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

Solution 1:

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 $2\leq a\leq {n-1}$

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:

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

Solution 2:

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