5
$\begingroup$

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?

$\endgroup$
10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.