Math 272b Notes; Spring 2005 Michael Hopkins

Math 272b Notes; Spring 2005 Michael Hopkins David Glasser February 9, 2005 1 Wed, 2/2/2005 11 Administrivia mjh@mathharvardedu Computing homotopy groups lets you classify all spaces Homotopy theory on general categories Get Hatcher s book on spectral sequences 12 Spectral Sequences We know that a pair of spaces A X gives you a long exact sequence in homology: H (A) H (X) H (X, A) We also learned that H (X, A) = H (X CA) which is usually H (X/A) (the CA is the cone on A) Homology is good for cofiber sequences, which are A X X CA But homotopy groups are not good for them On the other hand, we have a fiber sequence, or fibration, which is F where F = p 1 ( ) (where is the basepoint of homotopy in B) A fibration is a map with the homotopy p B 1

lifting property: A {0} A [0, 1] B That is, the diagonal map exists This is a generalization of covering space, where the diagonal map needs to be unique If it exists for all A, it s a Hurewicz fibration; if it exists for all simplicial complexes A, it s a Serre fibration We can get a long exact sequence of homotopy groups: π n F π n π n B π n 1 F (Have to be careful near n = 1 or so) Fibrations are good for homotopy groups Cofibrations are good for mapping p out of homology and fibrations are good for mapping into homotopy But the miracle is that fibrations aren t so bad for homology! Suppose I have a fibration (as above); let B be a CW-complex Suppose we understand B and H F, and we want H The miracle is that you can actually do this Let B 1 B 2 B, where B i+1 is one more cell (of dimension n i ) than B i Now let i = p 1 i (B i ) ( D, S) ( i+1, i ) (D ni+1, S ni ) (B i+1, B i ) This is a pullback fibration We want to inductively calculate the homology of i Suppose we start with B 0 as a basepoint So 0 = F, the inverse image of the single point, which is by definition its fiber xamine the LS H ( i ) H ( i+1 ) H ( i+1, i ) If we know the left and right one here, we can probably figure out the middle one Let s look at the pair (B i+1, B i ), and the pair ( i+1, i ) which sits over it The characteristic map of the cell we re attaching is in the last diagram (the bottom map, I think) The definition of fibration guarantees excision for the top map So we get an isomorphism H ( D, S) = H ( i+1, i ) Lemma: suppose we have a fibration B, and we have a map A [0, 1] B which is a homotopy 2

between two maps A B: X A [0, 1] B p(fibration) Now we look at just the beginning and the end of the homotopy: X 0 X X 1 A {0} A [0, 1] A {1} Here X 0 and X 1 are the inverse images of A {0} and A {1} respectively I think! Then the maps X 0 X and X 1 X are weak homotopy equivalents, which mean they induce isomorphisms on homotopy groups, which implies that they induce isomorphisms in homology groups (proved last term apparently) Didn t need to have A [0, 1], could have had any C containing A, so in A C A C B if A C is a weak homotopy equivalence, so is A C If we map a point: F D pt D ni+1 B The fiber (F ) is by definition the inverse image of the basepoint; here we re talking about an arbitrary point, which is a bit vague Then we have that F D is a weak homotopy equivalence (in a Hurewicz fibration, 3

it s actually a homotopy equivalence) So the lift D has the same homology as F What about S? S? S ni F S ni S ni [0, 1] B char map at 0 S ni {1} justthebasepoint (Hopkins stated he was too lazy to come up with a name for the? here) This implies that the homology of S is the homology of S ni F, so H ( D, S) = H (D ni+1 F, S ni F ) = H ni 1(F ) by the Kunneth formula Thus, H ( i+1, i ) = H ni 1(F ) So it is possibly to calculate the homology of xample time! Path-loop fibration: ΩX P X p X P X is the space of paths {γ : [0, 1] X γ(0) = } starting at the basepoint, and pγ = γ(1) Then ΩX = {γ : [0, 1] X γ(0) = γ(1) = }, the set of loops at This is a fibration (check!) Claim: P X is contractible This works because a kid can suck spaghetti into his/her mouth if it has one end in the mouth already Homotopy: h: P X [0, 1] P X defined by h(γ, s)(t) = γ(st) Zeno eating spaghetti The LS of homotopy groups tells us that π k ΩX π k P X π k X π k 1 ΩX is a LS The groups π P X are 0 because it is contractible So π k 1 ΩX = π k X This is like what a suspension does to homology Now let s find homology of ΩX Let s take X to be the sphere S n+1 So we have: ΩS n+1 P S n+1 p S n+1 4

The induction will be very easy, since there s just one step (basepoint plus n + 1-cell) Now S n+1 is the equivalent of B So B 0 is just the basepoint, and B 1 is the whole thing is P S n+1, and 0 is the fiber ΩS n+1 LS of homology: H ΩS n+1 H P S n+1 H (P S n+1, ΩS n+1 ) H 1 ΩS n+1 etc We see that the relative group there is isomorphic to H (D n+1 ΩS n+1, S n ΩS n+1 ) = H n 1 (ΩS n+1 ) So since P S n+1 is contractible, there are isomorphisms in the LS, so H ΩS n+1 = H +n (ΩS n+1 ) So it s periodic with period n H ΩS n+1 H +n ΩS n+1 (n 1) 0 H 1 = 0 (n 2) 0 H 2 = 0 0 Z H n = Z 1 0 H n+1 = 0 n Z H 2n = Z So H ΩS n+1 is Z when is a multiple of n and 0 otherwise 2 Fri, 2/4/2005 21 Spectral Sequences Last time, we discussed the technique, where given a fibration F p B over a CW-complex B, we can relate the homology of the three spaces To relate H B, H, and H F, we filter B by its skeleton We write n = p 1 (B (n) ) (the n-skeleton), and study H ( n+1, n ) in order to get H n+1 from H n Now, H ( n+1, n ) = H (B (n+1), B (n) ) H F This deals with C cell n+1(b) H F, the cell complex First let s explain an exact couple This is the traditional way to introduce spectral sequences; Hopkins says he learned it this way and then never thought about exact couples again it s only good for getting 5

into spectral sequences An exact couple is a long exact sequence A i A k j forming a long exact sequence A A A A What you do with an exact couple is form the derived exact couple: A i A A is just the image under i of A, ie A = i(a) er fix this up to attach them: k j A i A i A i A k j k j k j If we let d = jk, we have that d 2 = jkjk = j0k = 0 by exactness So let = ker d/ im d Given α A, we have that α = i(a) for some a Now let j (α) = j(a); we have to check that this actually is in more or less See that it s in ker d Next see that it doesn t depend on image of d Last, let k (x) = k(x), which is in the image of i because dx = 0 (Skipping some conversions between elements and classes Look this over!) Last, i = i restricted to A Fact: the new thing we get is an exact couple So we can actually keep deriving over and over This gives a sequence ( j, d j ) of pairs of -groups and differential maps, where each j+1 = H ( j ; d j ) This sequence of groups is a spectral sequence Here s the main source of examples Space filtered by subspaces (say, n-skeletons): = X 0 X n 1 X n X H (X n 1 ) H (X n ) H (X n, X n 1 ) 6

isn t quite an exact couple, but n H (X n 1 ) n H (X n ) n H (X n, X n 1 ) is So this can get us a spectral sequence! Simple example! Suppose X n is the n-skeleton of a CW-complex Then 1 is just C cell (X) (cellular chain complex) and d 1 is the cellular boundary map (check!) So 2 is just H X And A 2 = im(h X n H X n+1 ) = im(h X n H X) = H n X Interesting example, gives you the idea that taking d 1 is a good idea In general, we have that A k is im(h X n H X n+k ) Suppose for given = m, n, the image H m X n H m X n+k equals the image H m X n H m X Then for this group, i k is an isomorphism (where I think i k : H m X k H m X) If you work this out, you find eventually that for r >> 0, assuming the preceding stability condition, the r term is the image of H X n H X mod H X n 1 H X, or actually the direct sum of that quotient over all n (This is for fixed ) This means that we can think of the r (for large r) as the graded group associated to the filtration on H X by the image of H X n (We have X whose homology we want, which is filtered by the X i, and we look at the successive quotients, or something, and then maybe we get the homology of X) Now restrict to the case where we have a fibration F p B, where B is a CW-complex Let n = p 1 B (n) ; study the spectral sequence associated to this filtration Conflict of notation! Let s try again: Now restrict to the case where we have a fibration F X p B, where B is a CW-complex Let X n = p 1 B (n) ; study the spectral sequence associated to this filtration Let A 1 = n H X n, 1 = n H (X n, X n 1 ) which last class we showed was equal to H (B (n), B (n 1) ) H F = n Cn cell (B) H F = C cell (B) H F = C cell p (B) H q (F ); let the summand here be p,q We usually show this in a grid: [notebook F1] What do the differentials do? We should check that d 1 maps 1 p,q 1 p 1,q This gives us a grid of 2 7

where each dot is ker d 1 / im d 1 Now we go over two and up one (instead of just over one) So in general d r : p,q r p r,q+r 1 r Keep doing lots of kernel mod image things ventually this diagonal hits negative p, so the groups are zero, which means that everything is the kernel and the image is trivial, so taking homology doesn t change it, so there s a limit on p,q r for fixed p, q as r goes to infinity It turns out the groups p,q r assemble to form H p+q, which is the whole point Main theorem: when π 1 B acts trivially on H F (to be explained next time), then d 1 = d cell so that p,q 2 = H p (B; H q (F )) (When B is simply connected π 1 B is trivial so it obviously acts trivally on anything) Another example! Fibration S 1 S CP ; we know the homology of S 1 and S is contractible Let s find the homology of CP, which is connected (and simply connected, even) So the left hand side of the grid has two Zs in it [Figure F2] Now, the limit of every square is 0 since S is contractible Looking at d 2 and such we can somehow reason that the homology of CP is Z in even spots and 0 in odd spots Basically the only way to get rid of the Z in spot (0, 1) is to have a Z in spot (2, 0) or something Now suppose B is (n 1)-connected and look at ΩB P B B P B is contractible (spaghetti argument from Wednesday), so again the spectral sequences go to 0 So the columns before n are all 0 (except for the 0 column) So the only opportunity to zero out the bottom n 1 (or so) elements on the first column must be zero because they can t change after n and they can t change before n because the maps are coming from 0 In fact, the only possible non-vanishing groups have p at least n and q at least n 1, or p = 0 or q = 0 Or maybe not F3 So this implies that H ΩB = H 1 B for < 2n 2 This leads to an inductive proof of the Hurewicz Theorem, which involes the diagram π n B = π n 1 ΩB H n B =byss H n 1 ΩB by reducing it to Poincare s theorem that H 1 is the abelianization of π 1 8

3 Mon, 2/7/2005 31 Action of the fundamental group of B on the homology of F This is an extension of the way that the fundamental group acts on the fiber in a covering space, which is just a discrete set of points Take a loop in the base space: γ : [0, 1] B with γ(0) = γ(1) = and define a pullback; here s pullback notation: X S X S B Now in this case we have: X {0} X [0,1] X {1} {0} [0, 1] {1} B The maps on the top induce isomorphisms in H (because they are homotopy equivalences?) and the top left and right groups are just F So we have H F = H X [0,1] = H F and then the composed map from the first to the last which is the action of γ on H F If all of these actions are trivial, then we get a Serre sequence where s,t = H s (B; H t F ), and that converges to H s+t X In the cohomology version, we still assume that π 1 B acts trivially on H F Then we get a spectral sequence of H s (B; H t (F ; R)) (where R is some commutative ring) which converse to H s+t (X; R) Now we have that d r : s,t r s+r,t r+1 And guess what, we get a ring structure again Here, d r (xy) = d r (x)y + ( 1) x xd r (y) where the product is cup product 9

xamples! ΩS n+1 P S n+1 S n+1 where p finds the endpoint See diagram D1 We have e which is a Z generated by e; to make it go to zero (P S is contractible) we need there to be an a 1 at a certain spot, which makes this other thing exist for some reason Huh d r+1 (a 2 1) = 2a 1 d r+1 a 1 = 2a 1 e, so a 2 1 = 2a 2 This gives us that a n 1 = n!a n in the ring And a p a q = ( ) p+q p ap+q This is a divided power algebra on one generator Or another example S 1 S CP This is the one where it goes off to the right Diagram D2 Looking for cohomology You do something or other and get that H (CP ) = Z[x] You have to look at some stuff with the definition of the thing to figure out that this works 32 A Series of Really Amazing Theorems Due To Serre π 3 (S 3 ) is Z by the Hurewicz Theorem, but we don t know about any of the other homotopy groups of S 3 We can map this to the ilenberg-mclane group K(Z, 3) (by adding higher cells to kill off higher homotopy groups): S 3 K(Z, 3) (which induces an isomorphism on π 3 ) Aside: Any map can be made into a fibration Recall that given a map A X gives you that H (X, A) = H (X/A) only most of the time But we can always make it work by making A nice f X X s by doing a mapping cylinder thingy: let X be X disjoint union A [0, 1] mod (a, 0) f(a) and including A at the other end of the cylinder (the 1 end) So similarly H (X, a) = H ( X/A) = H (X CA) (cone on A) Now, if we look at A X X CA (X CA) CX; this ends up being homotopy equivalent to the suspension of A (ΣA), which is just the homology of A with the degree shifted Now let s go even another step, by adding the mapping cone of this map, to get ((X CA) CX) CX CA Barrett-Puppe sequence 10

Now we convert a map into a fibration: A à X fibration this is like mapping out of thepuppe sequence Let à A X[0,1] such that the path (element of X [0,1] ) starts at f(a) Then the map A à sends a (a, constant(a)) The map à X gets the final point of the path Then you get a fiber F = {(a, γ) γ(0) = f(a), γ(1) = } This gives a LS of homotopy groups, where π k F = π k+1 (X, A) Then we can get a dual Barratt-Puppe sequence Map F A X At each step you use the above method to replace F A by a fibration, and then take the fiber This gives you loopspace so you get ΩF ΩA ΩX F A X and so on This is backing up Note though that the new maps are the negative of the map induced by the loopspace functor So going back, we had an isomorphism (on π 3 ) induced by S 3 K(Z, 3) So we can do this X S 3 S3 K(Z, 3) This gives us an LS π n X π n S 3 π n K(Z, 3) π n 1 X π n 1 S 3 π n 1 K(Z, 3) But most of the groups with K are zero (except for n = 3) So we get (by playing with the LS) that π 0 X is the basepoint, π 1 X = 0, π 2 X = 0, π 3 X = 0, π n X = π n S 3 for n 4 So X has the same homotopy as S 3 except for the one we already knew, which was n = 3 Now, by Hurewicz, π 4 S 3 = π 4 X = H 4 X We can compute this by the Serre spectral sequence! So let s do this Though we re going to do cohomology first We use ΩK(Z, 3) X S 3 (we re backing up a step) Now, ΩK(Z, 3) = K(Z, 2) = CP, which we know the coho of See diagram D3 Since π 3 X = 0, H 3 X = 0, so H 3 X = 0 Which means that x has to go and hit e at d 3 So by Leibniz, d 3 x 2 = 2xe, and in general d 3 x n = nx n 1 e So when we re all done, 4 term (ie, 11

the page after d 3 ) is in the next bit of the diagram, so H 3+2(n+1) (X; Z) = Z/n, and all other H terms are zero You can try to remember universal coefficient formula, or just do it at the level of spectral sequences This eventually gives us that H 2n X = Z/n and that H odd = 0 In particular, H 4 X = Z 2 and so π 4 S 3 = Z 2 Amazingly powerful technique Next time, Serre classes 12