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Consider the classic map $F:\mathbb{RP}^2\rightarrow \mathbb{R}^4$ defined by $F[x,y,z]=(x^2-y^2,xy,xz,yz)$. This defines a smooth embedding of $\mathbb{RP}^2$ in $\mathbb{R}^4$. It is clearly a topological embedding.

Now, what is the best way to show such map is an immersion? We can compute $DF$ and note that the matrix will have rank 2, but is there an intuitive geometric way of showing that this topological embedding is actually an immersion?

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    oh, I see. I was confused because I thought $x, y, z$ referred to projective coordinates.2011-05-23

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The following is essentially the same thing as checking that the Jacobian has rank 2, but the reinterpretation might make it much easier.

Consider $S^2$ as a subset of $\mathbb C \times \mathbb R$ and identify $\mathbb C^2 = \mathbb R^4$.

Then consider the map $G: \mathbb C\times \mathbb R \to \mathbb C^2$ given by $G(z,r) = (z^2, rz)$. The restriction $G|_{S^2}$ of this map (almost) descends to your $F$. Now you can use that to check that $G$ is an immersion, which implies that $F$ is an immersion. But that's easy, because

$dG(z,r) = \begin{pmatrix} 2z & 0 \\ r & z \end{pmatrix}$

and $z$ and $r$ can't be 0 at the same time. (Note that $dG$ acts on vectors $(\zeta, \rho)$ with $\zeta \in \mathbb C$, $\rho\in \mathbb R$)

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    Reader, beware! What Sam writes is interesting and correct. However his notation, although elegant, is extremely susceptible to misinterpretation. In particular even if the determinant of the displayed matrix is nonzero, the linear mapping it represents is guaranteed *not* to be an isomorphism!2011-05-23