How to isometrically imbed the projective plane (identifying antipodal points of the unit sphere) in $\mathbb{R^5}$?
My textbook indicates that there is an isometric imbedding of the projective plane in $\mathbb{R}^6$ via
$$F(x,y,z)=(x^2/\sqrt{2}, y^2/\sqrt{2}, z^2/\sqrt{2}, xy, xz, yz).$$
But how to reduce the dimension from $6$ to $5$?
Check the norm of each point in that image, they're all on a sphere: since that image misses a point it lies (up to an homeomorphism) in $\mathbb{R}^5$.