22 have no common eigenvalue then the map P A 11 ,A 22 is non-singular, and if R is a number ring Q mod is finite, namely the norm of the resultant.
Proof. We write the matrix X as a vector, arranging the columns of X below each other. Then the linear transform P A 11 ,A 22 becomes If A 11 and A t 22 are diagonal matrices with diagonals (α 1 , · · · , α m 1 ) and (β 1 , · · · , β m 2 ) then P A 11 ,A 22 would be diagonal with diagonal entries β j − α i . If A 11 and A 22 are arbitrary then we can find B 1 ∈ GL m 1K and B 2 ∈ GL m 2K with such that the matrices B 1 A 11 B −1 1 and B 2 A t 22 B −1 2 are in Jordan block form. Let (α 1 , · · · , α m 1 ) be the diagonal entries of B 1 A 11 B −1 1 and is upper triangular. Its diagonal entries are β j − α i . Therefore if α i = β j for all i and j then P A 11 ,A 22 is non-singular. If R is a number ring then Im(P A 11 A 22 ) = P mod is a sublattice of maximal rank in Hom O K (P 2 , P 1 ) of index the norm N K/Q of the resultant. Therefore, Q mod = Hom(P 2 , P 1 )/P mod is finite.
Proof. (of proposition 1.7) If A 12 , A 12 ∈ Hom O K (P 2 , P 1 ) are such that the the endomorphisms are conjugate to each other in Aut O K (P ), then there exists B in Aut O K (P ) of the same block form such that Then the (2, 1)-entry of the product gives However, the map is non-singular from the previous lemma. Therefore we have B 21 = 0. Now the conditions on the other blocks of B become simpler: Equivalently, Corollary 1.12 If f (t) is monic polynomial of degree m with integer coefficient, and f (0) = ±1 without repeated roots then there are finitely many matrices A ∈ GL m (O K ) with characteristic polynomial f (t), where K is a number field.
Proof. We will proceed by induction on the number if irreducible factor of the characteristic polynomial. If the characteristic polynomial is irreducible we use the finiteness of class numbers and use corollary 1.4. Suppose we have proven the statement when the characteristic polynomial factors into n irreducible polynomials. Consider a matrix A whose characteristic polynomial factors into n + 1 irreducible polynomials. We can assume that A is of the form where A 11 has irreducible characteristic polynomial and the characteristic polynomial of A 22 factors into n irreducible polynomials. There are finitely many matrices having the same characteristic polynomial as A 11 and being non-conjugate to A 11 . Also, by assumption we can replace A 22 only by finitely many matrices in order to obtain new matrices non-conjugate to the initial one and with the same characteristic polynomial. To prove finiteness of matrices that are non-conjugate to A but have the same characteristic polynomial, we use that for fixed A 11 and A 22 there are only finitely many options for A 12 which follow from proposition 1.7.

Centralizers and homological Euler characteristics
Lemma 2.1 Let A 11 and A 22 be square matrices with coefficients in O K . Suppose that they have no common eigenvalues. And let Then the admissible matrices C are determined by the following properties: C 11 ∈ C(A 11 ), C 11 A 12 C −1 22 ≡ A 12 mod ImP A 11 A 22 . Also, the matrix C 12 is uniquely determined by C 11 and C 22 , and it is given by In particular, where C(A) denotes the centralizer of A.
Proof. We have The (2, 1)-entry of the products give However the map P A 22 A 11 : C 21 → C 21 A 11 − A 22 C 21 is non-singular by lemma 1.11. Therefore C 21 = 0. Considering the (1, 1)-entry and the (2, 2)-entry of the above matrix product we obtain that C 11 ∈ C(A 11 ) and C 22 ∈ C ( A 22 ). Fix C 11 and C 22 to be in the centralizers C(A 11 ) and C(A 22 ), respectively. Then use that A = CAC −1 in order to determine the block C 12 . C 11 ∈ C(A 11 ) and C 22 ∈ C(A 22 ). Then We have . Therefore A 12 and C 11 A 12 C −1 22 coincide as elements in Therefore C 11 × C 22 stabilizes A 12 as element of Q mod . Conversely, if C 11 × C 22 stabilizes A 12 as an element of Q mod we can find a unique C 12 such that And we can set . Then the equation We are going to define a map i A 12 : C(A) → C(A 11 ) × C(A 22 ) And let C be in the centralizer of A. Consider C as a block matrix of the same block form as A. Then by lemma 2.1 we have that C 21 = 0 and that C 11 ∈ C(A 11 ) and C 22 ∈ C(A 22 ). Let i A 12 sends C to C 11 × C 22 . Then i A 12 is a homomorphism.
Corollary 2.2 With the above definition of i A 12 we have that i A 12 is always injective and the index of Im(i A 12 ) in C 11 ×C 22 is equal to the number of elements in C 11 ×C 22orbit of A 12 inside Q mod .
Proof. Let C be in the centralizer of A. Let also be of the same block form as A. From lemma 2.1 we know that the blocks of C must satisfy C 11 ∈ C(A 11 ) and C 22 ∈ C(A 22 ) with the relation that A 12 and are congruent modulo the image of P A 11 A 22 . Equivalently, A 12 and C 11 A 12 C −1 Therefore the image Im(i A 12 ) consists of such C 11 and C 22 that the group element C 11 × C 22 fixes A 12 as an element of Q mod . Therefore the image Im(i A 12 ) coincides with the stabilizer of A 12 ∈ Q mod , and the index of Im(iA 12 ) in C(A 11 ) × C(A 22 ) corresponds to the C(A 11 ) × C(A 22 )-orbit of A 12 in Q mod .
Lemma 2.3 Let A 11 and A 22 be invertible matrices with coefficients in O K , or automorphisms of projective modules having no common eigenvalues. Let f 1 and f 2 be the characteristic polynomials if A 11 and A 22 , respectively. Then where the sum is taken over all non-conjugate torsion elements with fixed A 11 and A 22 , and R(f 1 , f 2 ) is the resultant of the two polynomials.
Proof. By Proposition 1.10, taking the sum over all non-conjugate matrices with fixed A 11 and A 22 is the same as varying A 12 through representatives of For a fixed A 12 by corollary 2.2, we have that the group C(A) is a finite index subgroup of C(A 11 ) × C(A 22 ), and the index is equal to the number of elements in the C(A 11 ) × C(A 22 )-orbit of A 12 in Q mod . Thus, Summing over all orbits, we obtain χ(C(A)) = #|Q mod | · χ(C(A 11 ))χ(C(A 22 )).
On the other hand, by lemma 1.11 We are going to define a resultant of k polynomials f 1 , . . . , f k , k ≥ 2 by Proposition 2.4 Let A 11 , A 22 , · · ·, A kk be invertible matrices with coefficients in O K such that A ii and A jj have no common eigenvalues for i = j. Let f i be the characteristic polynomial of A ii . Then where the sum is taken over all non-conjugate torsion elements A is a block-diagonal form such that the blocks on the diagonal are A 11 , . . ., A kk , and the blocks under the diagonal are zero, and R(f 1 , . . . , f k ) is the resultant of f 1 , . . . , f k defined above.
Note that all A's in the above sum have the same characteristic polynomial; namely Proof. We are going to prove the statement by induction on k. For k = 2 it is true by the previous lemma. Assume it is true for k − 1. Notice that Consider the matrices A in the sum as 2 × 2-block-matrices in the following way: with A 11 a block-triangular matrix with blocks on the diagonal A 11 , . . . A k−1,k−1 , and zero blocks under the diagonal. Let A ′ be another matrix in the above sum.
Consider it similarly as a 2 × 2-block matrix We can assume that either A 11 and A ′ 11 coincide, or that they are non-conjugate. We can make that assumption for the following reason. If they are conjugate say by a matrix B 11 instead of the representative A ′ one can take with the matrix I having the size of A kk . Then the new matrix A ′′ = BAB −1 will have A ′′ 11 = A 11 . Thus, we can separate the summation over non-conjugate matrices A into two summations. First fix A 11 and let A 12 very over all representatives where Q mod is the cokernel of the inclusion P A 11 A kk . Then sum over non-conjugate A 11 in block-diagonal form with A 11 , . . . , A k−1,k−1 on the diagonal, and zero under the diagonal. The summation over all The first summation leads to by the previous lemma because f 1 · . . . · f k−1 is the characteristic polynomial of any of the A 11 matrices. And by induction assumption We need e few more lemmas on the size of the centralizer.
Lemma 2.5 Let A and B be matrices in GL m O K , or Aut(P ) which are conjugate as elements of GL m K the C(A) and C(B) are comensurable.
Proof. The group C(A) is an arithmetic subgroup of C GLmK (A), which is conjugate to C GLmK (B). Therefore C GLmO K (A) and C GLmO K (B) are comeasurable.
Lemma 2.6 Let T n ∈ GL m (O K ) be an n-torsion matrix with with irreducible characteristic polynomial, or an n-torsion automorphisms. Then the centralizer C(T n ) contains O K [ξ n ] × , and is comensurable to it where ξ n = e 2πi/n .
Proof. The matrices in M at m,m K commuting with T n are precisely because this is the maximal abelian sub-Lie algebra commuting with T n ; it is of where R and O K(ξn) are comeasurable by the previous lemma. In this intersection R the invertible elements are Lemma 2.7 Let T n ∈ GL m Z be an n-torsion matrix with irreducible characteristic polynomial. Let T be a k × k-block matrix with blocks on the diagonal T n and the rest of the blocks being zero. Let R be the ring of endomorphisms that commute with T n . Then C(T ) ∼ = GL k R.
Proof. Then the matrices commuting with T are Among them the ones with integer coefficients are where R is a order in O K(ξn) isomorphic to the ring of matrices (not necessarily with unit determinant) with coefficients in O K commuting with T n And the invertible ones with integer coefficients are The following two propositions give bases for the proof of the vanishing results, namely theorem 0.1 and theorem 0.2.
Proposition 2.8 Let A be a torsion element of GL m O K . Then χ(C(A)) = 0 if and only if the set of eigenvalues of A is inside the set (a) {1, −1, i, −i, ξ 3 ,ξ 3 , ξ 6 ,ξ 6 }, and the multiplicity of 1 and −1 is at most 2 and the multiplicity of the rest of the roots of unity is at most 1 if K = Q; Proof. From theorem 1.5 we can assume that A is in block-triangular form with zero under the block-diagonal. We can also assume (by grouping similar blocks together) that the diagonal blocks have no common eigenvalues. Construct a new matrix A ′ having the same block diagonal as A and zeroes both above and below the diagonal. Then A and A ′ are conjugate in GL m K. By lemma 2.5 the Euler characteristic of A is a non-zero rational multiple of the Euler characteristic of A ′ . On the other hand, χ(A ′ ) can be expressed as the product of the Euler characteristics of the diagonal blocks. Thus, it is enough to examine all possible blocks that cannot be decomposed further. Such block B has eigenvalues n-th roots of unity repeated k times. Also, the centralizer of such a block is comeasurable with the centralizer of a matrix consisting of an n-torsion matrix T n sitting in each diagonal block where T n is an n-torsion matrix in GL l O K . By lemma 2.7 The number χ(O K [ξ n ] × ) will be zero if the units are infinitely many, because χ(Z) = 0.
Thus, we are left with the roots with multiplicities of +1 and −1 at most 2, and multiplicity of ±i, ± ξ 3 and ±ξ 6 at most 1. For K = Q(i), we can have only n-th roots of unity for n = 1, 2, 4 such that Also, we have that Thus, we are left with the roots with multiplicities at most 1. For K = Q(ξ 3 ), we can have only n-th roots of unity for n = 1, 2, 3, 6 such that Thus, Thus, we are left with the roots {1, −1, ξ 3 ,ξ 3 , ξ 6 ,ξ 6 } with multiplicities at most 1.
we can have only n-th roots of unity for n = 1, 2 such that because otherwise the units in the extension will be infinitely many. Also, we have that Thus, we are left with the roots {1, −1} with multiplicities at most 1. If K is not Q, not an imaginary quadratic extension of Q then for any n. Thus χ(C(A)) = 0 for such fields K.
For SL m O K we have a similar statement.
Proposition 2.9 Let A be a torsion element of SL m O K . Then χ(C(A)) = 0 if and only if the set of eigenvalues of A is inside the set (a) {1, −1, i, −i, ξ 3 ,ξ 3 , ξ 6 ,ξ 6 }, and the multiplicity of 1 is at most 2, the multiplicity of −1 is 0 or 2 and the multiplicity of the rest of the roots of unity is at most (e) χ(C(A)) = 0 always when K is not totally real and different from Q(i) and Q(ξ 3 ).
Proof. Then proof is similar to the one of the previous proposition. Given a torsion matrix A in SL m O K we can conjugate it with a matrix from GL m K to a matrix A ′ . Then C(A) and C(A ′ ) will be comeasurable. Consider A as a matrix in GL m O K . We can assume that A is of block-triangular form, using theorem 1.5. Then we can take A ′ to be of block-diagonal form with the same blocks on the diagonal as A. With another conjugation by a matrix from GL m K, we can modify A ′ so that blocks with identical eigenvalues will be represented by the same matrices. And finally, we combine the similar matrices into bigger block so that: (1) each new block is zero if it is not on the diagonal; (2) a block on the diagonal consists of smaller blocks such that on the diagonal we have smaller blocks repeated as many times as needed, and off the diagonal the smaller blocks are zero. The centralizer of such a matrix can be computed using lemma 2.6 and 2.7. We need a minor modification of lemma 2.6: We have that C GLmO K (T n ) is equal to the group of units in O K(ξn) while C SLmO K (T n ) is equal to the group Ker(N K(ξn)/K ) inside the units in O K(ξn) . For the vanishing of the Euler characteristic this does not have an effect since both groups have the same rank.
Thus, we are left with the roots with multiplicities of +1 and −1 at most 2, and multiplicity of ±i, ± ξ 3 and ±ξ 6 at most 1. For K = Q(i), we can have only n-th roots of unity for n = 1, 2, 4 such that Thus, we are left with the roots with multiplicities at most 1. For K = Q(ξ 3 ), we can have only n-th roots of unity for n = 1, 2, 3, 6 such that Thus, we are left with the roots we can have only n-th roots of unity for n = 1, 2 such that because otherwise the units in the extension will be infinitely many. Also, we have that We are left with the roots {1, −1} with multiplicities at most 1. Since −1 will lead to a determinant −1, we have that the only with the eigenvalues 1 with multiplicity 1, which gives that A is in SL 1 and for higher SL m the Euler characteristic will vanish. For K a totally real field different from Q, we have that there are infinitely many units and that ζ K (−1) = 0, which follows from the functional equation for the Dedekind zeta function. Also, ζ K(ξn) (−1) = 0 for n > 2. Thus the eigenvalues 1 or −1 might occur with multiplisity 2, and another root of 1 might occur with multiplicity at most 1. Suppose T n occurs as a block in A ′ , n > 2. Recall T n is an n-torsion block that has irreducible over K characteristic polynomial. By the modification of lemma 2.6 in the beginning of this proof we have that the centralizer of A ′ will contain a copy of the group of units α in O K such that In order to have non-vanishing Euler characteristic the rank of this group must be zero. This can happen only when the ranks of the groups of units in K and in K(ξ n ) coincide. This is the case when K(ξ n ) is an extension of degree 2 over K. Thus, the block T n is of size 2 × 2.
We are going to show that A ′ consists of at most block T n . Suppose that the diagonal of A ′ consists of T n 0 , T n 1 , . . ., T n l on the diagonal, where each T n i for i = 1 . . . l is n i -torsion block with irreducible characteristic polynomial of degree 2, and T n 0 is either empty, or(±1), or ±I 2 . Let The rank of the group is (l − 1) times the rank of units in K. In order to have nonvanishing Euler characteristic we must start with a zero rank group. Thus, l = 1 and A ′ = T n i for some i.
The rank of the group is l times the rank of units in K. In order to have nonvanishing Euler characteristic we must start with a zero rank group. Thus, l = 0 and A ′ = (±1). If T 0 = ±I 2 then the Euler characteristic of the centralizer of A ′ is isomorphic to the Euler characteristic of The rank of the abelian group is l times the rank of units in K. In order to have non-vanishing Euler characteristic we must start with a zero rank group. Thus, l = 0 and A ′ = ±I 2 . This completes the case when K is totally real.
In the rest of the cases we have that the group of units of K is infinite and that ζ K (−1) = 0. Then the Euler characteristic of the centralizer of A ′ always vanishes.
The following theorems are very useful for computational purposes. It expresses the homological Euler characteristic as a sum of very few terms. And each of the terms can be easily computed. We use the following notation: means that the square block A 11 through A ll are placed on the block-diagonal of A and the blocks of A outside the block-diagonal are zero blocks. Also, let and f i is the characteristic polynomial of A ii , and f i is a power of an irreducible polynomial. As a consequence of proposition 2.4 and 2.8 we obtain theorems 2.10, 2.11, 2.12 and 2.13.
Theorem 2.10 Let V be a finite dimensional representation of GL m Q. Then the homological Euler characteristic of GL n Z with coefficients in V is given by where the sum is taken over torsion matrices A consisting of blocks A 11 , . . . A mm on the block-diagonal and zero blocks off the diagonal. Also the matrices and the characteristic polynomial f i of A ii is a power of an irreducilbe polynomial, and f i and f j are relatively prime.
Theorem 2.11 Let V be a finite dimensional representation of GL m Q(i). Then the homological Euler characteristic of GL n (Z[i]) with coefficients in V is given by where the sum is taken over torsion matrices A consisting of blocks A 11 , . . . A mm on the block-diagonal and zero blocks off the diagonal. Also the matrices A ii are in the set {+1, −1, i, −i} and the characteristic polynomials Theorem 2.12 Let V be a finite dimensional representation of GL m Q(ξ 3 ). Then the homological Euler characteristic of GL n (Z[ξ 3 ]) with coefficients in V is given by where the sum is taken over torsion matrices A consisting of blocks A 11 , . . . A mm on the block-diagonal and zero blocks off the diagonal. Also the matrices A ii are in the set {+1, −1, ξ 3 ,ξ 3 , ξ 6 ,xi 6 } and the characteristic polynomials f i of A ii are relatively prime if i = j.
Theorem 2.13 Let P be a projective module of rank 2 over the ring of integers O K in a number field K = Q( √ −d) for d = 3, 4. Let V be a finite dimensional representation of GL 2 (Q( √ −d)) Then the homological Euler characteristic of Aut(P ) with coefficients in V is given by where n is the number of ways that P can be written as a direct sum of two projective modules, P = P 1 ⊕ P 2 counting also the order of the summands. In particular if P is free we obtain the homological Euler characteristic of GL 2 (O Q( √ −d) ) with coefficients in any representation.
Proof. (of theorems 2.10-2.13) Using proposition 2.4, we combine the torsion elements that have a common characteristic polynomial. Using proposition 2.8, we can take the sum only over those torsion elements that lead to a non-zero Euler characteristic of their centralizer. That leads to theorems 2.10-2.12.
For theorem 2.13 we need a little bit more. By proposition 2.8 part (d) we know that the only block-diagonal endomorphisms that will give contribution to the homological Euler characteristic is acting as 1 on P 1 and as −1 on P 2 . We have to take into account all decompositions of P into a direct sum of rank one projective modules. Note that Thus, the formula becomes In the rest of the section we will deal with general statements about the homological Euler characteristics of groups comensurable to GL . Let A be a torsion element of Γ. Then then the centralizer of A inside Γ, Thus, where q is a positive integer. Thus, vanishing of χ(C GLmO K (A)) implies vanishing of χ(C Γ (A)). We use a generalization of Kenneth Brown's formula (see section 10,and [B2]): where the summation is over all torsion elements in Γ up to conjugation, and V is a representation of Γ Let K = Q. Let also A ∈ GL m Z.
Then by proposition 2.9 part (a) we obtain that A multiplicity of the eigenvalues 1, −1 at most 2, and multiplicity of the eigenvalues i, −i, ξ 3 ,ξ 3 , ξ 6 ,ξ 6 at most 1. Thus, the dimension of A is at most 10. Therefore if the dimension of A is greater than 10 then the Euler characteristic of the centralizers of A will vanish and so will the homological Euler characteristic of a finite index subgroup Γ of GL m Z with coefficients in V , Suppose that χ(C(A)) = 0.
Then by proposition 2.9 part (b) we obtain that A multiplicity of the eigenvalues 1, −1, i, −i is at most 1. Thus, the dimension of A is at most 4. Therefore if the dimension of A is greater than 4 then the Euler characteristic of the centralizers of A will vanish and so will the homological Euler characteristic of a finite index Suppose that χ(C(A)) = 0.
Then by proposition 2.9 part (c) we obtain that A multiplicity of the eigenvalues 1, −1, ξ 3 ,ξ 3 , ξ 6 ,ξ 6 at most 1. Thus, the dimension of A is at most 6. Therefore if the dimension of A is greater than 6 then the Euler characteristic of the centralizers of A will vanish and so will the homological Euler characteristic of a finite index subgroup Γ of GL m (Z[ξ 3 ]) with coefficients in V , Suppose that χ(C(A)) = 0.
Then by proposition 2.9 part (d) we obtain that A multiplicity of the eigenvalues 1, −1, is at most 1. Thus, the dimension of A is at most 2. Therefore if the dimension of A is greater than 2 then the Euler characteristic of the centralizers of A will vanish and so will the homological Euler characteristic of a finite index subgroup Γ of And finally, for all other fields K by proposition 2.9 part (e), we have that always Therefore, the homological Euler characteristic of a finite index subgroup Γ of Similarly we obtain the vanishing result for finite index subgroups of SL m (O K ). Proof. (of theorem 0.2) Let Γ be a finite subgroup of SL m (O K ). Let A be a torsion element of Γ. Then then the centralizer of A inside Γ, where q is a positive integer. Thus, vanishing of χ(C SLmO K (A)) implies vanishing of χ(C Γ (A)). We use a generalization of Kenneth Brown's formula (see section 10,and [B2]): where the summation is over all torsion elements in Γ up to conjugation, and V is a representation of Γ Let K = Q. Let also A ∈ SL m Z.
Then by proposition 2.9 part (a) we obtain that A multiplicity of the eigenvalues 1, −1 at most 2, and multiplicity of the eigenvalues i, −i, ξ 3 ,ξ 3 , ξ 6 ,ξ 6 at most 1. Thus, the dimension of A is at most 10. Therefore if the dimension of A is greater than 10 then the Euler characteristic of the centralizers of A will vanish and so will the homological Euler characteristic of a finite index subgroup Γ of SL m Z with coefficients in V , Suppose that χ(C(A)) = 0.
Then by proposition 2.9 part (b) we obtain that A multiplicity of the eigenvalues 1, −1, i, −i is at most 1. Thus, the dimension of A is at most 4. Therefore if the dimension of A is greater than 4 then the Euler characteristic of the centralizers of A will vanish and so will the homological Euler characteristic of a finite index Suppose that χ(C(A)) = 0.
Then by proposition 2.9 part (c) we obtain that A multiplicity of the eigenvalues 1, −1, ξ 3 ,ξ 3 , ξ 6 ,ξ 6 at most 1. Thus, the dimension of A is at most 6. Therefore if the dimension of A is greater than 6 then the Euler characteristic of the centralizers of A will vanish and so will the homological Euler characteristic of a finite index subgroup Γ of SL m (Z[ξ 3 ]) with coefficients in V , Let K be totally real field.. Let also Suppose that χ(C(A)) = 0.
Then by proposition 2.9 part (d) we obtain that the dimension of A is at most 2. Therefore if the dimension of A is greater than 2 then the Euler characteristic of the centralizers of A will vanish and so will the homological Euler characteristic of And finally, for all other fields K by proposition 2.9 part (e), we have that always Therefore, the homological Euler characteristic of a finite index subgroup Γ of 3 Torsion elements of GL 2 Z and GL 3 Z This section deals with the computational task of finding all the torsion element up to conjugation in GL 2 Z and GL 3 Z.
Proposition 3.1 All torsion elements in GL 2 Z up to conjugation are listed in the table below together with their centralizers and Euler characteristic of the centralizers.
Proof. Let A be a torsion element in GL 2 Z. If the characteristic polynomial of A is irreducible over the rational numbers then its root belong to a quadratic extension of Q. The only such options occur if the roots are 3-rd, 4-th or 6-th root of unity. The corresponding rings of integers are unique factorization domains. Therefore by proposition 1.3 there is only one matrix T n up to conjugation having eigenvalues n-th roots of unity for n = 3, 4, or 6. This gives the torsion elements in part (d), (e) and (f).If the characteristic polynomial of A is reducible then its roots are 1 or −1. If both roots are 1 or both are −1 then we have only one representative for each case which is given in part (a) and (b). It remains to consider the case when the eigenvalues of A are 1 and −1. By theorem 1.5 we can assume that A is upper triangular. What remains to be done is to determine which elements in the upper right corner of A give conjugate elements. Consider P A 11 A 22 with A 11 = 1 and A 22 = −1. Then the matrix representing P A 11 A 11 is Thus, 1 a 0 −1 and 1 b 0 −1 are conjugate if and only if a ≡ b mod 2. Thus we obtain the two torsion elements in (c1) and (c2) corresponding to even a and odd a.
For GL 3 Z we find all the torsion elements up to conjugation.
Proposition 3.2 All torsion elements in GL 3 Z up to conjugation are listed in the table below together with their centralizers and Euler characteristic of the centralizers.
Proof. If A is a torsion element in GL 3 Z then its eigenvalues are roots of 1. If λ is an eigenvalue then all of its Galois conjugates are eigenvalues. If λ is an n-th root of 1 then all its Galois conjugates are all the primitive n-th roots of 1. Their number is ϕ(n) and we must have at must 3 of them. The inequality ϕ(n) ≤ 3 has solutions n = 1, 2, 3, 4, 6. In all of these cases we have ϕ(n) ≤ 2.That is, Q(λ) is at most quadratic extension of Q Thus, the remaining eigenvalue must be rational. Therefore the matrix A must have an eigenvalue +1 or −1. If all the eigenvalues are 1 then the matrix is either the identity or it contains a non-trivial Jordan block, which cannot be of finite order. That gives case (a).Minus that matrix gives case (b). If A the characteristic polynomial of A has a root at −1 and a double root at 1 then we need to use the material that we developed so far. Using theorem 1.6 we know that the matrix A can be conjugated to a blocktriangular matrix with a zero block under the diagonal. The diagonal block must be A 11 = I 2 and A 22 = −1. Then acts on M at 2,1 Z. Therefore P mod = 2M at 2,1 Z and Q mod = M at 2,1 (Z/2).
The centralizer of A 11 = I 2 , C(A 11 ) = GL 2 Z acts on Q mod and has two orbits: the zero vector and the non-zero vectors. Therefore we have exactly two non-conjugate matrices up to conjugation with diagonal blocks I 2 and −1. We assumed that below the diagonal the block is zero. Above the diagonal there are two cases leading to (c1) and (c2). In one of them we have zero above the diagonal, and in the other a non-zero representative of Q mod , for example [1, 0] t . By lemma 2.1 we obtain the centralizer in the case of (c1). For the case (c2), we have that the centralizer of A 22 = −1 is and −1 acts trivially on [1, 0] t ǫQ mod .
In case not all the eigenvalues of A are ±1, we must have either third, fourth or sixth root of 1. Together with them the third eigenvalue is either 1 or −1. That exhausts all cases for the eigenvalues of a torsion element in GL 3 Z.
Suppose the eigenvalues of A are primitive third roots of 1 and 1. We can assume that the matrix A is in block-triangular form. Let the blocks be For A 11 we could have picked any other matrix with characteristic polynomial t 2 + t + 1.
In Z[ξ 3 ] all ideals are principal. Now using corollary 1.4, we obtain that any two matrices with characteristic polynomial t 2 + t + 1 are conjugate to each other inside GL 2 Z. We assumed that A 21 = 0. For the simplification of A 12 we use Therefore, Then C(A 22 ) = ±1 acts on Q mod by exchanging the two nonzero elements. Thus there are two orbits on Q mod . A representative of the zero orbit is [0, 0] t and for the non-zero orbit [1, 0] t . Thus, we obtain the cases (e1) and (e2). By lemma 2.1 we obtain the centralizer of (e1). For the centralizer of (e2), we observe that −1 from C(A 22 ) and −I 2 from C(A 11 ) do change independently the element However, if they act simultaneously, they keep the element [1, 0] t fixed. This is an element of order 2 that fixes [1,0] t . An element of order 3 from C(A 11 ) must fix [1, 0] t . Thus, the centralizer is C 3 × C 2 . Considering −A we obtain the six torsion from (f1) and (f2).
If the eigenvalues of A are primitive third roots of 1 and −1 then we can again assume that A is in block-diagonal form with block A 11 = 0 1 −1 −1 and A 22 = −1.
Then P A 11 A 22 = I 2 ⊗ A 22 − A 11 ⊗ 1. Therefore, Then det P A 11 A 22 = 1 which means that Q mod = 0. Therefore, there is only one conjugacy class with such A 11 and A 22 , and we can take for A 12 any vector. So we take [0, 0] t . The centralizer of A is a product of two centralizers by lemma 2.1. That concludes case (g). Case (h) is taking −A in the previous case. Thus, we have exhausted all 3-torsions and all 6-torsions in GL 3 Z. Suppose we have a matrix A with eigenvalues i, −i and 1. (Taking the minus sign we obtain a matrix with eigenvalues i, −i and −1.) Using theorem 1.3, we can assume that A is in block-triangular form with zero below the diagonal block, and A 11 = 0 1 −1 0 and A 22 = 1. Then Therefore, which means that Q mod = Z/2. The module Q mod has no automorphisms therefore each of its elements correspond to a conjugacy class. Thus there are two classes: (i1) and (i2). The centralizer of (i1) can be computed via Lemma 2.1. For (i2) note that every element of C(A 11 ) or C(A 22 ) fixes the non-zero element of Q mod . Therefore the centralizer in the case of (i2) is isomorphic as an abstract group to the centralizer of (i1). Taking the minus sign we obtain (j1) and (j2).

Resultants
This section is computational. It deals with resultants and Euler characteristics of centralizers needed for computation of homological Euler characteristics. Let us recall the notation that we are using. By A = [A 11 , . . . A kk ] we mean a matrix A, whose diagonal blocks are A ii for i = 1, . . . k. We also assume that the characteristic polynomial f i of A ii is a power of an irreducible polynomial. Also, where R(f i , f j ) is the resultant of f i and f j . In the following statements we compute

|R(A)| · χ(A)
for various matrices A for the need of computation of the homological Euler characteristic of an arithmetic group over Z, Z[i] and Z[ξ 3 ] (see theorem 2.10).
From theorem 2.8 part (b) we have that the only torsion element in GL 2 Z[i] are these element whose Euler characteristic of their centralizer is not zero are the ones whose eigenvalues are different and belong to the set {±1. ± i}.

Lemma 4.4 This is the computation of |N Q(i)/Q (R(T ))| · χ(C(T )) for all torsion elements up to conjugation T in GL
Proof. If k is not congruent to l modulo 4 then It remain to compute R(A) in the two cases.
From theorem 2.8 part (b) we have that the only torsion element in GL 2 (Z[ξ 3 ]) are these elements whose Euler characteristic of their centralizer is not zero are the ones whose eigenvalues are different and belong to the set Lemma 4.5 This is the computation of |N Q(i)/Q (R(T ))| · χ(C(T )) for all torsion elements up to conjugation T in GL 2 Z[ξ 3 ] such that χ(C(T )) = 0. Let ξ 6 = e ])) = 1 9 , for k = 0, 1, 2.
Proof. If k is not congruent to l modulo 6 then It remain to compute R(A) in the two cases.

Homological Euler characteristic of GL m (O K ) with coefficients symmetric powers
In this section we compute the homological Euler characteristic of GL n (O K ) for K = Q, Q(i) and Q(ξ 3 ). For other imaginary quadratic fields the formula is given in theorem 2.13. And for all other fields the Euler characteristic vanishes (see theorem 0.1 part (c)).

Homological Euler characteristic of GL m Z.
Lemma 5.1 For the symmetric powers of the standard representation V 2 of GL 2 Z, we have: (a) Tr(I 2 |S n V 2 ) = n + 1 Proof. If λ 1 and λ 2 are the two eigenvalues of A acting on V 2 then Tr(A|S n V 2 ) = n i=0 λ i 1 λ n−i 2 .
Note that Tr(A|S n V 2 ) depends on the on the eigenvalues of A on V 2 not the conjugacy class that A belongs to. Also, Tr(−A|S n V 2 ) = (−1) n tr(A|S n V 2 ).

From
Tr(I 2 |S n V 2 ) = dim(S n V 2 ) = n + 1, we obtain part (a) and (b). If λ 1 = 1 and λ 2 = −1 then Tr(A|S n V 2 ) = 1 − 1 + 1 − . . . . and we have n + 1 summands. Thus Tr(A|S n V 2 ) is 1 if n is even and 0 if n is odd. This proves part (c). For part (d) the eigenvalues of T 3 on V 2 are ξ 3 and ξ −1 3 . Thus, Also the sum of three successive summands is zero, and the total number of summands is n + 1. Therefore, if n ≡ 0mod3 then Tr(T 3 |S n V 2 ) = ξ n 3 = 1.
This proves part (d). It also proves part (e) because −T 3 is conjugate to T 6 . Similarly, for the 4-torsion T 4 we have Also, every two consecutive summands add up to zero. Therefore if n is odd the trace is zero. If n is even we have which is 1 if n ≡ 0 mod4 and −1 if n ≡ 2 mod 4. This proves part (f).
Therefore we need only formulas for half of the torsion elements. One other remark: we need only to examine the trace for torsion elements whose Euler characteristic of the centralizer is not zero. For GL 3 Z these elements are I 3 and −I 3 .

Lemma 5.3
The trace of the torsion elements in GL 3 Z acting on the symmetric power of the standard representation are given by: We have omitted part (a) and (b) corresponding to I 3 and −I 3 because they give no contribution to the homological Euler characteristic since Proof. The above formulas can be derived from the following fact: let AǫGL k Z and

This proves part (c) and (d).
For part (e) we use lemma 5.1 part(d) and Similarly,

This proves part (e) and (h).
For part (f) we use lemma 5.1 part (e) and Tr([ This proves part (i) and (j).
For part (a) we use lemma 6.3 part(c) and Tr (

Homological Euler characteristic of GL m (Z[i])
Lemma 5.7 For the symmetric power of the standard representation of GL 2 (Z for k and l not congruent to each other modulo 4.
Theorem 5.8 The homological Euler characteristic of GL 2 (Z[i]) with coefficients the symmetric powers of the standard representation are given by Proof. If in the symmetric power n is not divisible by 4 then the central element [i, i] acts non-trivially on the cohomology Therefore the cohomology and the homological Euler characteristic vanish. If the symmetric power is 4n then we can use lemma 4.4 and 5.7, and we obtain
Theorem 5.10 The homological Euler characteristic of GL 2 Z[ξ 3 ] with coefficients the symmetric powers of the standard representation are given by Proof. If in the symmetric power n is not divisible by 6 then the central element [ξ 6 , ξ 6 ] acts non-trivially on the cohomlogy Therefore the cohomology and the homological Euler characteristic vanish. If the symmetric power is 6n then we can use lemma 4.5 and 5.9, and we obtain )Tr(A|S 6n V 2 ) = 6 · 1 36 + 6 · 1 12 + 3 · 1 9 = 1.
We are going to compute the homological Euler characteristic of Γ 1 (2, N ), Γ 1 (3, N ) and Γ 1 (4, N ) using the torsion elements in the groups. Recall Γ 1 (m, N ) is a subgroup of GL m Z stabilizing the covector [0, · · · , 0, 1] modulo N . The same notation is used in [G1]. Recall, also, the homologocal Euler characteristic of a group G with coefficients a finite dimensional representation V over a field k over is We are going to use a generalization of a formula due to K. Brown (see [B2]) where C consists of representatives of the torsion elements of G up to conjugation, and C(A) is the centralizer of A.In order to make use of this formula, we have to find the torsion elements of Γ 1 (m, N ) using the torsion elements in GL m Z. In a more general setting let G be a group and Γ be a subgroup. It is possible to find two elements of Γ which are conjugate to each other in G but not in Γ. In the case of Γ = Γ 1 (m, N ) and G = GL m Z, we need to find all element of Γ which are not conjugate to any of the others however they become conjugate when considered as elements of the bigger group G. To do that we construct a set N Γ G (A) with the property that the double quotient Γ\N Γ G (A)/C G (A) parametrizes the elements in Γ that are non-conjugate to each other but all become conjugate to A when considered as elements of G. The same set N Γ G (A) is used in K. Brown paper [B2].
Lemma 6.1 Let G be a group and Γ be a subgroup. Given an element A ∈ Γ, the set of elements in which are conjugate to A in G but not in Γ is parametrized by the elements of the double quotient Proof. Let A 1 and A 2 be two elements of Γ both conjugate to A in G and Conjugate to each other in Γ. Then there exist X 1 , X 2 in the bigger group G such that and, equivalently, (X −1 2 Y X 1 )A = A(X −1 2 Y X 1 ). We obtain that X −1 2 Y X 1 ǫC G (A). Therefore Y X 1 ∈ X 2 C G (A) and X 1 ∈ ΓX 2 C G (A) We obtain that X 1 and X 2 belong to the same double quotient in Conversely, suppose X 1 and X 2 belong to the same double quotient. We are going to show that both elements X 1 AX −1 1 and X 2 AX −1 2 belong to Γ, and also that they are conjugated to each other in Γ. By definition of N Γ G (A) we have that X 1 AX −1 1 and X 2 AX −1 2 belong to Γ. Since the two elements belong to the same double quotient, we have that X 2 = Y X 1 C, where Y ∈ Γ and C ∈ C G (A). Then Thus, the elements X 1 AX −1 1 and X 2 AX −1 2 are conjugate to each other in Γ.
6.1 Homological Euler characteristic of Γ 1 (m, N) Lemma 6.2 Let A be an k-torsion element of Γ 1 (m, N ). If k and N are relatively prime then A has an eigenvalue +1.
Proof. Let L be the field obtained by adjoining all eigenvalues of A to Q. Let p be a prime dividing N . And let p be a prime ideal in L sitting above p. Denote, also by µ(L) the roots of 1 in L. And let Then π maps µ(L) onto F q because there are no p-roots of 1 in L since p does not divide k. So the eigenvalues on A in L are mapped onto F q . However, 1 is an eigenvalue of A mod p in F q . Therefore 1 is an eigenvalue of A in L. By proposition 2.8 part (a) we know that for a torsion element A, we have χ(C(A)) = 0 only when the eigenvalues of A are among {1, −1, i, −i, ξ 3 ,ξ 3 , ξ 6 ,ξ 6 }. Lemma 6.3 Let A ∈ Γ 1 (m, N ) be a torsion element with χ(C(A)) = 0, let N be relatively prime with 2 and 3 and let f be the characteristic polynomial of A. Then +1 is a root of f and it has multiplicity 1 or 2.
(a) If the root +1 has multiplicity 1 then the set Let X ∈ N Γ G (A) and let B = XAX −1 . Then B ∈ Γ 1 (m, N ). We can write the matrices A, B and X in 2 × 2 block form with diagonal blocks A 11 , B 11 and X 11 of size (m − 1) × (m − 1) and A 22 , B 22 and X 22 of size 1 × 1. Then Consider the (2, 1)-block of the product. The left hand side is X 21 A 11 and the right hand side is 1 · X 21 mod N.
Examine the map P B 22 A 11 : X 21 → X 21 A 11 − B 22 X 21 with B 22 = 1. The eigenvalues of A 11 and B 22 are different. From lemma 1.11 follows that the eigenvalues of P B 22 A 11 are λ i − 1 where λ i runs through the eigenvalues of A 11 . From the previous lemma we have that λ i − 1 is invertible mod N . Thus, is non-singular mod N . Therefore, N ) is defined as the subgroup of GL m Z that sends [0, . . . , 0, 1] to [0, . . . , 0, a] modulo N for any a = 0 mod N.
On the other hand, Γ 0 (m, N ) lies inside the normalizer of Γ 1 (m, N ). Also, from the definition of N Γ G (A). We obtain that Let X be an element in Γ 0 (m, N ) of the same block form as before. Inside the quotient the element X is determined by Note that there are ϕ(N ) options for X 22 because it has to be invertible modulo N . From the block triangular theorem (theorem 2.3) we know that we can choose A so that A 21 = 0. If C is in the centralizer of A inside GL m Z, by lemma 3.2 we know that C 21 = 0. Therefore the centralizer of A modulo Γ 1 (m, N ) is determined by C 22 which could be +1 or −1. Therefore the double quotient consist of Therefore, the double quotient has one element. Assume that m > 2. By the block triangular theorem we can conjugate A to a matrix of the block form Note that the block A 11 is an (m − 2) × (m − 2) matrix and it does not have an eigenvalue 1. We can assume that A is of the above form. Let X be an element of N Γ G (A), and let B = XAX −1 . If we write this equation with respect to the block form of A we obtain: Consider the (3, 1)-block of the product. From the left hand side we obtain and from the right hand side we obtain 1 · X 31 . We need to examine the map We have to consider the map If λ k are the eigenvalues of A 11 then λ k − 1 are the eigenvalues of P B 33 A 11 from the previous lemma it follows that λ k − 1 is not divisible by p for any p dividing N . This, the map P B 33 A 11 is non-singular modulo N . We have We are going to show that the inclusion is in fact an equality. We are going to prove the following inclusions modulo N : The last inclusion always holds, not only modulo N . Modulo any prime number q relatively prime to N , we have where SL ± m (Z/qZ) is the group of m×m matrices with coefficients in Z/qZ whose determinant is ±1. Therefore, the only restriction on N Γ G (A) become apearant modulo N . From these inclusions it will follow that Therefore the double quotient will have one element. We have proved the first inclusion. The last inclusion follows directly from the definition of N Γ G (A). And the second to the last inclusion holds because H ′ is a subgroup of C GLmZ (A). Now we are going to prove the second inclusion. Consider A as a 2 × 2-block matrix be in the centralizer of A. Consider C as a 2 × 2 block matrix. Namely, with C 22 a 2 × 2 matrix. The admissible matrices C are such that C 11 A 12 C −1 22 and A 12 map to the same element in Q mod , where Q mod = M at n−2,2 Z/Im P A 11 ,I 2 , the matrix C 12 is determined uniquely from C 11 and C 22 . Let be a subgroup of C(A). We are going to show that are subgroups. In this setting C 12 is uniquely determined by C 11 ; and we are going to write it as C 12 (C 11 ).
Then we have that M · P −1 A 11 ,I 2 has integer entries. Therefore, Im (P A 11 ,I 2 ) ⊂ M · M at m−2,2 Z.
If C 22 is in Γ(2, M ) then Therefore A 12 C −1 22 ≡ A 12 mod Im (P A 11 ,I 2 ). Set Then However, we need H ′ for comparison of groups modulo N . Note that M and N are relatively prime because M = det(P A 11 I 2 ) which has prime factors only 2 and 3. And by assumption N is relatively prime to 2 and 3 Then Γ(2, M ) ≡ SL 2 Z mod N.
We need to show that N 31 ⊂ Γ 1 (n, N ) · H ′ modulo N . Let X ∈ N 31 . We have that X 32 and X 33 are relatively prime modulo N . Otherwise their common factor will divide det(X), which is ±1. Then we can find Y and Z such that Y X 33 − ZX 32 = 1. Let Then Let C be a corresponding element in H ′ with C 22 congruent to the above matrix modulo M and C 11 = I n−2 . Then and we are done.
Theorem 6.4 Let V be a representation of GL m Z, and let N be an integer relatively prime to 2 and 3. Then where A 1 and A 2 are block-diagonal matrices with zero blocks off the diagonal, and with block on the diagonal A ii varying through {−1, −I 2 , T 3 , T 4 , T 6 } and A ii and A jj have no common eigenvalues. We set Also ϕ(N ) is the Euler function of N , and ϕ 2 (N ) is the arithmetic function generated by Proof. Let A be a torsion element of such that χ(C Γ (A)) = 0.
By lemma 6.2 we have that 1 is an eigenvalue of A. Also, if χ(C Γ (A)) = 0 then the multiplicity of the eigenvalue 1 is at most 2. We are going to prove that if the eigenvalue has multiplicity 1 then and if the eigenvalue 1 has multiplicity 2 then Assume that we have proven these two formulas. From the generalization of Brown's formula, we have In the summation it is enough to sum over torsion elements A that have eigenvalue 1 either with multiplicity 1 or with multiplicity 2. First take the sum over elements A that become conjugate to each other in GL m Z. Let A has eigenvalue 1 with multiplicity 1. By lemma 6.3 where the sum is taken over all non-conjugate A ′′ in Γ 1 (m, N ) that become conjugate to A ′ in GL n Z. Now sum over the torsion elements A ′ in GL m Z that become conjugate to each other in GL m C. Similarly, in the case when the eigenvalue 1 has multiplicity 2, we have instead of ϕ(N ). Thus, it remains to prove that if the eigenvalue 1 has multiplicity 1 then χ(C Γ (A)) = 2 · χ(C GLmZ (A)), and if the eigenvalue 1 has multiplicity 2 then Suppose A has eigenvalue 1 with multiplicity one. We can assume that A is in block-triangular A 11 A 22 0 1 with A 22 = 1 and A 21 = 0. If C is a matrix in GL m Z commuting with A then C is if the same block form with C 21 = 0 and C 22 = ±1. Exactly one of the matrices C and −C belongs to Γ 1 (m, N ) because of the C 22 entry. Thus the centralizer of A in GL m Z contains twice as many element as the centralizer inside Γ(m, N ). Therefore, Suppose A has eigenvalue 1 with multiplicity two. We can assume that A is of the form If C commutes with A then it is of the same block type with The possibilities of C 22 are determined by relations modulo the image of Note that if χ(C(A)) = 0 then the eigenvalues of A are among {±1, ±i, ±ξ 3 }.
Then det(P A 11 ,I 2 ) has only prime factors 2 and 3. Therefore the conditions on C 22 imposed by P A 11 ,I 2 are independent from the condition CǫΓ 1 (n, N ), because that last condition is a congruence modulo N but N is relatively prime 2 and 3. Therefore Corollary 6.5 For N not divisible by 2 and 3 the homological Euler characteristics of Γ 1 (2, N ) with coefficients the symmetric powers of V 2 are given by where ϕ(N ) is the multiplicative Euler function generated by ϕ(p n ) = p n (1 − 1 p ) and ϕ 2 (N ) is the multiplicative function generated by Proof. From theorem 6.4, we have that Similarly, Theorem 6.6 The torsion elements of Γ 1 (3, N ), for N not divisible by 2 or 3, are given in the following table together with the Euler characteristic of their centralizers. In the table is the multiplicative function generated by For abbreviation set Γ = Γ 1 (3, N ) and G = GL 3 Z.
Proof. This table follows from the table in proposition 3.2 and from the previous two lemmas on the calculation of The only part that requires a proof given two elements A and B in Γ 1 (3, N ) such that both are conjugate in GL 3 Z to the same element form the above list, we have Indeed, if A and B are conjugate to each other in Γ 1 (3, N ) then we are done. If they are not conjugate to each other in Γ 1 (3, n) then the double quotient Γ\N Γ G (A)/C G (A) has more than one element. From lemma 5.3 we have that Corollary 6.7 For N not divisible by 2 and 3 the homological Euler characteristic of Γ 1 (3, N ) is given by where ϕ(N ) is the multiplicative Euler function generated by and ϕ 2 (N ) is the multiplicative function generated by Proof. Brown's formula gives the homological Euler characteristic as a sum of the usual (orbifold) Euler characteristic of the centralizers of the torsion elements. From the previous proposition we obtain χ h (Γ 1 (3, N )) = − 1 48 ϕ 2 (N ) − 1 16 ϕ 2 (N ) − 1 48 ϕ(N ) − − 1 16 ϕ(N )+ + 1 12 ϕ(N ) + 1 6 ϕ(N ) + 1 12 ϕ(N ) + 1 8 ϕ(N ) + 1 8 ϕ(N ) = = − 1 12 ϕ 2 (N ) + 1 2 ϕ(N ).

Homological Euler characteristic of
Lemma 6.9 Let A be an k-torsion element of Γ 1 (m, a). If 2k and a are relatively prime then A has an eigenvalue +1.
Proof. Let L be the field obtained by adjoining all eigenvalues of A to Q(i). Let p be a prime ideal dividing a. And let P be a prime ideal in L sitting above p. Denote, also by µ(L) the roots of 1 in L. And let where F q = L/P.
Let p be the rational prime sitting below p. Then π maps µ(L) onto F q because there are no p-roots of 1, in L since p does not divide k. So the eigenvalues on A in L are mapped onto F q . However, 1 is an eigenvalue of A mod p in F q . Therefore 1 is an eigenvalue of A in L. By proposition 2.8 part (b) we know that for a torsion element A, we have χ(C(A)) = 0 only when the eigenvalues of A are among Lemma 6.10 Let A ∈ Γ 1 (m, a) be a torsion element with χ(C(A)) = 0, let a be relatively prime with (1 + i), and let f be the characteristic polynomial of A. Then +1 is a root of f and it has multiplicity 1. And the set ).
Proof. In order to save some space set and Let X ∈ N Γ G (A) and let B = XAX −1 . Then B ∈ Γ. We can write the matrices A, B and X in 2 × 2 block form with diagonal blocks A 11 , B 11 and X 11 of size (m − 1) × (m − 1) and A 22 , B 22 and X 22 of size 1 × 1. Then are λ i − 1 where λ i runs through the eigenvalues of A 11 . From the previous lemma we have that λ i − 1 is invertible mod a. Thus, is non-singular mod a. Therefore, the element X is determined by Note that there are ϕ Z[i] (a) options for X 22 because it has to be invertible modulo a. From the block-triangular theorem (theorem 2.3) we know that we can choose A so that A 21 = 0. If C is in the centralizer of A inside GL m (Z[i]), by lemma 3.2 we know that C 21 = 0. Therefore the centralizer of A modulo Γ 1 (m, a) is determined by C 22 which could be +1, −1, i or −i. Therefore the double quotient consist of elements.
Theorem 6.11 Let V be a representation of GL m (Z[i]), and let a be an ideal relatively prime to (1 + i). Then is the arithmetic function defined on the ideals of Z[i] generated by ).
Proof. Let A be a torsion element of such that χ(C Γ (A)) = 0.
By lemma 6.9 we have that 1 is an eigenvalue of A. Also, if χ(C Γ (A)) = 0 then the multiplicity of the eigenvalue 1 is at most 1. We are going to prove that Assume that we have proven this formula. From the generalization of Brown's formula, we have In the summation it is enough to sum over torsion elements A that have eigenvalue 1 either with multiplicity 1. First take the sum over elements A that become conjugate to each other in GL m (Z[i]). Let A has eigenvalue 1 with multiplicity 1. By lemma 6.9 where the sum is taken over all non-conjugate A ′′ in Γ 1 (m, a) that become conjugate to A ′ in GL n (Z[i]).
We can assume that A is in block-triangular A 11 A 22 0 1 with A 22 = 1 and A 21 = 0. If C is a matrix in GL m (Z[i]) commuting with A then C is if the same block form with C 21 = 0 and C 22 = i k . Exactly one of the matrices C, −C, iC and −iC belongs to Γ 1 (m, a) because of the C 22 entry. Thus the centralizer of A in GL m (Z[i]) contains 4 times as many element as the centralizer inside Γ(m, a). Therefore, By the previous lemma we have that the number of non-conjugate matrices in Γ that become conjugate in G is ϕ Z[i] (a)/4. This proves the theorem.
Theorem 6.12 The homological Euler characteristic of Γ 1 (2, a) ⊂ GL 2 (Z[i]) is given by where ϕ Z[i] (a) is the multiplicative function defined on the ideals of Z[i], generated by ).
Proof. From theorem 6.11 and lemma 4.4 we have

Homological Euler characteristic of
Lemma 6.13 Let A be an k-torsion element of Γ 1 (m, a). If 6k and a are relatively prime then A has an eigenvalue +1.
Proof. Let L be the field obtained by adjoining all eigenvalues of A to Q(ξ 3 ). Let p be a prime ideal dividing a. And let P be a prime ideal in L sitting above p. Denote, also by µ(L) the roots of 1 in L. And let where F q = L/P.
Let p be the rational prime sitting below p. Then π maps µ(L) onto F q because there are no p-roots of 1, in L since p does not divide k. So the eigenvalues on A in L are mapped onto F q . However, 1 is an eigenvalue of A mod p in F q . Therefore 1 is an eigenvalue of A in L. By proposition 2.8 part (b) we know that for a torsion element A, we have χ(C(A)) = 0 only when the eigenvalues of A are among {1, −1, ξ 3 , −ξ 3 , ξ 6 , −ξ 6 }.
Lemma 6.14 Let AǫΓ 1 (m, a) be a torsion element with χ(C(A)) = 0, let a be relatively prime with (1 − ξ 3 ), and let f be the characteristic polynomial of A. Then +1 is a root of f and it has multiplicity 1. And the set ).
Proof. In order to save some space set and Let XǫN Γ G (A) and let B = XAX −1 . Then BǫΓ. We can write the matrices A, B and X in 2×2 block form with diagonal blocks A 11 , B 11 and X 11 of size (m−1)×(m−1) and A 22 , B 22 and X 22 of size 1 × 1. Then Consider the (2, 1)-block of the product. The left hand side is X 21 A 11 and the right hand side is 1 · X 21 mod a. are λ i − 1 where λ i runs through the eigenvalues of A 11 . From the previous lemma we have that λ i − 1 is invertible mod a. Thus, is non-singular mod a. Therefore, Therefore, On the other hand, Γ 0 (m, a) lies inside the normalizer of Γ 1 (m, a). Also, Let X be an element in Γ 0 (m, a) of the same block form as before. Inside the quotient Γ 1 (m, a)\Γ 0 (m, a) the element X is determined by Note that there are ϕ Z[ξ 3 ] (a) options for X 22 because it has to be invertible modulo a. From the block-triangular theorem (theorem 2.3) we know that we can choose A so that A 21 = 0. If C is in the centralizer of A inside GL m (Z[ξ 3 ]), by lemma 3.2 we know that C 21 = 0. Therefore the centralizer of A modulo Γ 1 (m, a) is determined by C 22 which could be +1, −1, i or −i. Therefore the double quotient consist of elements.
Proof. Let A be a torsion element of such that χ(C Γ (A)) = 0. Then By lemma 6.9 we have that 1 is an eigenvalue of A. Also, if then the multiplicity of the eigenvalue 1 is at most 1. We are going to prove that Assume that we have proven this formula. From the generalization of Brown's formula, we have In the summation it is enough to sum over torsion elements A that have eigenvalue 1 either with multiplicity 1. First take the sum over elements A that become conjugate to each other in GL m (Z[ξ 3 ]). Let A has eigenvalue 1 with multiplicity 1. By lemma 6.9 where the sum is taken over all non-conjugate A ′′ in Γ 1 (m, a) that become conjugate to A ′ in GL n (Z[ξ 3 ]). We can assume that A is in block-triangular with A 22 = 1 and A 21 = 0. If C is a matrix in GL m (Z[ξ 3 ]) commuting with A then C is if the same block form with C 21 = 0 and C 22 = i k . Exactly one of the matrices ξ k 6 C, k = 0, 1, . . . 5 belongs to Γ 1 (m, a) because of the C 22 entry. Thus the centralizer of A in GL m (Z[ξ 3 ]) contains 6 times as many element as the centralizer inside Γ(m, a). Therefore, χ(C Γ (A)) = 6 · χ(C GLmZ[ξ 3 ] (A)).
By the previous lemma we have that the number of non-conjugate matrices in Γ that become conjugate in G is ϕ Z[ξ 3 ] (a)/6. This proves the theorem.
Theorem 6.16 The homological Euler characteristic of Γ 1 (2, a) ⊂ GL 2 (Z[ξ 3 ]) is given by where ϕ Z[ξ 3 ] (a) is the multiplicative function defined on the ideals of Z[ξ 3 ], generated by 7 Dedekind zeta fuction at −1 and χ h (SL 2 (O K )) for totally real number fields In this section we examine the homological Euler characteristic of SL 2 (O K ) for K a totally real number field. At the end of the section we consider two examples for the fields Q and Q( √ 3). The relation to the Dedekind zeta function is We obtain the following theorem.
Theorem 7.1 Let K be a totally real number field. Then where the first summation is taken over all roots ξ of 1 such that [K( where the sum is taken over the torsion elements of SL 2 (O K counted up to conjugation. Note that besides ±I 2 the rest of the torsion elements have irreducible characteristic polynomial. Their eigenvalues are ξ and ξ −1 , where ξ is a root of 1 such that [K(ξ) : K] = 2. The centralizers of I 2 and of −I 2 are the entire group. Therefore the Euler characteristic of their centralizer gives two copies of ζ K (−1).
We need to examine closely the rest of the torsion elements A in SL 2 (O K . We have that the eigenvalues of A are ξ and ξ −1 , where For that reason, in the theorem we sum over such ξ. By proposition 1.7 we have that the non-conjugate matrices with the same characteristic polynomial as A are parametrized by where the above set denotes the set of ideal classes in O K [ξ] that are free as O Kmodules. For that reason, in the theorem, we are summing over ideal classes from this set. Fix such a torsion matrix A. Let I be the corresponding ideal class from Cl(O K [ξ]/O K ), and ξ and ξ −1 be its eigenvalues. We are going to define R I . Consider the matrices with coefficients in K that commute with A. Denote denote the collection by C M at 2 K (A). We have that C M at 2 K (A) ∼ = K(ξ), By sending A to ξ, this isomorphism ψ is an isomorphism of K-algebras. Consider the ring of matrices with coefficients in O K that commute with A. Denote it by C M at 2 O K (A). Define R I = ψ(C M at 2 O K (A)). Then for the centralizers in GL 2 O K and SL 2 O K we have ). The second equality holds because the determinant of a matrix B in the centralizer C GL 2 O K (A) corresponds to the norm N K(ξ)/K (β), where β = ψ(B). Thus, det(B) = 1 is equivalent to N K(ξ)/K (β) = 1. The last equality holds only when β is a root of 1 because the ranks of the groups of units in K and in K(ξ) coincide. In Brown's formula we need to compute χ(C(A)), which we have done Proposition 1.3 classifies non-conjugate matrices in GL 2 (O K ). In order to pass to SL 2 (O K ), we use lemma 6.1. It gives that the non-conjugate matrices in SL 2 (O K ) that become conjugate to A in GL 2 (O K ) are parametrized by is parametrized by the determinant of the matrices, namely, by O × K . The group by which we quotient from the right is Its determinant leads to N K(ξ)/K (R × I ). Thus, the double quotient is isomorphic to O × K /N K(ξ)/K (R × I ), which gives the last ingredient in the theorem. Thus, we have where the 1/2 occurs because we are summing over all roots of unity ξ in , while when we are summing over the torsion matrices A we count the eigenvalues ξ and ξ −1 at the same time.
Corollary 7.2 Let K be a totally real number field. Suppose that for each root of 1 ξ such that [K(ξ) : K] = 2, we have that calO K [ξ] is integrally closed. Then where the sum is taken over all roots of 1 ξ such that [K(ξ) : K] = 2, and where all the maps are norm maps.
Proof. If R is a number ring we have that K 0 (R) = Cl(R) ⊕ Z and K 1 (R) = R × . If O K [ξ] is integrally closed then it coincides with O K(ξ) . Then the ideal classes in O K [ξ] that are free O K -modules are precisely Ker(K 0 (O K(ξ) ) → K 0 (O K )).
In the remaining portion of the section we give a simple method for determining what R I is. And we end the section with two examples.
Let A be a matrix in SL 2 (O K ) that has an irreducible over K characteristic polynomial. Let ξ and ξ −1 be its eigenvalues. By theorem 1.3 we have a correspondence between the ideal classes in O K [ξ] that are free as O K -modules. We give a way of determining what is the order R I . Let C M at 2,2 O K (A) be the ring of matrices with coefficient in O K that commute with A. Let C M at 2,2 K (A) be the ring (actually field) of matrices that commute with A. Consider that O K -isomorphism ψ : C M at 2,2 K (A) → K(ξ) that send A to ξ. Then R I is defined by Every element λ in O K(ξ) can be written as a linear function f (t) with coefficients in K, so that λ = f (ξ).
If the norm is (1+ √ 5)/2 then all the constants a, b, c, and d must be zero. Therefore (1 + √ 5)/2 is not in the image of the norm map and the cokernel of N is The kernel of N is Z[ξ 10 ] × tors = C 10 . Thus the contribution towards the homological Euler characteristic is 1 2 · 8 · 4 10 = 8 5 .
If ξ = ξ k 6 for k = 1, 2, 4, 5 then which is integrally closed. We are going to show that it has class number 1. First, we need to examine the units in the ring. Let N be the norm map from Q( √ 5, √ −3) to Q( √ 5). We try to solve N (α) = 1 + √ 5 2 .
We are going to use the formula for the leading coefficient of the Dedekind zeta function at 1. It is given by 2 r 1 (2π) r 2 hR w |d| , where r 1 and r 2 are the number of the real and the complex imbeddings if the field h is the cless number, R is the regulator, w is the number of roots of unity in the field, and d is the discriminant of the field.
which is integrally closed and of class number 1. Let N be the norm map from Q( √ 5, i) to Q( √ 5). We want to solve N (α) = 1 + √ 5 2 .
Therefore (1 + √ 5)/2 is not in the image of the norm map. Thus, the cokernel of N acting on the units in Q( √ 5, i) is isomorphic to C 2 × C 2 . And the kernel is isomprphic to C 4 . Then the contribution to the homological Euler characteristic is 1 2 · 2 · 4 4 = 1. We obtain that ζ Q(

Generalization of Brown's formula
The main result of this section is to prove a generalization of Brown's formula relating the homological Euler characteristic to the Euler characteristics of certain centralizers. Given an arithmetic group Γ and a finite dimensional representation V over a field K of characteristic zero. We prove the following theorem.
Theorem 8.1 The homological Euler characteristic of Γ with coefficients in V is given by where the sum is taken over all torsion elements of Γ counted up to conjugation, C(A)) is the centralizer of A inside Γ and Tr(A −1 |V ) is the trace of the action of A −1 on the finite dimensional vector space V .
Before we start with the proof we recall what is complete Euler characteristic, and what are some of its properties. For more detailed treatment of generalized Euler characteristic we refer to Bass' article [Ba] and Brown's book [B1].
Let R be a unital non necessarily commutative ring. Let F be a free R-module. And let f : F → F be an endomorphism of R-modules. We can choose a basis for F and then f can be written as a matrix (a ij ). We define the trace of f to be whereā is the projection of a from R to T (R) = R/[R, R], and [R, R] is the additive group generated by ab − ba for a, bǫR. Note that [R, R] is not an ideal so T (R) is only an additive group. The trace can be extended to endomorphisms of a finitely generated projective moduleS P . Such a module can be imbedded in a free module of finite rank F . Denote by i the imbedding, and by π the projection from F to P . The the trace of f : P → P if defined by Tr R (f ) := Tr R (i • f • π).
Note that i • f • π is an endomorphism of F . The trace can be also extended to modules M that admit finite length projective resolution by finitely generated projective modules. Then an endomorphism Let Γ be an arithmetic group. The relation between E(Γ), χ(Γ) and χ h (Γ) is the following χ(Γ) = c (I) , where the sum is taken over all conjugacy classes. In fact this sum is finite which follows from Brown's theorem Then a (A) = χ(C(A)).
We generalize the complete Euler characteristic in the following way. For a finite dimensional representation V of Γ, let E(Γ, V ) = Tr QΓ (id V ).
In order to pass to homological Euler characteristic we use a lemma due to Chiswell (see [Ch], lemma 12) at a time. Using infinitely many of them would lead to a contradiction. Another remark is that if any of the summation is finite , we can use the finite number. Let We are going to define an inclusion i : V R → F, and a projection π : F → V R .
Let v be an element of V . We define And extend it by R-linearity to V R . Let where rǫR and vǫV Then π • i = id V R .
Note that F is a free R-module of finite rank and V R is a projective module. By the definition of the trace, we have Tr R (id V R ) = Tr R (i • π) = (A) x (A) x Tr(A −1 |V ) · (A).
In order to find the right regularization of x (A) /x, we compare this result to the the Brown formula for the trivial representation. We have a natural inclusion QΓ → R.
Then this inclusion induces equality between Tr QΓ (id Q ) and Tr R (id Q R ). Using Brown's formula, we obtain that We also have that χ(C(A)) = 0 for non-torsion elements A. Thus, the formula for the trace over R leads to E(Γ, V ) = A: torsion χ(C(A))Tr(A −1 |V ) · (A).
Using Chiswell's lemma we obtain the formula for the homological Euler characteristic.
Proof. (second proof of theorem 8.1) This proof follows more colosely the paper of Brown [B2]. And it turn out that the formula we need is a consequence of a more general formula of Brown For a finite length chain C • of QΓ-modules that admit generalized Euler characteristic, define Define e(Γ, C • ) = c (I) , where I is the identity element in Γ. We have that e(Γ, Q) = χ(Γ).
The general formula that Brown obtains is that c (A) coincides with the coefficient next to (I) in Tr QΓ (M A ), where M A : (C • ) A → (C • ) A is multiplication by A. Now, let X be a contractable CW complex on which Γ act properly and discontinuously. One can assume also that a cell is mapped to itself by an element of Γ then the entire cell is fixed by that element pointwise. One can achieve that by subdivision of the cells. Then E(Γ, V ) = E(Γ, C • (X, V )), where C • (X, V ) is the cochain complex of X with coefficients in V . Let f ∈ C i (X, V ) and A ∈ Γ. let also σ be an i-th cell and v ∈ V so that f (σ) = v. Then A acts on f by (A · f )(Aσ) = A · v. Since we want to find the trace we consider the fixed cells inder the action of A. Consider a the following basis for C i (X, V ). Let v 1 , . . . , v n be a basis for V . Let f σ,v i be the function that sends the cell σ to the vector v i , and sends all other cell to zero. Consider the contribution of f σ,v i to the trace. We have that A · f σ,v i maps σ to a ii · v i , where a ii is a constant. But A · f σ,v i maps Aσ to Av i . Thus, we have a contribution to the trace if Aσ = σ.
Consider the space X A of cells which are fixed under the action of A. On X A the centralizer C(A) acts propersly and discontinuously. Also X A is contractable. Thus c (A) = e(C(A), C(X A , V )).