I'm hoping to understand the motivation for certain parts of the definition of a derivative operator $\nabla$ on a manifold $M$. In Wald's General Relativity, two clauses of the definition are:

  1. Commutativity with contraction: For all tensors $\mathit{A} \in \mathscr{T}(k,l)$, $\nabla_{d}(\mathcal{A}^{a_1...c...a_k}_{b_1...c...b_l}) = \nabla_{d}\mathcal{A}^{a_1...c...a_k}_{b_1...c...b_l}$. (the parenthesis indicates contraction)

  2. Consistency with the notion of tangent vectors as directional derivatives on scalar fields: for all functions $f \in \mathscr{F}:M \rightarrow \mathbb{R}$ and all tangent vectors $t^a \in V_p$, it is required that $t(f)=t^a\nabla_af$

What is the point of having derivative operators commute with contraction (where we sum over the vectors and dual vectors for some (i,j) slot in the tensor)? What theorems are not possible to prove if this commutativity isn't stipulated? The so-called "Leibnitz rule" for derivative operators on manifolds corresponds to the simple product rule in elementary calculus; does this commutativity-with-contraction rule correspond to something simple as well?

My second question is about the notation of clause 2. We know that the $t^a$ are tangent vectors in the tangent space $V_p$ at point $p$ on the manifold. Then what is $t$ supposed to be?

  • $\begingroup$ Wald's notation is a bit goofy in the second clause. He means that $t(f)$ is the directional derivative of f along the direction of $t^{a}$. $\endgroup$ Aug 17, 2014 at 16:04

2 Answers 2


Since an $(k,\ell)$ tensor can be viewed as linear combinations of pertinent tensor products of $(1,0)$ tensors (=vectors) and $(0,1)$ tensors (=covectors), and by linearity and Leibniz rule for the covariant derivative, it is enough to consider the latter.

  1. The condition (1) in manifestly covariant notation then reduces to $$ X[\eta(Y)]~\stackrel{(2)}{=}~\nabla_X (\eta(Y)) ~=~ (\nabla_X \eta)(Y) +\eta (\nabla_X Y), \tag{1} $$ where $X,Y$ are vectors and $\eta$ is a covector. Here $\eta(Y)$ denotes the contraction of $\eta\otimes Y$. The first equality in eq. (1) follows from condition (2) below. In local coordinates $\eta(Y)=\eta_{\mu}X^{\mu}$ and $\nabla_X=X^{\mu}\nabla_{\mu}$. Condition (1) ensures, in plain words, that the Christoffel symbols$^1$ used to covariantly differentiate a vector must be related to the Christoffel symbols used to covariantly differentiate a covector.

  2. The condition (2) in manifestly covariant notation reads $$ \nabla_X f ~=~ X[f], \tag{2} $$ where $f$ is a $(0,0)$ tensor (=scalar function). In local components with $X=X^{\mu}\partial_{\mu}$, we have $X[f]~=~X^{\mu}\partial_{\mu}f$. Condition (2) ensures that the covariant differentiation of scalars corresponds to partial/directional differentiation.


$^1$ It is convenient to call $\Gamma^{\lambda}_{\mu\nu}$ Christoffel symbols even if the tangent-space connection $\nabla$ is not torsionfree nor compatible with a metric.

  • $\begingroup$ Note added for purists: It should be stressed that the notion and defining properties of a covariant derivative/connection $\nabla$ (and interplay with contractions) make sense for manifolds $M$ that doesn't have a distinguished metric tensor $g$. $\endgroup$
    – Qmechanic
    Aug 18, 2014 at 14:41
  • $\begingroup$ I have very much liked the observation you make at point 1 ("contravariant Christoffel is the same as covariant Christoffel" because of commutativity with contractions). This is something that I have never found on textbooks and is quite illuminating. $\endgroup$ Sep 29, 2016 at 9:19

For $1.$, you have, by applying the Leibnitz rule for covariant derivatives :

$\nabla_{d}(\mathcal{A}^{a_1...c...a_k}_{b_1...c...b_l}) \\= \nabla_{d}(\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l} \delta_{c'}^{c}) \\= \nabla_{d}(\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l} g^{ca}g_{ac'}) \\=(\nabla_{d}\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l}) \delta_{c'}^{c} + \mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l} (\nabla_{d} g^{ca})g_{ac'} + \mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l} g^{ca} (\nabla_{d}g_{ac'})$

The covariant derivative of the metric tensor is zero, then $\nabla_{d} g^{ca} = \nabla_{d}g_{ac'}=0 $:

$\nabla_{d}(\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l} \delta_{c'}^{c}) = (\nabla_{d}\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l}) \delta_{c'}^{c}$

So, covariant derivatives and contractions commute.


Your Answer

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

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