Questions tagged [cobordism]
Cobordism is a fundamental equivalence relation on the class of compact manifolds of the same dimension, set up using the concept of the boundary of a manifold.
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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^...
6
votes
0
answers
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Triple data for Pontrjagin dual of the Spin bordism group
It is said that the Pontrjagin dual of the 3-dimensional Spin bordism group of $BG$ for $G$ a finite group,
$$
\text{Hom}(Ω^{spin}_3(BG),\mathbb{R/Z}),
$$
can be expressed by triples of cochains $$(w, ...
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votes
1
answer
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The structure of complex cobordism cohomology of the Eilenberg-Maclane spectrum
Let $MU$ be the complex bordism spectrum and let $H\mathbb{Z}$ be the Eilenberg-Maclane spectrum.
Is it know what the structure of the complex cobordism cohomology $MU^{*}(H\mathbb{Z})$ is?
EDIT: ...
5
votes
0
answers
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views
Linear Independence & Integral Homology Cobordism Group
The set of integral homology spheres up to integral homology cobordism forms
an abelian group with the operation induced by the connected sum. This group is called integral homology cobordism group ...
8
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answers
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Relating bordism generators in d and d+2 dimensions --- an explicit example
This is an attempt to make my relation between bordism invariants in $d$
and $d+2$ dimensions, following a previous attempt more explicit. This counts as a different question, since some more specific ...
2
votes
1
answer
141
views
Extending the maps between the bordism groups: NONE exsistence of a certain kind of extended group
Let $M^d$ be a nontrivial bordism generator for the bordism group
$$
\Omega_d^G= \mathbb{Z}_n,
$$
suppose $G$ (like O, SO, Spin, etc) specify the group structure of the boridsm group. The $\mathbb{Z}...
8
votes
0
answers
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views
"Gerbes" in the cobordism theory
In a lecture I attended today, I heard the use of gerbes in the cobordism theory.
Previously, I use cobordism theory, but I never encounter the term "gerbes" when I work on bordism or cobordism group ...
4
votes
0
answers
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views
Relating bordism invairants in $d$ and $d+2$ dimensions
Are there some relationship between mapping the bordism invairants of eq.1 and eq.2?
$$\Omega_{O}^{d}(B(PSU(2^n)\rtimes\mathbb{Z}_2)) \tag{eq.1}$$
and
$$\Omega_{O}^{d+2}(K(\mathbb{Z}/{2^n},2)) \...
5
votes
1
answer
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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 ...
9
votes
2
answers
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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 ...
5
votes
1
answer
189
views
Relating bordism groups of different dimensions
Let
$M_d$
be a $d$-manifold generator of a subgroup of bordism group
$$
\Omega_d^{G},
$$
or further generalization
$$
\Omega_d^{G}(K(\mathcal{G},n+1)),
$$
which $G$ is the given structure ...
3
votes
1
answer
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views
Bordism invariants vanishes in a lifted twisted $Pin^- \times Spin$-structure
It looks to me that the bordism group
$$\Omega_3^{SO} (B(O(2) \times SO(3))) \tag{1}$$
(whose Pontryagin dual for the manifold generator) contains at least a nontrivial invariant:
$$
w_1(O(2))\big(...
4
votes
1
answer
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views
(Co)bordism invariant of Eilenberg–MacLane space becomes vanished
Consider a (co)bordism invariant
$$
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$$
obtained from
$$
\Omega^5_{O}(K(\mathbb{Z}/2,2)).
$$
Here $u \in H^2(K(\mathbb{Z}/2,2),\mathbb{Z}_2)$. The $K(\mathbb{Z}/2,2)$ is ...
4
votes
1
answer
381
views
null-bordant vs null-homologous sub-manifolds of $\infty$-d spaces/CW complexes
$\require{AMScd}$
Preliminaries: Let $\Sigma$ be a closed manifold, $X$ be a CW complex and $f:\Sigma \to X$ be a map. We say that the pair $(\Sigma,f)$ is null-homologous (over $\mathbb{Z}_2$) if $...
5
votes
0
answers
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views
Does the theorem that genera vanishing on even-dim complex projective bundles are elliptic also apply for integral-valued genera?
Ochanine proved in this paper that for genera taking values in $\mathbb{Q}$-algebras, vanishing on even-dimensional projective bundles is equivalent to being an elliptic genus (i.e. a specialization ...
7
votes
1
answer
382
views
Pontryagin square and $\frac{1}{2}(\mathcal{P}(x) -x^2) =x \cup_1 Sq^1 x$
The Pontryagin square, maps $x \in H^2({B}^2\mathbb{Z}_2,\mathbb{Z}_2)$ to $ \mathcal{P}(x) \in H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4)$. Precisely,
$$
\mathcal{P}(x)= x \cup x+ x \cup_1 2 Sq^1 x.
$$
...
4
votes
1
answer
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views
Sphere spectrum, Thom spectrum, and Madsen-Tillmann bordism spectrum
This is a following up question of Sphere spectrum, Character dual and Anderson dual.
What are the differences and the significances of the following:
(1). Homotopy classes of maps from a Thom ...
10
votes
2
answers
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views
Sphere spectrum, Character dual and Anderson dual
The homotopy groups of the sphere spectrum are the stable homotopy groups of spheres.
However, could you help me to appreciate the mathematical meanings of the following:
What is the significance of ...
3
votes
0
answers
119
views
Twisted spin cobordism v.s. KO theory in low dimensions
Based on the background info and this this webpage, here is a more advanced problem:
Question: If we consider a different more subtle twisted structure, like
$${\Omega_d^{(\mathrm{spin} \times G)/N}},...
8
votes
1
answer
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views
Spin cobordism v.s. KO theory in low or in any dimensions
It seems that from this webpage, the spin cobordism is equivalent to KO theory in low dimension.
If we denote the $p$-torsion part (mean $\mathbb{Z}_{p^n}$ for some $n$) $$\Omega_d(BG)_p.$$
...
9
votes
0
answers
129
views
Relating bordism groups of $\Omega_{d}^{Spin_c}$ and $\Omega_{d}^{(Spin \times SU(N))/\mathbb{Z}_2}$ to that of $U(N)$
I felt that the earlier question may be too challenging, so let me provide a different angle and more infos to tackle an easier and separate problem.
Let us consider a more explicit a short exact ...
6
votes
0
answers
122
views
Bordism groups and a short exact sequence
Let us consider a short exact sequence:
$$
1\to N\to G\to Q \to 1,
$$
where $N$, $Q$, and $G$ can be continuous Lie groups in general (or finite groups).
Suppose I have the data and the computations ...
3
votes
0
answers
169
views
Pairing the Arf with Stiefel-Whitney class
The Arf invariant is a nonsingular quadratic form over a field of characteristic 2.
The form that I looked at was:
$$
S(q)=|H^1(M^2,\mathbb{Z}_2)|^{-1/2} \sum_{x\in H^1(M^2,\mathbb{Z}_2)} \exp[\pi \;...
3
votes
0
answers
158
views
Cobordant of 5d manifolds, and the generalization of bordisms
Some of the 5-dimensional manifolds are (co)bordant via oriented cobordism.
For example, if I understand correctly, 5-dimensional Dold manifold and Wu manifold are manifolds which are cobordant to ...
6
votes
1
answer
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views
Stable normal framings of parallelizable manifolds
Suppose $M$ is a compact, connected, orientable manifold ($\dim M=m$) with trivial tangent bundle and let $j \colon M \to \mathbb R^n$ be an embedding. Suppose we choose a trivialization of $TM$. Then ...
3
votes
0
answers
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views
Reference Request: Equivariant Symplectic bordism
Non-equivariantly, symplectic bordism has been developed extensively by Ray, Gorbunov, and specially S. Kochman in this memoir: http://dx.doi.org/10.1090/memo/0496 Yet the coefficients ...
4
votes
1
answer
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views
Thom space, homotopy group and cohomology group
In Thom's 1952 paper, Thom showed that the Thom class, the Stiefel–Whitney classes, and the Steenrod operations were all related. He used these ideas to prove in the 1954 paper Quelques propriétés ...
10
votes
2
answers
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Is the boundary of a manifold topologically unique? [duplicate]
Let $X$ be a manifold without boundary and let $Y$ and $Z$ be two manifolds with boundary such that $X$ is homeomorphic to their interiors: $X \cong Y^\circ \cong Z^\circ$. Does it follow that $Y \...
5
votes
1
answer
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2-morphisms for Bord(n)
I am currently reading in Boundary Conditions for Topological Quantum Field Theories, Anomalies and Projective Modular Functors, and have a (I guess) pretty basic question for my understanding of the (...
14
votes
1
answer
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views
Is there a PL, or topological, bordism hypothesis?
The bordism hypothesis says that the $(\infty, n)$-category of smooth, framed $n$-bordisms, $(n-1)$-dimensional boundaries, and corners down to points, is freely generated symmetric monoidal with ...
2
votes
0
answers
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views
Does this condition imply symplectic birational cobordism?
From the definition of symplectic birationality given here (https://arxiv.org/pdf/0906.3265.pdf, Definition 2.1), two compact symplectic $2n$-manifolds $(M_{1},\omega_{1}),(M_{2},\omega_{2})$ are ...
11
votes
1
answer
381
views
Twisting bordism classes
Let $X$ be a reasonable topological space (I'd be happy to assume that $X$ is a smooth closed manifold) and let $f\colon M^n \rightarrow X$ be a continuous map from a smooth oriented $n$-manifold $M^n$...
5
votes
1
answer
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Relative Steenrod's problem
Thom's theorem states that for every homology class $\alpha \in H_{*}(X)$ there exists an integer $k = k(\alpha)$ such that the class $k\, \alpha$ comes from the fundamental class of an orientable ...
12
votes
0
answers
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Simply connected homology cobordisms
I'm looking for interesting examples of a homology 3-sphere $Y$ for which there exists a smooth, simply connected homology cobordism from $Y$ to itself (or simply to another homology 3-sphere $Y'$, ...
4
votes
0
answers
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views
Can the cobordism hypothesis be formulated as a statement about adjoint functors?
I would like to formulate the cobordism hypothesis for general tangential structure as a statement about adjoint $(\infty,1)$-functors.
For a space $Y$ with an action of $O(n)$ let $X=Y\times_{O(n)} ...
3
votes
0
answers
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Landweber Exact Functor Theorem for Cohomology
I have seen the Landweber exact functor theorem beeing used to retrieve cohomology theories, in particular singular cohomology and K-Theory.
However the statement of the theorem itself is always ...
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Is there a closed 5-manifold $M$ with $w_1(M)w_2(M)\ne 0$?
I'm trying to find generating manifolds for the cobordism group $\mathit{MO}_5(K(\mathbb Z/2, 2))\cong (\mathbb Z/2)^4$, which can be represented as the cobordism group of closed 5-manifolds $M$ ...
2
votes
1
answer
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Oriented Bordism Group and Un-Oriented Bordism Group of points $pt$
Do we know, or are there any References that list down complete oriented and unoriented Bordism Group $Ω_{n,O}(pt)$ and $Ω_{n,SO}(pt)$ of points $pt$ for dimensions $n=1,2,...,10$?
Here are some ...
7
votes
1
answer
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views
What is the growth rate of the number of unoriented cobordism classes?
Let $\Omega_n^O$ denote the abelian group of cobordism classes of closed, unoriented manifolds of dimension $n$, and let $d(n) := \lvert\Omega_n^O\rvert$. What are the asymptotics of $d(n)$?
It's ...
5
votes
1
answer
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views
Manifold bounded by Sp(2) realisable inside End(H^2)?
The Lie group $Sp(2) = \{A\in GL(2,\mathbb{H})\mid A^\dagger A = I \}$ has a variety of nice geometric aspects. One of which is that it is the boundary of the disk bundle $D(V)$ of the rank-1 ...
6
votes
1
answer
344
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More on categories of modules over the algebraic cobordism spectrum
I have the following questions on monoidal model structure(s) for the motivic stable homotopy category $SH(k)$ (where $k$ is a field); certainly, I am also interested in general statements concerning ...
16
votes
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answer
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Explicit cobordism between Wu manifold and Dold manifold P(1,2)?
The Wu manifold $SU(3)/SO(3)$ and the Dold manifold $P(1,2)$, the latter being defined as $(S^1\times \mathbb{C}P^2) / (p, x) \sim (-p, \overline{x})$, are cobordant because they are both generators ...
3
votes
1
answer
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Submanifold generator of Pontryagin-dual of the torsion subgroup of the $Pin^+$ bordism group (dimension 4th)
I am interested in figuring out the following submanifold generator of a $Pin^+$ Bordism or Cobordism group in the dimension 4, say for a $Pin^+$ cobordism group of the classifying space $BG=BSU(2)$:
...
8
votes
1
answer
1k
views
Cobordisms and Euler characteristics
I am trying to understand exactly which role the Euler characteristic plays in (smooth) cobordism theory, and especially why the answer seems to depend on the dimensions of the manifolds in question. ...
2
votes
1
answer
179
views
Cobordism/bordism group based on orbifolds with corners
We define a geometric homology group of a topological space $X$ as follows: the chain complex $C_{\bullet}$is freely generated by the maps $f$ from a compact oriented orbifold with corners $P$ to $X$, ...
7
votes
3
answers
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views
Is the bordism from disjoint union to connected sum universal for connected manifolds?
Let $M_1$ and $M_2$ be two oriented, connected, closed $n$-manifolds. It is known that the disjoint union $M_1 \sqcup M_2$ and the connected sum $M_1 \# M_2$ are cobordant, via a bordism $\Sigma_{M_1, ...
21
votes
2
answers
824
views
Does Spin cobordism vanish in dimension $4k-1$?
For the purposes of a remark in a paper in preparation, I would like to know if anyone can confirm that $\Omega^{spin}_{4k-1} = 0$.
In the Atiyah-Patodi-Singer paper, Spectral asymmetry and ...
9
votes
0
answers
249
views
Building examples of elements of $\Omega_4(\xi)$ via surgery theory: how to do it?
When computing special bordism groups, I often need to determine existence of (singular) smooth $4$-manifolds with fixed fundamental group and certain properties like the spin behaviour (i.e. being ...
4
votes
1
answer
298
views
Is equivariant oriented cobordism finite?
It is known that for $n \not\equiv 0 \mod 4$, the oriented cobordism ring $MSO_n$ is finite. That is, for oriented n-dimensional manifold $Y$, there exists $m\in \mathbb{N}$, such that $mY$ bounds.
...
7
votes
2
answers
657
views
Variants and Generalizations of Arf (-Brown-Kervaire) invariants
(1) I encounter the Arf invariants in Kirby-Taylor, Pin structures on low-dimensional manifolds. The form that I looked at was:
$$
S(q)=|H^1(M^2,\mathbb{Z}_2)|^{-1/2} \sum_{x\in H^1(M^2,\mathbb{Z}_2)} ...