Let $V$ be an $n$-dimensional vector space. Is the space of embeddings $$ \coprod_1^{k} V \to V $$ path connected for large enough $n$? Clearly $n=1$ is not enough, but I feel like $n=2$ is enough for $1$-connected. Does the space become highly connected as $n\to \infty$? This feels like it is equivalent to a question about the little disks operads, but I don't know how to frame it as such.
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2$\begingroup$ Do you mean embeddings as topological spaces? If so, do you mean proper embeddings? My guess is the answers are Yes and No, since you mentioned the little disks operad. $\endgroup$– Kevin WalkerCommented Jun 17, 2011 at 4:16
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$\begingroup$ Smooth embeddings as manifolds. Proper I'm not sure... $\endgroup$– Alan WilderCommented Jun 17, 2011 at 5:46
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$\begingroup$ No not proper, just smooth. I think the answer below is what I need. $\endgroup$– Alan WilderCommented Jun 17, 2011 at 6:01
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1 Answer
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No, it is not connected: for example, if $k=1$ it has two path components, given by the two orientations with which $V$ can be embedded into itself.
In general, it has the homotopy type of $F_k(V; O(n))$ the space of configurations of $k$ particles in $V$ with labels on the orthogonal group, which has $2^k$ path components given by the possible configurations of the orientations.
If you ask for the embedding of each $V$ to be orientation-preserving, then the space is path-connected for $n > 1$ by Tilman's argument (as $SO(n)$ is connected).
answered Jun 17, 2011 at 6:19
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1$\begingroup$ Thanks for the clarification. To take the last statement further, if I demand that the embeddings preserve a framing of $V$ do I get a homotopy equivalence to $F_k$ with unlabeled points, and so the highly connected as $n\to\infty$ result? $\endgroup$ Commented Jun 18, 2011 at 0:38
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