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I ran into the geometrisation conjecture a few days ago, and I started wondering how to prove Poincaré's conjecture. Let $M$ be a compact, simply connected, $3$-manifold. Clearly it is irreducible since it is simply connected. How would show that the whole $M$ carries a spherical geometry ?

Have a nice day,

Selim

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2 Answers 2

up vote 6 down vote accepted

Poincaré's conjecture follows from Perelman's proof on the Thurston Elliptization Conjecture. To put it simply, Thurston's Geometrization Conjecture claims that if you have a closed prime orientable $3$-manifold than you can cut it along a suitable collection of embedded tori so that each of the pieces you are left with can be endowed with a "nice" geometry. This is in a sense analogous to the Uniformization of closed surfaces, with the difference that in the $2$-dimensional case you don't have to cut anything (and that there you have just 3 geometries of constant curvature, whereas in 3 dimensions you get 8 geometries).

The Elliptization Conjecture is a part of this big program, which claims that a closed $3$-manifold with finite fundamental group is spherical, which means it can be endowed with the elliptic geometry (riemannian metric of constant curvature $1$); then it is a standard theorem in metric geometry that a space with such a geometry has the sphere as its universal cover. (Have a look here). This Elliptization Conjecture was proven by Perelman in 2003 using the so called Ricci flow, i.e. techniques from analysis of PDEs.

Coming back to your question, if you have your $3$-manifold $M$, it satisfies the hypothesis of the theorems I quoted; in particular its fundamental group is trivial, thus finite. So $M$ is covered by $S^3$. But $M$ is simply connected and this implies it is isomorphic to every cover, so $M \cong S^3$.

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This is a fantastic lay-person article on Thurston's program by Erica Klarreich

https://simonsfoundation.org/features/science-news/getting-into-shapes-from-hyperbolic-geometry-to-cube-complexes-and-back/

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