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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeNov 15th 2013

After scanning a bunch of literature, my favorite survey of the Adams spectral sequence is now this gem here:

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeNov 17th 2013
• (edited Nov 17th 2013)

I have added a section Motivation from Hurewicz theorem and Serre spectral sequence, closely following Wilson’s section 1.1.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeNov 17th 2013
• (edited Nov 17th 2013)

I have started now a section

with the general perspective as in Lurie’s “Higher Algebra”, following the helpful outline in Dylan Wilson’s note.

I still need to fill in some of those technical conditions.

One question:

at the very end of the “Chromtatic Homotopy Theory“-lecture we learn that the image of J is just the $E(1)$-localization of the spheres. So in view of Dylan Wilson’s prop. 1.3.4 we want to know the “core” of $\pi_0(K(1))$. What is it?

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeJan 15th 2015

I don’t have time to look at Wilson’s survey right now but maybe you can answer this quickly. The “Motivation” section at Adams spectral sequence says that if $X$ is a finite spectrum, then it has finitely many nontrivial cohomology classes, which we can use to define a map into a finite wedge of EM spectra, and then we iterate this process on the fiber of this map. I can’t see any reason why that fiber should again be a finite spectrum, though. Hatcher uses instead the assumption of being “finite type”, which I don’t fully understand yet. Is that what one has to do?

• CommentRowNumber5.
• CommentAuthorKarol Szumiło
• CommentTimeJan 15th 2015
• (edited Jan 15th 2015)

Maybe I’m missing something, but if we had a fiber sequence of spectra of the form $F \to X \to \prod H \mathbb{F}_2$ with both $X$ and $F$ finite, then this would imply that $\prod H \mathbb{F}_2$ is also finite since the sequence is also a cofiber sequence.

• CommentRowNumber6.
• CommentAuthorMike Shulman
• CommentTimeJan 16th 2015

That sounds right to me. But I guess $H \mathbb{F}_2$ is finite type and hence so is the fiber?

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeJan 16th 2015

Maybe I had this wrong, you’re saying it must be “spectrum of finite type” instead of “finite spectrum”? I’d need to check and to remind myself. But don’t have time to do this right now.

• CommentRowNumber8.
• CommentAuthorKarol Szumiło
• CommentTimeJan 16th 2015
• (edited Jan 16th 2015)

$H \mathbb{F}_2$ is finite type (since it’s cohomology is finite type, I’m not sure if there is a more direct argument) so this works with spectra of finite type. However, I think we have to be a little careful because a spectrum of finite type may have infinitely many nontrivial cohomology classes and we need to observe that the corresponding infinite product of EM spectra is still of finite type.

• CommentRowNumber9.
• CommentAuthorDylan Wilson
• CommentTimeJan 17th 2015

Hey- that’s probably from a typo in my notes. I meant to say “spectrum of finite type” (which means its cohomology is finitely generated in each degree, but could exist in infinitely many degrees) not “finite spectrum”. Now you’re mapping into a locally finite wedge of EM spectra (so that has finite type), and this property persists to the fiber.

A belated thanks for the praise of my notes- I’m glad you found them useful! Let me know if you find other typos, though I’m not sure I can fix them on Hiro’s page.

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeJan 17th 2015

Thanks, I have changed in the entry, “finite spectrum” to “spectrum of finite type”.

(I really have no time for this right now, even though I ought to. Sorry. But with all you experts around here, please all feel invited to hit “edit” and hack away on the entry.)

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeJan 17th 2015

By “locally finite wedge” you mean a wedge with only finitely many summands at each suspension level?

I am, of course, trying to figure out how much of this can be done in HoTT, where we have to do without AC and therefore weird things happen like not every module over a field is free and not every free module is projective. However, the finite AC is always true, so I have some hopes that things may work out somehow.

• CommentRowNumber12.
• CommentAuthorDylan Wilson
• CommentTimeJan 17th 2015

@Mike yes.

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeJan 28th 2016

Just in case you are watching the logs and are wondering: I am editing little bits at Adams spectral sequence and various related entries with the goal of making it all eventually more comprehensive. But not quite there yet.

• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeJan 30th 2016

Miller 81 gave a nicely streamlined perspective exhibiting $E$-Adams resolutions in direct analogy with injective resolutions. This is the perspective that the “coctalos” lectures (Hopkins 99) adopted. A lot of details involved have nicely been spelled out by Aramian. I’ll be adopting this into the $n$Lab entries.

I have added to Adams resolution a new subsection Via injective resolutions. So far this contains all the relevant definitions and statement of all the relevant fact, but no proofs yet.

• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeMar 8th 2016

I am in the process of spelling out full details in the section Via injective resolutions. Not done yet, but need to interrupt now.

• CommentRowNumber16.
• CommentAuthorUrs
• CommentTimeMar 16th 2016

I have further been expanding the section Via injective resolutions – The first page and Hopf algebroid structure.

Not done yet. Need to interrupt now to get some lunch.

• CommentRowNumber17.
• CommentAuthorUrs
• CommentTimeMar 17th 2016

I have now added content to The second page and homological co-algebra.

• CommentRowNumber18.
• CommentAuthorUrs
• CommentTimeMay 20th 2016
• (edited May 20th 2016)

I have expanded in the section on $E$-Adams resolution by adding (some more infrastructure and then) the example (here) of the “associated inverse sequence” for the “normalized $E$-Adams resolution”, showing that this is the actual thing that Adams and Bousfield originally considered.

Then I have added the statement (here) that the tower spectral sequences induced from a) the associated Adams tower and b) the “associated inverse sequence” of an $E$-Adams resolution are isomorphic.

This connects the classical construction of the $E$-Adams spectral sequences (Adams, Bousfield, Ravenel, …) to the construction in Hopkins99 (the coctalos lecture notes).

What I still don’t have is complete proof that the latter is indeed isomorphic to the spectral sequence of the Tot-tower of the corresponding cosimplicial spectrum. I see where the pieces want to go, but I haven’t sorted out a rigorous proof yet. Probably it’s obvious and I am being dense.

• CommentRowNumber19.
• CommentAuthorUrs
• CommentTimeJun 30th 2016
• (edited Jun 30th 2016)

I have reworked the first part at Adams spectral sequence a little, up to and including the section “The second page”.

I have

• moved the discussion of $E$-Adams resolutions following Miller and Hopkins down to the end of the classical discussion, and instead replaced it with a section (here) that gives the direct construction of the $E$-Adams spectral sequence right away, the one from Adams 74, III, theorem 15.1.

• then I adapted the proof of the form of the $\mathcal{E}_1$-page accordingly (now this theorem and this lemma)

• made explict in all these arguments where one uses that $Ho(Spectra)$ has the structure of a tensor triangulated category;

• added a quick introduction to the relevant spectral sequences of filtered spectra (here), in order to have all the definitions and notation available in one place;

• changed, for sake of conciseness, the mentioning of homotopy groups of mapping spectra to the more elementary construction of forming graded hom-groups $[X,-]_\bullet$ in the stable homotopy category

In the course of all this I have tried to harmonize and improve notation conventions.

• CommentRowNumber20.
• CommentAuthorUrs
• CommentTimeJul 11th 2016
• (edited Jul 11th 2016)

I have added a bunch of further details to the section “The first page”, up to and excluding its last subsection “Universal coefficient theorem”.

• CommentRowNumber21.
• CommentAuthorUrs
• CommentTimeJul 12th 2016
• (edited Jul 12th 2016)

I have finally spelled out full proof (here) of the “Adams universal coefficient theorem” needed for identifying the first page of the $E$-Adams spectral sequence, namely the statement that for $E$ a homotopy ring spectrum such that $E_\bullet(E)$ is flat over $\pi_\bullet(E)$ and $X$ a spectrum such that $E_\bullet(X)$ is projective over $\pi_\bullet(E)$, then

$[X, E \wedge Y]_\bullet \overset{\simeq}{\longrightarrow} Hom^\bullet_{E_\bullet(E)}( E_\bullet(X), E_\bullet(Y) )$

is an isomorphism, where on the right we have the graded hom of comodule homomorphisms over the dual $E$-Steenrod algebra.

The proof in Adams 74, p. 323 of this fact invokes the stronger statement that under these conditions and for $Z$ any homotopy $E$-module spectrum then also

$[X, Z]_\bullet \simeq Hom_{\pi_\bullet(E)}( E_\bullet(X), \pi_\bullet(Z) )$

is an isomorphism (where what is needed for the above is only the case that $Z$ is a free module spectrum $E \wedge Y$ ), but this stronger statement (which is really what Adams calles the UCT) holds only under much more restrictive conditions. The general proof of the above I am deducing from Schwede 12, chapter II, prop. 6.20, which however at the end is vague about how the conclusion via idempotent splitting in $Ho(Spectra)$ follows. (It looks like there may have been a copy-and-pasting mismatch in the compilation of that proof.) The relevant result is Bökstedt-Neeman 93, prop. 3.2, which says that in the stable homotopy category all idempotents split.

• CommentRowNumber22.
• CommentAuthorUrs
• CommentTimeJul 15th 2016
• (edited Jul 15th 2016)

I have finally brought in all the infrastructure such as to state (just state) in a self-contained fashion the two big convergence theorems by Bousfield, here

• CommentRowNumber23.
• CommentAuthorUrs
• CommentTimeJul 20th 2016
• (edited Jul 20th 2016)

I have finally found a reference with a proof that Bousfield’s original definition of E-nilpotent completion is equivalent to the one in terms of the Tot-tower of the cosimplicial spectrum:

it is proposition 2.14 in the (neat) article

As the authors notice on p. 9

This result is certainly not new, but we have included it for lack of a convenient reference.

Indeed.

I have correspondingly expanded the relevant paragraph in the entry here.

• CommentRowNumber24.
• CommentAuthorJon Beardsley
• CommentTimeJul 20th 2016

Oh fantastic!! I tell this to people all the time, but never know what to cite, and I guess I’ve just been too lazy to write it down myself.

• CommentRowNumber25.
• CommentAuthorUrs
• CommentTimeJul 20th 2016
• (edited Jul 20th 2016)

Yeah. At times the difference in sociology between the alg-top community and the string theory community is not that large. ;-)

Anyway, praise to Akhil, Niko and Justin for writing it out. Their article is a gem in many other ways, too.

• CommentRowNumber26.
• CommentAuthorUrs
• CommentTimeJul 21st 2016
• (edited Jul 21st 2016)

Jon, here is one thing that confuses me, maybe you’ll spot where I am making a mistake:

$CRing^{op} \underoverset {\underset{\pi_0}{\longleftarrow}}{\overset{}{\hookrightarrow}}{\bot} E_\infty Ring^{op}$

and hence

$(\pi_0)_! \;\colon\; Sh_\infty(E_\infty Ring^{op}) \longrightarrow Sh_\infty(CRing^{op})$

preserves not only $\infty$-colimits but also $\infty$-limits.

Let then $E$ be an $E_\infty$-ring and consider its 1-image factorization

$Spec(E) \longrightarrow Spec( \underset{\longleftarrow}{\lim}_n Tot^n(E^\bullet) ) \longrightarrow Spec(\mathbb{S})$

given by the formal dual of the totalization of its cosimplicial ring spectrum.

Now if $(\pi_0)_!$ preserves both $\infty$-limits and $\infty$-colimits, then this 1-image is preserved under $(\pi_0)_!$ to yield a 1-image factorization of $\pi_0(E) \to \mathbb{Z}$.

But the former yields a formal completion, while the latter in general does not. (?)

Specifically, let $A$ be a Noetherian commutative ring, $I \subset A$ an ideal and $E \coloneqq H (A/I)$ is EM-spectrum of the quotient ring. Then theorem 4.4 in Carlsson’s arXiv:0707.2585 says that the 1-image factorization in the $\infty$-topos yields on $\pi_0$ the $I$-adic completion. But the 1-image factorization of the image of $E \to \mathbb{S}$ under $(\pi_0)_!$ does not.

So that looks like a contradiction. Where is my mistake?

• CommentRowNumber27.
• CommentAuthorJon Beardsley
• CommentTimeJul 21st 2016
• (edited Jul 21st 2016)

Hey Urs,

Not sure I can be of that much help here. I think my “answer” to your question is just going to result in more questions, so feel free to not bother and to ask someone else. The topos notation (e.g. $(\pi_0)_!$) kind of confuses me, so if it’s okay with you I’m just going to phrase my response in terms of spectra and $\pi_0$ and so forth.

One question I have is that Gunnar’s theorem seems to relate $A$ and $A/I$ (or their associated EM-spectra), but you want to relate $H(A/I)$ to $\mathbb{S}$? Are you assuming that $A/I\cong \mathbb{Z}$ here? Also, Gunnar’s theorem, in my opinion, is really poorly written. Is $M_B^\wedge$ the homotopical completion at $B$ of $M$ thought of as a module spectrum over $H(A)$? I can’t tell what’s a ring and what’s the EM-spectrum of a ring here… and I feel like this is confusing matters.

But so, let’s see. I guess Gunnar is saying the following: if you have a map of commutative discrete rings $A\to A/I=B$, and a $B$-module $M$ regarded as an $A$-module, then we can do two things: complete $M$ at $B$, which is isomorphic to $M$ as an $A$-module, and take its EM-spectrum OR we can look at $H(M)$ as an $H(A)$ module and complete it homotopically at $H(B)=H(A/I)$, and his theorem is saying that these two things are equivalent? I.e. $H(M)_{H(B)}^\wedge\simeq H(M_I^\wedge)$? Am I reading this correctly?

• CommentRowNumber28.
• CommentAuthorDavid_Corfield
• CommentTimeNov 28th 2018
• (edited Nov 28th 2018)

At Adams resolution what is meant by the equivalence to the final term in Example 1.2?

$S = \Sigma^\infty S^n \simeq \Sigma^n \mathbb{N}$

• CommentRowNumber29.
• CommentAuthorAli Caglayan
• CommentTimeNov 28th 2018

I think that should really be the sphere spectrum $\mathbb{S}$?

• CommentRowNumber30.
• CommentAuthorUrs
• CommentTimeDec 10th 2020

fixing some trivial typos

• CommentRowNumber31.
• CommentAuthorUrs
• CommentTimeDec 10th 2020

Only now that this thread is bumbed up do I see #28 from over two years ago. Have fixed that typo now (here).

Also, only now do I see Jon’s kind reply in #27. I hope to look into this again..