1

We define a relation in the sphere by identifying the antipodal points, the quotient space obtained is the projective plane $\mathbb{P}^2$. Also, the quotient map $\pi:\mathbb{S}^2\longrightarrow\mathbb{P}^2$ turns out to be a double covering of the projective plane. Given an isometry $T$ in $\mathbb{P}^2$ we have a lifting $\widetilde{T}:\mathbb{S}^2\longrightarrow\mathbb{S}^2$ such as $T\circ\pi=\pi\circ\widetilde{T}$.

My first question is: is this lifting $\widetilde{T}$ unique? And if it is not, can we make it unique? How? Well, that have been three questions.

My second question is: Can we use the previous to show that $\widetilde{T}$ is an isometry on the sphere? How?

My third question is: If we show that $\widetilde{T}$ is an isometry, how this help us to get all the isometries in $\mathbb{P}^2$?

The next part of the problem is about the geodesics in $\mathbb{P}^2$, Any hint about it? Maybe using the quotient map $\pi$?

Kyor
  • 61
  • A hint for your first question: Let $a: S^2 \to S^2$ be the antipodal map. Given a lifting $\tilde{T}$, consider the map $a \circ \tilde{T}$. Do you see why this is also a lifting of $T$? (Hint: $\pi \circ a = \pi$.) – Phillip Andreae May 21 '14 at 14:31
  • For a somewhat related question, see here, which explains that the isometry group of $S^2$ is $O(3)$ (i.e., restrictions to $S^3$ of orthogonal linear transformations of $\mathbb{R}^3$). – Phillip Andreae May 21 '14 at 14:35
  • Yes, I see that the map $a\circ\widetilde{T}$ is also a lifting of $T$. So it´s not unique. Now, the next part is how to make it unique? I think I heard about fixing a point, what do you know about that? – Kyor May 22 '14 at 02:08

0 Answers0