# Covariant derivative vs Ordinary derivative

How correct is the statement: Covariant derivative, denoted by $$\nabla_u v$$ = Ordinary derivarive (denoted by $$u(v)$$) - Normal components, also called second fundamental form.

If the former statement is correct, is it correct both for intrinsic and extrinsic case?

• Since you're doing stuff with covariant derivatives, I should mention that I literally just wrote a program which generates the equations for covariant derivatives for LaTeX and UnicodeMath in Python. Looking at the results could help you learn more about how covariant derivatives work. It can also help you write covariant derivative equations for stackexchange questions and math in word documents. Nov 8, 2022 at 3:51

IMHO there is no need to introduce lots of definitions when teaching/learning basics in GR, or in differential geometry, when everything comes quite naturally from simple laws of derivation.

You can define the "covariant derivatives" as the components of the gradient of scalar/vector/tensor fields.

The gradient of a tensor field can be written as

$$\nabla \mathbb{T} = \mathbf{b}^i \otimes \dfrac{\partial \mathbb{T}}{\partial q^i}$$

being $$\mathbf{b}^i = g^{ik} \mathbf{b}_k$$ the vectors of the reciprocal basis of the natural basis $$\{ \mathbf{b}_k \}_k$$ induced by the coordinates $$q^k$$, $$\mathbf{b}_k = \frac{\partial \mathbf{r}}{\partial q^k}$$.

I'll provide here three examples:

1. the gradient of a scalar field $$f(q^k)$$,

$$\nabla f = \mathbf{b}^i \dfrac{\partial f}{\partial q^i}$$

and thus the covariant derivatives are $$\nabla_i f= \frac{\partial f}{\partial q^i}$$.

2. the gradient of a vector field, written as the linear combination of the vector of the natural basis, $$\mathbf{v}(q^k) = v^k(q^j) \mathbf{b}_k(q^j)$$,

$$\nabla \mathbf{v} = \mathbf{b}^i \otimes \dfrac{\partial \mathbf{v}}{\partial q^i} = \mathbf{b}^i \otimes \dfrac{\partial }{\partial q^i} (v^k \mathbf{b}_k) =$$
$$\qquad \qquad \qquad \quad=\mathbf{b}^i \otimes \left[ \mathbf{b}_k \dfrac{\partial v^k}{\partial q^i} + v^k \dfrac{\partial \mathbf{b}_k}{\partial q^i} \right] =$$
$$\qquad \qquad \qquad \quad=\mathbf{b}^i \otimes \left[ \mathbf{b}_k \dfrac{\partial v^k}{\partial q^i} + v^k \Gamma^{\ell}_{ik} \mathbf{b}_{\ell} \right] = \mathbf{b}^i \otimes \mathbf{b}_k \left[ \dfrac{\partial v^k}{\partial q^i} + v^{\ell} \Gamma^{k}_{i \ell} \right]$$

and thus the covariant derivatives of a vector field are $$\nabla_i v^k = \dfrac{\partial v^k}{\partial q^i} + v^{\ell} \Gamma^{k}_{i \ell}$$; here I introduced Christoffel symbols as the components in the natural basis of the derivatives of the vectors of the natural basis itself w.r.t. the coordinates;

3. gradient of a $$2^{nd}$$-order tensor field reads (I won't write the full computation here, do it as an exercise)

$$\nabla \mathbb{T} = \mathbf{b}^i \otimes \mathbf{b}_k \otimes \mathbf{b}_m \underbrace{\left[ \dfrac{\partial T^{km}}{\partial q^i} + T^{k \ell} \Gamma^{m}_{i \ell} + T^{m \ell} \Gamma^{k}_{i \ell} \right]}_{\nabla_i T^{km} }$$