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Consider the projective space $\mathbb R P^2$, constructed as the quotient space of $S^2$ with the equivalence relation where we identify opposite points (or alternatively, the space of all lines in $\mathbb R^3$ through $0$). I now want to show that the mapping

$$f: \mathbb R P^2 \to \mathbb R^4, (x : y : z) \mapsto (x^2 - y^2, x y, x z, y z)$$

is an injective immersion, in order to prove that $\mathbb R P^2$ is a submanifold of $\mathbb R^4$.

Now I've found similar threads about this topic like this and this, but the part where I'm struggling is actually not showing that $f$ is an immersion, but that $f$ is injective, which I couldn't find anywhere.

I tried to start with an image point $(a, b, c, d)$ of $f$ and tried to show that this already determines the equivalence class of any $(x, y, z) \in S^2$ with $f(x, y, z) = (a, b, c, d)$, but no matter which components of that equation I tried to add and subtract, I couldn't get anywhere. I was thinking that maybe there is an easy argument to it that I've just missed? Or it's really just a flat calculation that I can't seem to put together. Any help would be appreciated.

moran
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1 Answers1

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Note: Here I'm viewing $\Bbb RP^2 $ as a quotient of $\Bbb R^3 \setminus 0$ instead of as a quotient of $S^2$, to go from the former (triples $(x,y,z)$ with the relation $(x,y,z) \sim (rx,ry,rz)$) to the latter ($S^2$ with antipodal points identified) you need to take either of the elements of norm 1 from a given equivalence class.

If either of $xz$ or $xy$ are nonzero, then we know that $x\ne 0$ and can use $(x:y:z) = (1:y/x:z/x) = (1:yz/xz:yz/xy)$.

If $x=0$ (indicated by $xy=xz=0$) but $yz \ne 0$, then we can use $(x:y:z) = (0:1:y/z) = (0:1:y^2/yz)$, where $y^2$ is known since $x^2-y^2 = -y^2$.

If all three values $xy = yz = xz$ then there are three possible points: $(1:0:0), (0:1:0), (0:0:1)$, which can be differentiated by observing whether $x^2-y^2$ is positive, negative, or zero.

Rolf Hoyer
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