Let $S^m$ be the $m$-sphere and $$F(S^m,2)/\mathbb{Z}_2=\{(a,b)\mid a,b\in S^m, a\neq b\}/(a,b)\sim (b,a)$$ be the $2$-nd unordered configuration space on $S^m$. Why $F(S^m,2)/\mathbb{Z}_2$ is homotopy equivalent to $\mathbb{R}P^m$?

Note that $F(S^m,2)/\mathbb{Z}_2$ is obtained by the action of $\mathbb{Z}_2$ on $F(S^m,2)$ by permuting coordinates $(a,b)\to (b,a)$. But $\mathbb{R}P^m$ is obtained by the action of $\mathbb{Z}_2$ on $S^m$ by the antipodal map. Although $F(S^m,2)\cong TS^m \simeq S^m$, the actions of $\mathbb{Z}_2$ are different.

Added Question: Is $F(S^m,2)/\mathbb{Z}_2$ homeomorphic to $\mathbb{R}P^m\times \mathbb{R}^m$?

  • $\begingroup$ Yes. I also have this question. $\endgroup$ – Shiquan Jul 11 '15 at 2:26
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    $\begingroup$ $F(S^m, 2) \cong S^m \times S^m \setminus \overline{\Delta}$ (where $\overline{\Delta}$ is the antidiagonal $\{(a,-a) \in S^m \times S^m\}$) is homeomorphic to the total space of the tangent bundle, $TS^m$. The homeomorphism is essentially orthogonal projection. Exploit this description. $\endgroup$ – user98602 Jul 11 '15 at 2:43
  • $\begingroup$ Thanks. Does this homeomorphism imply that after quotienting the $\mathbb{Z}_2$-action by reversing coordinates, we still can get a homeomorphism $F(S^m,2)/\mathbb{Z}_2\cong (S^m\times S^m\setminus \bar\Delta )/\mathbb{Z}_2$? $\endgroup$ – Shiquan Sep 5 '15 at 15:23
  • $\begingroup$ Hi Mike, I am still confused. Could you explain more in detail? Thanks! $\endgroup$ – Shiquan Sep 6 '15 at 7:26

Observe that:

  1. $F(S^m,2)/\mathbb{Z}_2$ is the space of all unordered pairs of points on $S^m$.

  2. $\mathbb{R}P^m$ is the space of all unordered pairs of antipodal points on $S^m$.

The first space deformation retracts onto the second. Specifically, given an unordered pair $\{a,b\}$ of points on $S^m$ that are not antipodal, let $C$ be the great circle containing them. Then we can move $a$ and $b$ directly away from each other along $C$ until they are antipodal, which defines the required deformation retraction.


Define $i:S^m\to F(S^m,2)$ by $i(a)=(a,-a)$. This map is equivariant with respect to the $\mathbb{Z}_2$-actions on both sides, and I claim it actually realizes $S^m$ as a $\mathbb{Z}_2$-equivariant deformation retract of $F(S^m,2)$. Indeed, given a point $(a,b)\in F(S^m,2)$, we can continuously move $a$ and $b$ in opposite directions from each other along the great circle joining them until they are antipodal, and every stage of this homotopy is $\mathbb{Z}_2$-equivariant.

An alternative description of the induced deformation retraction on the quotients is the following. Think of $\mathbb{R}P^m$ as the set of lines through $0$ in $\mathbb{R}^{m+1}$, and think of $F(S^m,2)/\mathbb{Z}_2$ as the set of lines in $\mathbb{R}^{m+1}$ that intersect $S^m$ at two points. Given such a line, take the midpoint of its two intersections with $S^m$, and translate your line continuously so the midpoint moves in a straight line to the origin.

  • $\begingroup$ Thanks! Will $F(S^m,2)/\mathbb{Z}_2$ be homeomorphic to $\mathbb{R}P^m\times [0,1)$? $\endgroup$ – Shiquan Sep 5 '15 at 3:17

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