Lets take a vector field $X$ on some Riemannian manifold $(\mathcal{M},g)$ with Levi-Civita connection $\nabla$. Then, the components of its covariant derivative in coordinates are


Now, since $\nabla X$ is again tensor field, we can apply a second covariant derivative, which in coordinates yields


Now, instead of writing it like this, let us formally change the order in which the covariant detivative act, i.e. let us write


Now, mathematically speaking, the two terms on the right-hand side are ill-defined and do not make sense, since $\nabla$ is an operation acting on tensor fields and neither $\partial_{\alpha}X^{\beta}$ nor $\Gamma_{\alpha\delta}^{\beta}X^{\delta}$ are the components of a tensor field. However, if we treat these two terms as if they were rank (1,1) tensor fields, i.e. elements of $\Gamma^{\infty}(T\mathcal{M}\otimes T^{\ast}\mathcal{M})$, and use the standard formula for the connections, we will find the same and correct result for the components of $\nabla^{2}X$.

Now, this seem to work in general, i.e. for arbitrary rank tensors and an arbitrary amount of covariant derivatives (at least I never have seen a counter example).

My question, or lets say, my curiosity, is:

Why does the second "approach", in which we produce ill-defined terms in the steps in-between, work?

Is there any mathematical reason? For example, maybe one can extend the covariant derivative to more general "objects with indices" in a unique way, such that the steps in between become well-defined. For example, one can extend the covariant derivative to a map acting on tensor densities (i.e. sections of tensor product of a tensor bundle and a density bundle). Maybe there is a similar and more general notion which also includes objects like the partial derivative and the Christoffel symbols.

  • 2
    $\begingroup$ You know, I've used this types of calculations numerous times and never stopped to think about it like this. Good question. $\endgroup$
    – K.defaoite
    Commented Jan 29, 2023 at 15:07
  • 1
    $\begingroup$ @peek-a-boo Some help, maybe? :) $\endgroup$
    – K.defaoite
    Commented Jan 29, 2023 at 15:08
  • $\begingroup$ The second way is technically not correct, but the formula you get by writing both covariant derivatives in terms of partial derivatives and Christoffel symbols is independent of the order you do the substitution. $\endgroup$
    – Deane
    Commented Jan 29, 2023 at 23:31

3 Answers 3


The notion that Christoffel symbols are not tensor fields is not entirely true, or at least somewhat deceptive. If we choose a set of local coordinates $x^\alpha$, then the corresponding Chrisotoffel symbols $\Gamma^\alpha{}_{\beta\gamma}$ represent a perfectly well-defined local tensor field. Issues only arise when we try to interpret the objects involved as global, coordinate-independent objects, since a different set of coordinates will not generally result in the same tensor field, even on their common domain.

In this particular computation, we are fixing a coordinate system at the start, and finding an expression for $\nabla\nabla X$ in terms of the Christoffel symbols of those coordinates, so we can interpret expressions like $\nabla_\alpha(\Gamma^{\beta}{}_{\gamma\delta})$ as a covariant derivative of a tensor field without issue.


As an alternative point of view, let me point out the following: In your second approach, if you keep the term $(\partial_{\alpha}X^{\beta}+\Gamma_{\alpha\gamma}^{\beta}X^{\gamma})$ in brackets and use linearity only once there is no more covariant derivative, everything stays perfectly well-defined:


Now, as soon as you arrive here, you can use linearity, since $\partial_{\gamma}$ acts on every $C^{\infty}$-function and the terms with Christoffel symbols can also be clearly separated.

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    $\begingroup$ This is exactly the way I see this $\endgroup$
    – Didier
    Commented Jan 30, 2023 at 7:39

The way I have understood this is that in the abstract index notation, $ \nabla_a \nabla_b Z = \nabla^2 Z$. That is, $\nabla F$ (where F is a tensor) is the tensor that takes a vector field $X$ and enough vector and covectors fields to fill the inputs of $F$ and spits out $(\nabla_X F)(...)$. Therefore $(\nabla F)(...,X):=(\nabla_X F) (...)$, and clearly after we fix $F$ we get a tensor by linearity in the direction argument. This means that if we define $\nabla^2_{X,Y} F := (\nabla (\nabla F))(...,Y,X)$ [where we have swapped the indices under the convention that the last vector field is the vector field being differentiated along], we in fact get a tensor field that in coordinates is $\nabla_a (\nabla_b F)$. To see this, realize that the coordinate expression says "Give me a vector field for the slot $a$, and I'll take the covariant derivative of the tensor (\nabla F). Then, fill the slots of this new tensor". This is to be contrasted with taking the derivative of the quantity obtained after already filling in the slots (that is, $\nabla ^2_{X,Y} F \ne \nabla_X (\nabla_Y F)$! Most starkingly, latter isn't even linear in $Y$, but the former is.).

Your computation is correct because it effectively interprets the equation $\nabla_a \nabla_b X^{\beta}$ as taking a covariant derivative of a $(1,1)$ tensor, obtaining a $(1,2)$ as a result (one covariant component for the outer derivative, and both another covariant component and a contravariant component for the arguments of covariant derivative of $\nabla_b X^{\beta}$. The upside is that you can interpret the inner covariant derivative as taking in some inputs and then outpouring a tensor in terms of the Christoffel Symbols, without any problems, because still have a multilinear expression that takes in the right amount of slots and so can use the expression for the outer covariant derivatives as applied to the inner one.

Note: My source for the definition of the second covariant derivatives is John Lee's Introduction to Riemannian Manifolds. Other sources give the same definition, but often interpret it differently or dont explain as well what the distinction between it and $\nabla_X \nabla_Y$ is. Unfortunately the book doesn't explain the coordinate expression as much.

  • $\begingroup$ This is not addressing the actual question $\endgroup$
    – Didier
    Commented May 17, 2023 at 8:47

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