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Results tagged with gt.geometric-topology
Search options questions only
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user 27004
Topology of cell complexes and manifolds, classification of manifolds (e.g. smoothing, surgery), low dimensional topology (e.g. knot theory, invariants of 4-manifolds), embedding theory, combinatorial and PL topology, geometric group theory, infinite dimensional topology (e.g. Hilbert cube manifolds, theory of retracts).
6
votes
1
answer
362
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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 …
- 10.5k
2
votes
1
answer
161
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
…
- 10.5k
0
votes
0
answers
259
views
Define a characteristic class on a simplicial complex (non-manifold)
Given a simplicial complex with only triangulation and only branching structure, is it enough to define Stiefel–Whitney class?
(Please provide Yes or No answers, and reasonings.)
Given a fixed tria …
- 10.5k
7
votes
1
answer
258
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 $M$.
$$\Omeg …
- 10.5k
9
votes
1
answer
566
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 differe …
- 10.5k
5
votes
1
answer
370
views
$\pi_{2n-1}(\operatorname{SO}(2n))$ element represents the tangent bundle $TS^{2n}$, not tor...
Question: Is the element $\alpha$ in $\pi_{2n-1}(\operatorname{SO}(2n))$ representing the tangent bundle $TS^{2n}$ of the sphere $S^{2n}$ indivisible and not torsion?
My understanding so far —
An $ …
- 10.5k
6
votes
0
answers
162
views
Can $X_4 \times S^1$, $X_4 \times I^1$, or $X_4 \times \mathbb{R}^1$ be a triangulable, PL o... [duplicate]
Question: If $X_4$ is a non-triangulable topological (TOP) manifold,
can $X_4 \times S^1$, $X_4 \times I^1$, or $X_4 \times \mathbb{R}^1$ be a triangulable manifold?
can $X_4 \times S^1$, $X_4 \time …
- 10.5k
6
votes
0
answers
208
views
If $X_d$ is a non-triangulable manifold, can $X_d \times T^k$, $X_d \times I^k$, or $X_d \ti...
If $X_d$ is a non-triangulable manifold, can $X_d \times T^k$, $X_d \times I^k$, or $X_d \times \mathbb{R}^k$ always be a triangulable manifold?
Let $X_d$ be a $d$-manifold which is NOT a triangula …
- 10.5k
4
votes
0
answers
158
views
Postnikov square explicitly on a simplicial complex
$\DeclareMathOperator\Z{\mathbb{Z}}$
Following Wikipedia, a Postnikov square is a certain cohomology operation from a first cohomology group $H^1$ to a third cohomology group $H^3$, introduced by Post …
- 10.5k
6
votes
0
answers
178
views
Visualize how the 5d Dold manifold and Wu manifold are cobordant via a 6d manifolds with bou...
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 oriente …
- 10.5k
4
votes
0
answers
272
views
Instanton numbers for diverse gauge bundles on diverse manifolds --- their relations to char...
It is standard (?) that the $SU(N)$ gauge theory has the instanton number $n$ quantized as $n \in \mathbb{Z}$
$$
n = { 1 \over 8\pi^2} \int_{\mathcal{M}_{4}} \text{tr} \left(F \wedge F\right) = {1 \o …
- 10.5k
3
votes
0
answers
182
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Cobordism theory of some weird space
Let $G=SU(3)$ and $N=SO(3)$, then $G/N= SU(3)/SO(3)$ = a 5-dimensional Wu manifold $W$.
The $W$ is a homogeneous space (also a quotient space), but not a group.
Previously, I am aware of the bordis …
- 10.5k
4
votes
0
answers
234
views
Any cobordism invariant made of "characteristic classes", on unorientable manifolds, must be...
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 Stie …
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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 ar …
- 10.5k
4
votes
2
answers
356
views
Chirality and Anti-Chirality of links in 3 and in 5 dimensions
We know there is a chiral knot which is a knot that is not equivalent to its mirror image. It is well known in the mathematical field of knot theory:
https://en.wikipedia.org/wiki/Chiral_knot
My fir …
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