# 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.

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:

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

• 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$g'_{r\phi_k}$. May 27, 2013 at 11:39 • 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. Oct 20, 2014 at 9:12 • @GiuseppeNegro Done. Thanks for the feedback. Nov 10, 2014 at 14:04 • It seems$g'_{11}=r^2$, no? Nov 13, 2018 at 14:17 • Can somebody explain more thoroughly why the off diagonal terms vanish? 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 $$r^{2} g(t/r) = \frac{4 (dt_{1}^{2} + \cdots + dt_{n}^{2})}{(\|t/r\|^{2} + 1)^{2}} = \frac{4r^{4} (dt_{1}^{2} + \cdots + dt_{n}^{2})}{(\|t\|^{2} + r^{2})^{2}}.$$