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Results tagged with gt.geometric-topology
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user 2349
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).
15
votes
Accepted
Does every orientable surface embed in $\mathbb{R}^{3}$
Expanding slightly on my comment above: here is how one can get the embedding theorem from the classification theorem.
The classification theorem for non-compact surfaces (theorem 3 in http://www.ams …
- 23.5k
3
votes
Can you explain a step in a proof about 2-sided surfaces in 3-manifolds?
We have to prove that $F$ separates some connected neighborhood. Choose a neighbourhood $U$ that can be contracted onto $F$ (e.g. a tubular one) and suppose $F$ does not separate $U$. Then there is a …
- 23.5k
3
votes
Accepted
Poincare duality
For any CW complex $X$ one defines a chain complex $C_*(X)$: choose an orientation of each cell; the group $C_n(X)$ is the free abelian group with a basis whose elements correspond to the $n$-cells of …
- 23.5k
6
votes
Accepted
Decide a manifold via its boundary
Yes, assuming the manifold compact. One way to see this is this: if we glue a disk along the boundary, we get a manifold which is simply-connected manifold by Seifert-van Kampen, and closed, hence a s …
- 23.5k
1
vote
Action of U(1) on a sphere bundle, non-vanishing vector fields on odd-dimensional manifolds
Here is a proof that a compact oriented manifold $M$ of dimension $n$ with zero Euler characteristic has a nowhere vanishing vector field. There a vector field $X$ on $M$ with all zeroes non-degenerat …
- 23.5k
7
votes
Covering spaces of surfaces
For an explicit example of a non-cyclic group $G$ acting freely on a surface $S$ take the union $X$ of the edges of the 3-cube $[-1,1]^3\subset\mathbb{R}^3$; set $S$ to be the boundary of a small neig …
- 23.5k
1
vote
homotopy type of complement of subspace arrangement
Here is a sketch of a solution. As Petya tells us, $\mathbf{R}^4\setminus M$ is invariant under the action of $\mathbf{Z}^4$ by integral translations, so it suffices to show that the image of $\mathbf …
- 23.5k
8
votes
1
answer
657
views
A conjecture of Montesinos
Not every orientable 3-manifold is a double cover of $S^3$ branched over a link. For example, the 3-torus isn't. However, in 1975 Montesinos conjectured (Surjery on links and double branched covers of …
- 23.5k
11
votes
2
answers
1k
views
Number of the Reidemeister moves needed to transform one diagram into another one
A recent question Random Reidemeister moves to unknot contains a link to the paper http://www.ams.org/journals/jams/2001-14-02/S0894-0347-01-00358-7/S0894-0347-01-00358-7.pdf, in which J. Hass and J. …
- 23.5k
14
votes
"Largest" finite-dimensional Lie subgroups of Diff(S^n), are they known?
The quotient of $GL_{n+1}(\mathbf{R})$ by the positive scalars acts on $S^n$; it has dimension $n^2+2n$, so for $n>1$ the answer to the first question is no. For $n=3$ an alternative proof would be as …
- 23.5k
17
votes
Classification problem for non-compact manifolds
Complementing Ryan's answer: as shown by McMillan (Transactions AMS 102, 373-382) there is a continuum of pairwise nonhomeomorphic contractible open subsets of $\mathbf{R}^3$. So classifying noncompac …
- 23.5k
7
votes
Twisted cohomology of the mapping class group
I think the answer is 0.
Indeed, $G=SL_2(\mathbb{Z})$ contains a normal $H=\mathbb{Z}/2\cong\{\pm I\}$. So if $V$ is set to be $\mathbb{Z}^2$ with the standard $G$-action, then the restriction of $H$ …
- 23.5k
7
votes
Diffeomorphism of 3-manifolds
About the elliptic case $S^3/G$: elliptic 3-manifolds are classified up to homeomorphism by their $\pi_1$'s, except for the lens spaces. For the lens spaces the simple homotopy type classification is …
- 23.5k
8
votes
1
answer
621
views
Cohomology map induced by the group actions on homogeneous vector bundles
Here is a topological question which seems quite elementary. The answer to this question may be useful e.g. in estimating the orders of the automorphism groups of some algebraic varieties and in compu …
- 23.5k
18
votes
1
answer
940
views
Do chains and cochains know the same thing about the manifold?
This question was inspired by Poincaré quasi-isomorphism
Let $M$ be a closed oriented $n$-manifold. The cap product with the fundamental class of $M$ induces an isomorphism $H^i(M,\mathbf{Z})\to H_{n …
- 23.5k