You Could Have Invented Spectral Sequences, Volume 53, Number 1

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 . “downstairs” to Cd−1,1 . In contrast, the elements of Zd,2 + Cd,1 are more special: their boundaries are Recall that we have been trying to find a natural boundaries of chains that come from Cd,1. Hence way of decomposing Hd into a direct sum. Just as Z + C we observed before that Cd is not a direct sum of d,2 d,1 + Cd,1 and Cd,2, we observe now that Zd/Bd is not a Bd,2 Cd,1 direct sum of Zd,1/Bd,1 and Zd,2/Bd,2 ; indeed, is a subspace of E1 , the subspace of elements Zd,2/Bd,2 by itself is already the entire homology d,2 group. But again we can use the same trick of mod- whose boundaries are boundaries of Cd,1-chains. ding out by the “downstairs part” and then direct Intuitively, the problem is that the associated summing with the “downstairs part” itself: graded complex only “sees” activity that is confined to a single horizontal level; everything above and Z Z + C Z ∩ C Z + C Z d d d,1 ⊕ d d,1 = d,2 d,1 ⊕ d,1 . below that level is chopped off. But in the original + ∩ + Bd Bd Cd,1 Bd Cd,1 Bd,2 Cd,1 Bd,1 complex, the boundary map ∂ may carry things down one or more levels (it cannot carry things up Now, the naïve hope would be that one or more levels because ∂ respects the filtration), ? Z + C (7) E1 d,2 d,1 and one must therefore correct for this inter-level d,2 B + C d,2 d,1 activity. and The Emergence of Spectral Sequences ? Z 1 d,1 The beautiful fact that makes the machinery of (8) Ed,1 , Bd,1 spectral sequences work is that both of the above corrections to the homology groups E1 can be re- and even that the numerators and denominators d,p in equations (7) and (8) are precisely the “cycles” garded as “homology groups of homology groups”! and “boundaries” in the definition (equation (5) Notice that ∂ induces a natural map—let us call 1 1 1 1 above) of E . For then expression (6) would give it ∂ —from Ed+1,2 to Ed,1, for all d, for the bound- d,p 1 us a direct sum decomposition of H . Unfortu- ary of any element in Ed+1,2 is a cycle that lies d 1 nately, in general, neither (7) nor (8) holds. Cor- in Cd,1, and thus it defines an element of Ed,1. The rections are needed. key claims (for n =2) are the following. 1 Claim 1. If we take E1 and mod out by the Let us first look “downstairs” at Ed,1. The • d,1 1 0 1 “cycles” of E are the cycles in E , and the image of ∂ , then we obtain Zd,1/Bd,1 . To see d,1 d,1 1 “boundaries” are the image I of the map this, just check that the image of ∂ gives all the boundaries that lie in C . 0 0 → 0 d,1 ∂ : Ed+1,1 Ed,1. 1 1 • Claim 2. The kernel of ∂ is a subspace of Ed,2 0 isomorphic to The space of cycles in Ed,1 is Zd,1 , which is the nu- Z + + C + merator in equation (8). However, the image I is d 1,2 d 1,1 . + not Bd,1, for Bd,1 is the part of Bd that lies in Cd,1, Bd+1,2 Cd+1,1

JANUARY 2006 NOTICES OF THE AMS 17 Again, simply check that the kernel consists r r r+1 from Ed+1,p+r to Ed,p, and E is defined to be the just of those elements whose boundary equals homology of (Er ,∂r ). a boundary of some element of C . d+1,1 The verification that, for general n, We can visualize these claims by drawing the fol- n Hd p Ed,p is a conceptually straightforward lowing diagram. generalization of the ideas we have already seen, but it is tedious so we omit the details. 00000 What Good Is All This? ∂1 ∂1 ∂1 ∂1 In analysis, the value of having a series approximation converging to a quantity of interest is fa- ··· 1 1 1 ··· Ed+1,2 Ed,2 Ed−1,2 miliar to every mathematician. Such an approximation is particularly valuable when just the first (9) ∂1 ∂1 ∂1 ∂1 couple of terms already capture most of the information. 1 1 1 Similar remarks apply to spectral sequences. ··· E + E E − ··· d 1,1 d,1 d 1,1 One common phenomenon is for a large number of the Er and/or the boundary maps ∂r to become ∂1 ∂1 ∂1 ∂1 d,p zero for small values of r. This causes the spectral sequence to stabilize or collapse rapidly, allowing 00000the homology to be computed relatively easily. We illustrate this by sketching the proof of Theorem 2 in a paper of Phil Hanlon [2]. This is far from a Diagram (9) is a collection of chain complexes; it’s “mainstream” application of spectral sequences, but just that the chain complexes do not run horizon- it has the great advantage of requiring very little tally as in diagram (4), but slant downwards at a background knowledge to follow. Readers who 45◦ angle, and each complex has just two nonzero know enough topology may wish instead to proceed terms. (Reminder: Our indexing convention is dif- directly to the standard examples that may be ferent from that of most authors, whose diagrams found in any number of textbooks. will therefore look “skewed” relative to diagram (9).) Let Q be a finite partially ordered set that is 2 If we now define Ed,p to be the homology, i.e., ranked—i.e., every maximal totally ordered subset has the same number of elements, so that every el- 1 1 → 1 def ker ∂ : E E − − ement can be assigned a rank (namely, a natural (10) E2 = H (E1 ) = d,p d 1,p 1 , d,p d d,p 1 1 → 1 im ∂ : Ed+1,p+1 Ed,p number indicating its position in any maximal totally ordered subset containing it)—and that is then the content of Claim 1 and Claim 2 is that equipped with an order-reversing involution 2 ⊕ 2 → ∗ Ed,1 Ed,2 is (finally!) the correct homology of our x x . Let original filtered complex. γ ={α1,α2,...,αt } where For the case n =2, this completes the story. The (11) 0 1 2 sequence of terms E ,E ,E is the spectral se- α1 <α2 < ···<αt quence of our filtered complex when n =2. We be a totally ordered subset of Q. We say that γ is may regard E1 as giving a first-order approxima- ∗ tion of the desired homology, and E2 as giving a isotropic if αi = αj for all i and j. ˆ second-order approximation—which, when n =2, Now adjoin a minimum element 0 and a maxi- ˆ is not just an approximation but the true answer. mum element 1 to Q, and consider the family of What if n>2? The definitions (2), (5), and (10) all totally ordered subsets of the resulting par- still make sense, but now E2 will not in general give tially ordered set. These form an abstract simpli- the true homology, because E2 only takes into ac- cial complex ∆, and we can consider its simplicial count interactions between adjacent levels in dia- homology groups Hd. We can also restrict attention gram (4), but ∂ can potentially carry things down to the isotropic totally ordered subsets; these form two or more levels. Therefore we need to consider a subcomplex ∆0 , which has its own homology 3 4 n 0 further terms E ,E ,... ,E . For example, to define groups Hd . E3, we can check that ∂ induces a natural map— Hanlon’s Theorem 2 says that if Q is Cohen- 2 2 2 call it ∂ —from Ed+1,p+2 to Ed,p, for all d and p. One Macaulay and its maximal totally ordered subsets 0 ≤ obtains a diagram similar to diagram (9), except have m elements, then Hd =0if 0 d

18 NOTICES OF THE AMS VOLUME 53, NUMBER 1 5 knowing that Q is Cohen-Macaulay gives us information about Hd. 4 0 In order to deduce something about Hd from the information we have about Hd, 3 we seek a relationship between Hd 0 p and Hd . Given γ as in equation (11), Han- lon’s key idea is to let ρ(γ) be the rank 2 of αi, where i is maximal subject to the ∗ condition that αi = αj for some j>i. Then ρ(γ)=0if and only if γ is isotropic, 1 but more importantly, applying the boundary map can clearly only decrease ρ, 0 so ρ induces a filtration on ∆. Specifi- 10 9 8 7 6 5 4 3 2 1 0 cally, we obtain the pth level of the filtration by restricting to those γ such that d ρ(γ) ≤ p. Therefore we obtain a spectral sequence! This gives a relationship be- Why the Adjective “Spectral”? 0 1 tween Hd = Ed,0 and the limit Hd of the spectral se- A question that often comes up is where the term quence. “spectral” comes from. The adjective is due to The heart of Hanlon’s proof is to analyze E1. He Leray, but he apparently never published an ex- shows that E1 =0 except possibly for certain d,p planation of why he chose the word. John McCleary pairs (d,p). For instance, when m =10, E1 =0 d,p (personal communication) and others have specu- except possibly for the pairs (d,p) marked by dots lated that since Leray was an analyst, he may have in the diagram to the right. viewed the data in each term of a spectral sequence If you imagine the 45◦ boundary maps, then as playing a role that the eigenvalues, revealed one you can see that some potentially complicated at a time, have for an operator. If any reader has things may be happening for d ≥ 5=m/2, but for 2 1 better information, I would be glad to hear it. m/2 >d, Ed,p will be isomorphic to Ed,p for all p. In fact, Er E1 for all r ≥ 1 when m/2 >d; the d,p d,p Acknowledgments boundary maps slant more and more as r increases, This paper started life as some lecture notes ac- but this makes no difference. Therefore just by com- companying a talk I gave while I was a general puting E1, we have computed the full homology member of the Mathematical Sciences Research In- group for certain values of d. In particular, H0 H d d stitute in the spring of 1997, supported in part by for m/2 >d. It turns out that the Cohen-Macaulay a National Science Foundation postdoctoral fel- condition easily implies that H =0for m/2 >d, d lowship. Many thanks go to Art Duval, who went so this completes the proof. through the notes with a fine-toothed comb and A Glimpse Beyond provided detailed suggestions for improving the ex- When our simplifying assumptions are dropped, a position, as well as repeatedly encouraging me to lot of complications can arise. Over an arbitrary publish a proper writeup. I thank John McCleary commutative ring, equation (3) need not hold; not for pointing out references [3] and [5] to me; the every short exact sequence splits, so there may be latter is another exposition of spectral sequences extension issues. When our finiteness conditions that is targeted at the beginner. Finally, I thank Phil are relaxed, one may need to consider Er for arbi- Hanlon for helping me understand spectral se- trarily large r, and the spectral sequence may not quences in the first place. He deserves at least as converge. Even if it does converge, it may not con- much credit as I do for any merits of the approach verge to the desired homology. So in many appli- expounded in this paper. cations, life is not as easy as it may have seemed from the above discussion; nevertheless, our sim- References plified setting can still be thought of as the “ideal” [1] DAVID EISENBUD, Commutative Algebra With a View To- situation, of which more realistic situations are ward Algebraic Geometry, Springer-Verlag, 1995. perturbations. [2] PHIL HANLON, A note on the homology of signed posets, We should also mention that spectral sequences J. Algebraic Combin. 5 (1996), 245–250. turn out to be such natural gadgets that they arise [3] JOHN MCCLEARY, A history of spectral sequences: Ori- not only from filtered complexes, but also from gins to 1953, in History of Topology, edited by Ioan M. double complexes, exact couples, etc. We cannot James, North Holland, 1999, 631–663. [4] , A User’s Guide to Spectral Sequences, 2nd ed., even begin to explore all these ramifications here, ——— Cambridge University Press, 2001. but hope that our tutorial will help you tackle the [5] BARRY MITCHELL, Spectral sequences for the layman, textbook treatments with more confidence. Amer. Math. Monthly 76 (1969), 599–605.

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