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?

  • 2
    $\begingroup$ For those more familiar with physical instead of mathematical notation: You mean $u^\kappa v_\kappa$ and $u^\kappa\nabla_\kappa v^\lambda=u^\kappa\partial_\kappa v^\lambda+u^\kappa\Gamma_{\kappa\xi}^\lambda v^\xi$? $\endgroup$ Nov 3, 2022 at 4:04
  • 3
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Nov 3, 2022 at 5:04

2 Answers 2


$u(v)$ 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}


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.