HOMOLOGICAL ALGEBRA ALLEN KNUTSON CONTENTS. Left nd right modules 2. Exct functors 2 3. Projective nd injective modules 4 4. Complexes nd homology 5 5. Mpping resolutions 7 6. Derived functors 8 7. Ext groups 8. Interlude: filtrtions on rings nd modules 3 9. Filtered complexes 3. The Grothendieck spectrl sequence 5 Fix commuttive bse ring k (the miniml choice being k = Z). Then k-lgebr R is ring with unit, plus mp k Z(R). In prticulr the bidditive multipliction R R R fctors through the tensor product R k R R. Herefter will denote k.. LEFT AND RIGHT MODULES Unless otherwise noted, R is noncommuttive ring with unit, nd we indicte left nd right R-modules by R M, M R. We cn mke ny left R-module into right R op -module. The lest structure our objects will hve is being k-module, i.e. being n belin group if k = Z. An (R, S)-bimodule R M S possesses two commuting ctions, or equivlently, is module over R S op. As such ny left R-module is nturlly (R, k)-bimodule... Hom. There is nturl left S-module structure on Hom R ( R M S, R N): (s f)(m) := f(ms) First we check tht s f is gin in Hom R (M, N): Then tht the ction of S is left ction: Dte: April 2, 27. (s f)(rm) = f(rms) = r f(ms) = r (s f)(m) (s t f)(m) = (t f)(ms) = f(mst) = (st f)(m)
2 ALLEN KNUTSON In prticulr, if M S is only right S-module, we write M := Hom k (M, k) nd get left S-module. The nturl mp M M is not usully : or onto. The mnemonic, for when k is field, is tht Hom R (M, N) is something like M N, nd the dul of right module is left module. Of course we re used to hving inner products on vector spces, nd now tht looks impossible. So we sk for n ntiutomorphism : R R i.e. n isomorphism R R op. Then we cn mke M into left R-module, nd consider homomorphisms M M. If we wnt the nturl function M M to be R-liner, then should be n ntiïnvolution. Of course the most fmilir cse is R commuttive nd the identity. Fun cse: k = C, = complex conjugtion, nd then these homomorphisms give sesquiliner forms. This extends to the quternionic cse in wy tht = Id doesn t. Another is g g for k[g]. If N = R N T is n (R, T)-bimodule, then the hom-spce is lso right module. If R is commuttive, we cn soup up ny R M to R M R, hence Hom R (M, N) is gin (left or right) R-module. But if we try to use n ntiinvolution, the ctions of R nd R op my not commute, so my not mke M into bimodule..2.. The most generl nturl tensor is R M S S S N T, which is then (R, T)-bimodule. Of course common cse is to strt with R commuttive nd R M, R N. A fun 2-ctegory: rings, bimodules, bimodule homomorphisms..3. Adjointness. Theorem.. Let R M S be bimodule, so we hve functors Then these re djoint. M S : Mod S Mod R : Hom R (M, ) Proof. Let A, B be left R-, S-modules respectively. Then we need to show Hom S (B, Hom R (M, A)) = HomR (M B, A) ( ( )) f m b f(b) m is bijective, nd nturl in A, B. 2. EXACT FUNCTORS An dditive ctegory is ctegory object in the monoidl ctegory of belin groups, i.e. the homsets re belin groups nd composition is bidditive. We ll ctully wnt better thing, tht ll homsets re k-modules. Let A B C be short exct sequence of some kind of modules, nd T n dditive functor from their module ctegory to nother (gin respecting the dditive structure so e.g. not Schur functors). Then T is left exct resp. right exct if T(A) T(B) T(C) is exct t T(A), T(B) resp. T(B), T(C), nd exct if both. Proposition 2.. An exct functor preserves exctness of rbitrry exct sequences (not just short exct).
HOMOLOGICAL ALGEBRA 3 Proof. Let A f B g C be prt of n exct sequence, nd T n exct functor. We cn fctor it s A A B C C, where A B C is short exct. Apply T to tht sequence to get T(A) T(A ) T(B) T(C ) T(C) since T preserves : ness nd ontoness. Now the kernel of T(B) T(C) is the kernel of T(B) T(C ), which is the imge of T(A ) T(B) by exctness t B. But T(A) T(A ) sys tht it s the sme s the imge of T(A) T(B). Lemm 2.2. Hom(Z, ) is left exct. Proof. We pply Hom(Z, ) to A B C, or relly to A B B/A, obtining Hom(Z, A) Hom(Z, B) Hom(Z, B/A) Since A B, the first mp is :. If φ : Z B is in the kernel of the second mp, tht mens Z B B/A is zero, i.e. the imge of Z lies in A. So we obtin n element of Hom(Z, A). The min exmples re Hom R (M, ) nd M S, s follows from the next theorem. We need the definition of n object X Obj(C) representing covrint functor F : C Set, nmely F = Hom C (X, ). Theorem 2.3. Let C hve n object R tht represents the forgetful functor. Then if L : C D : R re djoint functors, then the left djoint is right exct nd vice vers. Proof. We hve bijections Hom(X, Z) = Hom(X 2, Z) tht re nturl in Z. For Z = X the left side hs the identity, giving us n element of Hom(X 2, X ), nd for Z = X 2 the RHS does, giving us n element of Hom(X 2, X ); these turn out to be inverses. Corollry 2.4. If Hom(Z, A) Hom(Z, B) Hom(Z, C) is exct for ll Z, then A B C ws exct. Proof tht djointness = exctness. Strt with n exct sequence A B C in D (the is unimportnt). Apply R to get R(A) R(B) R(C) in C. To mke use of djointness, we need to mp into this, so Hom C out of Z to get Hom C (Z, R(A)) Hom C (Z, R(B)) Hom C (Z, R(C)) Adjointness sys tht we hve commuting digrm Hom D (L(Z), A) Hom D (L(Z), B) Hom D (L(Z), C) Hom C (Z, R(A)) Hom C (Z, R(B)) Hom C (Z, R(C)) with verticl isomorphisms. We lredy proved tht the top sequence is exct, hence the bottom one is, for ll Z. Tking the cse Z = R, we get A B C is exct. In fct using the Yoned lemm, Lemm 2.5 (Yoned). If two objects represent the sme functor, they re cnoniclly isomorphic. one cn dispense with the object representing the forgetful functor.
4 ALLEN KNUTSON 3. PROJECTIVE AND INJECTIVE MODULES Even though Hom(M, ) isn t left-exct for ll R M, it is for some, e.g. for R itself, nd more generlly for free modules. Similrly M is exct for M free. However, free is difficult property to mintin. One result we hve in this direction derives from Theorem 3.. Let M be finitely generted module over PID R. Then M is direct sum of cyclic modules R/r i R. nmely Corollry 3.2. If R is PID, then finitely generted submodule of R n is free. but more generlly, we will work with projective modules P: if given M N nd P N, there lwys exists lift P M, then P is projective. Theorem 3.3. () Free modules re projective. (2) If P = Q R, then the direct summnds Q, R re projective. (Even this lredy is flse for free modules: try M = R = Z/6.) (3) Any SES A B P with P projective, splits. I.e. ny mp B P must be projection A P P. (4) P is projective iff it s direct summnd of free module. (5) If every SES onto P splits, then P is projective. Proof. () The genertors of P go somewhere in N. By the ontoness, we cn lift those to M, obtining mp P M. (2) Given mp Q N we cn extend it to P (by on R), lift to P M, restrict to Q M. (3) Tke N = P i.e. P N the identity. Lift to get mp P B. Now check tht B is the sum of the imges of A nd P. (4) = Pick genertors of P to get surjection F P. Now pply the previous to ker F P. = (2). (5) If every mp B P splits, then tke B free, nd use (4). Another reson people like projectivity is tht it cn be checked loclly, i.e. fter M is projective R-module iff R[ ] R M is projective R[ ]-module for enough, but the sme sttement doesn t work with free. A free resolution of module M is n exct sequence F 3 F 2 F F M where the F i re free; we could tke F to be freely generted by M, then F freely by the kernel of F M, nd so on. Exmple. Let R = C[, b, c]/ b 2 c, nd M = R/, b. Then we hve free resolution R 2 [c] [ b] R 2 [] [b] R 2 [c] [ b] [ b] 2 R M
HOMOLOGICAL ALGEBRA 5 Theorem 3.4 (Hilbert bsis theorem). If M is module over F[x,..., x n ], then M hs free resolution with F i = i > n. Proof. Use Gröbner bsis for the module, nd... More generlly though, we ll use projective resolutions, which look the sme but F i need only be projective. 3.. Injective modules. All these notions cn be dulized by reversing the rrows, but they cn get kind of weird. Wht would n injective Z-module I be? Given ny M M, nd M I, there should be n extension M I. A Z-module M is divisible if for every m M nd there exists m with m = m. The principl exmple is Q/Z s Z-module. Theorem 3.5. () Injective modules re divisible. (2) If M is torsion-free nd divisible, it is injective. Proof. () Consider the mps R R m M, where the first is injective by. Then injectivity sys tht this should complete to R M, with m. (2)... this is rel pin, nd uses Zorn s lemm. In prticulr, it s much less obvious (but true) tht R Mod hs enough injectives to mke n injective resolution of ny module. φ n 4. COMPLEXES AND HOMOLOGY φ n φ A complex A n An A is sequence of mps, the differentils, in n dditive ctegory such tht ll composites φ i φ i+ =. Then (ssuming the ctegory hs quotients, s in module ctegories) we cn define the homology s H i := ker φ i /imge(φ i+ ). Under this definition, n exct sequence is complex with vnishing homology. There re obvious definitions of direct sums, subcomplexes, nd isomorphisms of complexes ( sequence of mps giving commuting squres). The two bsic kinds of complexes re complexes concentrted in single degree (with trivil differentil, thus isomorphic to their homology) nd two-step complexes k k with no homology. Theorem 4.. Let (A i ) be finite complex of finite-dimensionl vector spces over k. Then (A i ) is isomorphic to the direct sum of its homology (with trivil differentil) nd bunch of two-step complexes. Proof. If the lst mp is not onto, split A into imge(φ ) nd complement, then use induction on totl dimension of the complex. Now sy it is onto. Consider the subspces imge(φ 2 ) ker(φ ) A. Pick complements H nd T, nd split A A into ( A 2 imge(φ 2 ) ) ( H ) ( T = A )
6 ALLEN KNUTSON nd use induction on the length of the complex. Very often we will only cre bout the homology of complex, becuse the complex involves choices but the homology does not. So we wnt wy to prove tht two complexes hve the sme homology. Before defining isomorphisms, we define morphism of complexes s digrm A 2 A A = = B 2 B B Theorem 4.2. A morphism of complexes induces mp on homology. Proof. The mp A B tkes imge(a 2 A ) to imge(a 2 B ) through B 2 by the left squre, so lnds inside imge(b 2 B ). It s only well-defined up to ker(a A ), which lies inside ker(a B ), which mps to ker(b B ). So s long s we kill tht in the trget, we get well-defined mp H (A) H (B). We my s well do this here: Theorem 4.3. A SES of complexes induces long exct sequence on homology. Proof. A 2 A A B 2 b B b B C 2 c C c C We lredy hve the mps H i (A) H i (B) H i (C), nd need the mp H i (C) H i (A), sy i = in the bove. Strt with c C C. Then B C so b c, mpping to b B. Then by the commuting squre, b C. Hence it the imge of unique A. Wht if the originl c were the imge of some c 2 C 2? Then we could lift tht c 2 to b 2, whose imge b B would go the sme plce s b does (nmely c ). Hence b = b + for some unique. Now b b b, since b 2 b horizontlly. But is the unique element mpping to b, so must be the imge of. I.e. if we lift imge elements, we lnd inside imge elements, giving well-defined mp on homology. We would like nturl reson tht two complexes (A i ), (B i ) would hve the sme homology. More generlly, we would like nturl reson tht two mps f, g : (A i ) (B i ) would induce the sme mp on homology. Define chin homotopy from f to g to be collection of mps h i : A i B i+ (not mp of complexes) such tht f g = d B h + hd A.
HOMOLOGICAL ALGEBRA 7 (The RHS is something like commuttor; note tht the d nd h opertions re both odd degree, so it is nturl to hve + insted of in their commuttor ). Proposition 4.4. If f, g : (A i ) (B i ) hve chin homotopy h relting them, then they induce the sme mp on homology. Proof. Let i ker(a i A i ). We need to show tht imge(b i+ B i ) (f g)( i ) = (d B h + hd A )( i ) = d B (h( i )) + h(d A ( i )) = d B (h( i )) +. In topology, nturl source of such hs is from mps [, ] A B. Sometimes we hve mps (A i ) (B i ) (A i ), where one composite is the identity nd the other composite is only chin homotopic to the identity, e.g. in topology from the inclusion of deformtion retrct. Then the bove sys tht the homologies re isomorphic. 5. MAPPING RESOLUTIONS Sy we hve mp M N nd resolutions (i.e. exct complexes) of M, N: A 2 A A M b B b 2 B B N Proposition 5.. If (A i ) is projective resolution, then we cn extend the bove to mp of complexes, uniquely up to chin homotopy. Proof. Since B N nd A M N nd A projective, there exists mp f : A B mking commuting squre. We cn obviously replce A with imge(a A ) whose mp to M is the zero mp (since the top row is complex), nd replce M with, but we wnt to lso replce B with imge(b B ) nd N with. To do tht sfely, we need to know tht f(imge(a A )) imge(b B ). If we lredy hd the mp A B, we could use tht. Insted we use imge(a A ) = ker(a M) ker(a N) f ker(b N) = imge(b B ) Now we cn indeed rip off tht lst column nd repet the rgument. To see the uniqueness, let f, g be two such mps. Strt with h := (the mp M B ). Assume we hve h,..., h i stisfying f g = d B h+hd A, nd now we wnt h i : A i B i+ solving b i h i = g i f i h i i. A 2 A A M b B b 2 B B N Obviously we ll use projectivity to construct desired mps, which mens we need to know tht A i mps vi the RHS into the imge of B i+ B i = ker(b i B i ). So we
8 ALLEN KNUTSON compute: b i (g i f i h i i ) = b i g i b i f i b i h i i = b i g i b i f i (g i f i h i 2 i 2 ) i = b i g i b i f i g i i + f i i = (b i g i g i i ) (b i f i f i i ) = = 6. DERIVED FUNCTORS Let (A i ), (B i ) two resolutions of the sme object M. Then we get mps between the resolutions, unique up to chin homotopies A 2 A A M B 2 b B b B M A 2 A A M which would induce mps on homologies, except tht s silly. It s worth noting tht the composite mps nd identity mps must be chin homotopic to one nother (which would induce isomorphism of homologies, except tht s sillier). Now ditch the M, nd pply covrint right exct functor T : C D (soon, X ) everywhere: T(A 2 ) T( ) T(A ) T( ) T(A ) T(B 2 ) T(b ) T(B ) T(b ) T(B ) T(A 2 ) T( ) T(A ) T( ) T(A ) The result is two complexes no longer exct with morphisms between them, ech composite chin homotopic to the identity. Hence their homologies re isomorphic. To get n ctul cnonicl definition, we cn use the cnonicl free resolution of M (use every element to generte, etc.), even though it s enormous. (We ll just never compute tht wy.) Given A 2 A k M M b B b 2 B k N N the cnonicl free resolutions, we get nturl mp k M k N. Then since tht squre commutes, the mp tkes ker(k M M) ker(k N N), with which we cn define the next mp, nd so on. This shows tht the cnonicl free resolution is functoril, nd from there, tht the homology of the complex T(A 2 ) T( ) T(A ) T( ) T(A )
HOMOLOGICAL ALGEBRA 9 is functoril in M. These re clled the left derived functors L i T of the right exct functor T. Wht does right exctness get us? The complex T(A ) T(A ) T(M) is still exct t T(A ), i.e. T(A ) T(A ) isn t; its homology is T(M). So L T = T. Exmple. Let T = Z n : Ab Ab, where Ab = Z Mod denotes finitely generted belin groups. Let g = gcd(m, n). Then we cn resolve Z n s the th homology of Z m Z whose T is Z n m Z n with kernels Z n/g Z n nd imges Z n/g giving homology Z n/g Z g Define Tor i (M, N) := H i (M projective resolution of N). With this definition, it s quite nonobvious tht Tor i (M, N) = Tor i (N, M). The long exct sequence. Let T be right exct, nd A B C SES. Then the covrint functors L i T give us complexes (L i T)(A) (L i T)(B) (L i T)(C) but now we clim tht there is nturl mp (L i T)(C) (L i T)(A) gluing them together into long exct sequence. We know how to get those from short exct sequences of complexes. The L i T come from T(projective resolution). So we wnt to replce A B C by short exct sequence A B C of projective resolutions, then pply T, then get the LES. Such SES would necessrily split by the projectivity of the C. Tht s why to build it, we first pick projective resolutions (A i ) of A nd (C i ) of C, nd dd them together, giving us A 2 2 A A A A 2 C 2 A C A C B C 2 c 2 C c C c C [ n The middle line hs projective elements, nd some obvious mps c n ], but doesn t yet hve the mp A C B. We cn build tht mp from A A B, nd the projectivity of C C B. But if we just bng this cnonicl nd non-cnonicl choice together, the rightmost two squres won t commute.
ALLEN KNUTSON We ll need to spoil the block digonlity of build it, consider the sequences C 2 C C C A A B C [ ] n, with mp A c i to or from C i. To n We clim the second one is exct. (Since A i B is :, the kernel of A B is the sme s tht of A A. Since A A is onto, the imge of A B is the sme s tht of A B.) Now use proposition 5. to crete verticl mps (t i ) mking ll squres commute. Then if our middle sequence is defined s n ( ) n t n [ ] c n i t A n C n A n C n, A C B we get everything we wnt. (The ( ) n re becuse the t i re constructed s degree mps, but we use them s degree mps.) T contrvrint. Everything works the sme if we pply contrvrint functor T, except tht our complexes re now bckwrds, which flips our definition of chin homotopies. To help keep trck, we cll the homology of the complex cohomology, but the kernel mod imge definition is the sme. Now left exct functor T like Hom(A, ) gets right derived functors R i T. T left exct. Use n injective resolution insted. 7. Ext GROUPS 7.. Using projective resolution. Define Ext i (C, A) := (L i Hom(, A))(C), i.e. using projective resolution of C nd the contrvrint right exct functor Hom(, A). Stupid cse: C is projective, so hs resolution C C. Then we Hom(, A) the complex C nd find out Ext i> (C, A) =. Consider the ctegory of SES A B C with fixed A, C, nd chin isomorphisms of SES. To such n SES, we look t the Ext i (, A) functors, getting LES Hom(C, A) Hom(B, A) Hom(A, A) Ext (C, A) Then A mps to n element of Ext (C, A). In prticulr, tht element is iff Hom(B, A) Hom(A, A) hs A in the imge, iff the SES is split. So we might hope tht the element of Ext (C, A) mesures the nonsplitness in some more generl sense. Theorem 7.. The mp from isomorphism clsses of extensions of C by A to Ext (C, A) is welldefined nd bijective. Proof. First, well-defined. Given two isomorphic SES, pply Ext s bove. The third nd fourth verticl mps come from the third nd first in the SES, which re equlities, so we get commuting squre with verticl equlities. Hence the horizontl mps re the sme.
HOMOLOGICAL ALGEBRA For onto, let g Ext (C, A), nd mke SES K κ F C where F is free. It gives the Ext(, A)s Hom(K, A) Ext (C, A) Ext (F, A) = i.e. g is the imge of some ḡ Hom(K, A). Define B := (A F) / ([ ḡ κ ] K ) Then we clim tht A B C is exct, nd induces g Ext (C, A). The best wy to show injectivity is to sy tht we re defining different R-module structures on the set A C, nd figure out how to mke those into n belin group so tht the mp to Ext (C, A) is group homomorphism. Then we lredy figured out tht the kernel is. Exmple. There re two extensions of Z p by Z p : the trivil one nd Z p 2. Consider SES Z p p Z p 2 b Z p These, b live in Z p. By scling by b, we cn reduce to b =. But then we still hve Z p different extensions (with the sme group in the middle). 7... Computing Ext groups of finitely generted Z-modules. If C or A is direct sum, then the Hom groups nd mps will lso be direct sums, so the result will be direct sum. The indecomposble options re Z nd Z p k, but we cn hndle Z m just s esily. C = Z. This ws the stupid projective cse from before. So Hom(C, A) = A nd Ext (C, A) =. C = Z m, with projective resolution Z m Z. A = Z. After Homming, the sequence is Z m Z, with homology groups Hom(Z m, Z) =, Ext (Z m, Z) = Z m. m A = Z n. After Homming, the sequence is Z n Z n. Both Ext groups re Z gcd(m,n). 7.2. Using n injective resolution. Define Ext i (A, B) := (R i Hom(A, ))(B), from Hom(A, ) being covrint left exct. To compute it, we strt with n injective resolution then nd look t B b b B b 2 B B2 Hom(A, B ) Hom(A, B ) Hom(A, B 2 ) / Ext (A, B) := {φ : A B b φ = } b {φ : A B }... Given SES A B C, pick function σ : C B splitting B C. Then we get symmetric mp (c, c 2 ) σ(c + c 2 ) σ(c ) σ(c 2 ) ker(b C) = A from C C A, mking no reference to B....
2 ALLEN KNUTSON 7.3. Universl coefficient theorems. This is usully the first ppliction of Ext groups. Let (C, d ) be complex of free belin groups to which we pply Hom(, A). Then we cn compute the homology of the first, denoted H i (C), nd the homology of the second, denoted H i (C; A) nd clled cohomology. We get nturl mp H i (C; A) Hom(H i (C); A) nd the theorem is tht () it fits into cnonicl SES Ext (H n (C), A) H i (C; A) Hom(H i (C); A) which (2) noncnoniclly splits (this uses the freeness). The rel reson tht we only need Ext s is tht this is bout Z-modules, which hve very short resolutions; things re much more complicted over rings requiring longer projective resolutions. This tkes while to prove (see e.g. https://nctlb.org/nlb/show/universl+ coefficient+theorem) so let s look t slightly simpler cse, computing homology of C A in terms of tht of C. Then the nturl mp () fits into the SES H i (C) A H i (C A) H i (C) A H i (C A) Tor(H i (C), A) where Tor(, A) = Tor (, A) is the first derived functor of A, nd (2) this sequence noncnoniclly splits. 7.3.. The proof. Here s trick: from C we consider the subcomplexes Z, B of kernels nd imges, on ech of which the differentil restricts (to ). Then we get two SES of complexes: B Z H (C) ll boundries trivil nd Z C B where the ltter splits row-wise (but not with commuting squres) if ech C i is free. Hence it still splits row-wise if we pply some covrint (sy) functor T, so we gin hve SES T(Z ) T(C ) T(B ), inducing LES T(B i ) T(Z i ) H i (TC) T(B i ) which mkes it look like it cme from SES B Z C, though it doesn t. (These shifts show up in the derived ctegory picture.) This breks into SES coker ( T(B i ) T(Z i ) ) H i (TC) ker ( T(B i ) T(Z i ) ) To understnd the first term, when T is right exct (e.g. A), pply it to the first SES obtining n exct sequence letting us identify the first with T(H i (C)). For the third term, we wnt to get T(B i ) T(Z i ) T(H i (C)) L T(Z i ) L T(H i (C)) T(B i ) T(Z i )
HOMOLOGICAL ALGEBRA 3 nd hve the first group vnish... so we obtin... H i (C) A H i (C A) Tor(H i (C), A) 8. INTERLUDE: FILTRATIONS ON RINGS AND MODULES Let R be ring. A grdtion by n belin group A is direct sum decomposition R = A R such tht R R b R +b. If R is vector spce over n infinite field F R, then this is the sme dt s n ction of the group Hom(A, F ) on R by ring utomorphisms. A grded module M is one with M = A M, where R M b M +b. If A contins submonoid A + we might be more interested in A + -grdings, the most common cse being N-grdings. One nice thing bout N-grded rings R is tht there is mp R R, not just R R. The nturls re ordered, nd the reverse order is not isomorphic, llowing for two distinct notions: decresing filtrtion on ring R is n N-filtrtion R = F F F 2. One often dds the requirement tht i F i = {}. Note tht ech F i is n idel, nd one obvious wy to mke decresing filtrtion is to define F k = I k for some idel I; this is clled the I-dic filtrtion. From n N-grdtion one cn construct decresing filtrtion, F i := k i R k. Conversely, from decresing filtrtion one cn construct the ssocited grded ring i (F i /F i+ ). One composite is the identity, but the other usully is not. All these definitions extend nturlly to modules. If R = R is just field, then the ssocited grded of module is unnturlly isomorphic to the originl module. Similrly we cn define incresing filtrtions E E... nd perhps sk tht i E i = R. In prticulr E is subring, nd we cn gin spek of the ssocited grded. One reson to work with filtrtions is tht ny sub or quotient ring of filtered ring is gin filtered (unlike the grded cse). For exmple, TV is grded hence Ug is filtered. Then the PBW theorem is tht gr Ug = Sym g. Given n incresing filtrtion E on R, define the Rees ring E[t] s the R [t]-sublgebr of R[t] given by i N R it i. Then for λ R, E[t]/ t λ = R, wheres E[t]/ t = gr R. Geometriclly, one should consider R [t] s the ring of functions on line, nd its grdedness s n ction of the multiplictive group. Then ll the fibers re isomorphic except for the centrl one, which is different. There is version for decresing filtrtions s well. Let F[t] be the R[t]-sublgebr of R[t, t ] given by i Z F it i, where F i := F = R. Once gin, F[t]/ t λ is R for λ unit, nd gr R for λ = (s long s one gives t degree ). Finlly, there s version i N F it i clled the blowup lgebr for its use in lgebric geometry (most often considered for I-dic filtrtions). 9. FILTERED COMPLEXES Let R = k[d]/ d 2, considered s Z-grded lgebr with deg d =. Then grded module M is exctly complex, nd nn d (M)/dM its homology. Wht if, unrelted to this, M hs 2-step filtrtion M M, where M is submodule (i.e. subcomplex)? (We ll keep the subscripts for M s being complex.)
4 ALLEN KNUTSON If M were ctully grded (M = M M ), then we could compute its homology in ech degree nd dd them together. Question. Wht s the reltion between H(M) nd H(gr M)? Obviously if M = M k, then H(M) = M k but H(gr M) = gr M k, so we cn t compute H(M) from H(gr M), only t best some sort of gr H(M). Things cn be much worse though. Exmple: let ech M i hve bsis t i, dt i, with M i spnned by just dt i. Then M is exct, but M nd M/M ech hve trivil differentil nd -D homology in every degree. The bsic picture, if we ssume k is field so tht we cn split M s M M where M is not subcomplex, is where ll two-step composites give. M i M i M i 2 M i M i M i 2 The bottom row is complex in itself, with homology H(M ). If we mod it out, we only see the top row, with homology H(gr M). Wht we ve forgotten is the digonl mp. Proposition 9.. There s nturl mp d : H (M/M ) H (M ), nd filtrtion on H(M) such tht gr H(M) = ker d coker d. Proof. Let m + M i (M/M ) i be in ker d. Then dm, priori in M i, is ctully in M i. Since d(dm) =, dm ker d. Then it remins to check tht if we chnge m by dding dm, it doesn t chnge dm. The filtrtion on H(M) is derived from the filtrtion on M. It is then tiresome to correspond the two spces. Question. Wht if M hs longer filtrtion? Sy, M M M 2. Now in ddition to M i M i M i 2 M i M i M i 2 M 2 i M 2 i M 2 i 2 we d hve knight s-moves mps to the SSE. One cn gin define the Southest mp d, but now it is only differentil on H(gr M), so insted of ker coker one should tke its homology, H(H(gr M), d ). But we re not done: the SSE mps give us nother differentil d on there. Then the sttement is tht for some filtrtion on H(M). H(H(H(gr M), d ), d ) = gr H(M) You cn now imgine mthemticl object, clled pge of spectrl sequence, where we hve grid of modules nd collection of semi-southest differentils, whose homology gives the next pge.
HOMOLOGICAL ALGEBRA 5 9.. Double complexes. Consider commuting grid of modules, where every row nd column is complex, nd cll this double complex. Then we cn long the digonls (which will be finite sums if this grid is supported in the correct qudrnt, in cse we cre), but to mke the result complex, we hve to introduce signs on every other row. Now, however, this resulting single complex comes with filtrtions, by either the rows or the columns of the originl double complex. One nturl wy to obtin such double complex is to strt with single complex, nd resolve every element (sy, projectively). Proposition 9.2. Let M, be double complex whose verticl homology is supported in the th row, i.e. is the complex (H, ). Consider H s (rther trivil) double complex too, so the mp M H gives n isomorphism of verticl cohomology. Then the flttening of M lso hs mp to H, now inducing n isomorphism of horizontl cohomology.. THE GROTHENDIECK SPECTRAL SEQUENCE Let M be module nd S, T two right exct covrint functors (there re mny other versions of course). Then S T is lso right exct, nd we should be ble to compute L (S T)(M). Strt with projective resolution M M whose only homology is H = M. Apply T to obtin complex. Wht we wnt to do now is pply S s well, then tke the homology, giving L (S T)(M). But insted we resolve the complex T(M ) to double complex, whose flttening would give the sme homology, nmely (L T)(M). Now pply S, obtining double complex whose flttening would give the thing we wnt (L S T)(M). The benefit of hving done this in two steps is tht the double complex comes filtered. So if we tke verticl cohomology obtining (L S)(L T)(M) first, we cn use the filtered story from before to continue tking homologies, in the limit obtining gr of L (S T)(M).