I guess that the least possible dimension of a Lie group $G$ acting smoothly and transitively on a compact manifold $M$ is $\operatorname{dim}(M)$.

Is this correct and is there a ref?


This follows from Sard's theorem. If $G \times M \to M$ is smooth, then in particular $G \times \{pt\} \to M$ is smooth. If $\operatorname{dim} G < \operatorname{dim} M$, then Sard's theorem implies the image has measure zero, and in particular is not all of $M$.

Compactness is inessential here.

  • $\begingroup$ A lovely argument there! $\endgroup$ – Benjamin Aug 6 '15 at 17:28
  • $\begingroup$ While I think I've got what you mean, you could add a definition of $p, t, n$ and a bit more clarity. $\endgroup$ – Benjamin Aug 7 '15 at 0:23
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    $\begingroup$ pt means a point in $M$. I for some reason thought we had set $n = \dim M$ already. Sorry, fixed. $\endgroup$ – user98602 Aug 7 '15 at 0:24
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    $\begingroup$ @AndreasBlass: OK, here's a proof. A manifold with a transitive action by a Lie group $G$ is $G/\text{Stab}_x$, $x$ some point. The stabilizer is closed, hence a Lie subgroup. Both have trivial $\pi_2$, and as they're Lie groups and deformation retract onto their maximal compact subgroup, finitely generated abelian $\pi_1$. So if $M$ has a transitive action by $G$, we obtain an exact sequence $\pi_2(G) = 0 \to \pi_2(M) \to \pi_1(\text{Stab}_x)$. So pick a manifold $M$ with infinitely generated $\pi_2$, as is constructed for instance here... $\endgroup$ – user98602 Aug 7 '15 at 1:46
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    $\begingroup$ in dimension as small as 4. (I don't know if there's a closed 3-fold with infinitely generated $\pi_2$.) $\endgroup$ – user98602 Aug 7 '15 at 1:46

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