All Questions
23 questions
19
votes
5
answers
2k
views
References for Eilenberg-Zilber shuffle product
Most of the treatments I can find in the literature for the Eilenberg-Zilber shuffle product approach it from the point of view of simplicial sets (including the original Eilenberg-MacLane paper). I ...
- 5,509
10
votes
1
answer
637
views
Cell structures of Dold manifold and Wu manifold
In Dold's 1956 paper Erzeugende der Thomschen Algebra N, Dold studied the Dold manifold $P(m,n)=(S^m\times\mathbb{CP}^n)/\tau$ where $\tau$ acts as $-1$ on $S^m$ and a complex conjugation on $\mathbb{...
- 1,329
9
votes
2
answers
641
views
Künneth formulas/theorem for bordism groups and cobordisms?
We are familiar with Künneth theorem:
The Kunneth formula is given by $R$ as a ring, $M,M'$ as the R-modules, $X,X'$ are some chain complex. The Kunneth formula shows the cohomology of a chain ...
- 10.5k
9
votes
1
answer
777
views
Intuition for torsion of a chain complex and application to lens spaces
I have read a bit about the torsion of an acyclic complex. One of my concrete hopes was that I could understand why $L(7,1)$ and $L(7,2)$ are not homeomorphic - I am under the impression that ...
- 5,349
9
votes
1
answer
1k
views
How does the Lefschetz-Poincare dual torsion linking pairing on manifolds with boundary interact with the maps of the long exact sequence of the manifold-boundary pair?
I'm wondering if anyone can point me to a reference on how the various
Lefschetz-Poincare dual torsion pairings of a manifold with boundary fit
together.
To explain in more detail, consider a ...
- 5,509
7
votes
3
answers
314
views
Eilenberg-Zilber-type theorem for good fiber products?
My question is:
If $p\colon X \to B$, $q\colon Y \to B$ are proper submersions, is there a characterization of $H_*(X \times_B Y)$ in terms of $H_*(X)$, $H_*(Y)$, $H_*(B)$ that is simpler than the ...
- 1,589
7
votes
4
answers
685
views
Realizing complexes with bases as cellular complexes
This is a question a friend of mine asked me some time ago. I suspect the answer is "no" but can't prove it.
Every free complex of abelian groups is isomorphic to the reduced cellular complex of some ...
- 23.5k
7
votes
1
answer
470
views
Twisted spin bordism invariants in 5 dimensions
[Note]: My question will be a bit long. So, first, thank you for your careful reading, generous comments, helps and answers, in advance!
The spin $G$-bordism invariant can be twisted in the way that ...
- 10.5k
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 ...
- 2,223
6
votes
1
answer
974
views
Constructible sheaves and dg-modules
Let $M$ be a smooth manifold, $A_M$ the de Rham algebra of $M$, $D_{A_M}$ the derived category of the category of differential graded (dg) $A_M$-modules and $D^+_c(M)$ the bounded below constructible ...
- 23.5k
6
votes
0
answers
142
views
Pin cobordism v.s. "KO" theory in low or in any dimensions
Fact: The spin cobordism is equivalent to "KO" theory in low dimension if we only consider the 2-torsion.
This is related to a question and an answer supports the claim.
Here we denote the $p$-...
- 10.5k
6
votes
0
answers
224
views
Arf-Brown-Kervaire invariant and a surjective map $G \to Pin^-$
We know that the Arf-Brown-Kervaire (abk) invariant is a bordism invariant of
$$
\Omega_2^{Pin^-}(pt)=\mathbb{Z}/(8\mathbb{Z}),
$$
where the $\mathbb{Z}/(8\mathbb{Z})$ is generated by a 2-manifold $M^...
- 2,147
5
votes
1
answer
278
views
Manifold generators of O-bordism invariants
If I understand correctly, I can obtain the $O$-cobordism group of
$$
\Omega^{O}_3(BO(3))=(\mathbb{Z}/2\mathbb{Z})^4,
$$
The 3d cobordism invariants have 4 generators of mod 2 classes, are generated ...
- 2,147
5
votes
0
answers
225
views
Generalizing the formula between Wu class and the Steenrod square
I know that on the tangent bundle of $M^d$, the corresponding Wu class and the Steenrod square satisfy
$$
Sq^{d-j}(x_j)=u_{d-j} x_j, \text{ for any } x_j \in H^j(M^d;\mathbb Z_2) .
\tag{eq.1}$$
...
- 10.5k
5
votes
0
answers
161
views
Twisted spin-bordism invariant and a possible Postnikov square from $d=2$ to $d=5$
This is a follow up more advanced question following Twisted spin bordism invariants in 5 dimensions. We follow the definitions in the earlier post.
I had discussed my computation of
$$
\Omega_5^{...
- 10.5k
4
votes
0
answers
158
views
Postnikov square explicitly on a simplicial complex
$\DeclareMathOperator\Z{\mathbb{Z}}$
Following Wikipedia, a Postnikov square is a certain cohomology operation from a first cohomology group $H^1$ to a third cohomology group $H^3$, introduced by ...
- 10.5k
4
votes
0
answers
181
views
Borromean Lines of three $\mathbb{R}^1$ in $\mathbb{R}^3$ and analogous Milnor link invariants
It is know that Borromean rings can be detected by Milnor invariant
$$
\bar{\mu}(\gamma_1,\gamma_2,\gamma_3)=
\# (\Sigma_1 \cap \Sigma_2 \cap \Sigma_3)-\frac{1}{2}\sum_{I,J,K}\epsilon_{IJK}
\sum_{\...
- 2,147
4
votes
0
answers
290
views
Generalized Postnikov square
Following Wikipedia (https://en.wikipedia.org/wiki/Postnikov_square), a Postnikov square is a certain cohomology operation from a first cohomology group $H^1$ to a third cohomology group $H^3$, ...
- 1,329
4
votes
0
answers
241
views
Non-spin 5-manifold and $2^2$-Bockstein homomorphism
The $2^2$-Bockstein is $\beta_4$ is associated to
$$0\to\mathbb{Z}/2\to\mathbb{Z}/{8}\to\mathbb{Z}/{4}\to 0,$$
(The $2^n$-Bockstein homomorphism
$$\beta_{2^n}:H^*(-,\mathbb{Z}/{2^n})\to H^{*+1}(-,\...
- 10.5k
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 $(5-d)$-...
- 10.5k
3
votes
1
answer
224
views
Homology modules and symmetry
Let $B$ be a cellular (simplicial, semi-simplicial etc) complex having $\mathbb{Z}^n$-symmetry (that is, there is a free action of $\mathbb{Z}^n$ on $B$, commuting with the boundary operators) and let ...
- 93
0
votes
1
answer
284
views
Künneth formula and induced map in homologies
Let $X,Y,Z$ be smooth connected manifolds and $f \colon X \times Y \rightarrow Z$ a smooth map. Suppose that we have $H_{*}(X \times Y; \mathbb{Z})$ is isomorphic to $\bigoplus_{p+q=*}(H_{p}(X; \...
- 369
0
votes
0
answers
260
views
Another definition of singular homology
The singular homology is defined via standard simplex. Now if I propose another definition of singular homology groups, based on arbitrary simplex, as follows:
Let $X$ be a topological space. A $n$-...
- 781