Currently, I started reading a book "Riemannian Geometry, Peter Petersen". In this book, Example 3 on page 3 explains the Euclidean sphere of radius $r$. $$S^n(r) = \{x \in R^{n+1} : |x| = r\}$$ The authors said that the metric induced from the embedding $j:S^n(r) \to R^{n+1}$ is the canonical metric on $S^n(r)$. The embedding is given by an inclusion.
I want to find the metric on the sphere induced from the given map. Since the embedding is given, we can construct a Riemmanian metric on $S^n(r)$ by pulling back the canonical metric on Euclidean space. $$g_{S^n(r)}(v,w) = g_{R^n}(Dj(v),Dj(w))$$ for $v,w \in T_pS^n.$
Since $j$ is an inclusion map, $Dj=Id$, so I think $$g_{S^n(r)}(v,w) = g_{R^n}(Dj(v),Dj(w)) = g_{R^n}(v,w).$$ Then, is the metric $g_{s^n(r)}$ the same as the metric $g_{R^n}$?
Edit: For $v,w \in T_pS^n$, $$g_{S^n(r)}(v,w) = g_{R^n}(Dj(v),Dj(w)) = g_{R^n}(v,w)$$ as mentioned above. Take an orthonormal basis $\{e_1, e_2, \cdots, e_n\}$. Then $v= \sum v^i e_i$ and $w= \sum w^i e_i$ for some $v^i$ and $w^i$. I want to calculate $$g_{S^n(r)}(v,w).$$ At first, I thought $$g_{S^n(r)}(v,w) = g_{R^n}(v,w) = v^i w^i$$ because $g_{R^n}$ is an identity matrix. However, I caught my mistake that $v=v^ie_i$ and $w=w^i e_i$ in the Euclidean space $R^{n+1}$. I cannot proceed further. How can I evaluate $g_{S^n(r)}(v,w)$ correctly?
Petersen recommends using Lee's Introduction to Smooth Manifolds in the preface, so let's just unravel his definitions to see what we get for the differential of a smooth map between smooth manifolds $F : M \to N$. Take coordinate charts $(U, \phi)$ around $p$ and $(V, \psi)$ around $F(p)$ and note that for any $v \in T_pM$ we have, for $f \in C^\infty(U)$:
\begin{align} \text DF(v)(f) &= v(f\circ F) = \sum_{i=1}^m v_i \frac{\partial}{\partial x^i} f\circ F \circ \phi^{-1}\big|_{x = \phi(p)} \\ &=\sum_{i=1}^m v_i \frac{\partial}{\partial x^i} [(f \circ \psi^{-1}) \circ(\psi \circ F \circ \phi^{-1})]\big|_{x = \phi(p)} \\ &=\sum_{i=1}^m v_i \sum_{j = 1}^n\frac{\partial}{\partial y^j} (f \circ \psi^{-1})\big |_{y = \psi(F(p))} \cdot \frac{\partial [\psi \circ F \circ \phi^{-1}]_j}{\partial x^i}\big|_{x = \phi(p)} \\ &=\sum_{j = 1}^n\left(\sum_{i=1}^m v_i \frac{\partial [\psi \circ F \circ \phi^{-1}]_j}{\partial x^i}\big|_{x = \phi(p)}\right)\frac{\partial}{\partial y^j} (f \circ \psi^{-1})\big |_{y = \psi(F(p))} \end{align} In particular, on basis elements: $$ \text DF\left(\frac{\partial}{\partial x^i}\right)(f) = \sum_{j = 1}^n\frac{\partial [\psi \circ F \circ \phi^{-1}]_j}{\partial x^i}\big|_{x = \phi(p)}\frac{\partial}{\partial y^j} (f \circ \psi^{-1})\big |_{y = \psi(F(p))} $$ since we want to find the components of the induced metric, it suffices to evaluate it on the tensor product of basis elements \begin{align} g_{M}\left(\frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}\right) &=\sum_{k,r = 1}^n\frac{\partial [\psi \circ F \circ \phi^{-1}]_k}{\partial x^i}\big|_{x = \phi(p)}\frac{\partial [\psi \circ F \circ \phi^{-1}]_r}{\partial x^j}\big|_{x = \phi(p)}g_N\left(\frac{\partial}{\partial y^k} (f \circ \psi^{-1})\big |_{y = \psi(F(p))},\frac{\partial}{\partial y^r} (f \circ \psi^{-1})\big |_{y = \psi(F(p))} \right) \\ &=\sum_{k,r = 1}^n\frac{\partial [\psi \circ F \circ \phi^{-1}]_k}{\partial x^i}\big|_{x = \phi(p)}\frac{\partial [\psi \circ F \circ \phi^{-1}]_r}{\partial x^j}\big|_{x = \phi(p)}[g_N]_{kr} \end{align} where $[g]_{kr}$ are the local coordinate components of the metric $g$. In particular, we can take the canonical metric on $N = \Bbb R^n$ to reduce $[g_N]_{kr} = \delta_{kr}$ if $F$ is the embedding $S^{n-1}\hookrightarrow \Bbb R^n$. Moreover, the chart $\psi$ is given by the identity and so the above equation yields, for your preferred parametrization $\phi$ of the sphere: \begin{align} [g_{S^{n-1}}]_{ij} &=\sum_{k = 1}^n\frac{\partial [\phi^{-1}]_k}{\partial x^i}\big|_{x = \phi(p)}\frac{\partial [\phi^{-1}]_k}{\partial x^j}\big|_{x = \phi(p)} \end{align}