I want to further explain why I care about this issue.

Consider $\{x_{n+1}>0 \}$ \begin{equation*} \left\{\begin{aligned} & y_1=x_1\\ &\vdots \\ &y_n=x_n\\ &y_{n+1}=x_{n+1}-(r^2-\sum^{n}_{i=1}x_i^2)^\frac{1}{2}. \end{aligned}\right. \end{equation*} We have \begin{equation*} \begin{aligned} & \begin{pmatrix} \frac{\partial}{\partial y_1}\\ \vdots\\ \frac{\partial}{\partial y_{n+1}} \end{pmatrix}=\begin{pmatrix} I&-\begin{pmatrix} \frac{x_1}{(r^2-\sum^n_{i=1} x_i^2)^\frac{1}{2}}\\ \vdots\\ \frac{x_n}{(r^2-\sum^n_{i=1} x_i^2)^\frac{1}{2}}\\ \end{pmatrix}\\ 0&1 \end{pmatrix}\begin{pmatrix} \frac{\partial}{\partial x_1}\\ \vdots\\ \frac{\partial}{\partial x_{n+1}} \end{pmatrix}=\begin{pmatrix} I&-\begin{pmatrix} \frac{x_1}{x_{n+1}}\\ \vdots\\ \frac{x_n}{x_{n+1}}\\ \end{pmatrix}\\ 0&1 \end{pmatrix}\begin{pmatrix} \frac{\partial}{\partial x_1}\\ \vdots\\ \frac{\partial}{\partial x_{n+1}} \end{pmatrix}, \end{aligned} \end{equation*} i.e. for $i\le n$, $\frac{\partial}{\partial y_i} = \frac{\partial}{\partial x_i} - \frac{x_i}{x_{n+1}} \frac{\partial}{\partial x_{n+1}}$(this formula told us the embedding of $\frac{\partial}{\partial y_i}$ in $\mathbb{R}^{n+1}$) and ${i=n+1}$, $\frac{\partial}{\partial y_{n+1}} =\frac{\partial}{\partial x_{n+1}} $, $\{ \frac{\partial}{\partial x_{i}} \}_{i=1,\cdots,n+1}$ is the standard orthogonal basis in Euclidean space $\mathbb{R}^{n+1}$, where the second equation is due to we restrict $y_{n+1}\equiv 0$.

For $1\le i,j\le n$, we have $g_{ij} = \delta_{ij} + \frac{x_ix_j}{x_{n+1}^2}$ and $g^{ij} = \delta_{ij} - \frac{x_ix_j}{r^2}$, moreover $$ \Gamma_{ij}^k = \left(\frac{x_ix_j}{x_{n+1}^2}+\delta_{ij}\right)\frac{x_k}{r^2},$$ where we used formula $\Gamma_{ij}^k = \frac{1}{2}g^{kl}\left(\frac{\partial g_{il}}{\partial y_j} + \frac{\partial g_{jl}}{\partial y_i} - \frac{\partial g_{ij}}{\partial y_l}\right)$, and $\{ \frac{\partial}{\partial x_{i}} \}_{i=1,\cdots,n+1}$ is the standard orthogonal basis in Euclidean space $\mathbb{R}^{n+1}$. By use the $\Gamma$, we can also compute curvature.

Using the comments of Willie Wong, we have $$(\vec{e}_{n+1})^\top=\vec{e}_{n+1}-\langle \vec{e}_{n+1},\vec{n}\rangle \vec{n}$$ where $\vec{n}=\frac{\sum^{n+1}_{i=1} x_{i}\vec{e}_i}{r}$. This will give the correct $\Gamma^k_{ij}$.

I want to thank Willie Wong again for answering my confusion (which has been tormenting me for several days).

I want to further explain why I care about this issue. I want to compute curvature by embedding.

Consider $\{x_{n+1}>0 \}$ \begin{equation*} \left\{\begin{aligned} & y_1=x_1\\ &\vdots \\ &y_n=x_n\\ &y_{n+1}=x_{n+1}-(r^2-\sum^{n}_{i=1}x_i^2)^\frac{1}{2}. \end{aligned}\right. \end{equation*} We have \begin{equation*} \begin{aligned} & \begin{pmatrix} \frac{\partial}{\partial y_1}\\ \vdots\\ \frac{\partial}{\partial y_{n+1}} \end{pmatrix}=\begin{pmatrix} I&-\begin{pmatrix} \frac{x_1}{(r^2-\sum^n_{i=1} x_i^2)^\frac{1}{2}}\\ \vdots\\ \frac{x_n}{(r^2-\sum^n_{i=1} x_i^2)^\frac{1}{2}}\\ \end{pmatrix}\\ 0&1 \end{pmatrix}\begin{pmatrix} \frac{\partial}{\partial x_1}\\ \vdots\\ \frac{\partial}{\partial x_{n+1}} \end{pmatrix}=\begin{pmatrix} I&-\begin{pmatrix} \frac{x_1}{x_{n+1}}\\ \vdots\\ \frac{x_n}{x_{n+1}}\\ \end{pmatrix}\\ 0&1 \end{pmatrix}\begin{pmatrix} \frac{\partial}{\partial x_1}\\ \vdots\\ \frac{\partial}{\partial x_{n+1}} \end{pmatrix}, \end{aligned} \end{equation*} i.e. for $i\le n$, $\frac{\partial}{\partial y_i} = \frac{\partial}{\partial x_i} - \frac{x_i}{x_{n+1}} \frac{\partial}{\partial x_{n+1}}$(this formula told us the embedding of $\frac{\partial}{\partial y_i}$ in $\mathbb{R}^{n+1}$) and ${i=n+1}$, $\frac{\partial}{\partial y_{n+1}} =\frac{\partial}{\partial x_{n+1}} $, $\{ \frac{\partial}{\partial x_{i}} \}_{i=1,\cdots,n+1}$ is the standard orthogonal basis in Euclidean space $\mathbb{R}^{n+1}$, where the second equation is due to we restrict $y_{n+1}\equiv 0$（The idea is from O' Neill's book P16 Proposition 28）.

For $1\le i,j\le n$, we have $g_{ij} = \delta_{ij} + \frac{x_ix_j}{x_{n+1}^2}$ and $g^{ij} = \delta_{ij} - \frac{x_ix_j}{r^2}$, moreover $$ \Gamma_{ij}^k = \left(\frac{x_ix_j}{x_{n+1}^2}+\delta_{ij}\right)\frac{x_k}{r^2},$$ where we used formula $\Gamma_{ij}^k = \frac{1}{2}g^{kl}\left(\frac{\partial g_{il}}{\partial y_j} + \frac{\partial g_{jl}}{\partial y_i} - \frac{\partial g_{ij}}{\partial y_l}\right)$, and $\{ \frac{\partial}{\partial x_{i}} \}_{i=1,\cdots,n+1}$ is the standard orthogonal basis in Euclidean space $\mathbb{R}^{n+1}$. By use the $\Gamma$, we can also compute curvature.

Using the comments of Willie Wong, we have $$(\vec{e}_{n+1})^\top=\vec{e}_{n+1}-\langle \vec{e}_{n+1},\vec{n}\rangle \vec{n}$$ where $\vec{n}=\frac{\sum^{n+1}_{i=1} x_{i}\vec{e}_i}{r}$. This will give the correct $\Gamma^k_{ij}$.

I want to thank Willie Wong again for answering my confusion (which has been tormenting me for several days).

I want to further explain why I care about this issue.

Consider $\{x_{n+1}>0 \}$ \begin{equation*} \left\{\begin{aligned} & y_1=x_1\\ &\vdots \\ &y_n=x_n\\ &y_{n+1}=x_{n+1}-(r^2-\sum^{n}_{i=1}x_i^2)^\frac{1}{2}. \end{aligned}\right. \end{equation*} We have \begin{equation*} \begin{aligned} & \begin{pmatrix} \frac{\partial}{\partial y_1}\\ \vdots\\ \frac{\partial}{\partial y_{n+1}} \end{pmatrix}=\begin{pmatrix} I&-\begin{pmatrix} \frac{x_1}{(r^2-\sum^n_{i=1} x_i^2)^\frac{1}{2}}\\ \vdots\\ \frac{x_n}{(r^2-\sum^n_{i=1} x_i^2)^\frac{1}{2}}\\ \end{pmatrix}\\ 0&1 \end{pmatrix}\begin{pmatrix} \frac{\partial}{\partial x_1}\\ \vdots\\ \frac{\partial}{\partial x_{n+1}} \end{pmatrix}=\begin{pmatrix} I&-\begin{pmatrix} \frac{x_1}{x_{n+1}}\\ \vdots\\ \frac{x_n}{x_{n+1}}\\ \end{pmatrix}\\ 0&1 \end{pmatrix}\begin{pmatrix} \frac{\partial}{\partial x_1}\\ \vdots\\ \frac{\partial}{\partial x_{n+1}} \end{pmatrix}, \end{aligned} \end{equation*} i.e. for $i\le n$, $\frac{\partial}{\partial y_i} = \frac{\partial}{\partial x_i} - \frac{x_i}{x_{n+1}} \frac{\partial}{\partial x_{n+1}}$(this formula told us the embedding of $\frac{\partial}{\partial y_i}$ in $\mathbb{R}^{n+1}$) and ${i=n+1}$, $\frac{\partial}{\partial y_{n+1}} =\frac{\partial}{\partial x_{n+1}} $, $\{ \frac{\partial}{\partial x_{i}} \}_{i=1,\cdots,n+1}$ is the standard orthogonal basis in Euclidean space $\mathbb{R}^{n+1}$, where the second equation is due to we restrict $y_{n+1}\equiv 0$.

For $1\le i,j\le n$, we have $g_{ij} = \delta_{ij} + \frac{x_ix_j}{x_{n+1}^2}$ and $g^{ij} = \delta_{ij} - \frac{x_ix_j}{r^2}$, moreover $$ \Gamma_{ij}^k = \left(\frac{x_ix_j}{x_{n+1}^2}+\delta_{ij}\right)\frac{x_k}{r^2},$$ where we used formula $\Gamma_{ij}^k = \frac{1}{2}g^{kl}\left(\frac{\partial g_{il}}{\partial y_j} + \frac{\partial g_{jl}}{\partial y_i} - \frac{\partial g_{ij}}{\partial y_l}\right)$, and $\{ \frac{\partial}{\partial x_{i}} \}_{i=1,\cdots,n+1}$ is the standard orthogonal basis in Euclidean space $\mathbb{R}^{n+1}$. By use the $\Gamma$, we can also compute curvature.

Using the comments of Willie Wong, we have $$(\vec{e}_{n+1})^\top=\vec{e}_{n+1}-\langle \vec{e}_{n+1},\vec{n}\rangle \vec{n}$$ where $\vec{n}=\frac{\sum^{n+1}_{i=1} x_{i}\vec{e}_i}{r}$. This will give the correct $\Gamma^k_{ij}$.

I want to further explain why I care about this issue.

Consider $\{x_{n+1}>0 \}$ \begin{equation*} \left\{\begin{aligned} & y_1=x_1\\ &\vdots \\ &y_n=x_n\\ &y_{n+1}=x_{n+1}-(r^2-\sum^{n}_{i=1}x_i^2)^\frac{1}{2}. \end{aligned}\right. \end{equation*} We have \begin{equation*} \begin{aligned} & \begin{pmatrix} \frac{\partial}{\partial y_1}\\ \vdots\\ \frac{\partial}{\partial y_{n+1}} \end{pmatrix}=\begin{pmatrix} I&-\begin{pmatrix} \frac{x_1}{(r^2-\sum^n_{i=1} x_i^2)^\frac{1}{2}}\\ \vdots\\ \frac{x_n}{(r^2-\sum^n_{i=1} x_i^2)^\frac{1}{2}}\\ \end{pmatrix}\\ 0&1 \end{pmatrix}\begin{pmatrix} \frac{\partial}{\partial x_1}\\ \vdots\\ \frac{\partial}{\partial x_{n+1}} \end{pmatrix}=\begin{pmatrix} I&-\begin{pmatrix} \frac{x_1}{x_{n+1}}\\ \vdots\\ \frac{x_n}{x_{n+1}}\\ \end{pmatrix}\\ 0&1 \end{pmatrix}\begin{pmatrix} \frac{\partial}{\partial x_1}\\ \vdots\\ \frac{\partial}{\partial x_{n+1}} \end{pmatrix}, \end{aligned} \end{equation*} i.e. for $i\le n$, $\frac{\partial}{\partial y_i} = \frac{\partial}{\partial x_i} - \frac{x_i}{x_{n+1}} \frac{\partial}{\partial x_{n+1}}$(this formula told us the embedding of $\frac{\partial}{\partial y_i}$ in $\mathbb{R}^{n+1}$) and ${i=n+1}$, $\frac{\partial}{\partial y_{n+1}} =\frac{\partial}{\partial x_{n+1}} $, $\{ \frac{\partial}{\partial x_{i}} \}_{i=1,\cdots,n+1}$ is the standard orthogonal basis in Euclidean space $\mathbb{R}^{n+1}$, where the second equation is due to we restrict $y_{n+1}\equiv 0$.

For $1\le i,j\le n$, we have $g_{ij} = \delta_{ij} + \frac{x_ix_j}{x_{n+1}^2}$ and $g^{ij} = \delta_{ij} - \frac{x_ix_j}{r^2}$, moreover $$ \Gamma_{ij}^k = \left(\frac{x_ix_j}{x_{n+1}^2}+\delta_{ij}\right)\frac{x_k}{r^2},$$ where we used formula $\Gamma_{ij}^k = \frac{1}{2}g^{kl}\left(\frac{\partial g_{il}}{\partial y_j} + \frac{\partial g_{jl}}{\partial y_i} - \frac{\partial g_{ij}}{\partial y_l}\right)$, and $\{ \frac{\partial}{\partial x_{i}} \}_{i=1,\cdots,n+1}$ is the standard orthogonal basis in Euclidean space $\mathbb{R}^{n+1}$. By use the $\Gamma$, we can also compute curvature.

Using the comments of Willie Wong, we have $$(\vec{e}_{n+1})^\top=\vec{e}_{n+1}-\langle \vec{e}_{n+1},\vec{n}\rangle \vec{n}$$ where $\vec{n}=\frac{\sum^{n+1}_{i=1} x_{i}\vec{e}_i}{r}$. This will give the correct $\Gamma^k_{ij}$.

I want to thank Willie Wong again for answering my confusion (which has been tormenting me for several days).