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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 …
mme's user avatar
  • 9,580
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 …
mme's user avatar
  • 9,580
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 …
mme's user avatar
  • 9,580
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 …
mme's user avatar
  • 9,580
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 …
mme's user avatar
  • 9,580