So in my relativity course, we recently learned about the covariant derivative. it is defined as:

\begin{equation} \nabla_{\mu}V^{\nu} = \partial_{\mu}V^{\nu}+\Gamma^{\nu}_{\mu,\lambda}V^{\lambda} \end{equation} where $V^{\mu}$ is a vector, and $\Gamma^{\nu}_{\mu,\lambda}$ is the connection, or Christoffel symbol. So recently I came accross: \begin{equation} \nabla_{\mathbf{e}'_{\beta}}\mathbf{e}'_{\alpha} = \Gamma^{\tau'}_{\alpha'\beta'}\mathbf{e}'_{\tau'} \end{equation} So I interpret this as the covariant derivative of $\mathbf{e}'_{\alpha}$ along $\mathbf{e}'_{\beta}$. However, because of the definition of this derivative, doesn't that mean $\mathbf{e}'_{\alpha}$ is a vector field? It is hard for me to understand the meaning of "a vector field of basis vectors." Also, given the index placement, wouldn't these be forms? Taking the covariant derivative of something with an index downstairs is confusing to me.

So my main question is, I cannot seem to calculate this result. So if we calculate $\nabla_{\mathbf{e}'_{\beta}}\mathbf{e}'_{\alpha}$ using the definition of the covariant derivative shown above, we get:

\begin{equation} \nabla_{\mathbf{e}'_{\beta}}\mathbf{e}'_{\alpha} = \left(\partial_{\mu}\mathbf{e}'_{\alpha}+\Gamma_{\mu\nu}^{\lambda}\mathbf{e}'_{\lambda}\right)\mathbf{e}'_{\beta} \end{equation}

So how the heck am I supposed to get the formula shown above? I can see that if I made $\mu = \alpha,\nu=\beta$ and $\lambda = \tau$, I would get the right form (as in how it looks) in the second term. But what about the first term? Again the index placement is weird. Any help would be appreciated!


2 Answers 2


In your first equation you gave the expression for the components of the covariant derivative of a contravariant vector field $V^\nu$. Your second equation is a bit different there you have the covariant derivative of a basis vector along a basis vector: we are dealing with vectors there.

Vectors and components (of vectors) are very very different objects: components are always related to a fixed (chosen) set of basis vectors, where vectors on the other hand are basis independent. That being said you can not just plug in vectors into your first equation. Your first equation does not hold for vectors it only holds for components. For the covariant derivative of a vector along a vector you would need: $$\nabla_\mathbf{v}\mathbf{u}=\nabla_{v^i\mathbf{e_i}}u^j\mathbf{e_j}=v^i\nabla_\mathbf{e_i}u^j\mathbf{e_j}=v^iu^j\nabla_\mathbf{e_i}\mathbf{e_j}+v^i\mathbf{e_j}\nabla_\mathbf{e_i}u^j=(v^iu^j \Gamma^k_{~ij}+v^i\frac{\partial u^k}{\partial x^i})\mathbf{e_k} \tag{1}.$$ In the last step we actually used that $$\nabla_{\mathbf{e_i}}\mathbf{e_j}=\Gamma^k_{~ij}\mathbf{e_k}\tag{2},$$ which is one definition of the affine connection: The basis vectors change if one moves from one point to another one in infinitesimal distance and the transformation that describes this change is the affine connection.

If one accepts the left hand side and the right hand side of (1) as a definition one can easily verity that (2) would be a consequence (set $\mathbf{v}=\mathbf{e_i} \Leftrightarrow v^i=1$ and $\mathbf{u}=\mathbf{e_j} \Leftrightarrow u^j=1$ but note that now $i$ and $j$ are not longer summation indices.) But equation (2) holds deeper meaning than just a trivial consequence of (1) since (2) is actually used to define (1). Equation (2) really describes the curvature of the space: in terms of the change of basis vectors.

So to your question about the basis vectors being vector fields: Yes of course they are. Differential-geometry or General relativity for that matter only become non trivial if this is the case: if the basis vectors where constant in space, the metric would be constant and all higher objects would vanish. All objects of GR are from a formal point of view fields: vector-, scalar- and even tensor-fields of the four dimensional space time. And even in "classical" differential-geometry this would be the case. The prime example: the basis vectors on the surface of a (2D) sphere change with position/angles, this is why from a mathematical point of the surface of a sphere has a non-vanishing curvature.

On your question of the "nature"/kind of equations (1) and (2): the covariant derivative of a component/vector is a tensor of rank 2.

  • $\begingroup$ Thanks so much for this response! This clears up all my confusions. $\endgroup$
    – user41178
    Sep 22, 2016 at 1:19

The notation $e_i$ is commonly used for frame fields (basis vector fields) as a more general notation than $\partial_\mu$. The connection forms, $\gamma^i{}_j$ (note that these indices are not tensorial) are defined relative to the frame fields such that $$ \nabla_v e_i = \gamma^j{}_i(v)e_j = \gamma^j{}_{ik}v^ke_j, $$ and we have $$ \nabla_{e_k}e_i = \gamma^j{}_{ik}e_j. $$ Here $\gamma^i{}_{jk}$ are the components of the connection forms. In a coordinate frame $\partial_\mu$ we have $\gamma^\mu{}_{\nu\sigma} = \Gamma^\mu_{\nu\sigma}$, the Christoffel symbols (but of course we could denote $\gamma^i{}_{jk}$ by $\Gamma^i{}_{jk}$ or $\Gamma^i_{jk}$; it is just a matter of convention). In fact starting with some postulates for the covariant derivative:

  1. $\nabla_{fu + gv}T = f\nabla_uT + g\nabla_vT$,

  2. $\nabla_v(T + Q) = \nabla_v T + \nabla_v Q$,

  3. $\nabla_v(f T) = v(f) T + f \nabla_v T$,

where $u$ and $v$ are vector fields, $f$ and $g$ functions, and $T$ and $Q$ are tensor fields, we can see that the connection forms must exist, and derive the formula for the covariant derivative of vectors \begin{align} \nabla_uv &= \nabla_{u^ie_i}v^je_j \\ &= u^i\partial_iv^j + u^i\gamma^k{}_j(e_i)v^j \\ &= u^i\partial_iv^k + \gamma^k{}_{ij}u^iv^j. \end{align} Formulas for more general tensors can further be derived from here.

To understand this it might be helpful to note that for frame fields we can select any linearly independent vector fields that span the tangent space at each point in its domain. A vector field $v$ then has components $v^i$ given by $v = v^ie_i$. Although an extremely helpful concept, the use of abstract index notation where we denote the vector itself by $v^i$ can sometimes muddle this. Similarly we can define dual frame fields on the cotangent bundle $\omega^i$ ($e^i$ is also commonly used, and $\omega^i{}_j$ is often used for the connection forms although I myself have never encountered $e^i{}_j$) by $\omega^i(e_j) = \delta^i_j$, and then a one-form $\alpha$ has components $\alpha_i$ given by $\alpha = \alpha_i\omega^i$. In this notation $$\alpha(v) = \alpha_i\omega^i(v^je_j) = \alpha_iv^j\delta^i_j = \alpha_iv^i,$$ i.e contraction over indices. More general tensors have components defined by the tensor products of a number of $e_i$:s and $\omega^i$:s.

If you want you can consider the frame vector fields defined in the common coordinate frame such that $e_i = e_i^\mu = e_i^\mu\partial_\mu$, but I personally don't find this very enlightening other than when we are transitioning to or from some coordinates for notation. Perhaps I should point out that the index $e_i$ is also not tensorial, but rather defines the tensorial indices $v^i$. This mixing of tensorial and non-tensorial indices may seem confusing at first, but in practice one quickly gets used to distinguishing between them.


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