# A Trying to understand covariant derivative on tensors

1. Jun 7, 2017

### Jonsson

Hello there,

Recently I encountered a type of covariant derivative problem that I never before encountered:

$$\nabla_\mu (k^\sigma \partial_\sigma l_\nu)$$

My goal: to evaluate this term

According to Carroll, the covariant derivative statisfies $\nabla_\mu ({T^\lambda}_{\lambda \rho}) = {{\nabla(T)_\mu}^\lambda}_{\lambda \rho}\ \ (\dagger)$ and also $\nabla(T \otimes S) = (\nabla T)\otimes S + T \otimes (\nabla S) \ \ (\ddagger)$. We know that $\partial \sigma$ forms a basis, so therefore $k^\sigma \partial_\sigma$ is a (1,0) tensor
$$\nabla_\mu (\underbrace{k^\sigma \partial_\sigma}_{\equiv T} l_\nu) = \nabla_\mu (T l_\nu) \stackrel{(\dagger)}{=} \nabla(T l)_{\mu \nu} = [\nabla(T l)]_{\mu \nu} \stackrel{(\ddagger)}{=} [\nabla(T) \otimes l + T\otimes \nabla(l)]_{\mu \nu}$$
But how do I continue from there? I want to distribute back the $\mu \nu$ and replace $\otimes$ with regular multiplication, but I dont know any algebra rules which applies, also my textbook doesn't help. I know that the tensors are multilinear, but its not quite what I need. How do I continue to evaluate $\nabla_\mu (k^\sigma \partial_\sigma l_\nu)$?

2. Jun 7, 2017

### Paul Colby

Is $k^\sigma\partial_\sigma$ a scalar operator when applied to the tensor $l_\nu$? I would think not?

3. Jun 7, 2017

### Jonsson

Great, I don't know. I am trying to understand how this all works. Would you like to explain some more?

4. Jun 7, 2017

### Paul Colby

This is the part I get in trouble with. The operator, $\partial_\sigma$, I read as regular partial differentiation. When applied to a scalar $\partial_\sigma \phi$ yields a 4-vector. In fact $\partial_\sigma = \nabla_\sigma$ when acting on a scalar. If I wrote $\nabla_\nu(k^\sigma\nabla_\sigma)l_\mu$ I would just distribute chain rule the way one normally does with covariant derivatives.

$\nabla_\mu(k^\sigma\nabla_\sigma)l_\nu = (\nabla_\mu k^\sigma)(\nabla_\sigma l_\nu)+k^\sigma (\nabla_\mu\nabla_\sigma l_\nu)$
Now if I replace $\nabla_\sigma = \partial_\sigma + \Gamma^._{\cdots}$ where the $\Gamma$'s are whatever that's needed to make things covariant, I could solve for the thing you are looking to evaluate.

5. Jun 7, 2017

### strangerep

The simplest way is to treat the argument as a (covariant) vector with index $\nu$ and apply the rules found in the "examples" section of this Wiki page. I.e., $$\nabla_\mu (k^\sigma \partial_\sigma l_\nu) ~=~\partial_\mu (k^\sigma \partial_\sigma l_\nu) ~-~ \Gamma^\alpha_{~\mu\nu} k^\sigma \partial_\sigma l_\alpha ~.$$If you're a masochist, you could in fact evaluate it using the Leibniz product rule as
$$\nabla_\mu (k^\sigma \partial_\sigma l_\nu) ~=~ k^\sigma \nabla_\mu ( \partial_\sigma l_\nu) ~+~ (\nabla_\mu k^\sigma) ( \partial_\sigma l_\nu) ~=~ \dots$$ This is indeed equivalent to "treating the argument as a (covariant) vector with index $\nu$".
(I've just now double-checked this, but if you want see my work you'll have to show me your attempt first. )

If you're a truly psychotic masochist, you could even treat it as the covariant derivative of a 3-index quantity, provided you follow the rules carefully, and preserve ordering when using the Leibniz rule.

HTH.

Last edited: Jun 8, 2017
6. Jun 7, 2017

### Orodruin

Staff Emeritus
Adding to this, I think it is worth pointing out that the argument is not a covariant vector. To make it covariant, you would need $k^\sigma\nabla_\sigma l_\nu$ instead. As written, I am very sceptical regarding the origins of the expression itself. Of course, it might be a term in a larger covariant expression.

Last edited: Jun 8, 2017
7. Jun 13, 2017

### stevendaryl

Staff Emeritus
There is a problem with the expression $k^\sigma \partial_\sigma l_\nu$. That isn't actually a tensor. This can be seen in two different ways:
1. $l_\nu$ is a one-form, and you can't apply a partial derivative to a one-form. Partial derivatives can only apply to scalar fields.
2. If you think of $\partial_\sigma$ as a basis vector, then $k^\sigma \partial_\sigma$ is just the vector $\vec{k}$ written in terms of components. So the expression $k^\sigma \partial_\sigma l_\nu$ is a vector acting on a one-form. How can a vector act on a one-form? Well, typically using covariant (or directional) derivatives: $k^\sigma \nabla_\sigma l_\nu$ makes sense, but $k^\sigma \partial_\sigma l_\nu$ doesn't.

8. Jun 13, 2017

### haushofer

Shouldn't the title of this thread read "trying to understand covariant derivatives on non-tensors"? :P