A Projective Plane Isometrically Imbedded in R⁵

This imbedding in R⁵ will derive from a mapping of a 2-sphere into R⁶.
Consider first the map H: R³ R⁶ given by

H(x,y,z)=(x²/2, y²/2, z²/2, xy, yz, zx)

The restriction of H to the 2-sphere S={v: ||v||=r} gives a map H\S: S R⁶. For simplicity we may as well take S to be the unit sphere, r=1.

Recall that the standard parametrization (cos(v) cos(u), cos(v) sin(u), sin(v)) of S has E(u,v)= cos²(v), F=0, G=1. Thus to show that H\S is an isometry, at least locally, it suffices to verify (by hand or machine computation) that the parametrization

x(u,v) = H(cos(v) cos(u), cos(v) sin(u), sin(v))

also has E(u,v)=cos²(v), F=0, G=1.

To convert H\S into a mapping h: P R⁶ of the projective plane P, it's enough to check that H(p)=H(-p). Then, formally, h sends each point {p,-p} of P to H(p)=H(-p).

In fact, h is one-to-one. To show this, prove (easily) that for points p,q of S:

If H(p) = H(q), then q is either p or -p.

The preceding facts show that h isometrically imbeds the projective plane in R⁶. But we can do better:

Clearly, any hyperplane in R⁶ is geometrically equivalent to R⁵. So to cut h down to R⁵, we need only check that the image of h lies in a hyperplane.
Let k=(1,1,1,0,0,0). Then it's immediate from the definition of H that, for every point (x,y,z) in the unit sphere S,

H(x,y,z) . k = 1/2,

This implies that the image of h lies in a hyperplane orthogonal to k.

Note. This isometric imbedding is a variant of projective geometry's classical imbedding of the projective plane P² into 5-dimensional projective space P⁵.

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