You Could Have Invented Spectral Sequences Timothy Y. Chow Introduction Fools rush in where angels fear to tread, so my The subject of spectral sequences has a reputation goal below is to make you, the reader, feel that you for being difficult for the beginner. Even G. W. could have invented spectral sequences (on a very Whitehead (quoted in John McCleary [4]) once re- good day, to be sure!). I assume familiarity with ho- marked, “The machinery of spectral sequences, mology groups, but little more. Everything here is stemming from the algebraic work of Lyndon and known to the cognoscenti, but my hope is to make Koszul, seemed complicated and obscure to many the ideas accessible to more than the lucky few who topologists.” are able to have the right conversation with the right Why is this? David Eisenbud [1] suggests an ex- expert at the right time. planation: “The subject of spectral sequences is el- Readers who are interested in the history of ementary, but the notion of the spectral sequence spectral sequences and how they were in fact in- of a double complex involves so many objects and vented should read [3], which gives a definitive ac- indices that it seems at first repulsive.” I have count. heard others make similar complaints about the proliferation of subscripts and superscripts. Simplifying Assumptions My own explanation, however, is that spectral se- Throughout, we work over a field. All chain groups quences are often not taught in a way that explains are finite-dimensional, and all filtrations (explained how one might have come up with the definition below) have only finitely many levels. In the “real in the first place. For example, John McCleary’s ex- world”, these assumptions may fail, but the es- cellent text [4] says, “The user, however, needs to sential ideas are easier to grasp in this simpler get acquainted with the manipulation of these gad- context. gets without the formidable issue of their origins.” Without an understanding of where spectral se- Graded Complexes quences come from, one naturally finds them mys- Chain complexes that occur “in nature” often come terious. Conversely, if one does see where they with extra structure in addition to the boundary come from, the notation should not be a stum- map. Certain kinds of extra structure are particu- bling block. larly common, so it makes sense to find a system- Timothy Y. Chow is a member of the research staff at the atic method for exploiting such features. Then we Center for Communications Research in Princeton, New do not have to reinvent the wheel each time we want Jersey. His email address is [email protected]. to compute a homology group. JANUARY 2006 NOTICES OF THE AMS 15 Here is a simple example. Suppose we have a for all d and p. (Warning: There exist different in- chain complex dexing conventions for spectral sequences; most 0 ∂ ∂ ∂ ∂ authors write E where q = d − p is called the ···−→ C −→ C −→ C −→··· p,q d+1 d d−1 complementary degree. The indexing convention I that is “graded”, i.e., each Cd splits into a direct sum use here is the one that I feel is clearest pedagog- n ically.) Then C = C n d d,p 0 p=1 (3) Cd Ed,p. p=1 and moreover the boundary map ∂ respects the The nice thing about this direct sum decomposi- ⊆ grading in the sense that ∂Cd,p Cd−1,p for all d tion is that the boundary map ∂ naturally induces and p. Then the grading allows us to break up the a map computation of the homology into smaller pieces: n n 0 0 → 0 simply compute the homology in each grade in- ∂ : Ed,p Ed−1,p dependently and then sum them all up to obtain p=1 p=1 the homology of the original complex. such that ∂0E0 ⊆ E0 for all d and p. The rea- Unfortunately, in practice we are not always so d,p d−1,p lucky as to have a grading on our complex. What son is that two elements of Cd,p that differ by an we frequently have instead is a filtered complex, i.e., element of Cd,p−1 get mapped to elements of each Cd comes equipped with a nested sequence Cd−1,p that differ by an element of of submodules ∂Cd,p−1 ⊆ Cd−1,p−1, by equation (1). Therefore we obtain a graded complex that splits 0=Cd,0 ⊆ Cd,1 ⊆ Cd,2 ⊆···⊆Cd,n = Cd up into n pieces: and the boundary map respects the filtration in the sense that (4) 0 0 0 0 (1) ∂C ⊆ C − ··· ∂→ 0 ∂→ 0 ∂→ 0 ∂→··· d,p d 1,p Ed+1,n Ed,n Ed−1,n 0 0 0 0 ··· ∂→ 0 ∂→ 0 ∂→ 0 ∂→··· for all d and p. (Note: The index p is called the fil- Ed+1,n−1 Ed,n−1 Ed−1,n−1 . tration degree. Here it has a natural meaning only . ≤ ≤ . if 0 p n, but throughout this paper, we some- ∂0 ∂0 ∂0 ∂0 ··· → E0 → E0 → E0 →··· times allow indices to “go out of bounds,” with the d+1,1 d,1 d−1,1 understanding that the objects in question are 1 Now let us define Ed,p to be the pth graded piece zero in that case. For example, Cd,−1 =0.) Although a filtered complex is not quite the of the homology of this complex: same as a graded complex, it is similar enough that we might wonder if a similar “divide and con- ker ∂0 : E0 → E0 1 def 0 d,p d−1,p quer” strategy works here. For example, is there a (5) E = Hd (E ) = d,p d,p im ∂0 : E0 → E0 natural way to break up the homology groups of a d+1,p d,p filtered complex into a direct sum? The answer (For those comfortable with relative homology, turns out to be yes, but the situation is surprisingly note that E1 is just the relative homology group complicated. As we shall now see, the analysis d,p leads directly to the concept of a spectral sequence. Hd(Cp,Cp−1).) Still thinking naïvely, we might hope Let us begin by trying naïvely to “reduce” this that n problem to the previously solved problem of graded 1 (6) Ed,p complexes. To do this we need to express each Cd p=1 as a direct sum. Now, Cd is certainly not a direct sum of the Cd,p; indeed, Cd,n is already all of Cd. is the homology of our original complex. Unfortu- However, because C is a finite-dimensional vector d nately, this is too simple to be true. Although each space (recall the assumptions we made at the out- 0 0 term in the the complex ( p Ed,p,∂ )—known as set), we can obtain a space isomorphic to Cd by the associated graded complex of our original fil- modding out by any subspace U and then direct tered complex (Cd,∂)—is isomorphic to the corre- summing with U; that is to say, C (C /U) ⊕ U . d d sponding term in our original complex, this does In particular, we can take U = C − . Then we can d,n 1 not guarantee that the two complexes will be iso- iterate this process to break U itself down into a morphic as chain complexes. So although E1 direct sum, and continue all the way down. More p d,p formally, define does indeed give the homology of the associated graded complex, it may not give the homology of 0 def (2) Ed,p = Cd,p/Cd,p−1 the original complex. 16 NOTICES OF THE AMS VOLUME 53, NUMBER 1 Analyzing the Discrepancy and while this contains I, it may also contain other This is a little disappointing, but let’s not give up things. Specifically, the map ∂ may carry some el- just yet. The associated graded complex is so closely ements x ∈ Cd+1 down from “upstairs” to “down- related to the original complex that even if its ho- stairs,” whereas I only captures boundaries of el- mology isn’t exactly what we want, it ought to be ements that were already downstairs to begin with. 1 a reasonably good approximation. Let’s carefully Therefore, Zd,1/Bd,1 is a quotient of Ed,1. 1 examine the discrepancy to see if we can fix the Now let us look “upstairs” at Ed,2. In this case, 1 problem. the space of “boundaries” of Ed,2 is Bd,2 + Cd,1, Moreover, to keep things as simple as possible, which is the denominator in equation (7). How- let us begin by considering the case n =2. Then the ever, the space of “cycles” in this case is the ker- array in diagram (4) has only two levels, which we nel K of the map shall call the “upstairs” (p =2) and “downstairs” 0 0 → 0 (p =1) levels. ∂ : Ed,2 Ed−1,2, The homology group Hd that we really want is which, by definition of E0, is the map Zd/Bd , where Zd is the space of cycles in Cd and Bd is the space of boundaries in Cd. Since Cd is fil- C C − ∂0 : d,2 → d 1,2 . tered, there is also a natural filtration on Zd and Bd : Cd,1 Cd−1,1 0 = Zd,0 ⊆ Zd,1 ⊆ Zd,2 = Zd Thus we see that K contains not only chains that and ∂ sends to zero but also any chains that ∂ sends 0 = Bd,0 ⊆ Bd,1 ⊆ Bd,2 = Bd .