7
$\begingroup$

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 $\mathbb{R}P^m\times \mathbb{R}^m$?

$\endgroup$
  • $\begingroup$ Yes. I also have this question. $\endgroup$ – Shiquan Jul 11 '15 at 2:26
  • 1
    $\begingroup$ $F(S^m, 2) \cong S^m \times S^m \setminus \overline{\Delta}$ (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
5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.