# Differential operator in Levi-Civita connection

I want to use the Levi-Civita connection to find geodesics on a manifold $$M$$.
In the german wikipedia article (https://de.wikipedia.org/wiki/Levi-Civita-Zusammenhang) the L-C connection is defined as following: $$\nabla_X^{\mathbb{R}^n}Y = \sum_{i=1}^n \nabla_Xf_i \frac{\partial}{\partial x_i}$$ Where $$X$$ and $$Y$$ are vector fields in the euclidean room $$\mathbb{R}^{n}$$ with $$Y = \sum _{i=1}^n f_{i}{\frac {\partial}{\partial x_{i}}}$$.

Where $$\nabla_Xf_i$$ is the directional derivative in direction of $$X$$. Which as far as I know can be expressed as the inner product between the gradient of $$f$$ and the unit vector in $$X$$ direction. $$\nabla_Xf_i = \nabla f_i \cdot \frac{X}{||X||}$$

If $$M\subset \mathbb R^n$$ is a subdivision of the $$\mathbb{R} ^{n}$$ and $$X,Y$$ are vector fields on $$M$$ then $$\nabla_{X}^{\mathbb {R} ^{n}}Y$$ is a vector field defined on $$M$$, but whose images are stored in the tangential space of the $$\mathbb{R}^n$$, do not necessarily lie in the tangential space of $$M$$. But for each $$x\epsilon M$$ you can use the orthogonal projection $$p:T_{x}\mathbb{R} ^{n} \rightarrow T_{x}M$$ and then define:

$$\nabla_X Y = p(\nabla_X^{\mathbb{R}^{n}}Y) = p(\sum_{i=1}^n \nabla_Xf_i \frac{\partial}{\partial x_i})$$

The geodesic equation is normally defined as: $$\nabla_{\dot \gamma}\dot \gamma = 0$$ which is a ODE which could be solved.

But rigth now I am confused what $$\frac{\partial}{\partial x_i}$$ means since it is not applied to anything at all. I am having a hard time setting up the ODE because of this term.
Sorry if this question is weird. I only just started differntial geometry and initially come from an other field.

• The gradient is expressed in terms of the natural basis vectors, which for Euclidean for $\mathbb{R}^n$ space are the basis $\partial_n$. This condition can obviously extend to representing the gradient in terms of physical basis vectors would one need to (for example if the gradient is to be expressed in curvilinear coordinates).
– user284001
Commented Sep 10, 2020 at 14:04

The $$\frac{\partial}{\partial x^{i}}$$ represents the chosen basis of the tangent space at a point. In your case, Y could act on some function g as (omitting the $$\Sigma$$, using Einstein summation notation) : Y(g) = $$f^{i}\frac{\partial g}{\partial x^{i}}$$ and that gives you a number at that point which the tangent space is around. Because the partial is linear and acts on scalars in the usual way (i.e. it doesn't), this gives basis for a vector space. So if we use the ideas from Hawking and Ellis (1973) 'The large scale structure of space-time' we have the definition $$\nabla_{X}Y$$ as the covariant derivative of Y in the direction of X at p. So let's work with this and let's choose an arbitrary basis, {$$E_{i}$$}. Then let $$Y = Y^{i}E_{i}$$ and let $$X = X^{i}E_{i}$$ (capital Y is a little more natural for me for components of Y). The covariant derivative of Y is $$\nabla Y = \nabla_{i}Y^{j}E^{i}\otimes E_{j}$$ where I use the fact that $$\nabla_{i}$$ with the lowered index is the natural use of the covariant derivative, because we can calculate partials, e.g. $$\nabla_{i}G_{j} = \frac{\partial}{\partial x^{i}}G_{j} - \Gamma^{k}_{ij}G_{k}$$. We let $$E^{i}$$ be the basis for the cotangent space so that <$$E^{i},E_{j}$$> = $$\delta^{i}_{j}$$ then projecting $$\nabla Y$$ in to X we have $$$$ = $$X^{k}\nabla_{i}Y^{j} E_{j}$$ = $$X^{k}\nabla_{i}Y^{j}\delta^{i}_{k} E_{j}$$ = $$X^{i}\nabla_{i}Y^{j} E_{j}$$ = $$\nabla_{X}Y$$, and as you would in a normal vector equation, solve the components of the equation. As we're on a manifold, our natural basis is the one of partials, subbing $$\frac{\partial}{\partial x^{i}}$$ for $$E_{i}$$. This should give you the ODE.