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Let $X^A$ be Cartesian coordinates on $\mathbb R^{n+1}$ and $x^\mu$ be an arbitrary coordinates on an $n$-dimensional sphere embedded in $\mathbb R^{n+1}$ in a usual way. Then $x^\mu, r = \sqrt{X_A X^A}$ would be coordinates on $\mathbb R^{n+1}$.

How can I show that in this coordinates the metric on $\mathbb R^{n+1}$ takes the form $$ ds^2 = r^2 h_{\mu\nu}(x) dx^\mu dx^\nu + dr^2, $$ where $h_{\mu\nu}$ is the metric tensor on a unit sphere induced by embedding?

OK, I can write $$ ds^2 = \eta_{AB} dX^A dX^B = \eta_{AB} \left( \frac{\partial X^A}{\partial x^\mu} dx^\mu + \frac{\partial X^A}{\partial r} dr \right) \left( \frac{\partial X^B}{\partial x^\nu} dx^\nu + \frac{\partial X^B}{\partial r} dr \right). $$ $g_{\mu\nu} = \eta_{AB} \frac{\partial X^A}{\partial x^\mu} \frac{\partial X^B}{\partial x^\nu}$ ($\eta_{AB} = diag(1,\dots,1)$) is the induced metric on a sphere $r = const$. $g_{\mu\nu}|_{r=1} = h_{\mu\nu}$. $g_{\mu\nu} = r^2 h_{\mu\nu}$ follows from dimensional considerations (but it would be nice to get a formal proof).

I'm stuck with the remaining parts of the tensor ($g_{\mu(r)} = g_{(r) \mu} \overset{?}{=} 0$ and $g_{(r)(r)} \overset{?}{=} 1$). I can differentiate $X_A X^A = r^2$ by $x^\mu$ and $r$ to get $$ X_A \frac{\partial X^A}{\partial x^\mu} = 0, \qquad X_A \frac{\partial X^A}{\partial r} = -r $$ but partial derivatives I need come in contraction with $X_A$ so it seems of no use...

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$g_{(r)\mu}$ vanishes because $\partial_r$ is orthogonal to $\partial_\mu$ (therefore to the whole tangent space to a sphere). This is because a hyperboloid is a level set and its gradient is orthogonal to it. Its gradient is $\sharp_\eta d (X^2) = \eta^{AB} 2X_B \partial_A = 2 X^A \partial_A = 2 r \partial_r$. So $\partial_r$ is proportional to the gradient to the hyperboloid and therefore is orthogonal to it's tangent hyperplanes.

And as for the last component: $$ g_{(r)(r)} = \eta(\partial_r, \partial_r) = \eta( \frac1r X^A \partial_A, \frac1r X^B \partial_B ) = \frac1{r^2} \eta(\partial_A, \partial_B) X^A X^B = -1. $$