# unordered configuration space over spheres and Euclidean spaces

For a topological space $X$, let $B(X,k)$ be the $k$-th unordered configuration space. Then

$$B(\mathbb{R}^n,2)\simeq \mathbb{R}P^{n-1},$$ $$B(S^n,2)\simeq \mathbb{R}P^n.$$ Hence

$(*)$ $$B(\mathbb{R}^{n+1},2)\simeq B(S^n,2).$$

Inspired by $(*)$, are there any relations between $B(\mathbb{R}^{n+1},k)$ and $B(S^n,k)$ for general $k\geq 3$ or for some particular $k=2^i$?

• Is it possible you meant to ask for relations between $B(\mathbb{R}^{n+1},k)$ and $B(S^n,k)$? – Mark Grant Sep 2 '15 at 7:28
• Yes. I want that there were some relations – QSR Sep 2 '15 at 7:41
• What happens if we take ordered configuration space instead? – user43326 Sep 2 '15 at 7:56
• If we take ordered configuration space instead, as a matter of fact we have a up-to-homotopy fibration (by combining two Fadell-Neuwirth) $S^n \setminus \mbox{$(k-1)$points}\rightarrow F(S^n,k)\rightarrow F(R^{n+1},k)$. It is possible that the map $F(S^n,k)\rightarrow F(R^{n+1},k)$ is homotopic to the one induced by the inclusion of $S^n$ in $R^{n+1}$, so $\Sigma _{n+1}$-equivariant, so induces the map from $B(S^n,k)$ to $B(R^{n+1},k)$. When $k=2$, the fiber is $S^n \setminus \mbox{ two points}$, which is contractible, so you get the homotopy equivalence. – user43326 Sep 2 '15 at 10:37

Put $$U(n,k)=\{(x_1,\dotsc,x_k)\in F(\mathbb{R}^{n+1},k): \sum_ix_i=0 \text{ and } \max(\|x_1\|,\dotsc,\|x_k\|)=1\},$$ and $V(n,k)=U(n,k)/\Sigma_k\subset B(\mathbb{R}^{n+1},k)$. Then $V(n,k)$ is a strong deformation retract of $B(\mathbb{R}^{n+1},k)$, and there is an obvious inclusion $i\:B(S^n,k)\to V(n,k)$ which is a homeomorphism for $k=2$ but not for $k>2$. I don't think that you can say much more than that. Already there is a big difference when $n=1$ and $k>2$.