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?

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    – Laff70
    Commented Nov 8, 2022 at 3:51

1 Answer 1


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)

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


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