Hopf algebroids and the structure of MU (MU)

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1 Hopf algebroids and the structure of MU (MU) Vitaly Lorman July 1, 2012 Note: These are my notes on sections B.3 and B.4 of Doug Ravenel s Orange Book (Nilpotence and Periodicity in Stable Homotopy Theory). If you have comments or notice any mistakes, please me. 1 Motivation: The dual Steenrod algebra We are going to motivate the definition of Hopf algebroids from the point of view of topology. Let s begin with some preliminaries about ring spectra. Definition: A ring spectrum E is a spectrum equipped with a map η : S 0 E, called the unit map. and m : E E E, called the multiplication map, such that some diagrams commute. The Hopf algebroids the we are going to be interested in are of the form (π (E), E E), where E is a ring spectrum with an extra property that we will discuss later. So to motivate this discussion, let s begin by analyzing the dual Steenrod algebra. It is a Hopf algebra, which means that it has the following structure maps: µ : A A A (product) η : Z/p A : A A A ɛ : A Z/p c : A A which fit into some commutative diagrams. (unit) (coproduct) (augmentation) (conjugation) We may identify Z/p with π (H), where H is the mod p Eilenberg-MacLane spectrum. Also, A = H H = π (H H). Since H H is a wedge of suspensions of H, we may also identify A A = H H π (H) H H with π (H H H) 1

2 In fact, the above homomorphisms on homotopy are induced by maps on spectra. H is a ring spectrum, and we let m : H H H denote the multiplication and η : S 0 H denote the unit map. 1 H m : H H H H H (product) η 1 H : H = S 0 H H H (unit) 1 H η 1 H : H H = H S 0 H H H H (coproduct) m : H H H T : H H H H (augmentation) (conjugation) Note that we could have also used the map η 1 H : S 0 H H H to induce a map Z/p A ; this would also be a plausible candidate for the unit, but it turns out that when our ring spectrum is H, the two maps are the same on homotopy. However, for an arbitrary ring spectrum E, the maps 1 E η and η 1 E will induce different maps, which we will denote by η L and η R, the left and right units respectively. We could also define a Hopf algebra over Z/p as a cogroup object in the category of Z/p-algebras. That is, for any Z/p-algebra, C, we have that Hom(A, C) is a group, with structure maps induced by the Hopf algebra structure maps in the following way. The coproduct induces the group multiplication: Hom(A, C) Hom(A, C) Hom(A A, C) Hom(A, C) where the map on the right is, i.e. precomposition with the coproduct. The augmentation ɛ picks out the identity morphism: A Z/p C where the map on the left is ɛ and the map on the right is the unit for C (since C is a Z/p-algebra. The conjugation c induces the map that sends each morphism f to its inverse: A A C where the map on the left is c and the map on the right is f. product maps are what give A the structure of a Z/p-algebra. Finally, the unit and 2 Hopf algebroids Now we will attempt to carry out the above constructions for an arbitrary ring spectrum E. We need one condition: Definition: A ring spectrum E is said to be flat if E E is a wedge of suspensions of E. Note that if E is flat, then E E = π (E E) is a flat π (E)-module. (The details of this need to be worked out) As we have seen, H is flat. Also, MU is flat. However, 2

3 the integral Eilenberg-MacLane spectrum is not flat. We use flatness for the following isomorphism: π (E E E) = E (E) π (E) E (E) (Ravenel sketches a proof of this in the Green Book. I will write this up later.) Because of the left and right units, E (E) will not be a Hopf algebra. It will be a generalization of this. Recall that we may think of a group as a category with one object in which all morphisms are isomorphisms. We think of the morphisms as elements of the group, and composition of morphisms gives us multiplication. We generalize as follows. Definition: A groupoid is a small category in which every morphism is an isomorphism. We have a set O of objects and a set M of morphisms. We can think of a groupoid as a map of sets (O, M) together with the following structure. There is a map O M which assigns to each object its identity morphism, a map M M which assigns to each morphism its inverse, and two maps M O assigning to a morphism its source and target. Composition is a map D M where D = {(m 1, m 2 ) M M : target(m 1 ) = source(m 2 )} These maps should satisfy the appropriate conditions. Definition: A Hopf algebroid over K is a pair (S, Σ) of graded commutative K-algebras with unit which is a cogroupoid object in the category of K-algebras. That is, given any K-algebra C, the pair of sets (Hom(S, C),Hom(Σ, C)) has a natural groupoid structure. More explicitly, we have : Σ Σ S Σ c : Σ Σ ɛ : Σ Σ η R : S Σ η L : S Σ (coproduct inducing composition) (conjugation inducing inverse) (augmentation inducing identity morphism) (right unit inducing target) (left unit inducing source) satisfying conditions corresponding to the groupoid structure. Using the left and right units, we may regard Σ as a left and right S-module, and in this way we define the tensor product Σ S Σ given above. Why is Σ S Σ the appropriate target for (and not just Σ k Σ?)? Well, induces composition of morphisms in the associted groupoids, and composition of morphisms is a map M M D M Notice that D is exactly the difference kernel between the two maps from M M to M O M, where the first map sends (m 1, m 2 ) to (m 1, target(m 1 ), m 2 ) and the second map sends (m 1, m 2 ) to (m 1, source(m 2 ), m 2 ). 3

4 The left unit η L induces the map that sends a morphism to its source (by precomposition with η L ). Similarly, the right unit η R induces the map that sends a morphism to its target. So if m 1 and m 2 are composable elements of Hom(Σ, C) Hom(Σ, C), the target of m 1 is the source of m 2 : that is, m 1 η R = m 2 η L Given such a pair of morphisms, we get an element of Hom(Σ, C) as follows. We have a map Hom(Σ, C) Hom(Σ, C) Hom(Σ Σ, C C) which we then compose with the multiplication µ C on C to get something in Hom(Σ Σ, C). We claim that this composition (*) Σ Σ C C C sends elements of the form σ 1 η R (s) σ 2 σ 1 η L (s)σ 2 to zero for all s S, σ 1, σ 2 Σ. Indeed, we have and m 1 m 2 (σ 1 η R (s) σ 2 = m 1 (σ 1 η R (s)) m 2 (σ 2 ) = m 1 (σ 1 )m 1 η R (s) m 2 (σ 2 ) = m 1 (σ 1 )m 2 η L (s) m 2 (σ 2 ) since m 1 η R = m 2 η L m 1 m 2 (σ 1 η L (s)σ 2 ) = m 1 (σ 1 ) m 2 η L (s)m 2 (σ 2 ) Applying µ C, both terms come out to be m 1 (σ 1 )m 2 η L (s)m 2 (σ 2 ) so their difference maps to zero. Thus, the above composition (*) factors through Σ Σ modulo the submodule generated by all such elements. But this quotient is exactly the tensor product Σ S Σ. Thus, we have map Σ S Σ C and so we define our coproduct to have Σ S Σ as its target to get the map Σ Σ S Σ C as the composition of the morphisms m 1 and m 2 and the product on the groupoid. As suggested above, the reason we are interested in Hopf algebroids is the following theorem. Theorem: If E is a flat homotopy commutative ring spectrum, then (π (E), E (E)) is a Hopf algebroid over Z. If E is p-local, then it is a Hopf algebroid over Z (p). Definition: A left comodule over a Hopf algebroid (S, Σ) is a left S-module M together with a left S-linear map ψ : M Σ S M 4

5 which is counitary and coassociative. (see Ravenel for the diagrams) A right comodule is defined similarly. Proposition: If E is as in the theorem above, then for any X, E (X) is a left comodule over (π (E), E (E)) with the homomorphism ψ induced by the map 1 E η 1 X : E X = E S 0 X E E X 3 The structure of MU (MU) Recall that L, the Lazard ring, is isomorphic to π (MU) is isomorphic to Z[x 1, x 2,...] where dim(x i ) = 2i. Also, as we proved sometime last year, H (MU) = Z[b 1, b 2,...] with dim(b i ) = 2i. Milnor and Novikov proved (Ravenel does this in the Green/Red Book) the following theorem. Theorem: The generators x i may be chosen in such a way that the Hurewicz map h : π (MU) H (MU) is given by h(x i ) = pb i +decomposables if i = p k 1 for some prime p, and h(x i ) = b i +decomposables otherwise. Landweber and Novikov originally computed the structure of MU (MU). Note that π (MU) Q = Z[m 1, m 2,..] with dim(m i ) = 2i. (actually, m i turns out to be the cobordism class of CP n divided by n) Theorem: As a ring, MU (MU) = MU [b 1, b 2,...] with dim(b i ) = 2. The coproduct is given by ( ) i+1 b i b j (b i ) = where b 0 = 1. (b n ) may be found be expanding the right hand side and finding the terms in dimension 2n, of which there are finitely many. The left unit η L is the standard inclusion MU MU (MU) = MU [b 1, b 2,...] The right unit on MU (MU) Q is given by η R (m i ) = ( m i c(b j ) ) i+1 where m 0 = 1 and c is the conjugation. The conjugation c is given by c(m n ) = η R (m n ) and ( ) i+1 c(b i ) b j = 1 Note that we have only defined the right unit on π (MU) Q. I guess that it induces the right unit on π (MU) via precomposition with the natural map π (MU) π (MU) Q. 5

6 Notice that the coproduct sends each b n to a polynomial in the b i b j with coefficients in Z rather than in π (MU). If we let B = Z[b 1, b 2,...], then this means that B is a Hopf algebra over Z. The right unit gives a B-comodule structure on π (MU) and it turns out that MU (MU) = π (MU) B. This is a general property for Hopf algebroids; we say that MU (MU) is split. It corresponds to the following property for groupoids: Definition: A groupoid is split if it is obtained in the following way. let X be a set acted on by a group G. Regard the elements of X as objects of a category C whose morphism set is G X. The sources and target of a morphism (g, x) are x and g(x), respectively. The composite of (g, x) and (g, g(x)) is (g g, x). Definition: A Hopf algebroid (S, Σ) over K is split if Σ = S B, where B is a Hopf algebra over which S is a comodule algebra. When (S, Σ) is split, we have Hom(Σ, C) = Hom(S B, C) = Hom(S, C) Hom(B, C) Since B is a Hopf algebra, Hom(B, C) is a group, and an action of Hom(B, C) on Hom(S, C) is a map Hom(S, C) Hom(B, C) = Hom(Σ, C) Hom(S, C) given by precomposing a morphism Σ C with the right unit map. In our situation, let R be a commutative ring. Then we can identify the set Hom(π (MU), R) with the set of formal group laws over R, which we will denote by FGL(R). We may regard the Hopf algebra B as a ring of integer valued functions on Γ: given γ = a i x i+1 think of b i as the function which assigns to γ the coefficient a i of x i+1 in its power series expansion. Proposition: Let B be the Hopf algebra Z[b 1, b 2,...] with the coproduct given above. Then for a commutative ring R, Hom(B, R) is the group Γ R of power series of the form x + b 1 x 2 + b 2 x with coefficients over R under functional composition. Proof: Given a morphism γ : B R, we send it to the element γ(b i )x i+1 in Γ R. It is not hard to see that this is one to one and onto. It remains to check that it is a group homomorphism. The group structure on B is induced by the coproduct. First, we claim that ( ) i+1 b i b j x j+1 (b i )x i+1 = 6

7 Indeed, notice that on the left and right hand sides, the degree of the coefficient of x N+1 is always 2N. If we set x = 1 in the left and right hand sides, the degree 2N term will be exactly the coefficient of x N+1, and indeed after we set x = 1, the left and right hand side are equal. Thus, (f g)( ( b i x i+1 )) = (f g)( (b i )x i+1 ) = (f g)( ( ) i+1 b i b j x j+1 = ( ) i+1 f(b i ) g(b j )x j+1 After applying the multiplication map R R R, we get ( ) i+1 f(b i ) g(b j )x j+1 Which is exactly the usual composition of the power series f(b i )x i+1 and g(b j )x j+1 which are the elements of Γ R associated to f and g. Thus, we have shown that the group Hom(B, R) is isomorphic to Γ R. It remains to show that Hom(B, R) acts on Hom(π (MU), R) in the same way that Γ R acts on formal group laws over R. That is, given a formal group law F over R represented by a map θ : π (MU) R and a map γ Hom(B, R) with associated power series f(x) = γ(b i)x i+1 Γ R, we act on F in one of two ways: 1) Apply the usual action of Γ R on F to get F f (x, y) = f(f (f 1 (x), f 1 (y))). 2) Apply the action of Hom(B, R) on Hom(π (MU), R) as described for any split Hopf algebroid above. That is, take the pair (θ, γ) Hom(π (MU), R) Hom(B, R) = Hom(π (MU) B, R) = Hom(MU (MU), R) and precompose with η R : π (MU) Hom(MU (MU)) to get a map (θ γ)(η R ) : π (MU) R and take the formal group law represented by this map. We claim that the formal group laws obtained in 1) and 2) are identical, which is exactly the statement that the action of Hom(B, R) on Hom(π (MU), R) is the same as the action of Γ R on the set of formal group laws over R. Recall that any formal group law over a Q-algebra has a unique logarithm. That is, if F is such a formal group law, there is a unique power series log F (x) = x +... such that log F (F (x, y)) = log F (x) + log F (y). 7

8 Note on the uniqueness of the logarithm: If φ and ψ are two logarithms for a formal group law F over a torsion-free ring R, that is, two isomorphisms between F and the additive formal group law, then ξ = ψ φ 1 will be a nontrivial automorphism of the additive formal group law. But this can t happen: if ξ(x) = a i x i+1 and then so that ξ(x + y) = ξ(x) + ξ(y) a i (x + y) i+1 = a i (x i+1 + y i+1 ) a i ((x + y) i+1 x i+1 y i+1 ) = 0 But since R is torsion-free, it is not possible for a i to kill the middle terms of the binomial expansion of (x+y) i+1, so for i 1, the a i must be zero, which means that ξ is the trivial automorphism and ψ = φ. Now, let s take our formal group laws over R and turn them into formal group laws over R Q. That is, given the two maps π (MU) R, turn them into maps π (MU) Q R Q. We claim that the two formal group laws defined by these maps have the same logarithms, and thus are the same. Why does this imply that the original formal group laws, before we rationalized, are the same??? Let s start by writing down the logarithm of the formal group law obtained in 1). We may rewrite the equation for F f as and we apply log F to get f 1 (F f (x, y)) = F (f 1 (x), f 1 (y)) log F (f 1 (F f (x, y))) = log F (F (f 1 (x), f 1 (y))) = log F (f 1 (x)) + log F (f 1 (y)) We also have We claim that This will imply that log F f (F f (x, y)) = log F f (x) + log F f (y) log F (f 1 (F f (x, y))) = log F f (F f (x, y)) log F f (x) = log F (f 1 (x)) which is a fact we need to proceed. So let s prove the claim. (Many thanks to Jeff for helping me with this.) 8

9 Check that log F f 1 and log F f are both power series whose linear term has coefficient 1. Notice that they are both isomorphisms between F f (x, y) and the additive formal group law. But the logarithm of F f is unique by the note above, so we must have that We conclude that log F (f 1 (F f (x, y))) = log F f (F f (x, y)) ( ) log F f (x) = log F (f 1 (x)) The logarithm for the universal formal group law is m ix i+1. Since θ is the map representing F, the logarithm for F will be θ(m i)x i+1. Recall that, the power series in Γ R corresponding to γ is f(x) = γ(b i )x i+1 Now, combining and from above, we have log F (x) = θ(m i )x i+1 log F f (x) = log F (f 1 (x)) ( ) log F f (x) = θ(m i )(f 1 (x)) i+1 We claim that f 1 (x) = θ(c(b j ))x j+1 To prove this, we compose f(f 1 (x)) = f( θ(c(b j ))x j+1 ) = ( ) i+1 γ(b i ) θ(c(b j ))x j+1 = (γ θ) ( ) i+1 b i c(b j )x j+1 Notice that the coefficient of x N+1 in the above power series has degree 2N so that we can read off the coefficient of x N+1 of the above power series by plugging in 1 and finding the degree 2N part of the resulting sum. But when we plug in x = 1, we get ( i+1 b i c(b j )) = 1 by the formula for the conjugation, c. Thus, we have ( ) i+1 b i c(b j )x j+1 = x 9

10 and (γ θ)(x) = x, so that the equality f 1 (x) = θ(c(b j ))x j+1 holds. Now we plug this into ( ) above: log F f (x) = θ(m i )( θ(c(b j ))x j+1 ) i+1 Now let s compute the logarithm of the formal group law that we got from 2). Recall that the formal group law given in 2) is represented by the map (θ γ)(η R ) : π (MU) R so if we apply this map to the coefficients of the logarithm of the universal formal group law, m i x i+1 we have (θ γ)(η R )( Now, compare this power series to the power series m i x i+1 ) = (θ γ)( η R (m i )x i+1 m i ( c(b j )x j+1 ) i+1 η R (m i )x i+1 ) Arguing as before, we see that in both power series the coefficient of x N+1 has degree 2N so that the two power series are the same if and only if they are the same when we plug in x = 1. But when we do this, we get η R (m i ) and m i ( c(b j ) ) i+1 which are equal by the formula for the right unit given above. Thus, (θ γ)(η R )( m i x i+1 ) = (θ γ)( η R (m i )x i+1 ) = (θ γ) ( m i ) i+1 c(b j )x j+1 = θ(m i )( θ(c(b j ))x j+1 ) i+1 10

11 which is exactly the logarithm we got for F f. Thus, the formal group laws in 1) and 2) have the same logarithms, so modulo some details they are the same. To conclude and rereiterate what has already been reiterated, the action of the group Γ R on the set of formal group laws over R is exactly the same as the action of the group Hom(B, R) on the set Hom(π (MU), R). 11