Sphere Embedding

A 4-sphere has Positive Curvature, with

$$R^2 = x^2+y^2+z^2+w^2$$ (1)

$$2x{dx\over dw} + 2y{dy\over dw} + 2z{dz\over dw} + 2w = 0.$$ (2)

Since

$${\bf r} \equiv x\hat{\bf x} + y\hat{\bf y} + z\hat{\bf z},$$ (3)

$$dw = - {x\,dx + y\,dy + z\,dz\over w} = -{{\bf r}\cdot d{\bf r}\over\sqrt{R^2-r^2}}.$$ (4)

To stay on the surface of the sphere,

$$ds^2 = dx^2+dy^2+dz^2+dw^2$$

$$= dx^2+dy^2+dz^2 + {r^2\,dr^2\over R^2-r^2}$$

$$= dr^2+r^2\,d\Omega^2 + {dr^2\over {R^2\over r^2} - 1}$$

$$= dr^2\left({1 + {1\over{R^2\over r^2} - 1}}\right)+ r^2\,d\Omega^2$$

$$= dr^2\left({{R^2\over r^2}\over{R^2\over r^2} - 1}\right)+ r^2\,d\Omega^2$$

$$= {dr^2\over 1 - {r^2\over R^2}} + r^2\,d\Omega^2.$$ (5)

With the addition of the so-called expansion parameter, this is the Robertson-Walker line element.