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.

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 WalkerJun 17 '11 at 4:16

$\begingroup$ Smooth embeddings as manifolds. Proper I'm not sure... $\endgroup$– Alan WilderJun 17 '11 at 5:46

$\begingroup$ No not proper, just smooth. I think the answer below is what I need. $\endgroup$– Alan WilderJun 17 '11 at 6:01
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 orientationpreserving, then the space is pathconnected for $n > 1$ by Tilman's argument (as $SO(n)$ is connected).

$\begingroup$ 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? $\endgroup$ Jun 18 '11 at 0:38
