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Considering a map $w$ from $ \mathbb{R}^3$ to $ \mathbb{R}^6$ defined as follows:

$$ w(x,y,z)= (x,y,z, \sqrt{2}xy,\sqrt{2}yz,\sqrt{2}zx )$$

in order to show that it defines an immersion of $S^2$ into $ \mathbb{R}^6$ I first proved that it defines an immersion from $ \mathbb{R}^3/{0}$ to $ \mathbb{R}^6$ by, first, computing the Jacobian matrix an proving that is has rank $3$ for all the points in $ \mathbb{R}^3/{0}$.

$$ J = \left(\begin{array}{cccc} 2x & 0& 0\\ 0 & 2y & 0\\ 0 & 0 & 2z \\ \sqrt{2}y & \sqrt{2}x & 0 \\ 0 & \sqrt{2}z & \sqrt{2}y \\ \sqrt{2}z & 0& \sqrt{2}x \end{array}\right) $$

It is also easy to verify that it is injective. After that, I ddefined a map that goes from $S^2$ to $ \mathbb{R}^6$, call it $ \psi$, that looks like:

$$ \psi(x,y,z) =(r^2\cos^2{\theta}\sin^2{\phi},r^2\sin^2{\theta}\sin^2{\phi},r^2\cos^2{\phi}, \sqrt{2}r^2\cos{\theta}\sin{\theta}\sin^2{\phi},\sqrt{2}r^2\sin{\theta}\cos{\phi}\sin{\phi},\sqrt{2}r^2\cos{\theta}\cos{\phi}\sin{\phi}) $$

Where I basically substituded the usual definition for spherical coordinates of $x$ . $y$ and $z$ in $w(x,y,z)$.I did the Jacobian matrix of this and since $(0,0,0)$ is not included then the rank of this Jacobian is also always $3$, therefore its an immersion. Is this attempt correct?

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It depends on what you mean by "since $(0, 0, 0)$ is not included". What you would like to check is that that Jacobian is injective. (And then the product of Jacobian matrices is also injective.)

But note that this can be done much simpler: First, the map $w : \mathbb R^3 \to \mathbb R^6$ is an immersion because already the projection onto the first three coordinates gives an immersion, it is just the identity map. Perhaps you saw this, and this is what you meant by "it is easy to verify". Second, because $S^2 \subset \mathbb R^3$ is a submanifold, the inclusion map $S^2 \to \mathbb R^3$ is an immersion. (Its differential is just the inclusion map on tangent spaces.) Therefore, the composition $S^2 \to \mathbb R^3 \to \mathbb R^6$ is also an immersion.

One way to prove that $S^2 \subset \mathbb R^3$ is a submanifold, is to define charts (spherical coordinates) and check that these define embeddings (are injective with injective differential). This is almost the same as what you have done. But it is not necessary to do that, if we are allowed to use that $S^2$ is a submanifold.

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  • $\begingroup$ So I must show, using the same procedure, that a map $ S^2 \longrightarrow \mathbb{R}^3$ is an immersion and then the composition $ S^2 \longrightarrow \mathbb{R}^3 \longrightarrow \mathbb{R}^6$ will be an immersion, right? And that map would be: $(r , \theta , \phi) \longrightarrow (r\cos{\theta}\sin{\phi},r\sin{\theta}\sin{\phi},r\cos(\phi))$ correct? $\endgroup$ Commented Mar 10, 2021 at 17:38
  • $\begingroup$ @MicrosoftBruh Yes, yes and no: Yes, you have to show that it is an immersion. The composition of immersions is an immersion, so then you are done. But there is no $r$, $r$ is fixed and equal to $1$. The coordinates for $S^2$ are just $\phi$ and $\theta$. $\endgroup$ Commented Mar 10, 2021 at 19:23
  • $\begingroup$ What map can I use that goes from: $S^2 \longrightarrow \mathbb{R}^3$ ? So that then I can use the Jacobian matrix to prove that this first part is an immersion. $\endgroup$ Commented Mar 11, 2021 at 9:44
  • $\begingroup$ The map that sends $(\theta, \phi)$ to $(\cos(\theta) \sin(\phi), \sin(\theta) \sin(\phi), \cos(\phi))$. (spherical coordinates) $\endgroup$ Commented Mar 11, 2021 at 13:53

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