The metric of a sphere and of a closed universe

A sphere is special because it is homogeneous and isotropic within its own surface: a two-dimensional creature living on it finds every point and every direction equivalent. That is exactly the property our universe has on large scales. So the curved spaces cosmology needs are spheres (positive curvature), flat space, and their negatively curved relative.

Take the sphere $x^2 + y^2 + z^2 = R^2$ in ordinary three-dimensional space, and label points with the usual polar angles: $\theta$ measured down from the $z$-axis and $\phi$ around it. Consider two points that differ by $d\theta$ and $d\phi$, and change one angle at a time:

• Change $\theta$: you move along a great circle (a meridian) of radius $R$, so $ds = R\,d\theta$.
• Change $\phi$: you move around a circle of latitude, whose radius is $R\sin\theta$ — zero at the poles, largest at the equator — so $ds = R\sin\theta\,d\phi$.

The two displacements are perpendicular, one along a meridian and one along a circle of latitude. Pythagoras then gives the metric:

Now the key move: forget the three-dimensional space. A two-dimensional creature needs only the coordinates $\theta, \phi$ and this formula to do all of its geometry. Gauss's "inner properties" are all here.

Guth also shows a brute-force route that needs no pictures. Write $x = R\sin\theta\cos\phi$, $y = R\sin\theta\sin\phi$, $z = R\cos\theta$, use the chain rule for $dx$, $dy$, $dz$, and add $dx^2 + dy^2 + dz^2$. After several uses of $\cos^2 + \sin^2 = 1$, the same metric comes out.

A closed universe has three space dimensions, so we need the three-dimensional analogue: the 3-sphere, the set of points

in a four-dimensional Euclidean space. Nobody can picture four dimensions, and Guth's advice is: usually don't; the equations are enough. One crutch helps, though. Pretend the fourth coordinate $w$ is time. At each "moment" $w$, the slice through the 3-sphere is an ordinary 2-sphere of radius $\sqrt{R^2 - w^2}$: it appears as a point at $w = -R$, grows to radius $R$ at $w = 0$, and shrinks back to a point at $w = +R$.

To give the 3-sphere coordinates, add a new angle $\psi$ measured from the $w$-axis, just as $\theta$ was measured from the $z$-axis:

Then multiply the 2-sphere formulas for $x, y, z$ by $\sin\psi$. The recipe repeats in any number of dimensions: add an axis, add an angle measured from it, and multiply the old coordinates by its sine. The angles range over $0 \le \psi \le \pi$, $0 \le \theta \le \pi$ and $0 \le \phi < 2\pi$.

Now find the metric as before. Changing $\psi$ alone moves along a circle of radius $R$: $ds = R\,d\psi$. Changing $\theta$ and $\phi$ alone moves on a 2-sphere of radius $R\sin\psi$. And the two kinds of displacement are perpendicular. Changing $\theta$ or $\phi$ leaves $w$ unchanged, so the $w$ part of the four-dimensional dot product vanishes. In the remaining $(x, y, z)$ directions, changing $\psi$ rescales all three by a common factor — a radial move — while changing $\theta$ or $\phi$ moves tangentially along a sphere. So once again the squares add:

This, Guth says, is precisely the space of a closed universe: homogeneous and isotropic. (Unlike flat space, it is also finite in volume.)

In curved space there is no single natural coordinate system, and coordinates are not distances. A useful alternative to $\psi$ is

(using $du = \cos\psi\,d\psi = \sqrt{1-u^2}\,d\psi$). Two warnings come with it. First, each value of $u$ below 1 labels two points, one on each "hemisphere" in $w$; $u = 1$ is the equator between them. Second, near the origin ($u \ll 1$) the factor $1/(1-u^2)$ is almost 1, and the metric becomes the ordinary flat metric in polar coordinates — the space is locally Euclidean, as Gauss's theorem promises. Isotropy about the origin is obvious in this form; homogeneity is still true but hidden, because moving the origin corresponds to a rotation in the four-dimensional space.

• On a 2-sphere, varying $\theta$ gives $ds = R\,d\theta$ and varying $\phi$ gives $ds = R\sin\theta\,d\phi$; they are perpendicular, so $ds^2 = R^2(d\theta^2 + \sin^2\theta\,d\phi^2)$.
• The 3-sphere $x^2 + y^2 + z^2 + w^2 = R^2$ gets a third angle $\psi$, and $ds^2 = R^2[d\psi^2 + \sin^2\psi(d\theta^2 + \sin^2\theta\,d\phi^2)]$: the space of a closed universe.
• With $u = \sin\psi$ the metric becomes $R^2[du^2/(1-u^2) + u^2(\ldots)]$, visibly flat near the origin.
• Once the metric is known, the embedding space can be forgotten: the geometry is fully inside the metric.