# A $k$-component link defines a map $T^{k}\rightarrow \mathrm{Conf}_{k} S^{3}$. Does the homotopy type capture Milnor's invariants?

A $k$-component link defines a map $T^{k}\rightarrow \mathrm{Conf}_{k} S^{3}$. 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 $T^{2}\rightarrow \mathrm{Conf}_{2} R^{3}$, which captures linking number.
• $k=3$, Melvin et al. proved it does.
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