5
$\begingroup$

The question is about understanding Adams spectral sequence intuitively and some of the meanings of its relations.

In Adams spectral sequence,

$$E_2^{s,t}=\text{Ext}_{\mathcal{A}}^{s,t}(H^*(MTG), \mathbb{Z}/2\mathbb{Z}) \Rightarrow (\pi_{t-s} MTG)_2^\wedge. \tag{eq.1} $$

  • $E_2^{s,t}$ means the $E_2$ page info in the spectral sequence.

  • $Ext^{s,t}_{\mathcal{A}} (A,B)$ means Ext functor, the group of extensions of length $s$: $$A[t] \to C_1 \to ... C_s \to B$$ as modules over Steenrod algebra $\mathcal{A}$ (not only for abealian groups), where $[t]$ denotes shift in the grading.

  • $MTG$ is the Madsen-Tillman spectrum. Pontryagin-Thom isomorphism provides a relation between the bordism groups of manifolds with stable tangential structure and homotopy groups of the Madsen-Tillman spectrum $MTG$ (which is a close cousin of the more usual Thom spectrum $MG$) associated to tangential structure $G$.

  • $( ... )_2^\wedge$ is the 2-completion applied to $\pi_n MTG$ (apply it to abelian groups). In general, a $p$-completion (prime $p$) of an Abelian group $G$ means we take limit lim${}_{n\to \infty} A/(p^n A)$. Practically we take $p$-torsion part of $A$ plus free part where $\mathbb{Z}$ is represented as $p$-adic integers $\mathbb{Z}_{(p)}$.


Question:

  1. Is my understanding above correct, precisely?

  2. How do we establish the relation in (eq.1) from $E_2^{s,t}=\text{Ext}_{\mathcal{A}}^{s,t}(H^*(MTG) \mathbb{Z}/2\mathbb{Z})$ intuitively?

  3. How do we deduce the relation in (eq.1) $$E_2^{s,t} \Rightarrow (\pi_{t-s} MTG)_2^\wedge,$$ and how to we know the $E_2$ page info is complete? (How do we know if we need all the prime $p$-completion, instead of only $2$-completion?


For some background info, one can check these Ref 1, 2, and citations therein.

$\endgroup$
6
$\begingroup$

Your questions seem very broad, and would probably be better served by a textbook reference. I'll try to give some perspectives that might help.

Given two spectra $X$ and $Y$, we would compute the mapping space $\operatorname{Map}(X,Y)$. The Adams spectral sequence is sort of a way to split this calculation into an "algebraic" part and a "topological" part, the idea being the algebraic calculation should be routine.

In general, spaces and spectra are hard to work with (compared with algebra), so instead of studying spectrum maps $X \to Y$, we apply a cohomology theory $E$ and consider maps of $E^* E$-modules $E^* Y \to E^* X$. There is a homomorphism $$[X, Y] \to \operatorname{Hom}_{E^* E}(E^* Y, E^* X),$$ but this is in general neither a monomorphism or an epimorphism.

We can do slightly better: instead of approximating the category of spectra by just $E^*E$-modules, we can approximate it by the (still algebraic) derived category of $E^* E$-modules, and consider $$\operatorname{Ext}_{E^* E}^*(E^* Y, E^* X)$$ as our approximation. Topologically, this ought to take into account some of the "primary attaching data".

But this homological algebra approximation still doesn't give us $[X,Y]$, for a couple of reasons. One is that we haven't factored in all of the topology in $X$ and $Y$. This will be reflected in the differentials in the Adams spectral sequence, and this is generally very hard, since in general there will be infinitely many differentials and there is no algorithm to compute all of them.

Another reason, as you alluded to, is that our algebraic approximation depends on how much the theory $E$ sees in the category of spectra. It turns out that if $E = H\mathbb{F}_p$, then we cannot see any of the $q$-local information for any prime $q \neq p$. On the other hand, if $E = MU$, then we do see "enough".

So the intuition is that the $E_2$-page of the Adams spectral sequence is the best approximation to the category of spectra by the derived category of $E^* E$-modules.

To answer your questions rigorously, you have to know how the Adams spectral sequence is constructed. There are different ways to think about it, but a particular nice approach is via descent. Let me specialize to the case $X = \mathbb{S}$ for simplicity.

The functor $E \wedge -$ from spectra to $E$-module spectra is part of an adjunction, and we can form the monadic descent object $$Y \leftarrow E^{\wedge \bullet + 1} \wedge Y.$$ This is a cosimplicial spectrum whose totalization is $$\operatorname{Tot}(E^{\wedge \bullet + 1} \wedge Y) =: Y_E,$$ the $E$-localization of $Y$. This is why the $E$-based Adams spectral sequence can only see $E$-local information.

The Adams spectral sequence is the Bousfield-Kan spectral sequence for the cosimplicial spectrum $E^{\wedge \bullet + 1} \wedge Y$. Observe that we have $$E^{\wedge \bullet + 1} \wedge Y \simeq (E \wedge E) \wedge_E \cdots \wedge_E (E \wedge E) \wedge_E (E \wedge Y),$$ so the $E_1$-page looks like $$(E_* E)^{\otimes s} \otimes_{E_*} (E_* Y),$$ which is the standard bar construction for $E_* Y$ in $E_* E$-comodules. Thus it should come as no surpise that the $E_2$-page is given by some standard functor in homological algebra. Specializing to $H\mathbb{F}_2$ and $Y = MTG$, dualizing (since we want the cohomology ASS), and doing some homological algebra, we get $$E_2 = \operatorname{Ext}_{H^*H}^{s,t}(H^* MTG, \mathbb{F}_2) \Rightarrow \pi_{t-s} MTG^\wedge_2,$$ which is the form of the Adams spectral sequence you wanted.

$\endgroup$
  • $\begingroup$ thanks +1, this is very helpful $\endgroup$ – wonderich Aug 20 '18 at 18:05
  • $\begingroup$ thanks again for the answer - accepted for now $\endgroup$ – wonderich Sep 3 '18 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.