Compute Christoffel symbols of sphere by embedding

Question

In his answer V. Semeria, starts by taking $$(y_1,\dots,y_{n+1})=\left(x_1,\dots,x_n,\sum_{i=1}^{n+1}x_i^2 -R^2\right)$$ Write $(\vec{e}_1,\dots,\vec{e}_{n+1})$ the canonical basis of $\mathbb{R}^{n+1}$. For all $i\leq n$, it is easy to obtain $$ \partial y_i = \frac{\partial}{\partial y_i} = \vec{e}_i - \frac{x_i}{x_{n+1}}\vec{e}_{n+1} $$ Differentiate the $\partial y_i$ in euclidean $\mathbb{R}^{n+1}$, $$ \tilde{\nabla}_{\partial y_i}\partial y_j = \left(-\frac{x_ix_j}{x_{n+1}^3}-\frac{\delta_{ij}}{x_{n+1}}\right) \vec{e}_{n+1} .$$ To compute $(\tilde{\nabla}_{\partial y_i}\partial y_j)^\top$ where $\top$ is the projection onto the tangent space of the sphere, He obtain $$ \tilde{\nabla}_{\partial y_i}\partial y_j = \left(-\frac{x_ix_j}{x_{n+1}^3}-\frac{\delta_{ij}}{x_{n+1}}\right) \vec{e}_{n+1}, $$ and after we just need to compute $(\vec{e}_{n+1})^\top$, that is $$ \begin{split} \sum^n_{i=1}\frac{\langle \vec{e}_{n+1},\partial y_i\rangle}{|\vec{e}_{n+1}||\partial y_i|} |\vec{e}_{n+1}|\frac{\partial y_i}{|\partial y_i|} & =\sum^n_{i=1}\frac{\langle \vec{e}_{n+1},\partial y_i\rangle}{|\partial y_i|^2} {\partial y_i}\\ &=\sum^n_{i=1}\frac{\langle \vec{e}_{n+1},\vec{e}_i - \frac{x_i}{x_{n+1}}\vec{e}_{n+1}\rangle}{|\vec{e}_i - \frac{x_i}{x_{n+1}}\vec{e}_{n+1}|^2} {\partial y_i}\\ &=\sum^n_{i=1} \frac{-\frac{x_i}{x_{n+1}}}{1^2+(\frac{x_i}{x_{n+1}})^2}\partial y_i\\ & =\sum^n_{i=1} -\frac{x_i x_{n+1}}{x_i^2+x^2_{n+1}}\partial y_i. \end{split} $$ Then, we obtain $$ \nabla_{\partial y_i}\partial y_j = \left(\frac{x_ix_j}{x_{n+1}^2}+\delta_{ij}\right)\sum_{k=1}^n\frac{x_k}{(x^2_k+x^2_{n+1})}\partial y_k, $$ i.e. $$ \Gamma_{ij}^k = \left(\frac{x_ix_j}{x_{n+1}^2}+\delta_{ij}\right)\frac{x_k}{(x^2_k+x^2_{n+1})}. $$ But this result is not consistent with the following: $$ \Gamma_{ij}^k = \frac{1}{2}g^{kl}\left(\frac{\partial g_{il}}{\partial y_j} + \frac{\partial g_{jl}}{\partial y_i} - \frac{\partial g_{ij}}{\partial y_l}\right)= \left(\frac{x_ix_j}{x_{n+1}^2}+\delta_{ij}\right)\frac{x_k}{R^2} $$ where

• $g_{ij} = \delta_{ij} + \frac{x_ix_j}{x_{n+1}^2}$ and
• $g^{ij} = \delta_{ij} - \frac{x_ix_j}{R^2}$.

These two methods should yield the same $\Gamma^k_{ij}$, where did I go wrong?

Answer 1

Gauss's Therorema Egregium describes a way of computing the Christoffel symbols on a hypersurface $X$ in $\newcommand{\bR}{\mathbb{R}}$ using the second fundamental form.

Up to a sign, the second fundamental form can be identified with shape operator, i.e., the differential of the Gauss map

$$ G: X\to S^{n-1}, \;\;x \mapsto \nu(x), $$ where $\nu(-)$ is a unit normal vector field along $X$. (An orientation on $X$ uniquely determines $\nu$.)

When $X$ is the unit sphere in $\bR^n$ the Gauss map is the identity map and its differential at a point $x\in S^{n-1}$ is the identity map $T_xS^{n-1}\to T_xS^{n-1}$ This leads to a fast computation of the Christoffel symbols. For details see the equalities (4.2.10), (4.2.12) and Example 4.2.21 of these notes.

Answer 2

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).