I obtained the following $g_{ij}$ for this metric. Do they make sense? I have the following problem:
Let $\varphi:\mathbb{R^n}\rightarrow \mathbb{R}^{n+k}$ be such that $(M,\varphi)$ is a differentiable manifold, where $M=\varphi(\mathbb{R}^n)\subseteq\mathbb{R}^{n+k}$. In other words, $\varphi$ is a global parametrization for $M$. Let $\langle,\rangle_{\varphi(0)}$ be the inner product in $T_{\varphi(0)}M$ such that $\{\frac{\partial}{\partial x_1}\vert_{\varphi(0)},\ldots,\frac{\partial}{\partial x_n}\vert_{\varphi(0)}\}$ is an orthonormal basis. Now, for $a\in \mathbb{R}^n$, we consider $L_a:M\rightarrow M$, defined by $L_a(\varphi(x))=\varphi(x+a)$. We choose $\langle,\rangle_{\varphi(a)}$ an inner product in $T_{\varphi(a)}M$ such that
$$DL_a(\varphi(0)):T_{\varphi(0)}M\rightarrow T_{\varphi(a)}M$$
is an euclidean isometry. Find the components $g_{ij}$ of $(M,\langle,\rangle)$.
My attempt:
The first thing I did was calculating the localization $f$ of $L_a$:
$$x\overset{\varphi}{\longmapsto}\varphi(x)\overset{L_a}{\longmapsto}\varphi(x+a)\overset{\varphi^{-1}}{\longmapsto}x+a$$
So $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is defined by $f(x)=x+a$. Thus, the jacobian matrix of $f$ is the identity $I$ and it doesn't depend on the point $\varphi(x)$ nor $a$. Now I want to calculate $$g_{ij}=\langle\frac{\partial}{\partial x_i}\vert_{\varphi(x)},\frac{\partial}{\partial x_j}\vert_{\varphi(x)}\rangle_{\varphi(x)}$$
As we know that
$$DL_x(\varphi(0)):T_{\varphi(0)}M\rightarrow T_{\varphi(x)}M$$
is an euclidean isometry, then:
$$\langle u,v\rangle_{\varphi(0)}=\langle DL_x(\varphi(0))(u),DL_x(\varphi(0))(v)\rangle_{\varphi(x)}$$
for all $u,v\in T_{\varphi(0)}M$.
But the jacobian of the localization of the previous differential is the identity, so $DL_x(\varphi(0))(\frac{\partial}{\partial x_i}\vert_{\varphi(0)})=\frac{\partial}{\partial x_i}\vert_{\varphi(x)}$, and we can conclude that:
$$g_{ij}=\langle\frac{\partial}{\partial x_i}\vert_{\varphi(x)},\frac{\partial}{\partial x_j}\vert_{\varphi(x)}\rangle_{\varphi(x)}=\langle\frac{\partial}{\partial x_i}\vert_{\varphi(0)},\frac{\partial}{\partial x_j}\vert_{\varphi(0)}\rangle_{\varphi(0)}=\delta_{ij}$$
But this sounds strange for me. Any help or suggestions are welcome. Thanks!
 A: I think that that is exactly the answer. You are prescribing a scalar product in the tangent space at point $\varphi(0)$ whose local expression is $\delta_{ij}$.
Then, you are applying a transitive family of diffeomorphisms $L_a$ and prescribing in every point the scalar product that these diffeomorphisms induce. Since in local coordinates they are translations, what you obtain is logical and coherent. The differential map is $dL_a=Id$ in local coordinates, so the coordinate expression for the induced scalar product is $\delta_{ij}$.
A: This seems unnecessarily confusing to me. Since $\varphi^{-1}$ is a global chart, aren’t you just looking at its pullback of the standard metric on $\Bbb R^n$?
The pullback of the metric by the function $f\colon M\to\Bbb R^n$ is given by the formula
$$f^*\langle \cdot,\cdot \rangle (v,w) = \langle Df_p(v),Df_p(w)\rangle \quad\text{for tangent vectors }v,w\in T_pM.$$
Here $\langle\cdot,\cdot\rangle$ is the standard inner product on $\Bbb R^n$. In particular, if you work in the chart $f$ on $M$, by definition you will have $g_{ij}=\delta_{ij}$ for all $i,j$ at every point.
