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49 votes
4 answers
7k views

Elegant proof that any closed, oriented 3-manifold is the boundary of some oriented 4-manifold?

I'm looking for an elegant proof that any closed, oriented $3$-manifold $M$ is the boundary of some oriented $4$-manifold $B$.
Kevin Wray's user avatar
  • 1,709
33 votes
1 answer
1k views

Nilpotence of the stable Hopf map via framed cobordism

The Pontryagin-Thom construction shows that the stable homotopy groups of spheres are the same as the groups of stably framed manifolds up to cobordism. Specifically the Hopf map corresponds to the ...
Noah Snyder's user avatar
  • 28.1k
18 votes
1 answer
872 views

Oriented cobordism classes represented by rational homology spheres

Any homology sphere is stably parallelizable, hence nullcobordant. However, rational homology spheres need not be nullcobordant, as the example of the Wu manifold shows, which generates $\text{torsion}...
Jens Reinhold's user avatar
12 votes
0 answers
266 views

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 $Y'$, ...
Adam Levine's user avatar
11 votes
1 answer
388 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 $M^n$...
Sarah's user avatar
  • 113
11 votes
1 answer
454 views

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 $M$ ...
Arun Debray's user avatar
  • 6,881
10 votes
2 answers
757 views

Embedded (framed) cobordisms

[The title initially was "Actions of gauge groups on framed cobordisms. This has been changed.] This question is a follow-up to my answer to When is a submanifold of $\mathbf R^n$ given by global ...
algori's user avatar
  • 23.5k
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 ...
wonderich's user avatar
  • 10.5k
9 votes
1 answer
588 views

Oriented bordism in higher dimensions (e.g. $12 \leq d \leq 28$)

The classification of oriented compact smooth manifolds up to oriented cobordism is one of the landmarks of 20th century topology. The techniques used there form the part of the foundations of ...
wonderich's user avatar
  • 10.5k
9 votes
1 answer
322 views

Torsion and nilpotence of framed manifolds under the Pontryagin-Thom map

The framed Pontryagin Thom construction produces a graded ring isomorphism from the framed cobordism ring $\Omega^{fr}_*$ to the stable ring of homotopy groups of spheres $\pi_*(\mathbb{S})$. I have a ...
João Lobo Fernandes's user avatar
9 votes
1 answer
305 views

Topological Spin manifolds in dimension 4

In his ICM Adress at Nice (Proceedings of the International Congress of Mathematicians Nice, September, 1970, Gauthier-Villars, editeur, Paris 6 e ,1971, Volume 2, pp. 133-163.), Robion Kirby ...
Nicolas Boerger's user avatar
9 votes
0 answers
287 views

Rational cobordism classes of manifolds fibered over spheres

Let us fix positive integers $k, m$. Let $A^k_{4m} \subset \Omega^{\text{SO}}_{4m} \otimes \mathbb Q$ be the subgroup generated by oriented cobordism classes of manifolds fibered over $S^k$. The ...
Jens Reinhold's user avatar
8 votes
1 answer
1k views

Cobordism Theory of Topological Manifolds

Unfortunately, due to my ignorance, my present knowledge is limited to the cobordism Theory of Differentiable Manifolds. Cobordism Theory for DIFF/Differentiable/smooth manifolds However, there are ...
wonderich's user avatar
  • 10.5k
8 votes
1 answer
557 views

Relation between 16 $\mathbf{CP}^2$ and $\overline{K3}$

In bordism theory and algebraic topology, 4d spin bordism group is generated by $K3$ surface, while 4d $SO$ bordism group generated by $\mathbf{CP}^2$. $K3$'s 4-manifold signature is $- 16$ and $\...
zeta's user avatar
  • 447
8 votes
0 answers
219 views

Computation of the 3-dimensional $\mathbb{Z}/m$-equivariant spin cobordism group (with possibly non-empty fixed-point set)?

$\newcommand{\odd}{\mathrm{odd}}\newcommand{\ev}{\mathrm{ev}}$Consider tuples of the form $(Y,\mathfrak{s},\widehat{\sigma})$ where: $Y$ is a closed oriented 3-manifold, $\mathfrak{s}$ is a spin ...
Ian Montague's user avatar
8 votes
0 answers
125 views

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 ...
annie marie cœur's user avatar
8 votes
0 answers
198 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 ...
wonderich's user avatar
  • 10.5k
7 votes
1 answer
260 views

Bordism for oriented triangulable manifolds without smooth differentiable structures

We know the bordism group for oriented smooth differentiable structures such as $\Omega_d^{SO}$ that requires the special orthogonal group structure on the tangent bundle $TM$ of manifold $M$. $$\...
wonderich's user avatar
  • 10.5k
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 ...
wonderich's user avatar
  • 10.5k
7 votes
1 answer
413 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, $$ \mathcal{P}(x)= x \cup x+ x \cup_1 2 Sq^1 x. $$ ...
annie marie cœur's user avatar
6 votes
1 answer
416 views

Do $h$-cobordism groups arise from a 'Thom-like' spectrum?

Not thinking about $h$-cobordism, one usually defines a cobordism between manifolds, realizes it is an equivalence relation, chooses an appropriate class of structured manifolds (framed, unoriented, ...
Matthew Niemiro's user avatar
6 votes
1 answer
373 views

Fivebrane bordism $\Omega_d^{\mathrm{Fivebrane}}$

$\newcommand{\Fr}{\mathrm{Fr}}\newcommand{\Fivebrane}{\mathrm{Fivebrane}}\newcommand{\String}{\mathrm{String}}\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SO{SO}\DeclareMathOperator\GL{GL}$What ...
wonderich's user avatar
  • 10.5k
6 votes
0 answers
209 views

"Inclusion" between higher categories of framed bordisms?

Let $\mathrm{Bord}_n$ be the bordism $(\infty, n)$-category of unoriented manifolds. It can be viewed as an $(\infty, n+1)$-category whose $n+1$-morphisms are equivalences. If $n$ is large enough, ...
Daniel Bruegmann's user avatar
6 votes
0 answers
184 views

Visualize how the 5d Dold manifold and Wu manifold are cobordant via a 6d manifolds with boundaries

Are there simple intuitions and arguments to visualize why the following two 5-manifolds are cobordant to each other with the oriented structures? (They can be two boundaries of 6-dimensional oriented ...
wonderich's user avatar
  • 10.5k
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 $p$-...
wonderich's user avatar
  • 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^...
annie marie cœur's user avatar
6 votes
0 answers
84 views

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, ...
wonderich's user avatar
  • 10.5k
5 votes
1 answer
652 views

Spin bordism group of classifying space $BG$ with a finite Abelian $G$

The spin bordism group for the classifying space $BG$ of group $G$ can be denoted as $\Omega^{Spin}_d(BG)$. For example, $\Omega^{Spin}_d(pt)$ are computed by Anderson-Brown-Peterson (D. W. Anderson, ...
miss-tery's user avatar
  • 755
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 ...
annie marie cœur's user avatar
5 votes
1 answer
192 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 ...
wonderich's user avatar
  • 10.5k
5 votes
1 answer
516 views

Computation of $\Omega^{Pin-}_{d}(B\mathbb{Z}_2)$ and Smith isomorphism

question: I am looking for the literature with the result or the computation of Pin- bordism group: $\Omega^{Pin-}_{d}(B\mathbb{Z}_2)$. Can someone point out some useful ways to do this or any helpful ...
wonderich's user avatar
  • 10.5k
5 votes
0 answers
349 views

Are there alternate descriptions of `elementary cobordisms'?

Let $M^d$, $N^d$ be cobordant $d$-manifolds. Then $M^d \sqcup \bar{N}^d = \partial W^{d+1}$ for some $(d+1)$-manifold $W$. This cobordism can be implemented via an elementary set of 'moves' called ...
Joe's user avatar
  • 545
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^{...
wonderich's user avatar
  • 10.5k
5 votes
0 answers
150 views

Categorification-like statement in the cobordism group?

Suppose we consider a $d$-cobordism group classifying manifolds with $H$-structure and with a classifying space $BG$ of a group $G$, written as $$ \Omega^{d}_{H}(BG)= \mathbb{Z}_m \oplus \dots, $$ ...
wonderich's user avatar
  • 10.5k
5 votes
0 answers
67 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 ...
xir's user avatar
  • 2,044
5 votes
0 answers
155 views

Low dimensional homotopy fibration TOP(M) -> TOP(int(M))

In the thesis of Nancy Cardim she proves that for $M$ a topological manifold of dim $\geq 5$ with connected boundary, there exists a homotopy fiber sequence $C(\partial M)\rightarrow TOP(M) \...
Anonymous's user avatar
4 votes
1 answer
354 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 ...
annie marie cœur's user avatar
4 votes
1 answer
193 views

Version of pseudo-isotopy $\neq$ isotopy for $(n+1)$-framings

Let $M$ be a closed $n$-manifold and $\varphi$ be a self-diffeomorphisms of $M$. There is a bordism from $M$ to itself given by $M\times [0,1]$ with the identification $M \cong M \times \{0\}$ induced ...
Daniel Bruegmann's user avatar
4 votes
1 answer
637 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 ...
wonderich's user avatar
  • 10.5k
4 votes
1 answer
291 views

Lower dimensional Pin cobordisms

I'm studying Pin cobordism groups of a point for some low dimensions. I found a general result by Anderson, Brown, Peterson in Theorem 5.1 of their paper "Pin cobordism and related topics" http://...
Zitao Wang's user avatar
4 votes
1 answer
432 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 $...
Julian Chaidez's user avatar
4 votes
0 answers
243 views

Any cobordism invariant made of "characteristic classes", on unorientable manifolds, must be a mod 2 class?

For any cobordism invariant (or simply bordism invariant) quantity $\omega$ that satisfy the conditions: $\omega$ can be fully decomposed from the cup product of characteristic classes (such as ...
wonderich's user avatar
  • 10.5k
3 votes
1 answer
205 views

Prove or disprove that there exists no $G$ structure with its bordism group $\Omega_1^{G} =\mathbb{Z}/N$ for $N>2$

It can be found that there are the following bordism group $\Omega_0^{G}$ at $d=0$ and 1 dimensions by requiring $G$ structure for $d$-manifolds: $$ \Omega_0^{SO} = \mathbb{Z} , \quad \Omega_1^{SO} = ...
annie marie cœur's user avatar
3 votes
1 answer
139 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(...
annie marie cœur's user avatar
3 votes
1 answer
135 views

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 $BG=BSU(2)$: ...
annie marie cœur's user avatar
3 votes
0 answers
122 views

Is there a framed nullbordism of $T^4$ with an action of $T^4$ that extends the self-action?

Under the identification of the stable homotopy groups with the (stably) framed bordism groups, it is well known that $\eta\in\pi_1\mathbb{S}$ is represented by $S^1$ with its Lie group framing. ...
kiran's user avatar
  • 2,052
3 votes
0 answers
194 views

The first Stiefel Whitney class v.s. fermion eta invariant v.s. spin structure v.s. $H^1(M,\mathbb{Z}_2)$

$H^1(M,\mathbb{Z}_2)$ specifies the 1st cohomology class of manifold $M$ (can be regarded as spacetime) with $\mathbb{Z}_2$ coefficient, it is often to see that we say the 1st Stiefel Whitney class $$...
annie marie cœur's user avatar
3 votes
0 answers
170 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 \;...
wonderich's user avatar
  • 10.5k
2 votes
1 answer
165 views

string bordism group and framed bordism group for $d \leq 6$ and $d \geq 7$

Why do the string bordism group and the framed bordism group coincide the same in dimensions lower than 7 ($d = 0,1,2,3,4,5, 6$)? Why do the string bordism group and the framed bordism group differ ...
wonderich's user avatar
  • 10.5k
2 votes
1 answer
142 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}...
annie marie cœur's user avatar