What is the metric on the $n$-sphere in stereographic projection coordinates? The metric on the $n$-sphere is the metric induced from the ambient Euclidean
metric. Find the metric, $d\Omega^2_n$, on the $n$-sphere and the volume form, $\Omega_{S_n}$ , of $S^n$ in terms of the stereographic coordinates on $U_N =S^n − (0, . . . , 0, 1)$. 
The stereographic projection coorinates $u_j$ are given by 
$u_j=\frac{x_j}{1-x_{n+1}}$ for $j=1,\dots,n$.
We also have $x_j=\frac{2u_j}{1+\sum_{k=1}^nu_k^2}$ and $x_{n+1}=\frac{1-\sum_{k=1}^nu_k^2}{1+\sum_{k=1}^nu_k^2}$
I took differentials and put them into the expression for the Euclidean metric put things are getting messy so I'm not sure if it's right. I also have to find the components of the Levi-Civitta connection, curvature tensor, Ricci tensor.
 A: Let me present the calculation using a slightly different, in my opinion a bit clearer, notation:

*

*(Definition of stereographic coordinates) Let $\newcommand{\bs}[1]{\boldsymbol{#1}}\newcommand{\bfx}{\mathbf{x}}\newcommand{\bfu}{\mathbf{u}}(\bfx,x_{n+1})\in S^n$ be an element of the unit $n$-sphere, where $\bfx\in\mathbb R^n$, and define the stereographic coordinates via the map $\varphi:S^n\to\mathbb R^n$ such that
$$\bfu\equiv\varphi(\bfx) \equiv \frac{\bfx}{1-x_{n+1}}.$$
Note that I'm assuming unit radius here. This map is a diffeomorphism on $S^n\setminus\{N\}$ (the sphere without the north pole), and its inverse $\phi:\mathbb R^n\to S^n$ defines the parametrisation:
$$\phi(\bfu) = \left(\frac{2\bfu}{1+u^2}, \frac{1-u^2}{1+u^2}\right),$$
where $u^2\equiv\|\bfu\|^2\equiv\sum_{i=1}^n u_i^2$.


*(Overall objective) We want to compute the metric on $\mathbb R^n$ that corresponds to the Euclidean flat metric on $S^n$.
In other words, let $\bs\xi,\bs \xi'\in T_p S^n$ be tangent vectors to the sphere, for some $p\in S^n$, and denote with $\langle \bs\xi,\bs\xi'\rangle_p$ the corresponding inner product defined in the usual (Euclidean) way. We want to exploit $\phi$ to see what this inner product/metric corresponds to in the parameter space.
More explicitly, this means to take $\bs\eta,\bs\eta'\in T_{\bfu}\mathbb R^n$ and compute their inner product using
$$\langle\bs\eta,\bs\eta'\rangle_{\bfu}
= \langle \mathrm d\phi(\bfu,\bs\eta), \mathrm d\phi(\bfu,\bs\eta')\rangle,$$
where the inner product in the RHS is here just the standard Euclidean product in the $\mathbb R^{n+1}$ space in which $S^n$ is embedded, and $\mathrm d\phi(\bfu,\bs\eta)\equiv\mathrm d\phi(\bfu)(\bs\eta)\equiv\partial_{\bs\eta}\phi(\bfu)$ denotes the differential of $\phi$ at the point $\bfu$ in the direction $\bs\eta$.


*(Calculation of differential of the parametrisation) Consider then two tangent vectors in the parameter space, $\bs\eta,\bs\eta'\in T_{\bfu}\mathbb R^n\simeq\mathbb R^n$, and observe that $\mathrm d\phi:T\mathbb R^n\simeq\mathbb R^n\times\mathbb R^n\to TS^n$, where the differential can be represented as
$$\mathrm d\phi(\bfu):\mathbb R^n\to T_\bfu S^n:
\bs\eta\mapsto \sum_i \eta_i \partial_i \phi(\bfu), \\
\partial_1\phi(\bfu) =
\frac{1}{(1+u^2)^2} \left(
2(1+u^2)-(2u_1)^2,
-4 u_1 u_2, ...
, -4 u_1 u_n, -4u_1
\right),
$$
and thus more generally,
$$
\partial_i\phi(\bfu)
= \frac{2}{1+u^2} \mathbf e_i
- \frac{4 u_i}{(1+u^2)^2}
(\bfu,1), \\
\mathrm d\phi(\bfu,\bs\eta)
= \frac{2\bs\eta}{1+u^2}
- \frac{4\langle\bfu,\bs\eta\rangle}{(1+u^2)^2} (\bfu,1),
$$
where $\langle\bfu,\bs\eta\rangle\equiv\sum_{i=1}^n u_i \eta_i$, and $\bs\eta\equiv\sum_{i=1}^n \eta_i \mathbf e_i$.


*(Final result) From the above, a brief calculation shows that only the first factor in the expression of $\mathrm d\phi(\bfu,\bs\eta)$ contributes, and the result is:
$$\langle\bs\eta,\bs\eta'\rangle_{\bfu}
= \frac{4}{(1+u^2)^2} \langle\bs\eta,\bs\eta'\rangle
\equiv \frac{4}{(1+u^2)^2} \sum_{i=1}^n \eta_i \eta'_i.$$
The end result is thus that the metric remains Euclidean, up to a (position-dependent) multiplicative factor. We thus say that the metric is conformal.
This result is also often expressed concisely saying that the metric is
$$\frac{4}{(1+u^2)^2}\sum_{i=1}^n {\mathrm du}_i^2.$$
