# Compute Christoffel symbols of sphere by embedding

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?

• There's an obvious typo in the answer you are quoting: you probably want $y_{n+1} = \sqrt{R^2 - \sum (x_i)^2}$. // Secondly, the formula $$(e_{n+1})^T ='' \sum \frac{\langle e_{n+1}, \partial y_i\rangle}{|\partial y_i|^2} \partial y_i$$ only holds if you know that $\partial y_i$ are mutually orthogonal. In this case this is far from being true. Commented Apr 17 at 7:02
• To compute the projection to the sphere, it is better to write $$X^T = X - \langle X, n\rangle n$$ where $n$ is the unit normal $$n = \frac{1}{R} \left(\sqrt{R^2 - \sum (x_j)^2} e_{n+1} + \sum x_i e_i \right)$$ Commented Apr 17 at 7:17
• Thank you very much for your comments, which made me suddenly realize. Commented Apr 17 at 13:13
• I think $(y_1,\dots,y_{n+1})=(x_1,\dots,x_n,\sum_{i=1}^{n+1}x_i^2 -R^2)$ is not a typo, If we restrict $y_{n+1}\equiv 0$ it is a local coordinate system of $S^n$. Commented Apr 17 at 13:54
• That's wrong. Fixing $y_{n+1} \equiv 0$ the solution set is $S^{n-1}$ realized as the equator in $S^n$; but then this is an $n-1$ dimensional manifold and using $(x_1, \ldots, x_n)$ the coordinates are over-specified. // The local coordinates should be $(x_1, \ldots, x_n)$; and notationally what you wrote as $\partial y_i$ should really be the pushforward of $\partial x_i$ Commented Apr 17 at 23:01

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.

• Maybe the question is where Measure32 went wrong. I suppose he went wrong by not reading your lecture notes, which are excellent. Commented Apr 17 at 17:34

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).