TITLE: What is the metric tensor on the n-sphere (hypersphere)?
QUESTION [37 upvotes]: I am considering the unit sphere (but an extension to one of radius $r$ would be appreciated) centered at the origin. Any coordinate system will do, though the standard angular one (with 1 radial and $n-1$ angular coordinates) would be preferable. 
I know that on the 2-sphere we have $ds^2 = d\theta^2+\sin^2(\theta)d\phi^2$ (in spherical coordinates) but I'm not sure how this generalizes to $n$ dimensions.
Added note: If anything can be discovered only about the determinant of the tensor (when presented in matrix form), that would also be quite helpful.

REPLY [14 votes]: $\newcommand{\Reals}{\mathbf{R}}$For posterity: Fix $r > 0$, and let $S^{n}(r)$ denote the sphere of radius $r$ centered at the origin in $\Reals^{n+1}$. Stereographic projection from the north pole $N = (0, \dots, 0, 1)$ on the unit sphere $S^{n} = S^{n}(1)$ defines a diffeomorphism $\Pi_{N}:S^{n} \setminus \{N\} \to \Reals^{n}$ given in Cartesian coordinates by
\begin{align*}
\Pi_{N}(x_{1}, \dots, x_{n}, x_{n+1}) &= \frac{1}{1 - x_{n+1}}(x_{1}, \dots, x_{n}), \\
\Pi_{N}^{-1}(t_{1}, \dots, t_{n}) &= \frac{(2t_{1}, \dots, 2t_{n}, \|t\|^{2} - 1)}{\|t\|^{2} + 1}.
\end{align*}
In these coordinates, the induced (round) metric on the unit sphere is well-known (and easily checked) to be conformally-Euclidean:
$$
g(t) = \frac{4 (dt_{1}^{2} + \cdots + dt_{n}^{2})}{(\|t\|^{2} + 1)^{2}}.
$$
Stereographic projection from the north pole $(0, \dots, 0, r)$ of $S^{n}(r)$ is given by the scaled mapping $x \mapsto t = r\Pi_{N}(x/r)$, whose inverse is $t \mapsto x = r\Pi_{N}^{-1}(t/r)$, i.e.,
\begin{align*}
r\Pi_{N}(x_{1}/r, \dots, x_{n}/r, x_{n+1}/r) &= \frac{1}{r - x_{n+1}}(x_{1}, \dots, x_{n}), \\
r\Pi_{N}^{-1}(t_{1}/r, \dots, t_{n}/r) &= \frac{\bigl(2t_{1}, \dots, 2t_{n}, r(\|t/r\|^{2} - 1)\bigr)}{\|t/r\|^{2} + 1}.
\end{align*}
The induced metric in these coordinates is consequently
$$
r^{2} g(t/r) = \frac{4 (dt_{1}^{2} + \cdots + dt_{n}^{2})}{(\|t/r\|^{2} + 1)^{2}}
= \frac{4r^{4} (dt_{1}^{2} + \cdots + dt_{n}^{2})}{(\|t\|^{2} + r^{2})^{2}}.
$$