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?

  • $\begingroup$ 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. $\endgroup$ – Manuel May 23 '11 at 14:50
  • $\begingroup$ I don't understand how you've defined the embedding. What are $x, y, z$? $\endgroup$ – Qiaochu Yuan May 23 '11 at 15:35
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    $\begingroup$ $(x,y,z)\in S^2$ and $[x,y,z]=\pi(x,y,z)$ where $\pi$ is the quotient map. $\endgroup$ – Manuel May 23 '11 at 16:18
  • $\begingroup$ oh, I see. I was confused because I thought $x, y, z$ referred to projective coordinates. $\endgroup$ – Qiaochu Yuan May 23 '11 at 17:51

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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    $\begingroup$ 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! $\endgroup$ – Georges Elencwajg May 23 '11 at 21:50

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