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Results tagged with gt.geometric-topology
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user 1450
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).
3
votes
Why is complex projective space triangulable?
Although I am more than a decade late to the discussion, here is an alternate approach to an explicit triangulation of $\mathbb{C}P^n$.
(1) A simplicial set $X$ is a natural generalization of a simpli …
- 56.6k
8
votes
Accepted
Does Lackenby's polynomial bound on knot moves imply polynomial mixing in "Quantum Money Fro...
Thanks to HJRW2 for the flattering invitation here, and I will give an answer, but it might be not all that deep. In fact I haven't been on MO much lately; maybe I should visit it more.
I don't see …
- 56.6k
29
votes
Accepted
Manifolds are paracompact
Theorem: A countable atlas of charts for a Hausdorff $n$-manifold $M$ can be refined to a locally finite atlas. In fact, each chart only needs to be trimmed.
Proof: Let $U_1,U_2,\ldots$ be the chart …
- 56.6k
12
votes
Singular chains in Spivak's Calculus on Manifolds
It's clearly an elementary oversight that, on the other hand, doesn't matter for the real development of the material. Yes, the chain $c$ ought to be at least $C^1$ for the pullback to be continuous …
- 56.6k
9
votes
Accepted
Generalization of Moise's theorem
In fact it helps immensely that $M$, $X$, and $Y$ are all Riemannian, so much so that the question is both true and not at all a generalization of Moise's theorem. Instead, you are looking for a smo …
- 56.6k
32
votes
Accepted
Why is Casson's invariant worth studying?
Wikipedia's description of the Casson invariant gives the first important reason to study it. As an invariant that comes from the $\text{SU}(2)$ representation variety of $\pi_1(M)$, it reveals in pa …
- 56.6k
22
votes
Accepted
Riemannian metrics on non-paracompact manifolds
On the contrary, the long line does not have a Riemannian metric. Every countable subset of the long line has a least upper bound, so if it were Riemannian then a geodesic ray in the long direction w …
- 56.6k
1
vote
Orientation of a "glued"-manifold
I agree that the difficulty in the question is that you are relying on the homological definition of an orientation of a manifold. As Ryan implies in the comments, the solution is undergraduate-level …
- 56.6k
8
votes
Conformal embedding of Riemann surfaces into 3-space
I have thought about this question before, but at the moment I can't remember links or references. Nonetheless, many years ago I thought of a sketch of an argument that should eventually work to prov …
- 56.6k
24
votes
Are there any very hard unknots?
There are really two questions here: (1) Can you an untangle any unknot with relatively little work, say a polynomial number of geometric moves of some kind? (2) Given a knot, can you quickly figure …
- 56.6k
10
votes
Utility of virtual knot theory?
My view of virtual knot theory: An R-matrix is a tensor with four indices that, in a natural sense, satisfies the Reidmeister relations. Actually it is enough to consider the third Reidemeister rela …
- 56.6k
1
vote
Accepted
Better term for a (simplicial) contractible plane continuum
In the end, we (Joel Kamnitzer, his student Bruce Fontaine, and I) agreed on the term "diskoid". In the abstract, the word "cactus" seemed too clever by half. When I actually wrote it into the paper …
4
votes
How indepenedent of a chosen metric is the box-counting dimension? Is there a non-integral d...
The so-called "box" dimension is best interpreted as a cheap approximation to Hausdorff dimension. Hausdorff dimension is much more robust, but it is not preserved by homeomorphisms. All Cantor sets …
- 56.6k
8
votes
Can all n-manifolds be obtained by gluing finitely many blocks?
It is not entirely clear in the question whether a "block" is a manifold with smooth boundary, or perhaps a manifold that is allowed to have ridges or more complicated corners. Let's assume that the …
- 56.6k
11
votes
Rugged manifold
To expand on Ryan's comment: A handle decomposition of a manifold is a restricted type of quotient of a disjoint union of handles. An $n$-dimensional $k$-handle is by definition the manifold $B^k \t …
- 56.6k