Let M be a differentiable manifold. I want to define the covariant derivative $\nabla$ of tensor fields over M, as a derivation on a suitable algebra A of tensors, that is:

1) $\nabla$ is linear over tensors of the same order.

2) $\nabla$ increases the covariant order of the tensor by one.

3) $\nabla$ satisfies the Leibniz rule for arbitrary tensors of the algebra A with respect to the tensor product.

My issue is with point 3. In Physics, usually one defines the covariant derivative of an arbitrary tensor by extending the covariant derivatives of vectors and covectors, requiring that it commutes with contraction and that it satisfies the Leibniz rule for the components. However, I want to work with the tensors themselves instead of just the components. If I do that, then the Leibniz rule doesn't seem to work. For example, say S is of order (1,0) (covariant) and T is of order (0,1) (contravariant), and that the covariant derivative stacks the additional covariant component at the beginning of the tensor. Writting the tensors in terms of a basis $\partial_{j}$ and its natural cobasis $dx^{j}$:

$$ S = S_{j}dx^{j} $$ $$ T = T^{j}\partial_{j} $$

Let $\Gamma^{i}_{j,k}$ be the connection coefficients defining the covariant derivative in that particular coordinate system. Then one would have:

$$ \nabla S = \partial_{i}(S_{j}) dx^{i} \otimes dx^{j} - S_{j}\Gamma^{j}_{k,l}dx^{l} \otimes dx^{k} $$

$$ \nabla T = \partial_{i}(T^{j}) dx^{i} \otimes \partial_{j} + T^{j}\Gamma^{k}_{j,l}dx^{l} \otimes \partial_{k} $$

But then $\nabla T \otimes S$ has summands of the form $dx^{\alpha} \otimes \partial_{\beta} \otimes dx^{\gamma}$ while $ T \otimes \nabla S$ has summands of the form $\partial_{\alpha} \otimes dx^{\beta} \otimes dx^{\gamma}$, so the two cannot be summed to form $\nabla (T \otimes S )$ unless some shuffling is applied to the factors (the total number of covariant and contravariant components is the same in both).

So my question is: is there a canonical way to order the covariant and contravariant parts in a product like this, so that $\nabla$ satisfies the Leibniz rule in the way I propose?

  • $\begingroup$ For example, if $X,Y$ are vector fields and $\alpha$ is a $1$-form, you are looking for something like $\nabla_Y(X\otimes\alpha)=\nabla_Y(X)\otimes\alpha+X\otimes\nabla_Y(\alpha)$? If you do not evaluate $\nabla(X\otimes\alpha)$ on a vector field $Y$, $\nabla_Y(X\otimes\alpha)$, I cannot imagine a global formula. $\endgroup$
    – CvZ
    Commented Sep 22, 2015 at 1:01
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    $\begingroup$ shuffling is always allowed: There are canonical isomorphism $V\otimes W \cong W\otimes V$. $\endgroup$
    – user99914
    Commented Sep 22, 2015 at 1:51
  • $\begingroup$ Thanks, yes I would have thought so. But is there a canonical/preferred isomorphism when you have more than two factors? If so, maybe this could provide a global formula without having to evaluate on a given vector field... $\endgroup$ Commented Sep 23, 2015 at 15:04

1 Answer 1


I think it's much more common to let the last index position be the one introduced by differentiation, not the first. But regardless of which convention you choose, you're never going to have a Leibniz rule for total covariant derivatives of the form $\nabla (T\otimes S) = \nabla T \otimes S + T\otimes \nabla S$. (Notice, for example, that if both $S$ and $T$ have positive rank, then the differentiation index in $\nabla T \otimes S $ is never the last one, and in $T\otimes \nabla S$ it's never the first one.)

The Leibniz rule for covariant derivatives of tensor fields applies to the covariant derivative in the direction of a vector field (or vector): $$ \nabla_V(S\otimes T) = \nabla_V S \otimes T + S \otimes \nabla_V T. $$ This is true whether you put the differentiated index last or first (or somewhere else).

  • $\begingroup$ Thanks for your answer. I see... so the covariant derivative along a given vector field $Y$ seems to be the more relevant object in this case (I would not call some operator a derivative unless it satisfies some form of the Leibniz rule). $\endgroup$ Commented Sep 23, 2015 at 15:01
  • $\begingroup$ In view of what you say, do you think it possible to define a more abstract covariant derivative in some sort of graded algebra of tensor fields, being related to the standard one via some suitable isomorphism? I was hoping to find something in the likes of the exterior derivative operator $d$, that is, some form of the covariant derivative which is actually a derivation (as defined in abstract algebra). $\endgroup$ Commented Sep 23, 2015 at 15:01
  • $\begingroup$ @JackLee: Can I treat with $\nabla \langle \omega, \eta\rangle $ like $\nabla_X \langle \omega, \eta\rangle $? i.e. I suppose that we have a vector field $X$ then I use cov. derivative properties then I remove $X$ because it was an arbitrary vector field. This tempt me to do similar trick for $\nabla (T\otimes S)$ but you say that is wrong!! What about $\nabla (\alpha\wedge \beta)$ and $\nabla_X (\alpha\wedge \beta)$? $\endgroup$
    – C.F.G
    Commented Feb 23, 2021 at 18:06
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    $\begingroup$ @C.F.G: It's still true that $\nabla \langle \omega, \eta \rangle$ is a tensor field that can be expressed in terms of $\omega$, $\eta$, $\nabla\omega$, and $\nabla\eta$. For example, if $\omega$ and $\eta$ are $1$-tensor fields, then in any coordinate system, you can write $\nabla\langle \omega, \eta \rangle = (\omega_{i;j}\eta^i + \omega^i \eta_{i;j})dx^j$ (using the summation convention). It's just that there isn't a good coordinate-free notation for this tensor field. $\endgroup$
    – Jack Lee
    Commented Feb 23, 2021 at 18:12
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    $\begingroup$ @C.F.G: Yes, unless you make a special definition for this situation, an expression like $\langle \nabla\omega,\eta\rangle$ doesn't make sense. Same goes for $(\nabla\alpha)\wedge \beta$. $\endgroup$
    – Jack Lee
    Commented Feb 23, 2021 at 18:29

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