At what points the application ceases to be an immersion?

Question

The application $f: S^2 \rightarrow\mathbb{R}^3$ given by $f (x, y, z) = (yz, xz, xy)$ induces an application of class $C^1$ of $\mathbb{R}\text{P}^2$ in $\mathbb{R}^3$.

But it is not an immersion in $6$ points, I am not able to find these points in which the differential is not injecting and its images are points of the coordinated axes?

Could someone give me a hint how to find them, please?

Answer 1

Your function $f$ has a natural extension $\widetilde{f}\colon \Bbb R^3 \to \Bbb R^3$, and we have no choice but to compute $${\rm d}\widetilde{f}_{(x,y,z)} = \begin{bmatrix} 0 & z & y \\ z & 0 & x \\ y & x & 0 \end{bmatrix}.$$However, the domain of the differential ${\rm d}f_{(x,y,z)}$ is the tangent space $T_{(x,y,z)}\Bbb S^2 = (x,y,z)^\perp$. Observe that $$\begin{bmatrix}0 & z & y \\ z & 0 & x \\ y & x & 0 \end{bmatrix}\begin{bmatrix} x \\ y \\ z\end{bmatrix} = \begin{bmatrix}2yz \\ 2xz \\ 2xy \end{bmatrix} \neq \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix},$$since $(x,y,z) \neq (0,0,0)$. So ${\rm rank}({\rm d}\widetilde{f}_{(x,y,z)}) = {\rm rank}({\rm d}f_{(x,y,z)}) + 1$, and ${\rm d}f_{(x,y,z)}$ will have full rank if an only if ${\rm d}\widetilde{f}_{(x,y,z)}$ has full rank, and the question becomes: what are the points $(x,y,z)\in \Bbb S^2$ for which this matrix does not have full-rank?

• Can it have rank $0$? No, because $x^2+y^2+z^2=1$ implies $(x,y,z) \neq (0,0,0)$, and so ${\rm d}f_{(x,y,z)} \neq 0$.

• Can it have rank $1$? This would mean that every order two subdeterminant vanishes. This also implies that $(x,y,z) = (0,0,0)$, by looking at the submatrices $\begin{bmatrix} 0 & \ast \\ \ast & 0\end{bmatrix}$, with $\ast \in \{x,y,z\}$.

• Can it have rank $2$? This means that $\det {\rm d}f_{(x,y,z)} = 0$. But this equals $$\det {\rm d}f_{(x,y,z)} = 0 + xyz + xyz - 0 - 0 - 0 = 2xyz.$$If $x = 0$, then $y^2+z^2 = 1$ gives us $4$ critical points. If $y = 0$, $x^2+z^2=1$ gives us $4$ more critical points. And from $z=0$, $x^2+y^2=1$ gives another $4$ critical points.

These $12$ critical points in $\Bbb S^2$ pass to the quotient as $6$ critical points in $\Bbb R P^2$.

Answer 2

Let's use spherical coordinates, which are safe away from the poles ($\sin \phi= 0$)

$$(x,y,z)=(\cos\theta\sin\phi, \sin\theta\sin\phi,\cos\phi)$$

Then writing $f$ as a function of $\theta$ and $\phi$, we have $$f(\theta,\phi) = \newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}} \bmat \sin\theta\sin\phi\cos\phi\\ \cos\theta\sin\phi\cos\phi\\ \sin\theta\cos\theta\sin^2\phi \emat = \frac{1}{2} \bmat \sin\theta\sin 2\phi\\ \cos\theta\sin 2\phi\\ \sin 2\theta\sin^2\phi\\ \emat $$

$$ df = \frac{1}{2} \bmat \cos\theta\sin2\phi & 2\sin\theta\cos2\phi\\ -\sin\theta \sin2\phi & 2\cos\theta\cos2\phi\\ 2\cos2\theta\sin^2\phi & \sin2\theta\sin 2\phi \emat. $$

The first $2\times 2$ minor is $$2\cos^2\theta\sin2\phi\cos 2 \phi +2\sin^2\theta\sin2\phi\cos 2\phi=2\sin 2\phi\cos 2\phi.$$

In order for $df$ to have low rank, this must be $0$, so either $\sin 2\phi=0$ or $\cos 2\phi =0$. Note that $\phi \in (0,\pi)$. Thus either $\phi = \pi/2$ or $\phi = \pi/4,3\pi/4$.

These correspond to three possibilities for $df$.

When $\phi = \pi/4$, $$ df = \frac{1}{2} \bmat \cos\theta & 0\\ -\sin\theta & 0\\ \cos2\theta & \sin2\theta \emat, $$ this has low rank when $\sin 2\theta = 0$, corresponding to $\theta = n\pi/2$, $n\in \Bbb{Z}$.

Similarly, when $\phi = 3\pi/4$, we end up with $$ df = \frac{1}{2} \bmat -\cos\theta & 0\\ \sin\theta & 0\\ \cos2\theta & -\sin2\theta \emat, $$ which also has low rank at $\theta = n\pi/2$.

Finally, when $\phi = \pi/2$, we have $$ df = \frac{1}{2} \bmat 0 & -2\sin\theta\\ 0 & -2\cos\theta\\ 2\cos2\theta & 0 \emat. $$

This has low rank if and only if $\cos 2\theta = 0$, corresponding to $\theta = \pi/4 + n\pi/2$.

Let $a = \frac{\sqrt{2}}{2}$ for notational convenience.

Translating back into rectangular coordinates, $f$ has low rank at the following twelve points: $(\pm a, 0, \pm a)$, $(0,\pm a, \pm a)$, and $(\pm a, \pm a, 0)$.

These descend to $6$ points where $f$ has low rank on the projective plane.

Since none of these are the points $(\pm 1,0,0)$ or $(0,\pm 1,0)$ we don't need to worry about the fact that we excluded the poles at the beginning, since by symmetry they can't be points where $f$ has low rank.