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41 votes
1 answer
6k views

Not all manifolds can be triangulated: In which dimensions?

I know that Ciprian Manolescu has settled the triangulation conjecture in the negative: Not all manifolds can be triangulated. I've only read secondary literature on this result, which did not detail ...
Joseph O'Rourke's user avatar
17 votes
1 answer
981 views

Smooth 4-manifolds with $E_8$ intersection form

Does there exist a closed orientable smooth 4-manifold $M$ whose intersection form is the $E_8$-form? Here by the intersection form I mean the $\mathbb{Z}$-valued bilinear form on $H^2(M;\mathbb{Z})/\...
Julia's user avatar
  • 193
8 votes
1 answer
686 views

Surgery along an embedded surface in a 4-manifold

Let $X$ be a 4-manifold and $\Sigma_g\subset X$ be an embedded closed orientable genus = $g$ surface. Suppose $\Sigma_g\subset X$ has a trivial closed normal bundle $N(\Sigma_g) = \Sigma_g\times D^2$. ...
Shinichiro Nakamura's user avatar
6 votes
1 answer
324 views

Computation of $\pi_1$ for a Mazur manifold and its boundary

If we attach a $4$-dimensional $1$-handle $D^1 \times D^3$ to a $4$-dimensional $0$-handle $B^4$, we obtain $S^1 \times D^3$. The null homologous knot in $S^1 \times S^2$ indicated in the picture ...
Max Schumann's user avatar
6 votes
1 answer
304 views

Stable torus that is not a torus [duplicate]

Let $M$ be a closed manifold such that $M\times \mathbb{S}^1$ is a torus. Is it true that $M$ is homeomorphic to a torus?
Anton Petrunin's user avatar