Covariant Derivative vs Vector Differentiation

Question

I am confused about the difference between a vector operating on another (like $u(v)$, used in Lie Brackets) and covariant derivatives ($\nabla_u v$). Can we not use Christoffel symbols in the vector derivative? Let us have a metric $ds^2 = g_{xx} dx^2 + g_{yy} dy^2$. Let us have two vectors in this system: $u = u^x e_x + u^y e_y$ and $v = v^x e_x + v^y e_y$. We assume that neither the coefficients nor the basis vectors remain constant. What should be $u(v)$ and $\nabla_u v$ in this case?

Answer 1

$u(v)$ can be a bit misguiding. Notice that $u$ does not act on $v$. Instead, we are really defining a new vector field $w$ whose action on some function $f$ is given by $w(f) = u(v(f))$. This might seem silly, but notice $u(v(f))$ means $u$ is acting on a function, not on a vector field. Vectors act on functions, not on vectors. Notice this definition does not make reference to any structure on the manifold apart from the vectors themselves (and the differential structure, if you want, but definitely not the metric nor the Christoffel symbols). In a coordinate basis, $w(f)$ is written as $$w(f) = w^{\mu} \frac{\partial}{\partial x^{\mu}} f = a^{\mu} \frac{\partial}{\partial x^{\mu}} \left(v^{\nu} \frac{\partial}{\partial x^{\nu}} f\right).$$ No Christoffel symbols are needed in here because all partial derivatives are acting on scalar functions. Hence, any covariant derivative would yield the very same result.

$\nabla_u v$ is a very different object. It is also a vector, so it is convenient for us to write it acting on a function $f$ to compare with the previous expression. In components, we have \begin{align} (\nabla_u v)^\mu &= u^\nu \nabla_\nu v^\mu, \\ &= u^\nu \frac{\partial}{\partial x^{\nu}} v^\mu + u^\nu \Gamma^{\mu}{}_{\nu \rho} v^{\rho}. \end{align} Therefore, \begin{align} (\nabla_u v)(f) &= (\nabla_u v)^\mu \frac{\partial}{\partial x^{\mu}} f, \\ &= u^\nu \left(\frac{\partial}{\partial x^{\nu}} v^\mu\right) \frac{\partial}{\partial x^{\mu}} f + u^\nu \Gamma^{\mu}{}_{\nu \rho} v^{\rho} \frac{\partial}{\partial x^{\mu}} f. \end{align}

Answer 2

You have to remember:

• the vector/tensor nature of the mathematical object you're dealing with, where its components are the coefficients of the linear combination of the objects of a vector or tensor basis;
• the definition of the vector/tensor operators.

As an example, with a set of coordinates $\{q^i\}$ inducing the natural basis $\mathbf{b}_i = \dfrac{\partial \mathbf{r}}{\partial q^i}$, and its reciprocal basis $\mathbf{b}^k = g^{ki} \mathbf{b}_i$, the gradient of a vector field (see below for details) reads

$\mathbf{\nabla \mathbf{v}} = \mathbf{b}^k \otimes\dfrac{\partial \mathbf{v}}{\partial q_k} = \mathbf{b}^k \otimes\dfrac{\partial}{\partial q_k} \left( v^i \mathbf{b}_i\right)$,

and exploiting the product rule for derivatives

$\mathbf{\nabla \mathbf{v}} = \mathbf{b}^k \otimes \left[ \dfrac{\partial v^i}{\partial q_k} \mathbf{b}_i + v^i \dfrac{\partial \mathbf{b}_i}{\partial q_k} \right] = \mathbf{b}^k \otimes \left[ \dfrac{\partial v^i}{\partial q_k} \mathbf{b}_i + v^i \Gamma_{ik}^{\ell} \mathbf{b}_{\ell} \right]$

and rearranging the dummy indices in the second term

$\nabla \mathbf{v} = \mathbf{b}^k \otimes \mathbf{b}_i \left[ \dfrac{\partial v^i}{\partial q_k} + v^{\ell} \Gamma_{\ell k}^{i} \right]$,

and in this expression you can recognize the covariant derivative of a vector field,

$\nabla_{/k} v^i = \dfrac{\partial v^i}{\partial q_k} + v^{\ell} \Gamma_{\ell k}^{i}$

that appears quite naturally just remembering the tensor nature of the field and applying the product rule for derivatives.

References. For some details about vector/tensor operators in differential calculus, take a look at here: https://basics.altervista.org/test/Math/tensor_calculus/all.html