[Math] To prove that the projection map $S^2 \to \mathbb RP^2 $ is a covering map via group action

algebraic-topology

I am reading Algebraic Topology and I got some problem in covering map. Please help me. Thnx in advance.

I want to show that the projection map $S^2 \to \mathbb RP^2 (\text{ real projective plane })$ is a covering map.

So I tried the action of $G=\mathbb Z_2$ on $X=S^2$ via $(1,x) \to x,(-1,x)\to -x \forall x\in S^2$

Now I want to show that for each $x\in S^2 \exists U\text{ ,nbd of } x, \text{ such that } (-1U)\cap U=\phi$

I think I am able to do this for any $x$ not on the great circle because in that case choosing an nbd $U$ which lies on the same hemisphere of $x$ will work. Am I right upto this point?

But I got stuck if $x$ lies in the great circle. I am not able to find a way how to choose $U$ in this case. Please help me.

I think thereafter it is obvious that the natural quotient map $X \to X/G$ is a covering map and it is indeed the projection map $S^2 \to \mathbb RP^2$ since $\mathbb RP^2$ is defined by identifying antipodal points.

Please rectify me if I am wrong a any point. I want to solve the problem by group action. And I think $\mathbb Z_2$ should work.

Please help me to complete the proof. Thnx agian.

Best Answer

There are three great circles - if we write $$S^2=\{(x,y,z) : x^2+y^2+z^2=1\} $$ the great circles are when $x=0$, $y=0$, or $z=0$. Given any point $(x,y,z)\in S^2$ we must have at least one non-zero coordinate - if $x=y=z=0$ then the point is not in $S^2$. Then you can apply your covering map using the great circle for one of your non-zero entries.

For example, if $(x_0,y_0,z_0)\in S^2$ and $x_0 >0 $ then let $$U_{(x_0,y_0,z_0)}=\{(x,y,z)\in S^2 : x>0\}$$. This upper hemisphere is an open neighbourhood for $(x_0,y_0,z_0)$. Then the lower hemisphere $$V_{(x_0,y_0,z_0)}= \{(x,y,z)\in S^2 : x<0\}$$ defines a neighbourhood of $-(x_0,y_0,z_0)$ and $U\cap V = \emptyset$.

A key point is that every point on the sphere will be on one side of some great circle. This is because every point will have at least one non-zero coordinate, and the three great circles are defined by $x=0$,$y=0$, and $z=0$. In the example above, we chose the great circle defined by $x=0$.

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