$\tilde{f}:\mathbb{P}^2(\mathbb{R})\rightarrow\mathbb{R}^6$ injective immersion

Question

I have to prove this one

Given $f:S^2\rightarrow\mathbb{R}^6$ defined by $f(x_1,x_2,x_3)=(x_1^2,x_2^2,x_3^2,x_1x_2,x_1x_3,x_2x_3)$, prove that:

1. $f$ is an immersion;
2. $f(-x_1,-x_2,-x_3)=f(x_1,x_2,x_3)$;
3. exists an injective immersion $\tilde{f}:\mathbb{P}^2(\mathbb{R})\rightarrow\mathbb{R}^6$ such that $\tilde{f}([x_1:x_2:x_3])=f(x_1,x_2,x_3)$, $\forall [x_1:x_2:x_3]\in\mathbb{P}^2(\mathbb{R})$.

I have this idea of the solution:

1. I compute the Jacobian matrix

$J_f(x_1,x_2,x_3)=\left( \begin{array}{ccc} 2x_1 & 0 & 0 \\ 0 & 2x_2 & 0 \\ 0 & 0 & 2x_3 \\ x_2 & x_1 & 0 \\ x_3 & 0 & x_1 \\ 0 & x_3 & x_2 \end{array} \right) $

and since $T_p S^2=\{v=(a,b,c)\in\mathbb{R}^3 : (x_1,x_2,x_3)\cdot(a,b,c)=0\}$, $\forall p=(x_1,x_2,x_3)\in S^2$, the expression $J_f(v)=0$ leads to the sistem

$\left\{\begin{array}{l} 2x_1a=0\\ 2x_2b=0\\ 2x_3c=0\\ ax_2+bx_1=0\\ ax_3+cx_1=0\\ bx_3+cx_2=0 \end{array}\right.$

that has unique solution $(a,b,c)=(0,0,0)$. So $\ker(f_*)=\{0\}$ for every $p\in S^2$, hence $f_*$ is injective and f is then an immersion.

2. $f(-x_1,-x_2,-x_3)=(x_1^2,x_2^2,x_3^2,x_1x_2,x_1x_3,x_2x_3)=f(x_1,x_2,x_3)$.
3. I consider the diagram

\begin{array}{ccc} & f & \\ S^2 & \rightarrow & \mathbb{R}^6 \\ \pi \searrow & & \nearrow \tilde{f} \\ & \mathbb{P}^2 (\mathbb{R})& \end{array}

where $\pi$ is the projection of $S^2$ in $S^2/\sim\cong\mathbb{P}^2(\mathbb{R})$ and since it is a local diffeomorphism, $\pi^{-1}$ is an immersion. Hence because of (1), $ \tilde{f}=f\circ\pi^{-1}$ is composition of immersion, then an immersion. Moreover it's injective because if we take $[x_1:x_2:x_3]\neq[y_1:y_2:y_3]$ we get $\tilde{f}([x_1:x_2:x_3])\neq\tilde{f}([y_1:y_2:y_3])$. Then we conclude that $\tilde{f}$ is an injective immersion.

Is everything all right?

Thanks for reading all this stuff

Answer 1

Points 1. and 2. are fine.

But for 3., take care: local diffeomorphisms usually don't have globally defined inverses. So, there is no such (globally defined) function $\pi^{-1}$.

Nevertheless, taking it on the level of differentials, $T_p\pi$ is indeed an isomorphism for each $p$, so we can take its inverse - maybe this was your intension, too - to conclude that $T_p\tilde f$ is injective, i.e. $\tilde f$ is an immersion.

About the injectivity: all you write is correct, but maybe you could detail it a bit more:

Assume $f(x_1,x_2,x_3)=f(y_1,y_2,y_3)$. Then $x_i^2=y_i^2$ so $|x_i|=|y_i|$ and the other $3$ coordinates ensure that signs of $x_1,x_2,x_3\,$ "change together" along $f$..