All Questions
Tagged with cobordism surgery-theory
14 questions
2
votes
0
answers
55
views
Tangential normal invariant isomorphism
Recently, I was reading the paper "Finite Group Actions on Kervaire Manifold" by Crowley, Hambolton. But I am having problem understanding a definition. Here it is,
In page 15-16 they are ...
4
votes
1
answer
230
views
Gluing a manifold along its boundary, via chain complexes
Given closed oriented $n$-manifolds $M, M', M''$ and bordisms $W, W'$ with $\partial W = M \sqcup - M'$ and $\partial W' = M' \sqcup - M''$, we can collar-glue them to obtain a bordism from $M$ to $M''...
- 898
2
votes
0
answers
117
views
About connected cobordism and surgery
I need to find various ways of performing two surgeries on a collection of circles so that the resulting 2-dimensional cobordism (the trace of the surgeries) is connected.
How can I find these ? up ...
- 119
5
votes
1
answer
173
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Is spin cobordism an invariant for surgery of codimension $q\ge3$?
Recall that a surgery of codimension $q$ on an $n$-manifold $X$ is a modification of $X$ of the following type. Let $\Sigma^{n-q}\subset X$ be a smoothly embedded $(n-q)$-sphere with a trivialized ...
- 10.5k
1
vote
0
answers
183
views
Mapping class group and surgery theory
Given a smooth manifold $M$ of dimension $n$ and a diffeomorphism $\phi: M \to M$, we can construct a smooth cobordism of dimension $(n+1)$ from $M$ to $M$ by gluing $M \times [0,1]$ with itself by $\...
- 5,230
5
votes
0
answers
350
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 ...
- 545
6
votes
0
answers
184
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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 ...
- 10.5k
9
votes
1
answer
305
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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 ...
- 2,630
2
votes
0
answers
102
views
Realizing an amalgamated product of groups by splitting a closed manifold along a codimension 1 submanifold
In the paper "A splitting theorem for manifolds" by S.E. Cappell,
https://www.maths.ed.ac.uk/~v1ranick/papers/capsplit.pdf
the following "inverse" of the Seifert-van Kampen theorem for closed ...
- 21
6
votes
1
answer
170
views
Diffeomorphism type of the added sphere in simply connected surgery
A classical result of simply connected surgery theory, is that if two normal maps $f:M_i\rightarrow X$, $i=0,1$ are normally cobordant and if the dimension of the manifolds is odd, there exists a ...
- 93
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
0
answers
257
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 ...
- 2,018
23
votes
2
answers
881
views
Vanishing of characteristic numbers vs vanishing of characteristic classes
A famous result by Thom states that Oriented Bordism classes are determined by characteristic numbers; specifically, two closed manifolds are orientedly bordant if and only if they have the same ...
- 521
1
vote
1
answer
129
views
Connected representant of a framed cobordism class (reference needed)
Let $N^n\subseteq M^m$ be a submanifold with a framing of the normal bundle, $2n<m$. Then $N^n$ is framed cobordant (in $M^m$) to something connected.
I believe it could be proved by directly ...
- 972