# Does it make sense to ask how the covariant derivative act on the partial derivative $\nabla_\mu ( \partial_\sigma)$? If so, what is the answer?

I want to find out how the covariant derivative acts on terms containing a partial derivative, e.g. $\nabla_\mu(k^\sigma\partial_\sigma l_\nu)$. But I don't know how to evaluate the terms of the form $\nabla_\mu(\partial_\sigma)$. If one writes $$\nabla_\mu(k^\sigma\partial_\sigma l_\nu) = \nabla_\mu(g^{\rho \sigma}k_\rho\partial_\sigma l_\nu) = g^{\rho \sigma}\nabla_\mu(k_\rho\partial_\sigma l_\nu) = g^{\rho \sigma}\left[ \nabla_\mu(k_\rho)\partial_\sigma l_\nu + k_\rho\nabla_\mu(\partial_\sigma )l_\nu + k_\rho\partial_\sigma\nabla_\mu( l_\nu) \right]$$ Problem: How to determine $\nabla_\mu(\partial_\sigma )$. How do I work it out, and understand whatever the answer, that it makes sense? Have I made a mistake?

EDIT: I add the context: suppose $k^a$ and $l^a$ are killing vector. Then I want to prove that the commutator $[k,l]_\alpha = k^\sigma\partial_\sigma l_\alpha - l^\sigma\partial_\sigma k_\alpha$ is a Killing vector. If you write out $\nabla_{(\mu}[k,l]_{\nu)}$, then you find these terms immediately.

• Covariant derivative is an invariant object, partial derivative only makes sense in a chart with coordinates. If you are doing a computation in coordinates you can rewrite the covariant derivative in terms of partial derivatives using Christoffel symbols, and manipulate the partial derivatives as in calculus. If you are doing an invariant computation there shouldn't be any partial derivatives in your expressions. – Conifold Jun 7 '17 at 22:52

If you want to use this for commutators, then either consider

$$\nabla_\sigma([k,l]^\mu)=\partial_\sigma[k,l]^\mu+\Gamma^\mu_{\sigma\kappa}[k,l]^\kappa=\partial_\sigma(k^\nu\partial_\nu l^\mu-l^\nu\partial_\nu k^\mu)+\Gamma^\mu_{\sigma\kappa}(k^\nu\partial_\nu l^\kappa-l^\nu\partial_\nu k^\kappa)...$$

and use this for further calculations, or consider that for any torsionless connection, we have $$[k,l]^\mu=k^\nu\partial_\nu l^\mu-l^\nu\partial_\nu k^\mu\equiv k^\nu\nabla_\nu l^\mu-l^\nu\nabla_\nu k^\mu,$$ and the latter expression contains only terms that are 'covariant'.

• If you can explain why torsionless connection implies we can replace partial by covariant derivatives, then I will accept this as the answer. Thank you ! – Mikkel Rev Jun 8 '17 at 12:20
• @MariusJonsson If you expand the covariant derivatives in terms of partials and connection coefficients (Christoffel-symbols), you will see that in the expression, the connection coefficients will cancel each other, but only if the connection is symmetric: $k^\nu(\partial_\nu l^\mu+\Gamma_{\nu\sigma}^\mu l^\sigma)-l^\nu(\partial_\nu k^\mu+\Gamma^\mu_{\nu\sigma} k^\sigma)=[k,l]^\mu+\Gamma^\mu_{\nu\sigma}k^\nu l^\sigma-\Gamma^\mu_{\nu\sigma}l^\nu k^\sigma$. If $\Gamma$ is symmetric in the lower indices, then the two expressions cancel each other. – Bence Racskó Jun 8 '17 at 14:32

My question would be why are you doing this?

The idea of the covariant derivative is that it maps a tensor to a tensor with one more lowered index, while satisfying a few other rules like the Liebniz rule.

But, if you have an object like $\partial_{a}\ell_{b}$, it is already, in general, not a tensor, and your mapping has a domain problem.

If you're trying to figure out the expression of some series of covariant derivatives in terms of partials and Christoffel symbols, you need to do something like:

$$\nabla_{a}\nabla_{b}v^{c} = \partial_{a}\left(\nabla_{b}v^{c}\right) - \Gamma_{ab}{}^{d}\nabla_{d}v^{c} + \Gamma_{ad}{}^{c}\nabla_{b}v^{d}\\$$

And then you can expand out the $\nabla v$ terms normally. But it doesn't mean anything to compute the covariant derivative of a partial, except in one case: when you are computing the derivative of a function. In that case, we have, for all $f$, $\nabla_{a}f = \partial_{a}f$ by definition, so it doesn't matter which one you use.

• Context: suppose $k^a$ and $l^a$ are killing vector. Then I want to prove that the commutator $[k,l]_\alpha = k^\sigma\partial_\sigma l_\alpha - l^\sigma\partial_\sigma k_\alpha$ is a Killing vector. If you write out $\nabla_{(\mu}[k,l]_{\nu)}$, then you find these terms immediately. Does that help? – Mikkel Rev Jun 7 '17 at 22:07
• @MariusJonsson: the commutator doesn't care whether you use covariant or partial derivatives, the symmetry of the lower two indices of the Christoffel symbols makes them equivalent. So you can do the calculation in covariant derivatives, and everything will be consistent. – Jerry Schirmer Jun 7 '17 at 22:13
• namely $k^{a}\nabla_{a}\left(\ell^{b}\nabla_{b}f\right) - \ell^{a}\nabla_{a}\left(k^{b}\nabla_{b}f\right) =k^{a}\partial{a}\left(\ell^{b}\partial_{b}f\right) - \ell^{a}\partial_{a}\left(k^{b}\partial_{b}f\right)$ for all functions $f$ – Jerry Schirmer Jun 7 '17 at 22:16

OP asks in the title (v2):

Does it make sense to ask how the covariant derivative acts on the partial derivative $\nabla_\mu ( \partial_\sigma)$?

I) Well, Yes, in a limited sense, if one is careful with the notation. Recall that a partial derivative $\partial_{\mu}$ may be interpreted as serving a double purpose in differential geometry: both as an actual derivative $\frac{\partial}{\partial x^{\mu}}$ acting on functions, or merely as a booking device, as a basis, that transforms correctly under change of local coordinates $x^{\mu}$. To make this distinction clear, let us introduce the notation $b_{\mu}$ for the latter role. Then we may e.g. write a vector field as $$X~=~X^{\mu}(x)b_{\mu}.\tag{1}$$ We can then reproduce covariant differentiation $$X^{\nu}_{;\mu}~=~\frac{\partial X^{\nu}}{\partial x^{\mu}} + \Gamma^{\nu}_{\mu\lambda} X^{\lambda}\tag{2}$$ via a trick: Introduce the formal first-order differential operator $$\nabla_{\mu}~=~\frac{\partial}{\partial x^{\mu}} + \Gamma^{\nu}_{\mu\lambda}b_{\nu} \frac{\partial}{\partial b_{\lambda}}.\tag{3}$$ Then $$\nabla_{\mu}X~\stackrel{(1)+(3)}{=}~\left(\frac{\partial}{\partial x^{\mu}} + \Gamma^{\nu}_{\mu\lambda}b_{\nu} \frac{\partial}{\partial b_{\lambda}}\right)(X^{\kappa}b_{\kappa})~=~\left(\frac{\partial X^{\nu}}{\partial x^{\mu}} + \Gamma^{\nu}_{\mu\lambda} X^{\lambda}\right)b_{\nu}~\stackrel{(2)}{=}~X^{\nu}_{;\mu}b_{\nu}\tag{4}$$

II) Similarly, the Lie bracket $$[X,Y]^{\nu} ~=~ X^{\mu}\frac{\partial Y^{\nu}}{\partial x^{\mu}} -Y^{\mu}\frac{\partial X^{\nu}}{\partial x^{\mu}}\tag{5}$$ of vector fields $$X~=~X^{\mu}b_{\mu} , \qquad Y~=~Y^{\mu}b_{\mu}, \tag{6}$$ can be reproduced by the Schouten-Nijenhuis bracket $$[X,Y]~=~X\left(\stackrel{\leftarrow}{\frac{\partial}{\partial b_{\mu}}}\stackrel{\rightarrow}{\frac{\partial}{\partial x^{\mu}}} - \stackrel{\leftarrow}{\frac{\partial}{\partial x^{\mu}}}\stackrel{\rightarrow}{\frac{\partial}{\partial b_{\mu}}} \right)Y. \tag{7}$$

III) Such formalism, which differentiates basis elements $b_{\mu}$, can be developed further in other areas of differential geometry.