Literally "together boundary", a cobordism is a relation between two compact manifolds stating that their disjoint union forms the boundary of a higher dimensional manifold. This defines a equivalence relation between compact manifolds that is very coarse: two manifolds may be cobordant but not ...

learn more… | top users | synonyms

2
votes
0answers
52 views

rho invariant of manifold

If $G$ is a finite group, then the rational oriented cobordism group $\Omega_{2k-1}^{Stop}(BG)\otimes{\mathbb Q}=0$, so if $N^{2k-1}$ is an orientable odd-dimensional Top manifold with fundamental ...
2
votes
1answer
40 views

On Steenrod's realization of cycles problem.

There is old problem of realization homology classes of (closed) manifold $M^n$ by fundamental classes of its submanifolds. Partially it was solved by René Thom in his "Quelques propriétés globales ...
-1
votes
0answers
33 views

Framed nullbordant and calculation of framing in coordinate chart

To prove the Hopf degree theorem (theorem 2.37) D. Freed in his notes proves the following lemma I have two questions about this lemma: (1) how to calculate the framing at time $s$? what is the ...
2
votes
1answer
26 views

What motivated trying to express the signature of a manifold as a linear combination of pontrjagin numbers?

I have tried reading a proof of the signature theorem but it is way beyond me, is there a way to motivate, in english, why anyone even started searching for such a formula? Why would one assume that ...
3
votes
0answers
20 views

surgery of type $(\lambda,n-\lambda)$ on manifold, h-cobordism theorem by Milnor

I am reading Milnor's Lectures on h-cobordism theorem, and I am stuck on Milnor's definition on surgery of type $(\lambda,n-\lambda)$ on manifold, where the definition following can be found on page ...
1
vote
1answer
18 views

Proving that for a multiplicative (B,f)-structure $\mathfrak{B}$ (or X-strcture, or B-structure), Thom spectrum $M\mathfrak{B}$is a ring spectrum.

I'm interested in filling the detail of the claim I made above. I'm following Kochman's notation (page 14 for a def.). Actually he never claims it, (he never spoke about ring spectra), but I think ...
1
vote
1answer
46 views

Building $MU$-spectrum via the definition of a $(B,f)$-structure

I want to construct the them spectrum $MU$ using the definition of spectrum associated to a $(B,f)$-structure. Here are the relevant definitions: A $(B,f)$-structure is a collection of strictly ...
0
votes
0answers
25 views

Homotopy sets for a pushout of spaces (Seifert-van-Kampen?)

my problem is the following: I have two bordisms $M : \Sigma_0 \to \Sigma_1$ and $M' : \Sigma_1 \to \Sigma_2$, so I can glue them along $\Sigma_1$ to get $M'\circ M$. The manifold $M'\circ M$ is the ...
1
vote
0answers
25 views

Cobordism Groups an the Pontryagin-Thom Construction

I am confused by the statement that $\Omega^\text{framed}_1(S^3) \cong \mathbb{Z}$ which I came across as an application of the Pontryagin-Thom construction for showing that $\pi_3(S^2) \cong \mathbb{...
0
votes
0answers
20 views

Why is the homomorphism $\Omega_* \to H_*(BSO, Q)$ well defined.

Let $\Omega_*$ be the oriented bordism ring of a point. The correspondence of the map in the title is defined by $(K_{M^{N-1}})_*[M^{n-1}]$ where $K_{M^{N-1}}$ is the classifying map of the tangent ...
2
votes
2answers
60 views

Why are bordism groups of a point nontrivial

My definitions of Bordism are from Tom Dieck's Algebraic topology book. Briefly, a singular manifold, $M \xrightarrow{f} pt $, for a closed smooth oriented manifold $(M, \omega)$ without boundary is ...
1
vote
1answer
34 views

What is the map from $H_j( \Sigma MSO(k)) \to H_{j-k}(BSO(k))$ on Tom Dieck page 537

I am reading Tom Dieck's page 537 and I am not sure what the vertical map that I put in the title is in the diagram in the bottom of the page. This map is labeled Thom Isomorphism. Here $MSO(k)$ is ...
0
votes
1answer
13 views

Why is the fibration $BSO(n-1) \to BSO(n)$, $n-1$ connected?

In other words I want to show that the induced map $\pi_k BSO(n-1) \to \pi_k BSO(n)$ is 0 for $k\leq n-1$. The fiber of this fibration is $S^{n-1}$. I am having trouble with the case $k=n-1$. I am ...
1
vote
1answer
89 views

Cobordism and h-cobordism

Is there a way to simply explain cobordism and h-cobordism? I am not looking for a math based explanation, but rather, just the main ideas behind the concepts.
3
votes
1answer
36 views

Is $\mathsf{nCob}$ bicomplete?

Let $\mathsf{nCob}$ be the category of $n$-cobordisms, whose objects are $(n-1)$-dimensional closed manifolds and morphisms are bordisms. Is this category bicomplete, or even finitely bicomplete? ...
2
votes
1answer
65 views

Proof of general Poincaré Conjecture $\dim > 5$

Given $M$ is simply connected, $\dim(M) > 5$ and homotopy equivalent to a sphere. Let $W := M - D_1 \cup D_2$ for two smoothly embedded disjoint disks. How does one see that $W$ is simply ...
3
votes
1answer
61 views

Set consisting of all unoriented cobordism classes of smooth closed $n$-manifolds can be made into additive group? [closed]

How do I see that the set $\mathfrak{N}_n$ consisting of all unoriented cobordism classes of smooth closed $n$-manifolds can be made into an additive group?
1
vote
1answer
46 views

Equivalency of h-cobordisms

I'm reading lectures on the h-cobordism theorem by Milnor and I have a little problem understanding some basic points. I can't understand this theorem , not the theorem , not the proof. I appreciate ...
3
votes
2answers
60 views

Oriented Bordism Abelian Group structure proof

I've started studying cobordism and I found difficult formalising the abelian group structure of the oriented bordism group $\Omega_n$. The main problem I encountered is dealing with the orientation ...
3
votes
1answer
51 views

Bordism between bordisms

I am reading about $2$-categories of bordisms, denoted $Cob_2(n)$. This paper by Lurie (see page 10) states that the objects of such a category are closed $(n-2)$-manifolds and and for $M, N \in Cob_2(...
4
votes
1answer
88 views

Todd genus as homomorphism from complex cobordism

It's well-known that the Todd genus/arithmetic genus $\chi(\mathcal{O}_X)$ (or probably preferably $\int_X \text{td}(T_X)$ so as to define it in purely terms of the complex structure) is a genus in ...
3
votes
2answers
68 views

Another clarification about Thom-Pontrjagin construction

This is the second part of the following solved question. [I'm following Bredon's Book]. After explaining the idea behind the "desired" bijection we want to build, Bredon start dealing with the well-...
5
votes
3answers
411 views

Examples of manifolds that are not boundaries

What are some examples of manifolds that do not have boundaries and are not boundaries of higher dimensional manifolds? Is any $n$-dimensional closed manifold a boundary of some $(n+1)$-dimensional ...
4
votes
1answer
86 views

Clarification about the Thom-Pontrjagin construction as explained in Bredon's book

In Bredon's book, at page 118-119, there is a little chapter about the Thom-Pontrjagin construction, and I'm trying to follow the reason depicted there. He starts with a map $f \colon R^{n+k}\to R^...
3
votes
1answer
67 views

Cut out characteristic Submanifold N ($w_1(M)=w_1$(Normalbundle of N in M)). Remainder M-N is orientable? Orientation Character or CW-Structure?

So I try to understand the following (which is taken from Dold, "Structure of the cobordism ring", Page 3/274, in the paragraph "1. La suite exacte de Wall."): https://eudml.org/doc/109581 ): Giving a ...
4
votes
2answers
83 views

Cobordism: Reference Request.

Where does one learn Cobordism theory - is there some canonical text or reference at the beginning graduate level? More general sources on modern differential topology (surgery etc.) are fine; any ...
1
vote
0answers
74 views

Almost complex structure gluing

Consider smooth complex manifold $X$ of complex dimension $n$ and its smooth submanifold $Z$ of codimension $k$. Denote the normal bundle of the inclusion $Z\subset X$ by $\nu.$ One can blow up $X$ ...
7
votes
1answer
57 views

Grassmanians and boudaries of manifolds

Let $M$ be a smooth, compact manifold without boundary. I will say that $M$ is a boundary when there is a smooth, compact manifold with boundary $W$ such that $\partial W=M$. After some lectures I ...
7
votes
1answer
124 views

Hurewicz map factors through bordism homology

I've read in multiple sources that the hurewicz map $h \colon \pi_n(X) \to H_n(X)$ factors through oriented bordism homology. I'm particularly interested in the injectivity of the map $h \colon \pi_n(...
1
vote
1answer
22 views

The trivial cobordism and orientations

I have just come across the definition of a cobordism between two closed oriented $n$-dimensional manifolds $M,M'$ as an oriented $(n+1)$-dimensional manifold $W$ with boundary such that $\partial W =...
3
votes
0answers
138 views

Uncertainties on the details of the Connor-Floyd isomorphism and Formal Group Laws

Let $\Omega^{\bullet}(-)$ be the complex cobordism cohomology. $\Omega^n(X) = \{ (M, f) | f: M \to X \}$ where cobordant maps are identified, $M$ and $X$ are smooth manifolds, and dim($M $) = $n$. ...
6
votes
2answers
134 views

Is $H^*(\mathbf{C} P^\infty)=R[X]$ or $R[[X]]$?

The first ring seems to be what one learns first: the underlying group is the cohomology of the total singular cochain complex $C^*(\mathbf{C} P^\infty)$, which is defined as $\oplus C^n(\mathbf{C} P^\...
3
votes
0answers
72 views

Why are homotopy spheres spin-cobordant in dimensions divisible by 4?

Manifolds are assumed to be smooth having dimensions $\geq5$. As usual, $\Omega^{Spin}_{*}$ denotes the spin bordism ring and $\Omega^{SO}_{*}$ the oriented bordism ring. Let $\Sigma^n$ be a closed ...
2
votes
0answers
49 views

Notation and shorthand of cobordism G=O, SO, U

This is really a basic question: From Wikipedia: "The basic examples are G = O for unoriented cobordism, G = SO for oriented cobordism, and G = U for complex cobordism using stably complex manifolds."...
1
vote
0answers
47 views

differential in AHSS for spin cobordism

According to these solutions, the differential $d_2: H_p(X,\Omega_1^{Spin})\rightarrow H_{p-2}(X,\Omega_2^{Spin})$ is the dual of $Sq^2$. Why? This MO post asks a similar question (but about $d_3$ in ...
2
votes
1answer
143 views

About existence of Morse functions

Let's consider 4-manifold $M$, $\partial M = \partial M_1 + \partial M_2 = S^1 \times S^2 + \mathbb{RP}^3$. Is it true that there exist a Morse function $$f\colon M^4 \to [0,1],\quad f^{-1}(0) = \...
1
vote
1answer
69 views

Differential topology question involving cobordism

Prove that if $X$ and $Z$ are cobordant in $Y$, then for every compact manifold $C$ in $Y$ with dimension complementary to $X$ and $Z$, $I_2(X, C) = I_2(Z, C)$. [HINT: Let $f$ be the restriction to $W$...
1
vote
1answer
57 views

Identity in Thom spaces.

Let $T$ be the one-point compattification, $E$ a real vector bundle, $\epsilon$ the trivial line bundle and $\Sigma$ the suspension operation. How can I prove that $$ T(\epsilon \oplus E) \simeq \...
1
vote
1answer
56 views

Bordism classes of real projective plane

I have to prove that doesn't exist a compact $3$-manifold such that $\partial X= \mathbb{R}P^2$. My book suggests to use Euler characteristic and define $Y = X \cup_{\mathbb{R}P^2} X$. What is $X \...
1
vote
0answers
136 views

If $Z$ is an oriented manifold with boundary such that $\partial Z=M$ where $M$ is a compact and oriented manifold then $\chi(M)=0\mod(2)$..

I need some help for showing the following result: Let $M$ be a compact ($\partial M= \emptyset$) and oriented manifold of dimension $n$ and $Z$ an oriented manifold with boundary such that $\...
3
votes
1answer
143 views

Bordism sanity check

Were framed cobordism and h-cobordism invented to use for different purposes? I have been slightly confused about all the different types of cobordism. Now I am wondering if h-cobordism was invented ...
1
vote
0answers
61 views

Cobordant to sphere or homotopy sphere

In Ranicki's notes (here) remark 6.19 distinguishes between cobordant to a sphere and cobordant to a homotopy sphere. Earlier in these notes right after example 1.6 he writes that homotopy equivalent ...
2
votes
1answer
382 views

On framed cobordism

If two manifolds are h-cobordant then their homotopy groups agree. The notion of framed cobordism is supposedly a weaker notion. How much weaker than h-cobordism is it? What can be said about two ...
4
votes
1answer
92 views

Axiomatizing oriented cobordism

According to the nLab entry for abstract cobordism categories, the natural way of axiomatizing the relation of two oriented manifolds being cobordant is the following: Definition 1 Two objects $M,...
3
votes
1answer
114 views

Suspension Operation on the Pontryagin-Thom Construction

I have a feeling that this is well-known: View the Pontryagin-Thom construction as the bijective correspondence between $[M,S^r]$ and the set of (appropriate equivalence classes of) framed ...
10
votes
1answer
228 views

Why is Lemma 6.3 of Milnor's Lectures on the h-cobordism Theorem True?

Milnor's statement is: "Let $M^r$ and $N^s$ be sub-manifolds of $V^{r+s}$ which are all smooth, compact, oriented and without boundary. If $p$ is a point of $M^r$ contained in an $r$-cell $U$,...
3
votes
0answers
78 views

How to do this surgery?

Let $L$ be a $0$-framed trivial knot in $S^3 \subset B^4$. Take $B^3 \subset B^4$ such that $B^3$ splits $B^4$ into two and $\partial B^3$ intersects $L$ only two points. Take a neighborhood $U$ of $...
1
vote
1answer
80 views

Surgery and boundary

Let $L$ be a framed link in $S^3$ with $m$ components and let $U$ be a closed regular neighborhood of $L$ in $S^3$. Let $B^4$ be a closed 4-ball bounded by $S^3$ so that $U \subset S^3$. Gluing $m$ ...
0
votes
1answer
129 views

Definition of a 4-cobordism with boundary

Definition of a 3-cobordism (in my context) is a pair $(M, \partial_{-} M, \partial_{+}M)$, where $M$ is a closed orientable topological 3-manifold and $\partial M$ is a disjoint union of $\partial_{-}...
1
vote
0answers
36 views

Cylindrical structure and homology

Let us consider a cobordism $(M, \partial_{-} M, \partial_{+}M)$, where $M$ is homeomorphic to $T \times I$, here $T$ is a torus $S^1 \times S^1$ and $I=[0, 1]$. I encountered the statement "...