A $k$-component link defines a map $T^k \rightarrow \operatorname{Conf}\_k S^3$. Does the homotopy type of this map capture the Milnor invariants?


Some special cases:


* $k=2$, no, it's null homologous, but you can look instead at the map $T^2 \rightarrow \operatorname{Conf}\_2 R^3$, which captures linking number.
* $k=3$, [Melvin et al.](https://arxiv.org/abs/0901.1612) proved it does.



A $k$-component link defines a map $T^k\rightarrow \operatorname{Conf}_k S^3$. Does the homotopy type capture Milnor's invariants?
It's been a while since I've thought about this but I think Koschorke answered much of your question back in 1997 "A generalization of Milnor's mu-invariants to higher-dimensional link maps" Topology 36 (1997), no 2. 301--324. Scanning through the paper I see he recovers many of the mu invariants but not all. He lists it as an open question (6.3) if the homotopy class of the map T^k --> C\_k R^3 is a complete link homotopy invariant of the link.


Related: 


Brian Munson put these Koschorke "linking maps" into the context of the Goodwillie calculus in a recent arXiv paper. I've wondered for a while if you could use these types of maps to create a direct construction of the Cohen-Wu correspondence between the homotopy groups of S^2 and their corresponding simplicial quotient object made from the brunnian braid groups.


