how to compute levi-civita connection on $S^2$ with coordinates in $\mathbb{R}^3$?

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

• "Hoe" ?_? ?_? ?_? Commented Jun 4, 2022 at 3:20
• oh , i didn't notice that. Commented Jun 4, 2022 at 3:22
• Hint. Commented Jun 4, 2022 at 4:05
• "$\nabla_{(x_1,y_1,z_1)}(x_2,y_2,z_2)$" 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 $(x_1,y_1,z_1)$. Commented Jul 20, 2023 at 15:33

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 $$S$$.

• I got the idea. 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

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