Definition for Covariant Derivative

Question

What is simple definition of the covariant derivative that looks like the definition of the derivative of a function in calculus?

definition of the derivative of a function in calculus is:

$$\frac {df(x)}{dx}=\lim_{\Delta x\to o}\frac {f(x+\Delta x)-f(x)}{\Delta x}$$

what about Covariant derivative, what is definition of the Covariant derivative of a function in calculus? $$\frac {{\mathcal D}f(x)}{dx}$$

Remark: ok lets say that:

$$\frac {{\mathcal D}f(x)}{dx}=\frac {df(x)} {dx} +\delta f(x)$$ where the covariant derivative is broken into two parts, the extrinsic normal component and the intrinsic covariant derivative component . Now, how do we define a simple definition for the intrinsic covariant derivative component $\delta f(x)$ (this small addition is the result of parallel translation)

Answer 1

There is a very intuitive way to understand the covariant derivative (for the Levi-Civita connection) for the special case of isometrically embedded submanifolds in $\mathbb R^n$. Roughly speaking, first take the usual derivative in $\mathbb R^n$, and then project the answer onto the tangent plane of the submanifold. The image of the projection is then the covariant derivative.

In fact, this is how the concept of covariant derivative if often introduced to undergraduates in their first differential geometry course. For instance, if you consider the sphere $S^2 \subset \mathbb R^3$, then it is easy (both analytically and visually) to prove that the tangent vector field to a great circle has zero covariant derivative and therefore that great circles must be geodesics.

Answer 2

One can motivate the covariant differentiation using only vector calculus. It works for an oversimplified case though (but since the OP doesn't accept either the definition via Ehresmann connection nor the vector bundle definition, may be it's justified.)

Consider new coordinates $y^i$ on $\mathbb{R}^n$ (e.g. spherical). We require orthonormality on that coordinate system, $$\mathrm{d}s^2=(u_i)^2\mathrm{d}y^i.$$ One has $\mathbf{x}=x^i(y)e_i$ and defines $$e'_j=\frac{\partial\mathbf{x}(y) }{\partial y^j}.$$ Then the metric is given by $$g_{ij}=e_i'\cdot e_j'.$$ If one considers a vector field $X:\mathbb{R}^n\to \mathbb{R}^n$ one can write $X=X^ie_i'$. If we now wish to differentiate $X$, we have to take into account the change of the components $X^i$ and of the basis $e'_j$, which is no longer rigid. That is $$\frac{\partial X }{\partial y^j}=\frac{\partial (X^i e'_i)}{\partial y^j}=\frac{\partial X^i}{\partial y^j}e_i'+X^i\frac{\partial e'_i}{\partial y^j}.$$ One can write $\frac{\partial e'_i}{\partial y^j}$ as linear combination of the $e_i'$ s, i.e. for some functions $\Gamma_{ij}^k$ $$ \frac{\partial e'_i}{\partial y^j}=\Gamma_{ij}^k e_k'.$$ Upon taking inner product of this equation with $e_l'$, one sees that these coefficients are given by $$\Gamma_{ij}^k=g^{kl}e_l'\cdot \frac{\partial e'_i}{\partial y^j}.$$

Now, in index notation, the covariant derivative of $X^i$ is given by the

$$\nabla_jX^i=\frac{\partial X^i}{\partial y^j}+\Gamma_{jk}^iX^k.$$ This is of the form $\frac {{\mathcal D}f(x)}{dx}=\frac {df(x)} {dx} +\delta f(x)$, but $f$ must be a vector field (or higher rank tensor), otherwise the covariant and ordinary derivatives concide.

Answer 3

$${\mathcal D}_{X} V=\lim_{\Delta x \to o}\frac {\Gamma(\gamma)^o_{\Delta x}V_{(\gamma)\Delta x}-V_{(\gamma)o}}{\Delta x},$$ where the ${\mathcal D}_{X}=\nabla_{X}$

Answer 4

Ok, you want to do the derivative quotient $ \frac {df}{dx} $, as you do it in $\mathbb R^n$ ( using norms, matrices, etc. when $n>1$.

The reason you need additional machinery is that , unlike the case of $\mathbb R^n$, vector spaces are not naturally-isomorphic. You use this natural isomorphism to declare a vector at $T_p \mathbb R^n$ is the same at each point of $ \mathbb R^n$. But in a manifold, Tangent Spaces , as vector spaces, are not naturally isomorphic. The connection provides a choice of isomorphism between tangent spaces at different points.

The old difference quotient then evaluates the difference of the function at a vector and that at it's (isomorphic) translate.