Guide to the Serre Spectral Sequence in 10 easy steps! Jane Tan January 2020 These notes are intended to serve as a practical and painless introduction to the Serre spectral sequence, based on a lecture by Andr´eHenriques. Through 10 step-by-step examples, we'll see that a lot can be deduced from just the co- homology of spheres and some well-known fibrations. There will be no proofs or even precise statements made; the aim is simply to get a feel for the calculations involved and pick up some basic properties of the spectral sequence along the way. To this end, instead of a definition, we begin by describing what a spectral sequence looks like. The data of a spectral sequence can be naturally organised into pages Er for r ≥ 1. Each page consists of a 2-dimensional lattice of objects, as well as maps of a specified bidegree between any two objects for which the degrees match up. These form a collection of chain complexes. The spectral sequence that we consider is first-quadrant. This means that if we view our groups as being located at points of the integer lattice on the Cartesian plane, everything below the horizontal axis or right of the vertical axis is zero. It will also be cohomologically graded, meaning differentials dr on Er have bidegree (r; 1 − r). s;t The group at position (s; t) on the rth page is denoted Er . Differentials on s;t s+r;t+1−r the rth page are therefore of the form Er ! Er . Here is a generic picture of E2 and E3 with dots representing groups and arrows representing differentials. 5 • • • • • • • • • • 5 • • • • • • • • • • 4 • • • • • • • • • • 4 • • • • • • • • • • 3 • • • • • • • • • • 3 • • • • • • • • • • 2 • • • • • • • • • • 2 • • • • • • • • • • 1 • • • • • • • • • • 1 • • • • • • • • • • 0 • • • • • • • • • • 0 • • • • • • • • • • 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 E2 E3 1 By convention, we have not drawn in any differentials that are necessarily the zero map. This includes, for instance, all differentials going out of rows 0, 1, and 2 of the E3 page which have codomain located below the horizontal axis. In our examples, the nonzero groups and differentials are usually sufficiently sparse to stack the pages into one picture. Let us now specialise to the (cohomological) Serre spectral sequence, which relates the cohomology of the total space of a fibration to the cohomology of the fiber and base spaces. One can compactly describe the essential features of the spectral sequence via a signature of the form s;t s t s+t E2 = H (B; H (F )) =) H (X) where F ! X ! B is the input fibration. This provides a formula for objects on the E2 page of the spectral sequence, and tells us that they `converge' in some sense to the cohomology of the total space. It is useful to note that when there is no torsion1, the formula simplifies to s;t s t E2 = H (B) ⊗ H (F ): (∗) We will focus on integral cohomology, so these entries, and indeed the entries on each page, are abelian groups. It is worth noting that there is also a homological version of the spectral sequence, although we make no further mention of it. In the steps that follow, we feed a series of fibrations into the Serre spectral sequence and see what we can extract. These have been chosen so that the starting point is minimal, the only assumed knowledge being H∗(Sn). Step 1: S1 ! S3 ! S2. (Working backwards to the Serre SS) Consider the quotient map from C2 − f0g to the complex projective line CP 1 = C [ f1g given by identifying (z; w) with (λz; λw) for any nonzero complex number λ, that is, λ 2 C×. If we view S3 as the unit sphere in C2, and CP 1 as S2, then this restricts to a map S3 ! S2 for which the fiber over each point is the space of complex numbers of unit norm, and hence can be identified with S1. This describes a fibration S1 ! S3 ! S2 known as the Hopf fibration. We already know the cohomology groups of each space involved in this case, so the idea in this first step is to work backwards from the answer to determine the associated Serre spectral sequence. From the formula given in (∗), we see that H∗(B) is `plotted' on the horizontal axis, and H∗(F ) on the vertical axis. For the Hopf fibration, this gives Zs at positions (0; 0), (0; 1), and (2; 0). We then fill in Z ⊗ Z =∼ Z at (2; 1). Due to all of the 0s on the axes, every other position will just contain 0. The picture so far is as follows, where we adopt the 1Specifically, the condition is that we require at least one of H∗(B) and H∗(F ) to be free and finitely generated. 2 convention of suppressing 0 groups, so any blank entry should be interpreted as 0, and as before do not draw in any differentials that are necessarily trivial. 1 Z Z 0 Z Z 0 1 2 There is no direct expression for groups on E3 or higher. Instead, we `turn the page' to see these groups by taking homology at every position. In order for this to make sense, note that all of the maps really are differentials satisfying 2 dr = 0. Here (and henceforth), we use dr to denote all differentials on the the Er page. We can specify which one we mean by the position of its domain or codomain when necessary, but do so only if it is not clear from context. s;t s;t Property 1. Er+1 = ker dr= im dr at Er . More precisely, s;t s;t s+r;t+1−r s−r;t−1+r s;t Er+1 = ker dr : Er ! Er = im dr : Er ! Er : Observe that if the differentials going in and out of a particular entry, say s;t s;t s;t Er , are both zero, then Er = Er+1. In particular, for any fixed position (s; t), since the differentials at (s; t) become longer on successive pages, there exists s;t s;t an ` such that E` = Er for any r > `. Taking ` = maxfs; tg + 1 would do. s;t Let the collection of stabilised groups be the entries E1 on the so-called E1 page. The Serre spectral converges to the cohomology of the total space X in n s;t the sense that H (X) comes from the E1 for which s + t = n. In general, we s;t say that the total degree of Er is s + t. Applying the above to our example, note that all differentials dr are 0 for r ≥ 3. This is because 0 groups remain the same when taking a quotient, so the four Zs that we have found indicate the only possible positions of nonzero groups in all higher pages. Then, it is easy to check that all differentials going in and out of any non-zero group land outside of the first quadrant. It follows ∼ that E3 = E1. 1 3 2 3 s;t Since H (S ) = H (S ) = 0, we deduce from convergence that each E1 = s;t E3 for which s + t = 1 or 2 must also be 0. This means that we need to kill the Zs at (0; 1) and (2; 0). Fortunately, we have just the right differential d2 : 0;1 2;0 E2 ! E2 between these groups. This is in fact the only possible (nonzero) differential on E2 as the domain or codomain vanish for all others. From 0;1 0;1 0 = E1 = E3 = ker d2= im d2; it follows that d2 must be injective to make ker(d2) = 0. Similarly, for 2;0 2;0 0 = E1 = E3 = ker d2= im d2 = Z= im d2 3 we need im d2 = Z, so d2 is also surjective. Hence, we find that d2 is an isomorphism. In the next picture, this isomorphism is shown in red, so the circled groups are the only ones that survive to infinity. 1 Z Z 0 Z Z 0 1 2 Turning to the E3 = E1 page, it is now possible to read off the cohomology of the total space S3 by assembling along the diagonals. In this case, we have n 3 M s;t H (S ) = E1 s+t=n 0;0 2;1 which gives a Z in dimension 0 from E1 , and Z in dimension 3 from E1 as expected. Since E3 = E1, we say that the spectral sequences collapses at E3. 3 2 1 Z 0 Z 0 1 2 3 ∗ 3 H (S ): Z 0 0 Z Step 2: S3 ! S7 ! S4 (Later pages) With the basic set-up in hand, we now essentially repeat Step 1 only this time working over the quaternions H. Specifically, if we view S7 as living in H2 and identify S4 with the quaternionic projective line HP 1 = H [ f1g, then one can analogously define a mapping S7 ! S4 with fiber S3, or the unit quaternions. Thus, there is also a quaternionic Hopf fibration S3 ! S7 ! S4. As before, we start by filling in the groups on the E2 page using the formula (∗). On the horizontal axis, which represents H∗(S4), we have a Z in positions (0; 0) and (0; 4), and all other entries are 0. We also have a Z on the vertical axis at (3; 0) which comes from H3(S3). Due to all of the 0s on the axes, the only other nontrivial entry is Z ⊗ Z at (4,3).