# embedding of $\mathbb{RP}^2$ in $\mathbb{R}^4$

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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Well, in this case it is easy to compute $DF$ on every coordinate chart and the identity 2x2 matrix will appear as a minor. But my question is a geometric one, I guess I wanted to ask if there is a intuitive geometric way to understand immersions. Maybe the question is confusing, sorry. –  Manuel May 23 '11 at 14:50
I don't understand how you've defined the embedding. What are $x, y, z$? –  Qiaochu Yuan May 23 '11 at 15:35
$(x,y,z)\in S^2$ and $[x,y,z]=\pi(x,y,z)$ where $\pi$ is the quotient map. –  Manuel May 23 '11 at 16:18
oh, I see. I was confused because I thought $x, y, z$ referred to projective coordinates. –  Qiaochu Yuan May 23 '11 at 17:51

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 $\zeta \in \mathbb C$, $\rho\in \mathbb R$)