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I'll first state the question as concisely as I can and then provide some motivation.

Consider two positive integers $m$ and $n$ such that $m < n+2$. Does there exist a topological group $G$ that is $n$-connected ($\pi_i(G) = 0$ for $0\le i \le n$) and that has a continuous action on a space $F$, which is an $m$-dimensional (say, topological) manifold, that extends the standard action of $GL(m)$ on $\mathbb{R}^m$? By extend, I mean that there is an inclusion $GL(m) \subset G$ such that the action of $G$ on $F$ restricts to the action of $GL(m)$ on $\mathbb{R}^m$ identified with an open subset of $F$.

[Note: In the original version of the question, I asked for $F$ to be a homogeneous space of $G$. I had the example of $G=\mathrm{Diff}(F)$ in mind, where both conditions are satisfied. Knowing that homotopy groups of diffeomorphism groups might be hard to come by, I tried to distill out the essential requirement, but was a bit too hasty. The $GL(m)$ condition is actually crucial for the extension of the normal bundle in the motivation given below.]

The question is motivated by the following negative remark in differential topology: Given a manifold $N$, not every closed, embedded submanifold $M \subseteq N$ can be represented as the zero set of a section of some vector bundle (even with a free choice of vector bundle). The obstruction is related to the normal bundle of $M$ in $N$ (cf. MO66732, MO78209). A natural way to generalize this submanifold representation question to get a positive answer is to introduce another variable into the problem: the topology of the fiber $F$ of the bundle over the ambient manifold $N$. Then $M$ could potentially be represented by the intersection locus of two sections of an $F$-bundle over $M$.

If we restrict our attention to a tubular neighborhood of $M$ in $N$, then the normal bundle already does the trick, with fiber $\mathbb{R}^m$, $m=\dim N - \dim M$. So the submanifold representation problem reduces to extending this bundle to some $F$-bundle over all of $N$, identifying $\mathbb{R}^m$ with a neighborhood in $F$ and also extending the appropriate sections along the way. Now, obstruction theory in algebraic topology has something to say about this. Namely, a sufficient condition (Hatcher, Cor.4.73, p.417) for such an extension to exist is that the relative cohomology groups $H^i(N,M;\pi_{i-1}BG) = 0$, for $0\le i\le n+2 = \dim N$, where we view an $F$-bundle as a fiber bundle with structure group $G$ and $BG$ is the corresponding classifying space.

Using the identity $\pi_{i-1} BG = \pi_{i-2} G$, shows that a positive answer to the above boxed question implies that the bundle extension problem can also be solved, if one first builds a principal $G$-bundle over $N$ and then constructs an associated bundle with fiber $F$. For the purposes of the submanifold representation question, it also has to be locally Euclidean (hence a manifold) and of dimension $m$.

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  • $\begingroup$ Well, there is the trivial example: Take $G = (\mathbb{R}^N,+)$, which is contractible and hence $n$-connected for all $n$. As long as $N\ge m$, the quotient $M^m = G/\mathbb{R}^{N-m}$ will do the job. It seems likely that you want some other properties of $G$ that you have not yet specified. $\endgroup$ Dec 7, 2013 at 10:38
  • $\begingroup$ @RobertBryant, that's an excellent point! I was a bit too hasty in abstracting the question. I've updated to question to require that $G$ actually acts on $F$ (not necessarily homogeneously) in a way that extends the action of $GL(m)$ on $\mathbb{R}^m$. Unfortunately, this seems to make the requirements rather more strict. :-/ $\endgroup$ Dec 7, 2013 at 12:09
  • $\begingroup$ A simple observation: $F$ must be non-orientable, since if it is orientable, then $G$, being path-connected, must act by orientation-preserving homeomorphisms, contrary to the fact that the action of $GL(m)$ on ${\mathbb R}^m$ includes orientation-reversing homeomorphisms. $\endgroup$ Sep 9, 2014 at 0:05
  • $\begingroup$ @AllenHatcher, thanks, that's a good point. I think it's sensible to relax the question by replacing $GL(m)$ with the connected component of its identity. But I do apologize for this generally poorly conceived and framed question. $\endgroup$ Sep 9, 2014 at 6:24
  • $\begingroup$ Igor, I think that you have some typos in the gray field. If indeed, then it'd be nice to fix them. $\endgroup$ Mar 8, 2015 at 1:34

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