27 votes
Accepted

Torsion in the Atiyah–Hirzebruch spectral sequence of a classifying space

Of course, in any spectral sequence $E_{r+1}$ is a subquotient of $E_r$ (the kernel of $d_r$ divided by the image of $d_r$). And in general new torsion can appear in the sense of torsion elements in $...
Tom Goodwillie's user avatar
20 votes
Accepted

Pullback and homology

This is not necessarily true. For example, there is a space $X$ constructed by attaching a 3-dimensional cell to $S^1 \vee S^2$, which serves as a standard counterexample to several questions. The map ...
Tyler Lawson's user avatar
20 votes
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Calculation of $H^{10}(K(\mathbb{Z}, 3); \mathbb{Z})$

I do not like naming a cohomology class $n$ because that deserves to be the name of an integer. I will use the name Hatcher does and call the generator of $H^2(K(\Bbb Z, 2); \Bbb Z)$ by the name "$a$"....
mme's user avatar
  • 8,988
19 votes
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Relating two different approaches to the Atiyah-Hirzebruch Spectral Sequence

For cohomology, this is theorem 3.3 in Maunder, C.R.F., The spectral sequence of an extraordinary cohomology theory, Proc. Camb. Philos. Soc. 59, 567-574 (1963). ZBL0116.14603. Theorem 3.3 If $...
Denis Nardin's user avatar
  • 16.1k
18 votes
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Persistence barcodes and spectral sequences

The answer to your question is no, nobody has used persistence to improve the algorithmic efficiency of computing differentials, although of course the relationship between persistence intervals of a ...
Vidit Nanda's user avatar
  • 15.2k
16 votes
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Sphere spectrum, Character dual and Anderson dual

The Anderson dualizing spectrum $I_\mathbf{Z}$ can be defined as follows. Consider the functor $X\mapsto \mathrm{Hom}(\pi_{-\ast} X,\mathbf{Q/Z})$ from the homotopy category of spectra to graded ...
skd's user avatar
  • 5,490
15 votes
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What is the relationship between spectral sequences and obstruction theory?

This is a partial answer, but every obstruction theory (in some precise sense) provides you with a spectral sequence (in fact several). Let me clarify what do I mean with obstruction theory. All this ...
Denis Nardin's user avatar
  • 16.1k
14 votes

cup product and Steenrod operations in Serre spectral sequence

The behavior of the Steenrod squaring operations in the Serre spectral sequence was determined by Araki and independently by Vázquez (whose article I cannot locate online). However, it's a little work ...
Tyler Lawson's user avatar
14 votes
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"Rotated" version of the Atiyah-Hirzebruch spectral sequence

Good question. I think the answer is yes. The unnamed spectral sequence is usually referred to as the isotropy spectral sequence. For a group $G$ acting on $X$ and an abelian group $A$ of ...
Mark Grant's user avatar
  • 34.6k
14 votes

To compare the total, base and fiber spaces of two fiber bundles

No. Consider the map from the fibre bundle $$B\mathbb{Z} \to BD_\infty \to B\mathbb{Z}/2$$ to $* \to * \to *$. Here $D_\infty = \mathbb{Z} \rtimes \mathbb{Z}/2$ is the infinite dihedral group. You ...
Oscar Randal-Williams's user avatar
13 votes
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cup product and Steenrod operations in Serre spectral sequence

1) No in general. A counterexample is the projective space bundle associated to a vector bundle. For a rank $n$ vector bundle, the fiber, $\mathbb C \mathbb P^{n-1}$, has cohomology ring $\mathbb Z[x]/...
Will Sawin's user avatar
  • 129k
13 votes

Multiplicative structure on spectral sequence

This is an expansion of John Rognes' answer. I have filled in a few details in Douady's seminare notes and noticed that one gets away with slightly weaker axioms. If there is already a reliable ...
Sebastian Goette's user avatar
13 votes

Why is it difficult to obtain the next differential in a spectral sequence?

Expanding on Tyler Lawson's comment, the point of a spectral sequence is often that we know what $E$ is concretely, and we want to use this to compute $A$. The issue is that if we want to explicitly ...
Dexter Chua's user avatar
13 votes
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Multiplicativity of the homology Atiyah-Hirzebruch spectral sequence for a ring spectrum

You can give a proof of multiplicativity by using that the smash product preserves connectivity. Here is a sketch proof. (EDIT: Denis Nardin pointed me towards this reference by Dugger. This ...
Tyler Lawson's user avatar
13 votes
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Zero differential in Serre spectral sequence for configuration spaces

I'll write $C_n$ for the configuration space, and $X_n$ for $\mathbb{R}^2$ with $n$ points removed. You are presumably thinking about the spectral sequence $$ E_2^{pq} = H^p(C_{n-1};H^q(X_{n-1})) \...
Neil Strickland's user avatar
12 votes
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Are there any cool applications of the generalized Atiyah-Hirzebruch(-Serre) spectral sequence?

A very nice generalized AHSS calculation that deserves to be better known is in Vershinin, V. V. and Gorbunov, V. G. Multiplicative spectra that do not have torsion in homology. (Russian) Mat. ...
Andy Baker's user avatar
12 votes

Grothendieck spectral sequence when one of the functors is contravariant

I think this case is actually not so obvious. The issue is that to derive $R\mathscr Hom$ in the first variable you would need to use a locally free resolution while $Rf_*$ being a covariant right ...
Sándor Kovács's user avatar
12 votes
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Oriented Bordism Group and Un-Oriented Bordism Group of points $pt$

Unoriented cobordism: can be read off from the structure of the unoriented cobordism ring (calculated in Thom's thesis): $\Omega_6^O = (\mathbb Z/2)^3$, $\Omega_7^O = \mathbb Z/2$, $\Omega_8^O = (\...
Arun Debray's user avatar
  • 6,676
12 votes
Accepted

Hodge Numbers and Leray Spectral Sequence

I don't think I defined the Hodge numbers in this way. Rather, the argument in Section 1 shows that the Hodge numbers agree with the dimensions of the terms in the $E_2$ page of the Leray spectral ...
Mark Gross's user avatar
11 votes
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Cohomology ring of a fiberwise join

What you wish to prove is not true. Namely, if $I=\operatorname{ker} f^*$ then $I^2\subseteq \operatorname{ker}(f\ast f)^*$ but the inclusion may be strict. The fibred join construction comes up in ...
Mark Grant's user avatar
  • 34.6k
11 votes

Pullback and homology

Here is a positive answer to a slightly different question. Call a map $X\to B$ "acyclic" if it induces an isomorphism in homology for every coefficient system on $B$. (If $B$ is simply connected ...
Tom Goodwillie's user avatar
11 votes

to compare cohomologies of fibers of two fiber bundles

No. Let $B'$ be any space, and take $E'=PB'$ and $F'=\Omega B$. The Kan-Thurston theorem gives a map $f\colon B\to B'$ such that $H^*(f;\mathbb{Q})$ is an isomorphism but $\Omega B$ is discrete, so $...
Neil Strickland's user avatar
11 votes
Accepted

The second stable homotopy group

I love this question! I've enjoyed thinking of it. Below, I show why the sequence splits always. Let $X\to Y=K(H_1(X,\mathbb{Z}/2),1)$ be the map inducing the identity in $H_1(-,\mathbb{Z}/2)$. By ...
Fernando Muro's user avatar
10 votes

Multiplicative structure on spectral sequence

As far as I know that 1954 paper of Massey is faulty, and you cannot get multiplicative spectral sequences just from such stucture on an exact couple. The best I know that you can do is to use Cartan-...
John Rognes's user avatar
  • 7,974
10 votes

Where does the primary obstruction of a fibration show up in its spectral sequence?

In the general case of integral coefficients and possibly non-trivial local coefficient system, let $\pi=\pi_1(B)$. A cocycle for the obstruction class is an element $o\in Hom_{\mathbb Z\pi}(C_{k+1}\...
user95545's user avatar
  • 451
10 votes

Why does strong convergence of the EMSS imply that Tot commutes with suspension spectrum?

I'm going to avoid the question and answer the edit. Hopkins has some results on this in his Oxford thesis which were announced without proof in his Asterisque paper of 1984. For a quite thorough ...
Rosona's user avatar
  • 101
10 votes
Accepted

Torsion in the integral cohomology of $BPU_{n}$

You may want to have a look at this paper: X. Gu. On the cohomology of classifying spaces of projective unitary groups. arXiv:1612.00506, (link to arXiv) The spectral sequence involving $BSU_n$ ...
Matthias Wendt's user avatar
10 votes

In the not necessarily abelian cat setting, is there a Grothendieck spectral sequence for computing the homotopy of a composition of derived functors?

Not in general, no. The problem is that animated functors play well with colimits, and homotopy groups play better with limits. However, if your functors $\mathcal{A}\xrightarrow{F}\mathcal{B}\...
Achim Krause's user avatar
  • 7,984
9 votes
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Reference request: cohomology of Eilenberg Maclane spaces with $p$-local groups

Here is a sketch proof. Step 1: For sensible spaces or spectra (connected, finite type) $X$, $H^*(X;\tau) $ will have exponent $p$ for all the coefficient groups $\tau$ you list exactly when the ...
Nicholas Kuhn's user avatar
9 votes

Sphere spectrum, Character dual and Anderson dual

One way to think of these spectra is in terms of the cohomology theories they define. In other words, if $E$ is a spectrum, what is $[E, I\mathbb Z]$? This is less of a description of what they are ...
Arun Debray's user avatar
  • 6,676

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