How to find the components of the induced metric on the graph of $f:\mathbb{R}^n\to\mathbb{R}$ Let $f:\mathbb{R}^n\to\mathbb{R}$ be a smooth function, let $M$ be the graph of $f$ defined by
$$M=\{(x,f(x))\in\mathbb{R}^{n+1}:x\in\mathbb{R}^n\},$$
and let $g$ be the metric on $M$ induced by the Euclidean metric on $\mathbb{R}^{n+1}$. In terms of the global coordinate chart $(x,f(x))\mapsto x$, we can write
$$g=g_{ij}dx^i\otimes dx^j$$
for $i,j=1,\ldots,n$. According to Lee's Introduction to Smooth Manifolds, $g_{ij}$ should be computed using
$$g_{ij}=\left<\partial_i,\partial_j\right>,$$
but it doesn't work that way in some other book. Instead, I found
$$g_{ij}=\left<\partial_i+f_i\partial_{n+1},\partial_j+f_j\partial_{n+1}\right>.$$
The subscripts on $f$ denote partial differentiation. Why's that? How could that be possible?
Thank you.
 A: Let $F \colon \Bbb R^n \to \Bbb R^{n+1}$ be the embedding given by $F(x) = (x,f(x))$.
Then the induced metric $g = F^* \langle \cdot, \cdot\rangle_{\Bbb R^{n}}$ on $\Bbb R^n$ is
$$
\forall x \in \Bbb R^n,\forall u,v \in T_x \Bbb R^n \simeq \Bbb R^n,\quad  g_x(u,v) = \langle dF(x)u, dF(x)v\rangle. 
$$
Identifying $\Bbb R^n$ with $\Bbb R^n \times \{0\} \subset \Bbb R^{n+1}$, it follows from direct computations that
$$
g_{ij}(x) = g_x(\partial_i,\partial_j) = \left\langle \partial_i + \partial_if(x)\partial_{n+1}, \partial_j + \partial_jf(x)\partial_{n+1}\right\rangle_{\Bbb R^{n+1}}, 
$$
so both formulations are right: the first one is the formulation in terms of the induced metric $g = F^*\langle \cdot,\cdot \rangle_{\Bbb R^{n+1}}$ on $\Bbb R^n$, the second one is in terms of the ambient metric $\langle \cdot,\cdot \rangle_{\Bbb R^{n+1}}$ on $\Bbb R^{n+1}$.
A: This answer is not original. @Didier actually made the first attempt to consider the map defined by $F(x) = (x,f(x))$, but people like me may want to see something more detailed.
First off, the formula
$$g_{ij}=\left<\partial_i+f_i\partial_{n+1},\partial_j+f_j\partial_{n+1}\right>\tag{1}$$
was found in Geometric Relativity written by Dan A. Lee. This is very different from the formula
$$g_{ij}=\left<\partial_i,\partial_j\right>,\tag{2}$$
which can be found in almost every textbook on manifolds. Lee didn't talk much about how (1) was related to (2), but if you have the same question I asked earlier, then you might be misunderstanding the meaning of the symbol $g$ used by Lee. Without doubt, $M$ is an embedded $n$-dimensional submanifold of $\mathbb{R}^{n+1}$ and is equipped with the induced metric $g:=\iota_M^*\overline{g}\ $ if the inclusion map of $M$ in $\mathbb{R}^{n+1}$ and the Euclidean metric on  $\mathbb{R}^{n+1}$ are denoted respectively by $\iota_M$ and $\overline{g}$. And with the help of a coordinate chart on $M$, you have every right to conclude that for every $p$ in the coordinate neighborhood (a subset of $M$),
$$g_{ij}(p)=g_p(\partial_i|_p,\partial_j|_p),$$
which is merely an alternative form of (2). And there is definitely nothing wrong about your idea. But the truth is, Lee is NOT talking about your $g$. He has, in fact, defined $g$ to be $F^*\overline{g}$. The map $F$ is sometimes called a parametrization of $M$, and if you want further information about it, I would suggest your going to page 111 and page 333 of Introduction to Smooth Manifolds by John M. Lee. Now you gotta understand that $p$ is roaming through $\mathbb{R}^n$ and we have
$$\begin{align}
g_{ij}(p)&=(F^*\overline{g})_p(\partial_i|_p,\partial_j|_p)\\
&=\overline{g}_{F(p)}(dF_p(\partial_i|_p),dF_p(\partial_j|_p)),\tag{3}
\end{align}$$
where we have used (2), the ultimate code and ethos you will never break in computing $g_{ij}$. Now, in order to deduce (1), you have to unravel (3), which begs the understanding that
$$dF_p(\partial_i|_p)=\sum_{j=1}^{n+1}\frac{\partial F^j}{\partial x^i}(p)\partial y^j|_{F(p)},$$
where $F^j$'s are the components of $F$ and $y^j$'s are coordinates on $\mathbb{R}^{n+1}$. The result should be clear to you now, and we won't expand on the algebra. Thank you.
