N-Complexes. Djalal Mirmohades. U.U.D.M. Project Report 2010:9. Department of Mathematics Uppsala University

Transcription

1 U.U.D.M. Project Report 2010:9 N-Complexes Djalal Mirmohaes Examensarbete i matematik, 30 hp Hanleare och examinator: Voloymyr Mazorchuk Maj 2010 Department of Mathematics Uppsala University

2

3 N COMPLEXES Djalal Mirmohaes ABSTRACT. Homological algebra mostly stuies complexes, having a ifferential satisfying 2 = 0. In this work we stuy some generalizations, with a focus on N complexes, where the ifferential instea satisfies N = 0. We investigate ifferent homologies an stuy the corresponing erive category. We also generalize the hom/tensor ajunction to N complexes base on a particular type of N:th root of unity. CONTENTS 1. Introuction 2 2. Homology for N complexes 3 3. Roots of unity 6 4. Hom/Tensor ajunction for sequences Application to N complexes The erive category Triangles in the erive category Projective resolutions Conclusion 22 Acknowlegements 22 References 23 1

4 2 N COMPLEXES 1. INTRODUCTION Definition 1.1. Given an abelian category A, we efine the category Seq(A), with objects being sequences in A:... 2 C 1 1 C 0 0 C an morphisms between them as commuting iagrams:... C 1 C 0 C 1... D 1 D 0 D 1... We call the objects of this category sequences. In fact Seq(A) is a functor category (functors to A), it is abelian since A is abelian. We follow the convention an, for a ring R, write Seq(R) instea of Seq(R Mo). We call the morphisms occurring in the objects of Seq(A) for ifferential an enote them by. Definition 1.2. We efine the category Com N (A) as the full subcategory of Seq(A) containing those object such that the composition of N consecutive ifferentials equals zero in A. The objects of this category is calle N complexes. The stuy of N complexes attracte some attention after the publication of [Kap] by M.M.Kapranov. Kapranov generalizes some of the tools of homological algebra for N complexes, inspire by their application in quantum groups. Since then a hanful of articles has been publishe in this area. See [Abr, D-V 1, D-V 2, D-VK, KW, Sit, Tik]. One thing this theory nees is the ability to turn a bicomplex into a N complex. In the classical theory (N = 2) this is achieve by a sign change in appropriate places in the bicomplex. Kapranov was intereste in C linear spaces so he use a complex primitive N:th root of unity to replace 1. We will investigate how this can generalize. Also, we stuy a few ways to efine homology for N complexes an calculate the erive category for some of the cases....

5 N COMPLEXES 3 2. HOMOLOGY FOR N COMPLEXES We want to efine a homology functor for N complexes. In the case of Com 2 (A) we have ker im an homology is efine as (1) ker /im In general, when N 2, the ifferential inuces the following poset of non-trivial subobjects where the arrows enote inclusion: (2) ker ker N 2 im ker 2 im 2 ker im im N 2 This iagram may be extene by intersections an sums since we have pullbacks an pushouts in abelian categories. We want to generalize the homology given in (1) to N complexes. A philosophy of throwing away as much information as possible may result in the homology (3) ker /(ker im ) In the same spirit, keeping as much information as possible may result in the homology (4) ker /im an in some sense, it oes preserve more information than say (5) ker /im when N > 2, but one shoul be aware that it is not possible to etermine (5) solely from (4). To see this, consier the following pair of 3 complexes of abelian groups:... Z/8Z 2 Z/8Z 2 Z/8Z 2 Z/8Z Z/4Z 0 2Z/4Z 0 2Z/4Z 0 2Z/4Z... Homology ker 2 /im 2 from (4) gives ientical results for both of them but homology ker 2 /im from (5) has non-isomorphic results.

6 4 N COMPLEXES Now, for N complexes an a + b N, ab > 0, efine the homology (6) H (a,b) i := {ker a /im b evaluate at position i} To see that can be extene to a functor Com N (A) A, efine another functor S (a,b) : Com N (A) Com 2 (A) that maps the N complex: to the following 2 complex:... C 1 C 0 C a C b b C 0 a C a b... an maps morphisms f : C D... C 1 C 0 C 1... to f 1... D 1 D 0 D C b C 0 f 0 C a f 1... f b... D b D 0 D 1... We are alreay familiar with the classical homology H (1,1) 0 which is functorial. Now observe that H (a,b) 0 = H (1,1) 0 S (a,b) Definition 2.1. Let T be the (invertible) translation functor that translates the complex one step to the left (position 0 moves to position 1). We now have H (a,b) i = H (1,1) 0 S (a,b) T i This means that H (a,b) i :s are functors an we may apply them to the morphism C : C TC: f 0 f a... C 1 C 0 C C 0 C 1 C 2... so that for each C we have a sequence:... H (a,b) 0 T 1 C H (a,b) 0 T 1 C H (a,b) 0 C which is equal to: H (a,b) 0 C H (a,b) 0 TC H (a,b) 1 C H (a,b) 1 C H (a,b) 0 C H (a,b) 0 C H (a,b) 1 C...

7 N COMPLEXES 5 This is in fact an M complex, where M = min(a, b). It is easy to verify that H (a,b) efines a functor H (a,b) : Com N (A) Com M (A) In the classical setting of Com 2 (A), H (1,1) maps 2 complexes to 1 complexes (where = 0) so this property may remain unnotice. But there are more connections between our homologies. Since the iagram in (2) is a poset of inclusions, it commutes, so we may consier it as a morphism of sequences: (7)... 0 im... im ker... ker 0... Since Seq(A) was abelian this morphism has a cokernel with objects being the homologies: H (1,) j i... i H (,1) j 0... Our point is however that there is, in a natural way, a morphism i : H (p,n p) j H (p+1,n p 1) j occurring as the ifferential above. So far, this was one only for egree j. The ifferential : C j C j+1 as a imension to iagram (7) by inucing the vertical morphisms in: (8) 0 im im N 2... im 0 0 im... im 2 im 0 where the rows are coming from the top row of (7). The same thing can be one for the bottom row of (7): (9) 0 ker ker 2... ker 0 0 ker... ker N ker 0 Inee, both of the iagrams (8) an (9) commute an are hence a bisequence (a sequence of sequences). Just like before, the vertical inclusion of (7) efines a morphism of bisequences from (8) to (9). Its cokernel reprouces our homologies, but now in a bisequence: (10) 0 H (1,) j i H (2,N 2) j i... i H (,1) j 0 0 H (1,) j+1 i... i H (N 2,2) j+1 0 i H (,1) j+1 0

8 6 N COMPLEXES Exten (10) vertically an efine a new sequence with objects at position n: G n = 2j+p=n H (p,n p) j an ifferential δ : G n G n+1 given by δ = i +. This homology was introuce by Kapranov in [Kap]. Kapranov shows that this is a functor G : Com N (A) Com (A) 3. ROOTS OF UNITY One construction in homological algebra looks like the following: Starting with a bi-aitive functor B : A A A, with = 1 or op, one extens it to a functor B : Com(A) Com(A) Com(A) in two steps: First one constructs a bicomplex by applying B object-wise. Next, one efines the objects of the final complex by taking the sum or prouct along the iagonal. To efine the ifferential in terms of the morphisms of the bicomplex, one introuces a sign change in appropriate places. However, to generalize this construction for N complexes, we nee some coefficient q that satisfies (11) (x q i y) = x N y N We o this in etails for the tensor an hom functor in the next section. Let R be a unital ring. In this section, we o not require R to be commutative. By the zero:th power of an element in R, we always mean the unit 1 R. Lemma 3.1. For q R the requirement (11) in R[x, y] is equivalent to the requirement that in R[x]: (12) (x q i ) = x N 1 Proof. We see (11) = (12) by evaluating (11) with y 1. For (11) = (12), it is enough to prove it in R[x](y) thanks to the injectivity of the embeing Define a map R[x][y] R[x](y) f : R[x] R[x](y) x x y

9 an calculate in R[x](y) (x q i y) = y N (x y qi) = y N f N COMPLEXES 7 = y N f(x N 1) = y N( (x) ) N 1 y ( ) (x q i ) = = x N y N Definition 3.2. An element q R is calle a binomial N:th root of unity if it satisfies equation (12) in R[x]. In [Kap], complex primitive N:th roots of unity are taken for binomial N:th roots of unity. This clearly works in C, but we want to investigate binomial roots of unity for rings in general. Lemma 3.3. Binomial N:th roots of unity are N:th roots of unity. Proof. Assume q R is a binomial N:th root of unity. Then (x q i ) = x N 1 = In particular it means that but then we have N k i x i q i = k = 0 q N = 1 + (q N 1) = 1 + (q 1) q i = 1 In [BP, p. 11] we fin the following Definition 3.4. An element q is sai to be a principal N:th root of unity if for j = 1, 2,...N 1. q ij = 0 Just like in the proof of lemma (3.3) we may use the case j = 1 to see that principal N:th roots of unity are N:th roots of unity. This means, in particular, that when N is a prime number, the cases j > 1 follows from the case j = 1 since integers j > 1 are invertible moulo N. Lemma 3.5. When N is prime, binomial N:th roots of unity are principal N:th roots of unity an vice versa. Proof. From the proof of lemma (3.3) we know that binomial N:th roots of unity satisfy the case j = 1 in the efinition of principal N:th roots of unity. As escribe above, it follows that they also satisfy cases j = 2,...N 1, hence they are principal N:th roots of unity.

10 8 N COMPLEXES To prove the other irection, consier that we have the ientity (x z i ) = x N 1 in Q[z]/(1+z+...+z )[x] since Q[z]/(1+z+...+z ) is the splitting fiel of x N 1. But the above equality must also hol in Z[z]/(1+z+...+z )[x] because of (the injectivity of) the embeing Z[z]/(1 + z z )[x] Q[z]/(1 + z z )[x] Now let q R be a principal N:th root of unity an efine the map f : Z[z]/(1 + z z )[x] R[x] z x q x it is well-efine since 1 + z z is mappe to zero. Then in R[x] (x q i ) = f ( ) (x z i ) = f(x N 1) = x N 1 However, when N is not prime, a primitive or principal N:th root of unity oes not nee to be a binomial N:th root of unity. To see this, let R = Z 2 [q]/(q 3 + 1) now q is a primitive an principal 6:th root of unity. But in Z 2 [q]/(q 3 + 1) 5 (x q i ) = x 6 + (q 2 q + 1)( x 4 + x 2 ) 1 x 6 1 To tackle this we introuce Definition 3.6. An element q in a ring R is calle a strong principal N:th root of unity if for any prime p that ivies N. p 1 q in/p = 0 Lemma 3.7. A strong principal N:th root of unity is a principal N:th root of unity. Proof. As escribe before, the notions coincie when N is prime. When N is not prime, assume q is a strong principal N:th root of unity an let 1 j N 1. Then q ij = q ic

11 N COMPLEXES 9 with c = gc(j, N). Now choose a prime p that ivies N/c an observe that ( N/pc 1 p 1 q ic = c q ic) q in/p = 0 Lemma 3.8. If q is a strong principal N:th root of unity then for any ivisor n of N, q n is a strong principal k:th root of unity, where k = N/n. Proof. This is an immeiate consequence of the fact that any prime ivisor of N/n is a prime ivisor of N. Lemma 3.9. Strong principal N:th roots of unity are binomial N:th roots of unity. Proof. We prove this by inuction on the number of prime factors in N. Lemma (3.7) with lemma (3.5) cover the case when N is prime. So assume we have proven this for N having n prime factors an let N have n + 1 prime factors. So let q be a strong principal N:th root of unity an p be a prime ivisor of N. Then (x q i ) = N/p 1 p 1 (x q jn/p q i ) q N/p is now a strong principal p:th root of unity by lemma (3.8) we use lemma (3.5) an (3.1) to get N/p 1 p 1 j=0 (x q jn/p q i ) = j=0 N/p 1 (x p q ip ) Now q p is a strong principal N p :th root of unity an N p has n prime factors so by the inuction hypothesis we have N/p 1 (x p q ip ) = x N 1 Now we have some tools to fin more examples of binomial N:th roots of unity: Definition The N:th cyclotomic polynomial is efine as the polynomial Φ N in Z[x] such that it has the following factorization in C[x]: Φ N (x) = j N (x e 2πij/N ) with N being the set of all positive integers less or equal to N that are relatively prime to N. We refer to [Rot] for a proof of the existence an other properties of the cyclotomic polynomials. Observe that, since the coefficients of the polynomials are integers, the polynomials can be realize in R[x].

12 10 N COMPLEXES Proposition Any root of the N:th cyclotomic polynomial in the ring R is a binomial N:th root of unity. Proof. Since for every prime ivisor p of N, the N:th cyclotomic polynomial ivies the polynomial p 1 x in/p = xn 1 x N/p 1 in Z[x] it also oes that after evaluation x q in R. So roots of the N:th cyclotomic polynomial are strong principal N:th roots of unity. By lemma (3.9) they are hence binomial N:th roots of unity. In [D-V 2, p. 7] the following axioms are introuce for the same purpose as ours: Definition An element q in a ring R is calle a istinguishe primitive N:th root of unity if q i = 0 an for every n = 1, 2,..., N 1 the sum is invertible in R. n 1 q i Proposition Distinguishe primitive N:th roots of unity are binomial N:th roots of unity. Proof. By lemma (3.9), it is enough to show that it is a strong principal N:th root of unity. So let p be any prime ivisor of N. Then 0 = q i = ( N/p 1 q p 1 i) q in/p We get the esire result by multiplying with the inverse of N/p 1 q i. It shoul be pointe out that there are binomial N:th roots of unity that are not istinguishe primitive N:th roots of unity. One example is the element i in the ring of Gaussian integers; it is a strong principal 4:th root of unity, but the sum 1 + i is not invertible. Lemma Every finite multiplicative subgroup of a fiel is cyclic. A proof of this lemma may be foun in [Gri, IV.1.6]. Proposition If the ring R is a commutative integral omain, the number of N:th roots of unity is a ivisor of N an all those roots are powers of some q R. Proof. The N:th roots of unity of the integral omain R form a group G uner multiplication since 1 G an for any q G we have q 1 = q G. But the group G is, in a natural way, a multiplicative subgroup of the fiel of fractions Q(R). Since the roots of unity are by efinition roots of

13 N COMPLEXES 11 the polynomial x N 1 there can be at most N roots. It follows from lemma (3.14) that the group is cyclic. A generator q of G then satisfies q G = 1 = q N, but then G must ivie N. The proposition gives some useful information since all binomial N:th roots of unity are N:th roots of unity. On the other han, if the ring has characteristic 0, we know something more: Proposition Let R be a commutative integral omain of characteristic zero an q a non-unit element of multiplicative orer g. Then q is a binomial N:th root of unity if an only if N = g. In other wors, the notion of binomial N:th root of unity coincies with the notion of primitive N:th root of unity for integral omains of characteristic zero. Proof. Since q has orer g, it is a N:th root of unity only when N = mg, m Z +. We begin by showing that q is a strong principal (hence binomial) g:th root of unity. Let p be any prime ivisor of g, then p 1 0 = q g 1 = (q g/p 1) q ig/p Since we are in a integral omain an q g/p 1 0 it follows that p 1 q ig/p = 0 It remains to show that q cannot be a binomial mg:th root of unity for m 2. So assume m 2, we have just shown that But then mg 1 g 1 (x q i ) = x g 1 (x q i ) = (x g 1) m x mg 1 since the characteristic is zero. Proposition Let R be a commutative integral omain of characteristic N. Then 1 is the only binomial N:th root of unity. Proof. First of all, since N 2 is the characteristic of a integral omain, N must be prime. In fact there are no other N:th roots of unity except for 1. To see this, embe R into the fiel of fractions Q(R) an observe that the only roots of the polynomial x N 1 is 1 since in Q(R)[x] the polynomial factors x N 1 = (x 1) N But this equation also shows that 1 is a binomial N:th root of unity.

14 12 N COMPLEXES 4. HOM/TENSOR ADJUNCTION FOR SEQUENCES Let R be a commutative ring with unity. We begin by generalizing the classical hom/tensor ajunction of Com 2 (R) making it epenent on coefficients from R but applie to the larger category Seq(R). We then show that by letting the coefficients epen on powers of a binomial N:th root of unity, they inuce a corresponing hom/tensor ajunction for Com N (R). Let k any function k, : Z Z R Definition 4.1. Define the k-hom functor Hom k (, ) : (Seq(R)) op Seq(R) Seq(R) on sequences as the sequence with objects Hom k (X, Y ) n = Hom(X i, Y j ) at position n an ifferential given by i+j=n : Hom(X i, Y j ) Hom k (X, Y ) i+j+1 f k i,j f + f On morphisms it is efine component-wise by the hom functor. Definition 4.2. Define the k-tensor prouct k : Seq(R) Seq(R) Seq(R) as the functor mapping a pair of sequences to the sequence with objects (X k Y ) n = (X i Y j ) at position n with ifferential given by i+j=n : X i Y j (X k Y ) i+j+1 x y k i+1,i+j+1 (x) y + x (y) efine on generators. The ifferential is well-efine because it originates from the bilinear map X i Y j (X k Y ) i+j+1 (x, y) k i+1,i+j+1 (x) y + x (y) On morphisms it is efine component-wise by the tensor prouct. Theorem 4.3. Given any sequence A in Seq(R) the functor A k to Hom k (A, ). Proof. Define the unit is left ajoint η : I Seq(R) Hom k (A, A k )

15 on the objects of a sequence as η Xn : X n Hom k (A, A k X) n = x i+j=n x i,j x i,j : A i A i X i+j a a x N COMPLEXES 13 i+j=n Hom(A i, k+m=j where x i,j is efine as the homomorphism (A k X m )) To see that η Xn efines a morphism in Seq(R) we must verify that it commutes with the ifferential: η X n X n Hom k (A, A k X) n X n+1 η Xn+1 Hom k (A, A k X) n+1 η Xn (x) = i+j=n = i+j=n = i+j=n = i+j=n+1 (x i,j ) k i,j x i,j + x i,j k i,j x i,j + k i+1,i+( i+j)+1 x i+1,j+1 + x i,j+1 x i,j = η Xn+1 (x) The calculation that η is a natural transformation is similar to the above except that all the k:s are zero. Define the counit ε : A k Hom k (A, ) I Seq(R) on the generators of the objects of the sequence as ε Xn : A n i Hom(A k, X m ) X n a k+m=i k+m=i f k,m f n i,n (a) ε Xn is well-efine since function application is bilinear. We show that it commutes with the ifferential: (A k Hom k (A, X)) n ε Xn X n (A k Hom k ε Xn+1 (A, X)) n+1 X n+1

16 14 N COMPLEXES ε Xn+1 (a k+m=i ε Xn+1 ( k n i+1,n+1 (a) ε Xn+1 ( k n i+1,n+1 (a) f k,m ) = k+m=i k+m=i f k,m + a f k,m + a k+m=i k+m=i ) (f k,m ) = ) ( k k,m f k,m + f k,m ) = k n i+1,n+1 f n+1 i,n+1 ((a)) k n+1 i,n+1 f n+1 i,n+1 (a) + f n i,n (a) = ( ε Xn a ) f k,m k+m=i It remains to show that the compositions (13) A k A k η A k Hom k (A, A k ) ε A k A k Hom k η Hom k (A, ) (A, ) Hom k (A, A k Hom k Hom k (A,ε) (A, )) Hom k (A, ) are the ientity natural transformations on A k an Hom k (A, ), respectively. To show the first equality, for a sequence X an a generator a x A n i X i from a summan of A k X at position n, we have (A k A k η X X) n (A k Hom k (A, A k ε A k X X)) n (A k X) n a x a x j,m x n i,n (a) j+m=i where x n i,n (a) = a x as before. For the secon equality, take a sequence X an a collection of homomorphisms f i,j Hom k (A, X) n i+j=n at position n where each f i,j Hom(A i, X j ), then Hom k (A, X) n i+j=n η Hom k (A,X) f i,j Hom k (A, A k Hom k (A, X)) n k+m=n ( i+j=n To see that ε Xm ( i+j=n f i,j ) f i,j ) k,m Hom k (A,ε X ) k+m=n Hom k (A, X) n ( ε Xm i+j=n k,m = f k,m we apply it to a A k : f i,j ) k,m

17 N COMPLEXES 15 ( ε Xm i+j=n ) ( f i,j (a) = ε X m a k,m i+j=n f i,j ) = f k,m (a) 4.1. Application to N complexes. Theorem 4.4. Let q R be a binomial N:th root of unity (efine in (3.2)) an k i,j = h(i) q j with h any function Z R. Then, for any N complex A, the functors A k an Hom k (A, ) map N complexes to N complexes. Proof. The construction of both functors begins with a bicomplex M; a commutative iagram with objects in R Mo: δ M 1, 1 δ M 0, 1 M 1, 1 δ M 1,0 δ M 0,0 M 1,0 δ M 1,1 δ M 0,1 M 1,1 Define a new sequence L in R Mo with objects L n = i+j=n M i,j an ifferential L i+j L i+j+1 efine on the components by q i+j δ : M i,j M i,j+1 M i+1,j To illustrate this, the composition of three consecutive ifferentials L 0 L 3 for the summan M 0,0 is the sum over all possible compositions in the iagram:

18 16 N COMPLEXES M 0,0 q 0 δ M 1,0 q 1 δ M 2,0 q 2 δ M 3,0 M 0,1 q 1 δ M 1,1 q 2 δ M 2,1 M 0,2 q 2 δ M 1,2 M 0,3 This choice of coefficients has the property that (by ignoring the inices of an δ) we may write the ifferential of L as simply q n δ : L n L n+1 This means the prouct of N consecutive ifferentials (starting at position k) can be written as the composition ( q k+ δ)... ( q k+1 δ) ( q k δ) This composition equals N δ N by lemma (3.1) since commutes with δ. It is obviously zero when both N an δ N equals zero. But we have more freeom in our choice of coefficients. To see this, fix an integer i an assume we multiply all the horizontal morphisms in column i with some fixe r i R. Then the ifferential restricte to the component M i,j becomes + r i q i+j δ : M i,j M i,j+1 M i+1,j We have seen that compositions of the ifferential L a L b, restricte an projecte between components M a c,c M b, can be written as a sum over all paths from M a c,c M b, in the bicomplex. Multiplying one column by a constant r i only affects paths passing through that column. Each such path is then multiplie by r i exactly once. In particular, if their sum was zero it remains zero. To prove the theorem for the functor A k set r i = h(i + 1) q. Then r i q i+j = h(i + 1) q i+j+1 = k i+1,i+j+1. The functor Hom k (A, ) iffers in a few ways. It is contravariant in the first argument an its objects are proucts instea of sums. We eal with the contravariance by simply reversing inices. We also observe that the same arguments can be one with proucts. The last ifference is solve by setting r i = h(i) q i then r i q i+j = h(i) q j = k i,j Corollary 4.5. If A k an Hom k (A, ) are efine as in the previous theorem they form a pair of ajoint functors for the subcategory of N complexes; Com N (R).

19 N COMPLEXES 17 Proof. Theorem (4.4) tells us that we may restrict the functors to the object class of Com N (R). But since Com N (R) is a full subcategory of Seq(R) the functors are well-efine on morphisms as well. In particular, the unit an counit may be restricte to our N complexes. They also satisfy the ajunction equalities (13) because they i that in Seq(R). 5. THE DERIVED CATEGORY Let A be an abelian category. Let T be the translation functor efine in (2.1). Given a sequence C in Seq(A) the ifferentials of C constitute a Seq(A) morphism C : C TC as escribe in section (2). To simplify, we may sometimes omit the inices, so say TC C will be written 2. Let B enote any aitive subcategory of Seq(A) that contains Com N (A). Definition 5.1. Let I be the inclusion functor Com N (A) B. Definition 5.2. For each sequence C we have the the following iagram of sequences in Seq(A) with ker N C ker N k C C C N T N C being a complex. We efine the functor [ker N ] : B Com N (A) C ker N C on sequences C. For morphisms, compose any f : C D in B with k C. The image of fk C lies in D but N fk C = T N (f) N k C = 0 hence fk C can be factore, with a unique f, through k D such that k D f = fk C : ker N C k C C N T N C [ker N ](f):=f ker N D f k D D N T N D T N (f) [ker N ](f) is efine as f. Note that [ker N ](f)[ker N ](g) also satisfy the property of [ker N ](fg) since the iagram commutes: ker N B k B B N T N B [ker N ](g) ker N C k C C g N T N C T N (g) [ker N ](f) ker N D f T N (f) k D D N T N D

20 18 N COMPLEXES Because of the uniqueness of [ker N ](fg), it is equal to [ker N ](f)[ker N ](g). This makes [ker N ] functorial. Proposition 5.3. (I, [ker N ]) is an ajoint pair. See [ML] for efinition of ajoint functors. Proof. We know that ker N is ientity on N complexes, then [ker N ]I = I ComN (A) because [ker N ] was just restriction on morphisms. The unit of the ajunction is the ientity natural transformation i : I ComN (A) [ker N ]I = I ComN (A) an the counit is the inclusion k efine in (5.2) k : I[ker N ] I B by construction k[ker N ]I(f) = fk for all f B so k is inee a natural transformation. We have k I I(i) = k I = i I since the inclusion of a complex into itself is the ientity. But also [ker N ](k) i [ker N ] = [ker N ](k) = i [ker N ] because k is monic an k [ker N ](k) = k i [ker N ]. Lemma 5.4. Assume gs = tf with s an t invertible. Then t 1 g = fs 1. Proof. t 1 g = t 1 gss 1 = t 1 tfs 1 = fs 1 Lemma 5.5. Given a pair of functors G, F between two categories C G F together with two natural transformations D an a pair of subclasses of morphisms such that with η : I D GF, ε : FG I C S Mor(C), T Mor(D) F(T ) S an G(S) T x Ob(C) : ε x S x Ob(D) : η x T then the localize categories C[S] an D[T ] are equivalent. Furthermore, this hols also if we arbitrarily change the irection of η or ε.

21 N COMPLEXES 19 Proof. Let Q : C C[S] an R : D D[T ] be the corresponing localizing functors. Since RG (QF) maps morphisms from S (T ) to isomorphisms we have an inuce G (F ) such that RG = G Q (QF = F R): C G D C F D Q C[S] R Q G D[T ] C[S] D[T ] F R We have a natural transformation Qε : QFG = F G Q Q. But Q is bijective on objects so for each object C of C[S] we have an isomorphism ε C : F G C C because ε consiste of morphisms from S. To show that ε efines a natural transformation F G I C[S] it is enough to check that it commutes with generators in C[S]. We alreay know that it commutes with generators in Mor(C). Form lemma (5.4) it follows that it then also commutes with the inverse generators in S. Lemma (5.4) also tells us that since ε is a natural transformation the isomorphisms ε 1 constitute an inverse natural transformation. Then εε 1 = i IC an ε 1 ε = i F G which proves the equivalence between I C an F G. Since the irection of ε i not matter the same argument can be applie to η, or to any of ε or η with the opposite irection. Definition 5.6. Given a functor H : X Y, we call a morphism f X an H-quasi-isomorphism if H(f) is an isomorphism. Definition 5.7. The erive category D H (X) is the category X localize with respect to H-quasi-isomorphisms. See [GM] for etails. Theorem 5.8. For any category C an functor H : Com N (A) C, we have an equivalence of categories D H[ker N ](B) D H (Com N (A)) Proof. The ajoint pair (I, [ker N ]) provie the functors an natural transformations for lemma (5.5). First we verify that the functors I an [ker N ] map quasi-isomorphisms to quasi-isomorphisms. Next we nee to see that the natural transformations i an k consist of quasi-isomorphisms. For i it is clear an for k it hols because [ker N ](k) = i. Corollary 5.9. D [ker N ](B) Com N (A) Proof. Let H = I ComN (A) in (5.8).

22 20 N COMPLEXES Definition Let D(A) enote the classical erive category D H (1,1)(Com 2 (A)). Corollary Let N 2. Consier the homology functor H : Com N (A) Com 1 (A) given by H = ker /(ker im ) then we have an equivalence Proof. Verify that H = H (1,1) [ker 2 ]. D H (Com N (A)) D(A) 5.1. Triangles in the erive category. Definition We efine the cone C(f) of a morphism f : C D in Seq(A) as the sequence C[1] D with ifferential ( ) C[1] 0 C(f) = f[1] We note that ( 2 C(f) = 2 ) ( ) C 0 2 f C + D f 2 = C 0 D 0 2 D D Proposition The cone construction commutes with [ker 2 ], namely C([ker 2 ](f)) = [ker 2 ](C(f)) for any f : C D. Proof. We have that C([ker 2 ]f) C(f) with ( ) [ker C([ker 2 ]f) = 2 ]C[1] 0 [ker 2 ]f[1] [ker 2 ]D then 2 C([ker 2 ]f) = 0 which implies C([ker 2 ]f) [ker 2 ](C(f)). To prove C([ker 2 ]f) [ker 2 ](C(f)) assume ( ( x 0 = y) 0) 2 C(f) then ( ( ) ( ) ( 2 C 0 x 2 0 D) 2 = C x 0 y 2 D y = 0) so x [ker 2 ]C, y [ker 2 ]D an ( x y) C([ker 2 ](f))

23 N COMPLEXES 21 D(A) is a triangulate category. But since D H (1,1)(B) is equivalent to D(A), we consier it triangulate via the equivalence. Triangles in D H (1,1)(B) are then by efinition commuting iagrams of the form IA If IB IC(f) X Y Z with the vertical morphisms being isomorphisms. The cone construction however also exists in D H (1,1)(B). Thanks to proposition (5.13) we know that the cone in D H (1,1)(B) also provies istinguishe triangles since the upper row in the iagram I[ker 2 ]A I[ker 2 ]f I[ker 2 ]B I[ker 2 ]i I[ker 2 ](C(f)) A k A f B k B i C(f) k C(f) is a triangle because [ker 2 ](C(f)) = C([ker 2 ]f) an the k:s are isomorphisms. 6. PROJECTIVE RESOLUTIONS Definition 6.1. P Com N (A) is sai to be a projective resolution of X A relative to a homology functor H if there is a morphism in Com N (A)... P 1 P X 0... which is mappe to an isomorphism by H. Lemma 6.2. Let X A be a non-projective object. A projective resolution of X relative to H (a,b) cannot be a (a + b 1)-complex. Proof. Let P be a projective resolution of X. We nee to fin a position where a+b 1 0. For the case a = 1 we have H (a,b) 0 (P ) = P 0 / b P b X but b P b = a+b 1 P b 0 since X was not projective. Now assume a > 1. At position 1 a the homology is H (a,b) 1 a (P ) = P 1 a/ b P 1 a b = 0...

24 22 N COMPLEXES so is epi. In the same way, b : P 1 a b P 1 a b : P 1 a P 1 a+b is epi since homology at position P 1 a+b is zero. Continue this n steps until 1 a + nb b. But like in the case of a = 1, b : P b P 0 is non-zero. This means in particular that 1+a nb : P 1 a+nb P 0 is non-zero an a+b 1 : P a+b 1 P 0 is non-zero. 7. CONCLUSION Say we use the homology H (a,b) [ker a+b ] for N complexes. From lemma (6.2) we see that we must have N a + b to allow for the existence of nontrivial projective resolutions. On the other han theorem (5.8) tells us that it is enough to take N = a + b when one is intereste in the erive category. Furthermore, the functor [ker a+b ] is ientity on (a + b) complexes. These facts suggests that the homology (10) introuce by Kapranov in [Kap] is the appropriate homology for N complexes. ACKNOWLEDGEMENTS I woul like to thank my avisor, professor Voloymyr Mazorchuk, for all the guiance an support.

25 N COMPLEXES 23 REFERENCES [Abr] V. ABRAMOV. On a grae q ifferential algebra. Journal of Nonlinear Mathematical Physics, Volume 13, Supplement [BP] D. BINI, V. Y. PAN. Polynomial an Matrix Computations: Volume 1: Funamental Algorithms. Birkhäuser Boston, [GM] S. I. GELFAND, Y. I. MANIN. Methos of Homological Algebra. Springer, [Gri] P. A. GRILLET. Abstract Algebra. Springer, [Kap] M. M. KAPRANOV. On the q analog of homological algebra. Preprint, Cornell University, 1991; q alg/96/ [D-V 1] M. DUBOIS-VIOLETTE. Generalize ifferential spaces with N = 0 an the q ifferential calculus. L.P.T.H.E.-ORSAY, [D-V 2] M. DUBOIS-VIOLETTE. Tensor prouct of N complexes an generalization of grae ifferential algebras. LPT-ORSAY 09/84, [D-VK] M. DUBOIS-VIOLETTE, R. KERNER. Universal q ifferential calculus an q analog of homological algebra. L.P.T.H.E.-ORSAY 96/48, [KW] C. KASSEL, M. WAMBST. Algèbre homologique es N complexes et homologie e Hochschil aux racines e l unité. I.R.M.A. - C.N.R.S., [ML] S. MACLANE Categories for the working mathematician. Springer Verlag, [Sit] A. SITARZ. On the tensor prouct construction for q ifferential algebras. Institut für Physik, Johannes-Gutenberg Universität Mainz, Germany, 2008 [Tik] A. TIKARADZE. Homological constructions on N complexes. Journal of Pure an Applie Algebra 176, [Rot] J. J. ROTMAN. Avance Moern Algebra. Prentice Hall, 2003.