Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
A three-manifold is a space that locally looks like Euclidean three-dimensional space
4
votes
Do we get a instanton $S^{3}$ if we do $1/n$ surgery on a knot in $S^{3}$?
In fact, if $C = \text{rank}(I_*(S^3_0(K)))$, we have $$f(k) = \text{rank}(I_*(S^3_{1/k}(K))) = kC.$$ This holds with any coefficient field.
Floer's exact sequence gives us an exact triangle relating …
4
votes
Accepted
Is there a way to calculate the Froyshov $h$-invariant for Seifert homology spheres?
No, not yet. The state of the art in computing instanton homology for Seifert spaces is in this paper. This is like knowing how to compute $\widehat{HF}(\Sigma)$, which is not enough: you also want to …
5
votes
Dual surfaces of a first cohomology class of a 3-manifold
$L(4,1)$ is a counterexample to your conjecture, taking $\alpha$ to be the nontrivial element of $H^1(L;\Bbb Z/2)$. Notice that this element squares to zero (the square is the same as the Bockstein, a …
2
votes
Accepted
Group of parallelizations of $M^3$ finitely generated?
I believe that for any finite $n$-complex $X$, the group $$[X, SO(n)]_*$$ is finitely generated. I will follow Igor Belegradek's approach. In fact I only think $H^*(X;\Bbb Z)$ finitely generated and m …
6
votes
Accepted
Possible orders of automorphisms for the Poincare homology sphere
Let $G$ act smoothly and orientably. The quotient $M/G$ is a spherical 3-orbifold. Therefore, by the elliptization theorem, it may be given a metric of constant curvature 1; pulling back, then $M$ is …