All Questions
39 questions
7
votes
1
answer
567
views
Long exact sequences for parametrized cohomology
I'm reading Michael Shulman's articles on cohomology in HoTT here and here, as well as Floris van Doorn's thesis here.
Given $E: Z \to \mathsf{Spectrum}$ a family of spectra over a homotopy type $Z$, ...
4
votes
1
answer
218
views
Poincaré dual of the Alexander dual of the fundamental class of a knot is given by a Seifert surface
Let $K\subset S^3$ be an oriented knot and let $F:\overline{B^2}\times K\rightarrow S^3$ be a thickening with self linking number $0$. I will denote $F(B^2\times K)$ by $(B^2\times K)$ for simplicity. ...
6
votes
1
answer
327
views
Steenrod squares in terms of chain maps
$\DeclareMathOperator\Sq{Sq}$The Steenrod squares $\Sq^i: H^n({-};\mathbb{F}_2) \to
H^{n+i}({-};\mathbb{F}_2)$ are fundamental cohomological
operations. By the Yoneda lemma, they induce a map between ...
10
votes
1
answer
418
views
Are all degree-1 cohomology operations Bocksteins?
I'm interested in cohomology operations (in ordinary cohomology)
$$H^i(-, G)\rightarrow H^{i+1}(-, H)\;,$$
that is, elements of
$$H^{i+1}(K(G, i), H)\;.$$
I know that $K(G, 1)=BG$, so for $i=1$, those ...
4
votes
0
answers
188
views
Multi-variable cohomology operations
Intuitively, cohomology operations are ways to locally compute a cocycle $\alpha\in H^i(X, G)$ from any cocycle $\beta\in H^j(X, H)$. Formally, they are in one-to-one correspondence with homotopy ...
9
votes
2
answers
1k
views
Hodge dual of de Rham cohomology and singular cohomology
We know that the de Rham cohomology is isomorphic to the singular cohomology, does the Hodge dual of differential forms induce a dual operation on de Rham cohomology, hence also on singular cohomology?...
13
votes
0
answers
864
views
A step in Toda's computation of a Cotor
I am trying to understand a proof from Toda's paper Cohomology of classifying spaces. The step I am stuck on is at page 96. Here is the setup.
We work with cohomology with $\mathbb{F}_2$ coefficients. ...
4
votes
0
answers
170
views
Relation between $Tor_{C^*(BG;K)}(K,K) $ and $K^G$?
Let $K$ be an algebraically closed field and $G$ a group.
Given a dg-algebra $A$, a left $A$-module $M$ and a right $A$-module $N$
let $Tor_A(M,N)$ denote the homology of the derived tensor product $M ...
7
votes
0
answers
270
views
Differentials in spectral sequences and Massey products
Consider a multiplicative spectral sequence such as the cohomological Serre spectral. It is known that differentials will satisfy a Leibniz rule. Is there a clean statement involving differentials and ...
5
votes
1
answer
416
views
triviality of homology with local coefficients
Let $X$ be a manifold or a CW-complex.
Let
$\pi: \tilde X\longrightarrow X$
be a covering map.
Let $\pi_1(X)$ be the fundamental group of $X$ and let $\rho: \pi_1(X)\longrightarrow O(n)$ be an ...
2
votes
1
answer
275
views
can the actions of fundamental groups annihilate homology?
Let $X$ be a path-connected manifold (or a CW complex).
Let $\pi_1(X)$ be the fundamental group of $X$.
Let $\pi: \tilde X\longrightarrow X$ be a covering map.
For each $m\geq 0$, let $C_m(\tilde X)$ ...
4
votes
1
answer
287
views
Conjugation action on relative homology
Let $G$ be a group and $K$ be a subgroup. Suppose $g \in G$ commutes with every element of $K$. Is it true that conjugation by $g$ will act trivially on $H_*(G,K)$?
4
votes
1
answer
482
views
Homology with local systems
Let $X$ be a connected topological space with abelian fundamental group. Let $\mathcal{L}$ be a $\mathbb{Z}$-valued local system on $X$.
Suppose that I know the full homology $H_*(X;\mathbb{Z})$. Are ...
2
votes
0
answers
53
views
Eilenberg–Zilber-type theorem for Map([n],A), where the degeneracy maps for [n] are forgotten
The following statement should be immediately implied by Eilenberg–Zilber theorem if the sequences $(i_0,\ldots,i_k)$ below are only monotone. But I need the strict monotone version which I believe to ...
5
votes
1
answer
185
views
Decompose $MT(E(d)\times_{\mathbb Z_2} SU(2))$ as the wedge sum or smash product of spectra
Consider the extension
$$1\to SU(2)\to X\to O\to1,$$
there are 4 possibilities for $X$:
$X=O\times SU(2)$ or $E\times_{\mathbb{Z}_2}SU(2)$ or $Pin^+\times_{\mathbb{Z}_2}SU(2)$ or $Pin^-\times_{\...
4
votes
0
answers
396
views
Eilenberg-Moore spectral Sequence calculation
I want to use the cohomology Eilenberg-Moore spectral sequence to calculate the cohomology of the fibre of the map
$$
S^{n} \to \Omega S^{n+1}.
$$
Question 1: Is anyone aware of any references for ...
9
votes
1
answer
993
views
Use of Steenrod's higher cup product and the graded-commutativity
In Steenrod's ``Products of Cocycles and Extensions of Mappings (1947),'' which derives [Theorem 5.1]
$$
\delta(u\cup_{i} v)=(-1)^{p+q-i}u\cup_{i-1}v+(-1)^{pq+p+q}v\cup_{i-1}u+\delta u\cup_{i}v+(-...
-2
votes
1
answer
89
views
Alternating property of H_2(T, Z)
Let us consider the torus $T = S^1_X \times S^1_Y$, where the former $S^1_X$ has the coordinate $X$, whereas the latter $S^1_Y$ has $Y$. We have an alternate property $dX \wedge dY = - dY \wedge dX \...
-1
votes
1
answer
163
views
Alternate property of H^2(T, Z) [closed]
Let us take $T = S^1_X \times S^1_Y$, which is a torus, where the former $S^1_X$ has the coordinate $X$, whereas the latter $S^1_Y$ has $Y$. If we consider the generator $dX \wedge dY \in H^2(T, {\Bbb ...
6
votes
0
answers
237
views
A Poincare dual spectral sequence for equivariant cohomology: extension to generalized cohomology?
This question is a follow-up to my previous question:
"Rotated" version of the Atiyah-Hirzebruch spectral sequence
In that question, I discussed two different spectral sequences for ...
11
votes
1
answer
553
views
Is there a kind of Poincare duality for Borel equivariant cohomology?
Let $G$ be a finite (or discrete) group, $M$ a $d$-dimensional manifold with smooth $G$-action (I am interested in the case where the action is not free, so $M/G$ is not a manifold). For an Abelian ...
3
votes
1
answer
266
views
Poincaré dual of the generators of $H^d(\mathbb{RP}^5,\mathbb{Z}_2)$
We know $H^d(\mathbb{RP}^5,\mathbb{Z}_2)=\mathbb{Z}_2$. So there are two classes of $\mathbb{Z}_2$ generators, trivial and nontrivial, for $d=0,1,2,3,4,5$.
Wha are the Poincaré dual $(5-d)$-...
16
votes
1
answer
808
views
"Rotated" version of the Atiyah-Hirzebruch spectral sequence
Let $G$ be a group, $X$ a topological space with $G$-action. For an Abelian group $A$, let $\mathcal{C}^n(X,A)$ be the group of $n$-cochains on $X$ with $A$ coefficients. We can treat this as a $G$-...
93
votes
3
answers
11k
views
What is homology anyway?
Disclaimer: I don't feel qualified to ask this question and yet it's been troubling me for some time now and I lost my patience and decided to ask to get some kind of answer. If there are any stupid ...
7
votes
0
answers
192
views
mod $p$ homology module of unordered configuration spaces of the projective plane
Let $M$ be a manifold and $k$ be a positive integer. Let $F(M,k)$ be the $k$-th ordered configuration space over $M$, consisting of all ordered $k$-tuples of distinct points in $M$. Let $\Sigma_k$ be ...
9
votes
1
answer
804
views
Known results in the Cohomology of finite groups
I am learning to compute cohomology of finite groups and came across this survey article http://www.ams.org/notices/199707/adem.pdf "Recent Developments in the cohomology of finite groups" by ...
21
votes
1
answer
2k
views
A spectral sequence for computing cohomology of a space from that of its strata
Let $X$ be a smooth complex variety (not necessarily compact) and let $D$ be a normal crossings divisors with components $D_1$, $D_2$, ..., $D_N$. For a set of indices $I$, let $D_I = \bigcap_{i \in I}...
3
votes
0
answers
104
views
Are there necessary and sufficient conditions for a chain complex $0 \to C_0 \to C_1 \to C_2 \to 0$ to be Poincare?
I am looking for necessary and/or sufficient conditions for the chain complex $0 \to C_0 \to C_1 \to C_2 \to 0$ over a principal ideal domain to be Poincare in the sense that $H_0 \cong H^2$, $H_1 \...
4
votes
0
answers
222
views
References for bilinear forms on chain complexes?
I am looking for references that include general results and theorems for bilinear forms defined on chain complexes. That is, bilinear forms $\langle \cdot , \cdot \rangle_i : C_i \times C_i \to \...
7
votes
1
answer
2k
views
generalized universal coefficient sequence
Take the familiar Universal Coefficient Theorem for ordinary homology with $\mathbb{Z}$-coefficients and ordinary cohomology with coefficients in some abelian group $A$:$$0\rightarrow \text{Ext}_\...
8
votes
3
answers
1k
views
Homological vs. cohomological dimension of a group/space
I have several related questions regarding homological vs. cohomological dimension of a space/group (this is not a duplicate of this).
The standard definition of the cohomological dimension $cd(X)$ ...
5
votes
1
answer
1k
views
LES for relative cohomology via sheaves
I was unable to find a suitable answer for the following question:
Once one learns that singular cohomology is the same as cohomology with coefficients in locally constant sheaf, it is natural to try ...
6
votes
1
answer
2k
views
cohomology version of Cartan-Leray spectral sequence that deduces cup product
On page 338, A User's Guide to Spectral Sequences. 2nd Edition, by John McCleary, Theorem 8.9, there is a Cartan-Leray spectral sequence for homology:
If $X$ is a connected pace on which the group $\...
12
votes
1
answer
2k
views
Cohomology ring of classifying space of spin group $\mathrm{BSpin}(n)$
$\DeclareMathOperator\BSpin{BSpin}\DeclareMathOperator\Pin{Pin}\DeclareMathOperator\BSO{BSO}\DeclareMathOperator\BO{BO}\DeclareMathOperator\BPin{BPin}$In the answer for question: Homology of ...
0
votes
1
answer
425
views
A generalization of cochain complex: quasi-cochain complex
It appears that we can generalize cochain complex to quasi-cochain complex, that still allow us to define cohomology.
Definition:
A quasi-cochain complex is a sequence of commutative monoids $M_n$ ...
25
votes
4
answers
6k
views
Singular Homology/Cohomology as a derived functor?
Hello,
Learning some Alg.geometry and Sheaf theory, I got used to the notion that cohomology arises naturally as a derived functor of some sort.
This has led me thinking, singular cohomology, from ...
8
votes
2
answers
2k
views
Splitting of the Universal Coefficients sequence
The really beautiful way to prove the Universal Coefficients theorem, to my taste,
is to use the fibration sequence $K(\mathbb{Z}, n) \to K(\mathbb{Z}, n) \to
K(\mathbb{Z}/k, n)$ (I'm using $\mathbb{...
34
votes
2
answers
5k
views
Example Wanted: When Does Čech Cohomology Fail to be the same as Derived Functor Cohomology?
I want to know exactly how derived functor cohomology and Cech cohomology can fail to be the same.
I started worrying about this from Dinakar Muthiah's answer to an MO question, and Brian Conrad's ...
20
votes
5
answers
2k
views
Equivalence of ordered and unordered cech cohomology.
Given a topological space X and a finite cover X = $\cup X_i$, one can define Cech cohomology of a sheaf of abelian groups F with respect to the cover $\{X_i\}$ in two different ways:
(Ordered): ...