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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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
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14 votes
3 answers
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Does every $SL_2\mathbb{C}$ representation of a closed oriented surface extend over a compact oriented three-manifold?

Let $F$ be a compact oriented surface and $\rho:\pi_1(F)\rightarrow SL_2\mathbb{C}$ be a representation. Does there exist a compact oriented three-manifold $M$ with $\partial M=F$ and a homomorphism $...
Charlie Frohman's user avatar
49 votes
4 answers
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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
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38 votes
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Explanation for the Thom-Pontryagin construction (and its generalisations)

In 1950, Pontryagin showed that the n-th framed cobordism group of smooth manifolds was equal to n-th stable homotopy group of spheres: $$ \lim_{k \to \infty} \pi_{n+k}(S^k) \cong \Omega_n^{\text{...
Sam Derbyshire's user avatar
8 votes
1 answer
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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 $n$) $$\Omega_d(BG)_p.$$ ...
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18 votes
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Homology theory represented by Madsen-Tillmann spectra

The generalized homology theory of the Thom spectrum $MO=\varinjlim\Sigma^nMTO_n$ is bordism theory:\begin{equation*}\pi_k(MO\wedge X)=\Omega^O_k(X)\end{equation*}These groups form the ring of (...
Alex Turzillo's user avatar
16 votes
1 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 ...
Taisong Jing's user avatar
13 votes
1 answer
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Why Lagrangian cobordism?

There are a good number of quantum topology papers in which a TQFT-like set-up is constructed as a functor to the category of vector spaces from some category of cobordisms which satisfy some "...
Daniel Moskovich's user avatar
11 votes
0 answers
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Madsen-Tillmann spectrum $MTE$ of the group $E$ which is defined in Freed-Hopkins's paper

In Freed-Hopkins's paper, the group $E(d)$ is defined to be the subgroup of $O(d)\times\mathbb{Z}_4$ consisting of the pairs $(A,j)$ such that $\det A=j^2$, where $\mathbb{Z}_4=\{\pm1,\pm\sqrt{-1}\}$ ...
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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 ...
wonderich's user avatar
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9 votes
2 answers
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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
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9 votes
1 answer
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Generators and relations for the 2-dimensional unoriented cobordism category

It is very well known in the field of TQFT that the 2-dimensional oriented cobordism category is generated by the disk and the pair of pants (each going in both directions), subject to a finite set of ...
Andi Bauer's user avatar
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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 ...
Riccardo's user avatar
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8 votes
1 answer
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Cobordism invariants: topological v.s. geometric

Some cobordism invariants are not cohomology classes. Such as the $\mathbb{Z}_{16}$-valued eta invariant $\eta$ of $\Omega_4^{Pin^+}$, the $\mathbb{Z}_8$-valued Arf-Brown-Kervaire invariant ABK of $\...
Borromean's user avatar
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7 votes
1 answer
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Is $MGL$ an $H\mathbb{Z}$-algebra?

Let $\mathrm{MGL}$ be the $\mathbb{P}^1$-ring spectrum over a field $k$ representing algebraic cobordism. Suppose, for simplicity, that $k$ is of characteristic 0. Let $H\mathbb{Z}$ be the motivic ...
Anandam Banerjee's user avatar
7 votes
1 answer
246 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
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7 votes
1 answer
444 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
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6 votes
0 answers
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Oriented cobordism group generated by mapping torus

The 4-dimensional oriented cobordism group of closed manifolds is $\Omega^{SO}_4=Z$, and we know that it cannot be generated by a mapping torus, since the Pontryagin for $p_1$ is zero for any mapping ...
Xiao-Gang Wen's user avatar
6 votes
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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 ...
wonderich's user avatar
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5 votes
0 answers
158 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
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4 votes
1 answer
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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
0 answers
287 views

"Naïve"cobordism?

The naive view of (say, complex) cobordism of a space $X$ is that it should be the group generated by continuous families of closed manifolds parametrized by $X$. Acting even more naively, one may try ...
მამუკა ჯიბლაძე's user avatar
4 votes
0 answers
71 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)) \...
annie marie cœur's user avatar
3 votes
2 answers
460 views

Does the Gysin map in $K$-theory respect bordism?

Let $X_1$ and $X_2$ be two closed spin$^c$ manifolds that are bordant via a spin$^c$ manifold-with-boundary $W$. Let $Z$ be a closed spin$^c$ manifold with $\dim Z=\dim X_1$ mod $2$. Let $$f_1:X_1\to ...
geometricK's user avatar
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0 votes
1 answer
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Relation between trivial tangent bundle $\Leftrightarrow$ certain characteristic classes of tangent bundle vanish [closed]

We know that framing structure means the trivialization of tangent bundle of manifold $M$. string structure means the trivialization of Stiefel-Whitney class $w_1$, $w_2$ and half of the first ...
zeta's user avatar
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