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This is reposted from MSE, but perhaps it is more appropriate to post it here. Let me know if I'm wrong.

Among the many unsuccessful attempts at solving the Poincaré conjecture, I'm wondering if there was an approach that went along the lines of showing that a closed simply connected $3$-manifold could be endowed with a Lie group structure. This might not be enough to solve it, since there are other compact $3$-dimensional Lie groups (if I'm not wrong), but it's tempting to think about since the only spheres which are Lie groups are of dimension $0, 1$ or $3$. This makes it feel like there is this special additional structure that can be used for the case $n=3$ as opposed to solving the generalized Poincaré conjecture, where no higher dimensional spheres are Lie groups.

So I'm wondering if someone tried this. If they did, what hurdle/subtlety did they find in the course of the attempt that prevented a solution? Did we learn something interesting from it?

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  • $\begingroup$ for reference: math.stackexchange.com/questions/3764770/… $\endgroup$
    – Carlo Beenakker
    Commented Jul 27, 2020 at 18:37
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    $\begingroup$ There are no other compact simply connected Lie groups of dimension $3$. $\endgroup$
    – abx
    Commented Jul 27, 2020 at 19:05
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    $\begingroup$ I don't really know why you would imagine it would be any easier to construct a Lie group structure on a simply connected closed 3-manifold than finding a diffeomorphism to $S^3$ and pulling back the group structure by that diffeomorphism. How do you imagine simple connectedness help you? $\endgroup$
    – mme
    Commented Jul 27, 2020 at 19:26
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    $\begingroup$ @MikeMiller Personally I don't know either, my knowledge of the subjects here is sparse. Perhaps this could lead to another question, can one find examples where "unconventional" properties of a manifold make it a Lie group? $\endgroup$
    – Paul Cusson
    Commented Jul 27, 2020 at 19:41
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    $\begingroup$ Tangentially you might find Stallings' paper how not to prove the Poincare conjecture a useful read $\endgroup$
    – Neal
    Commented Jul 27, 2020 at 20:28

1 Answer 1

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Thurston approaches 3-manifolds by cutting them up along various surfaces (one first cuts along spheres [Kneser-Milnor] and then along tori [Jaco-Shalen-Johannson]) into pieces which each admit a locally homogeneous geometric structure, modelled on a homogeneous space with an invariant Riemannian metric. A compact, simply connected manifold with such a structure is homogeneous. But a Lie group is homogeneous under its own left translations (or right translations), with many invariant Riemannian metrics (given by left, or right, translation of any positive definite inner product on any one tangent space. So Thurston's programme contains your programme as a special case.

Thanks to Allen Hatcher for correcting my description of the decomposition.

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