Suppose we have a Riemann sphere expressed as $S^2:=\{ (x,y,z)\in \mathbb{R}^3: x^2+y^2+z^2=1\}$ equipped with a standard inner product in $\mathbb{R}^3$ since $$ T_{(x_0,y_0,z_0)}S^2 = \{ (x,y,z) \in \mathbb{R}^3: xx_0+yy_0+zz_0=0 \} $$

In short, I want all differential calculations on $S^2$ to become the ones in $\mathbb{R}^3$. Suppose $\nabla$ is the corresponding Riemann connection, is there a general formula for $$ \nabla_{(x_1,y_1,z_1)}(x_2,y_2,z_2) $$

in a selected $T_{(x_0,y_0,z_0)}S^2$?

  • $\begingroup$ "Hoe" ?_? ?_? ?_? $\endgroup$ Commented Jun 4, 2022 at 3:20
  • $\begingroup$ oh , i didn't notice that. $\endgroup$ Commented Jun 4, 2022 at 3:22
  • 4
    $\begingroup$ Hint. $\endgroup$ Commented Jun 4, 2022 at 4:05
  • $\begingroup$ "$\nabla_{(x_1,y_1,z_1)}(x_2,y_2,z_2)$" is not defined. The second entry should not be a tangent vector, but a vector field. Meaning that $(x_2,y_2,z_2)$ must be a function of $(x_1,y_1,z_1)$. $\endgroup$
    – Didier
    Commented Jul 20, 2023 at 15:33

2 Answers 2


This is explained in a textbook or course on the differential geometry of curves and surfaces (e.g., Do Carmo's book).

First, you cannot take the covariant derivative of a single tangent vector $(x_2,y_2,z_2)$. You need a vector field. Second, a tangent vector field on $S$ is a map $V: S \rightarrow \mathbb{R}^3$ that happens to satisfy $V(x) \in T_xS$. Given a point $(x_0,y_0,z_0) \in S$, the covariant derivative of $V$ in the direction $w =(x_1,y_1,z_1)$ is the orthogonal projection of the directional derivative $D_wV(x_0,y_0,z_0)$ onto $T_{(x_0,y_0,z_0)}S$.

If $S$ is the unit sphere centered at the origin, then $$T_{(x_0,y_0,z_0)}S = \{ v \in \mathbb{R}^3\ :\ v\cdot x = 0 \}.$$ It is now straightforward to find an explicit formula for the covariant derivative of a tangent vector field on $S$.

  • $\begingroup$ I got the idea. $\endgroup$ Commented Jun 4, 2022 at 17:38

$\newcommand{\R}{\mathbb{R}}$ There is a general formula. If a manifold $M$ is given by a constraint $C(x) = 0$, considered as a map from $\R^n$ to $\R^k$ where $C'(x)$ (the Jacobian of $C(x)$) is of full rank then the Levi-Civita connection (for the embedded metric) is given by $$\nabla_XY = D_XY + \Gamma(X, Y)$$ where the vector fields $X$ and $Y$ are considered as functions from $M$ to $\R^n$ so we can take directional derivative $D_XY$. Here, $\Gamma(X, Y)$ is a Christoffel function, the operator version of Christoffel symbols. At $x\in M$, if $C^{(2)}$ is the Hessian ($\R^k$-valued bilinear form) of $C$ and $\xi, \eta\in T_xM$ then $$\Gamma(\xi, \eta) = C'(x)^T(C'(x)C'(x)^T)^{-1}C^{(2)}(x; \xi, \eta). $$ For the unit sphere in $\R^n$, if $C(x) = x^Tx-1$, then $C'(x)\omega = 2x^T\omega$ for $\omega \in \R^n$, $C^{(2)}(x, \xi, \eta) = 2\xi^T\eta$ for $\xi, \eta\in T_xS^{n-1}$ (deriving $2x^T\eta$ in direction $\xi$), hence $$\Gamma(\xi, \eta) = x\xi^T\eta.$$

If the metric is not constant, but given by an operator $g$ (so each $g(x)$ is a positive-definite operator and the pairing of two tangent vectors $\xi,\eta$ at $x$ is given by $\xi^T(g(x)\eta)$, the formula is given by the global version of the Christoffel symbol formula $$\Gamma(X, Y) = -(D_X\Pi)Y + \frac{1}{2}\Pi g^{-1}\{(D_Xg)Y + (D_Yg)X - \chi_g(X, Y)\} .$$ Here, $\chi_g(X, Y)$ is an index raising formula of $(D_Xg)Y$, ie $$\chi_g(X, Y).Z = ((D_Zg) X). Y$$ for three vector fields $X, Y, Z$, and $$\Pi(x) = I_n - g(x)^{-1}C'(x)^T\{C'(x)g(x)^{-1}C'(x)^T\}^{-1}C'(x)$$ is the projection from $\R^n$ to the tangent space $T_xM$ at $x$ compatible with the metric operator $g$, $D_X\Pi$ is the directional (Lie) derivative of $\Pi$ in direction $X$ (since $\Pi$ is an operator-valued function this makes sense) and it is easy to see $$-(D_X\Pi)Y = g(x)^{-1}C'(x)^T\{C'(x)g(x)^{-1}C'(x)^T\}^{-1}C^{(2)}(x; X, Y) $$ using the fact that $C'(x)Y(x) = 0$ for the vector field $Y$. In the references, [1] is classical, the formula for $\Gamma$ with constant $g=I_n$ is in section 4.9, derived using Lagrange multipliers. We derived the formula for nonconstant $g$ in [2] and apply this to several manifolds in Riemannian optimization, (Stiefel, positive definite\semidefinite matrices, flag manifolds). In [3] we give a derivation using Hamilton mechanics and apply it to rigid body mechanics, among other applications.

We can compute the curvature easily using this global $\Gamma$, similar to the way we compute the curvature using Christoffel symbols. For example, the curvature of the sphere at 3 tangent vectors $\xi, \eta, \phi$ at $x$ is given by $$R_{\xi\eta}\phi = (D_{\xi}\Gamma)(\eta, \phi) - (D_{\eta}\Gamma)(\xi, \phi) + \Gamma(\xi, \Gamma(\eta,\phi)) - \Gamma(\eta, \Gamma(\xi,\phi)) = \xi\eta^T\phi - \eta\xi^T\phi +x \xi^T(x\eta^T\phi) -x\eta^T(x\xi\phi)\\ =\xi\eta^T\phi - \eta\xi^T\phi $$ as $\xi^Tx = \eta^Tx = 0$ for tangent vectors. This is also proved in [3]. A derivation using local coordinates is a much more difficult computation with trigonometric functions. These formulas are very easy to implement numerically with numerical/automatic differentiation.

[1] A. Edelman, T. A. Arias, S. T. Smith, The geometry of algorithms with orthogonality constraints, SIAM J. Matrix Anal. Appl. 20 (1999) 303–35

[2] Nguyen, D. Operator-Valued Formulas for Riemannian Gradient and Hessian and Families of Tractable Metrics in Riemannian Optimization. J Optim Theory Appl 198, 135–164 (2023). https://doi.org/10.1007/s10957-023-02242-z. Read cube: https://rdcu.be/ddyWJ

[3] Nguyen, D. Geometry in global coordinates in mechanics and optimal transport https://doi.org/10.48550/arXiv.2307.10017

  • 1
    $\begingroup$ Sorry for the typo - I mean $$\Gamma(X, Y) = -(D_X\Pi)Y + \frac{1}{2}\Pi g^{-1}\{(D_Xg)Y + (D_Yg)X - \chi_g(X, Y)\} .$$ $\endgroup$ Commented Jul 20, 2023 at 14:39

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