What is the fundamental group of a 2-sphere with $x$ and $-x$ identified (i.e. $x=-x$)?

I thought about this in the following way: The 2-sphere with $x$ and $-x$ identified is just the half-sphere. Intuitively, I guess that the fundamental group is the same as for the original sphere $S^2$, i.e. the trivial group. Is this correct?


The sphere with antipodal points identified is called the real projective plane. In general, the fundamental group of a simply connected space modulo a properly discontinuous group action is the group. Here, you have $S^2$ module a $\mathbb{Z}/2$ action, so the fundamental group is $\mathbb{Z}/2$. So no, the fundamental group of the real projective plane is not the same as the fundamental group of its universal cover, the 2-sphere.

More intuitively, a path to an antipode becomes a loop which is not contractible, but traverse the path twice, and you get something that lifts to a loop in $\mathbb{S}^2$, which is contractible.

Similar remarks apply to the rotation group, which is a 3-sphere mod $\mathbb{Z}/2$, which has an interesting physical interpretation: It corresponds to somewhat surprising fact, that a rotation by 360º in three dimensional space is not actually indistinguishable from no rotation (for example, if the rigid body you rotate has some strings attaching to ambient room). But a rotation by 720º is.

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  • $\begingroup$ Does this correspond to figure 1 and 2 in the following document: damtp.cam.ac.uk/user/examples/D18S.pdf? $\endgroup$ – Quasar Nov 8 '17 at 20:20
  • $\begingroup$ @Quasar: yes, the rotation group is a 3-sphere mod Z/2 $\endgroup$ – ziggurism Nov 8 '17 at 20:22
  • $\begingroup$ Thank a lot! This makes it clear. $\endgroup$ – Quasar Nov 8 '17 at 20:23

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