Pullback metric of inversion map Let $I:\mathbb{R}^{n}\setminus\{0\} \longrightarrow \mathbb{R}^{n}\setminus\{0\}$ be the inversion map, i.e. $I(x)=\frac{x}{|x|^{2}}$ and equip $\mathbb{R}^{n}\setminus\{0\}$ with the standard euclidean metric $g_{eucl}=\langle\cdot,\cdot\rangle$. Show that for any $x\in \mathbb{R}^{n}\setminus\{0\}$ we have $I^{*}\langle\cdot,\cdot\rangle=\frac{1}{|x|^{4}}\langle\cdot,\cdot\rangle$ on the tangent space of $x$.
My work so far:
By definition, I have to show that for any $v,w\in T_{x}\left(\mathbb{R}^{n}\setminus\{0\}\right)$ we have $\langle d_{x}I(v),d_{x}I(w)\rangle = \frac{1}{|x|^{4}}\langle v,w\rangle$, where $d_{x}I$ is the differential of $I$ at $x$. Because we are in $\mathbb{R}^{n}\setminus\{0\}$, instead of choosing a curve $c$ that satisfies $c(0)=x$ and $c'(0)=v$ for $v\in T_{x}\left(\mathbb{R}^{n}\setminus\{0\}\right)$, I compute $d_{x}I$ by the product (or quotient) rule: $d_{x}I = \frac{1}{|x|^{4}}\left( |x|^{2}(\delta_{ij})_{i,j} -2(x_{i}x_{j})_{i,j}\right)\in \mathbb{R}^{n\times n}$. Now, it is no restriction if we set $v=e_{i}$ and $w=e_{j}$ to make the computation a little bit easier. Computing $\langle d_{x}I(e_{i}),d_{x}I(e_{j})\rangle$, I end up with an expression that containes $\frac{1}{|x|^{4}}\langle e_{i},e_{j} \rangle$ (which is good, I think) and something, that was not equal to zero (but it should). Can you give me some help with the last computation? I know it is probably an obvious mistake, but I don't see it.
 A: By definition of the pullback by $I$ of the euclidean metric, $(I^*\langle\cdot,\cdot\rangle)_x(v,w) = \langle\mathrm{d}I(x)v,\mathrm{d}I(x)w \rangle_{{I(x)}}$. The computation shows that $\mathrm{d}I(x): T_x\mathbb{R}^n \to T_{I(x)}\mathbb{R}^n$ is just, while canonically identifying these two tangent spaces to $\mathbb{R}^n$, equal to $$\mathrm{d}I(x)v = \frac{{v}}{\|x\|^2}-2\langle x,v\rangle \frac{x}{\|x\|^4}$$
Let $u$ and $v$ be two tangent vectors to $x$. Then
\begin{align}
(I^*g)_x(u,v) &= \left\langle \dfrac{u}{\|x\|^2}-2\frac{\langle u,x\rangle x}{\|x\|^4},\dfrac{v}{\|x\|^2}-2\frac{\langle v,x\rangle x}{\|x\|^4} \right\rangle \\
&=\frac{\langle u,v\rangle}{\|x\|^4} -2\dfrac{\langle u,x\rangle \langle x,v\rangle}{\|x\|^6} -2 \dfrac{\langle v, x \rangle\langle x ,u\rangle}{\|x\|^6} + 4 \dfrac{\langle u,x\rangle \langle v,x \rangle \|x\|^2}{\|x\|^8} \\
&= \dfrac{\langle u,v\rangle}{\|x\|^4} - 4 \dfrac{\langle x , u \rangle \langle x , v \rangle}{\|x\|^6} + 4 \dfrac{\langle x,u\rangle \langle x,v\rangle}{\|x\|^6}\\
&= \dfrac{\langle u,v\rangle}{\|x\|^4}
\end{align}
So you are done. Remark that you do not have to compute in any coordinate system or in any orthonormal basis.
