# What is the metric tensor on the n-sphere (hypersphere)?

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.

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I will define the metric of $S^{n-1}$ via pullback of the Euclidean metric on ${\mathbb{R}}^{n}$.

To start with we take $n$-dimension Cartesian co-ordinates:
$(x_1,x_2……x_n)$.
The metric here is $g_{ij }= \delta_{ij}$, where $δ$ is the Kronecker delta.

We specify the surface patches of $S^{n-1}$ by the parametrization $f$:
$$x_1=r{\cos{\phi_1}},$$

$$x_p=r{\cos{\phi_p}}{\Pi_{m=1}^{p-1}}{\sin{\phi_{m}}},$$

$$x_n=r{\prod_{m=1}^{n-1}}{\sin{\phi_{m}}},$$

Where $r$ is the radius of the hypersphere and the angles have the usual range.

We see that the pullback of the Euclidean metric $g’_{ab} = (f^*g)_{ab}$ is the metric tensor of the hypersphere. Its components are:

$$g’_{ab} = g_{ij} {\frac{\partial{x_i}}{\partial{\phi_a}}} {\frac{\partial{x_j}}{\partial{\phi_b}}} = {\frac{\partial{x_i}}{\partial{\phi_a}}}{\frac{\partial{x_i}}{\partial{\phi_b}}}$$

We get $2$ cases here:

i) $a>b$ or $b>a$, For these components one obtains a series of terms with alternating signs which vanishes, $g’_{ab}=0$ and thus all off-diagonal components of the tensor vanish.

ii) $a=b$,

$$g’_{11}=1$$

$$g’_{aa} ={r^2} \prod_{m=1}^{a-1} \sin^2{\phi_{m}}$$
where $2<a<{n-1}$

The determinant is very straightforward to calculate:

$$\det{(g’_{ab})} = {r^2} \prod_{m=1}^{n-1} g’_{mm}$$

Finally, we can write the metric of the hypersphere as:

$$g’ = {r^2} \, d\phi_{1}\otimes d\phi_{1} + {r^2} \sum_{a=2}^{n-1} \left( \prod_{m=1}^{a-1} \sin^2{\phi_{m}} \right) d\phi_{a} \otimes d\phi_{a}$$

$\newcommand{\Reals}{\mathbf{R}}$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
$$r^{2} g(t/r) = \frac{4 (dt_{1}^{2} + \cdots + dt_{n}^{2})}{(\|t/r\|^{2} + 1)^{2}} = \frac{4r^{4} (dt_{1}^{2} + \cdots + dt_{n}^{2})}{(\|t\|^{2} + r^{2})^{2}}.$$