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Let $M_1,M_2$ be two simply connected, connected, compact smooth manifolds without boundary and of the same dimension. Assume that $\mathfrak{X}(M_1)\cong \mathfrak{X}(M_2)$ as Lie algebras.

Question. Are $M_1$ and $M_2$ diffeomorphic?

This seems like a basic question but I did not find anything on Google.

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    $\begingroup$ For some context I can say that (abstract) diffeomorphism group determines compact manifold uniquely; of course, this has only very tangential relation to the question. $\endgroup$
    – Denis T
    Commented Dec 7, 2022 at 6:41

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The answer is yes, and the assumptions "simply connected" and "compact" are actually unnecessary.

In fact, it is possible to reconstruct any smooth manifold $M$, up to diffeomorphisms, by using the subalgebra $\mathfrak{X}_0(M)$ of vector fields with compact support. See

Shanks, M. E.; Pursell, Lyle E., The Lie algebra of a smooth manifold, Proc. Am. Math. Soc. 5, 468-472 (1954). ZBL0055.42105, JSTOR.

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    $\begingroup$ Can we reconstruct differentiable maps from maps of Lie algebras, or is this passing to Lie algebras fully faithful? $\endgroup$
    – Z. M
    Commented Dec 7, 2022 at 9:59
  • $\begingroup$ Surely, a isomorphism $\Phi$ of Lie Algebras (of vector fields) induces a diffeomorphism $\varphi$ between the two manifolds such that $\Phi = d \varphi$. $\endgroup$
    – Francesco Polizzi
    Commented Dec 7, 2022 at 10:14
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    $\begingroup$ I am asking slightly more generally – any maps, instead of isomorphisms. $\endgroup$
    – Z. M
    Commented Dec 7, 2022 at 10:25
  • $\begingroup$ Yes, I understand. I do not know, I should check the proof. $\endgroup$
    – Francesco Polizzi
    Commented Dec 7, 2022 at 10:35
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    $\begingroup$ @Z. M: It's not a functor, so it's not clear what your question would mean. $\endgroup$
    – Tom Goodwillie
    Commented Dec 7, 2022 at 12:52

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