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.


3 Answers 3


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 $\delta$ is the Kronecker delta.

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



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) $a=b$,


$$g'_{aa} ={r^2} \prod_{m=1}^{a-1} \sin^2{\phi_{m}}$$
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:

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

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    $\begingroup$ Great, thanks for the step-by-step explanation. Although what you've written is fairly clear, you can make it clearer by using the TeX environment, with the $ delimiters, like instead of g'rϕk, write $g'_{r\phi_k}$, which comes out as $g'_{r\phi_k}$. $\endgroup$ Commented May 27, 2013 at 11:39
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    $\begingroup$ This is a very good answer and very useful too, as it is not that easy to find this piece of information on the net. Proper formatting would make it great. $\endgroup$ Commented Oct 20, 2014 at 9:12
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    $\begingroup$ @GiuseppeNegro Done. Thanks for the feedback. $\endgroup$ Commented Nov 10, 2014 at 14:04
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    $\begingroup$ It seems $g'_{11}=r^2$, no? $\endgroup$
    – gamebm
    Commented Nov 13, 2018 at 14:17
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    $\begingroup$ Can somebody explain more thoroughly why the off diagonal terms vanish? $\endgroup$ Commented Apr 16, 2019 at 11:15

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


As asked in a comment by Ivan Burbano, here are more details about $g'_{ab}=0$ for $a\neq b$ in orange_soda's answer. Without loss of generality, assume $1\leq a<b\leq n-1$. Note that \begin{equation*} \dfrac{\partial x_p}{\partial \phi_a}=-r\delta_{pa}\sin\phi_a\prod_{m=1}^{p-1}\sin\phi_m+r\cos\phi_p\left(\prod_{m=1}^{p-1}\sin\phi_m\right)\left(\sum_{m=1}^{p-1}\dfrac{\delta_{ma}\cos\phi_m}{\sin\phi_m}\right),\quad \forall1<p<n \end{equation*} and \begin{equation*} \dfrac{\partial x_n}{\partial \phi_a}=r\dfrac{\cos\phi_a}{\sin\phi_a}\prod_{m=1}^{n-1}\sin\phi_m, \end{equation*} where $\delta_{ij}$ is the Kronecker delta and I used the product rule for more than two factors. Also, we can see that $\frac{\partial x_i}{\partial \phi_b}=0$ for all $i=1,\ldots,b-1$. Then, \begin{align*} g'_{ab}&=\sum_{p=b}^{n-1}\left[r\cos\phi_p\left(\prod_{m=1}^{p-1}\sin\phi_m\right)\left(\sum_{m=1}^{p-1}\dfrac{\delta_{ma}\cos\phi_m}{\sin\phi_m}\right)\right]\dfrac{\partial x_p}{\partial\phi_b}+r^2\dfrac{\cos\phi_a\cos\phi_b}{\sin\phi_a\sin\phi_b}\prod_{m=1}^{n-1}\sin^2\phi_m\\ &=r\dfrac{\cos\phi_a}{\sin\phi_a}\left[\sum_{p=b}^{n-1}\left(\prod_{m=1}^{p-1}\sin\phi_m\right)\cos\phi_b\dfrac{\partial x_p}{\partial\phi_b}+r\dfrac{\cos\phi_b}{\sin\phi_b}\prod_{m=1}^{n-1}\sin^2\phi_m\right]. \end{align*} Dividing the sum $\sum_{p=b}^{n-1}$ into $p=b$ and $\sum_{p=b+1}^{n-1}$, we obtain \begin{align*} g'_{ab}&=r^2\dfrac{\cos\phi_a}{\sin\phi_a}\left[-\sin\phi_b\cos\phi_b\prod_{m=1}^{b-1}\sin^2\phi_m+\dfrac{\cos\phi_b}{\sin\phi_b}\sum_{p=b+1}^{n-1}\cos^2\phi_p\left(\prod_{m=1}^{p-1}\sin^2\phi_m\right)+\dfrac{\cos\phi_b}{\sin\phi_b}\prod_{m=1}^{n-1}\sin^2\phi_m\right]\\ &=r^2\dfrac{\cos\phi_a\cos\phi_b}{\sin\phi_a\sin\phi_b}\left[-\prod_{m=1}^b\sin^2\phi_m+\sum_{p=b+1}^{n-1}\cos^2\phi_b\left(\prod_{m=1}^{p-1}\sin^2\phi_m\right)+\prod_{m=1}^{n-1}\sin^2\phi_m\right]. \end{align*} On the other hand, by $\sin^2\theta+\cos^2\theta=1$, we have \begin{align*} \sum_{p=b+1}^{n-1}\cos^2\phi_b\left(\prod_{m=1}^{p-1}\sin^2\phi_m\right)+\prod_{m=1}^{n-1}\sin^2\phi_m&=\sum_{p=b+1}^{n-2}\cos^2\phi_b\left(\prod_{m=1}^{p-1}\sin^2\phi_m\right)+\prod_{m=1}^{n-2}\sin^2\phi_m\\ &=\sum_{p=b+1}^{n-3}\cos^2\phi_b\left(\prod_{m=1}^{p-1}\sin^2\phi_m\right)+\prod_{m=1}^{n-3}\sin^2\phi_m\\ &\quad\vdots\\ &=\prod_{m=1}^b\sin^2\phi_m. \end{align*} Therefore, we conclude $g'_{ab}=0$.


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