Why does the projection $\pi:(a,b,c) \in S^2 \to[a,b,c] \in P^2$ have rank 2 everywhere?

Question

Consider the differentiable function $g: [a,b,c] \in \mathbb{P}^2 \to (bc,ac,ab) \in \mathbb{R}^3$ It is not an immersion because it has rank 2 everywhere except in 6 points. To see this consider the projection $\pi:(a,b,c) \in S^2 \to[a,b,c] \in P^2$ and the function $ G:(a,b,c)\in \mathbb{R}^3\to(bc,ac,ab)\in \mathbb{R}^3$. Since $G|_{S^2}=g\circ \pi$, for each $p \in S^2$ and $(p,\nu)\in T_pS^2, G_{*p}(p,\nu)=(G|_{S^2})_{*p}(p,\nu)=g_{*[p]}(\pi_{*p}(p,\nu))$. It can be shown that $\pi$ has rank 2 everywhere.

The points of $\mathbb{P}^2$ in which $g$ is not an immersion are the points $[p]$ coming from points $p \in S^2$ such that there exist a non-zero $(p,\nu)\in T_pS^2$and such that $G_{*p}(p,\nu)=0$....(**)

How can I prove that $\pi$ has rank 2 everywhere? I guess I have to find the jacobian of $\pi$ and I should find two linearly independent rows. Since it is a function that has as codomain the projective plane, I don't know how to proceed, can I take partial derivatives as if it where a function from $\mathbb{R}^3$ to $\mathbb{R}^3$? What is the analytical expresion of $\pi$ to do that?

As a side thing, where does the statement () come from?** In particular why do we require a non-zero $(p,\nu)\in T_pS^2$?

Answer 1

$\pi$ is a local diffeomorphism which shows that it has rank $2$ everywhere. In fact, the six sets $$U_j^\pm = \{(x_1,x_2,x_3) \in S^2 \mid (-1)^{\pm 1} x_j > 0 \}$$ with $j = 1,2,3$ are open and cover $S^2$. They are mapped by $\pi$ diffeomorphically onto the three open sets $$V_j = \{ [x_1,x_2,x_3] \mid x_j \ne 0\} \subset \mathbb P^2.$$

$g$ is a immersion at $p$ iff $dg_{[p]}$ has maximal rank (i.e. rank $2$). Write $h = G \mid_{S^2}$. We have $dh_p = dg_{[p]}\circ d\pi_p$. (I prefer to write $dh_p$ instead of $h_{*p}$; this notation is somewhat unusual.) Since $d\pi_p$ is an isomorphism, the rank of $dh_p$ agrees with the rank of $dg_{[p]}$ and it suffices to determine all $p$ such that $dh_p$ has rank $2$. This means that it maps no non-zero vector of $T_pS^2$ to $0$. Therefore $g$ is not an immersion at $[p]$ iff there exist a non-zero $(p,\nu)\in T_pS^2$and such that $dh_p(p,\nu)=0$.

Answer 2

Suppose that $M$ and $N$ are $n$-manifolds. Suppose that $f : M \to N$ is smooth. Suppose that $p$ is a point of $M$ and $q = f(p)$ is a point of $N$. Suppose that $U \subset M$ and $V \subset N$ are small neighbourhoods of $p$ and $q$. Suppose that $\alpha : U \to \mathbb{R}^n$ and $\beta : V \to \mathbb{R}^n$ are charts in the smooth atlas with $\alpha(p) = 0$ and $\beta(q) = 0$. The rank of $Df$ at $p$ is defined to be the rank of $D(\beta \circ f \circ \alpha^{-1})$ at zero.

It is thus a computation to determine the rank of (the covering map) $\pi$ at any point (say $(1, 0, 0)$). Since $\mathrm{SO}(3)$ acts nicely on both $S^2$ and $P^2$ you can deduce the result everywhere else.