Contravariant derivative tensor problem

[rafal_]

Tensor $$\left ( \begin{array}{cc} 1\\1 \end{array} \right )$$ is $$T=T^\alpha_\beta \omega^\beta \otimes \vec e_\alpha$$Why is contravariant derivative tensor $$\left (\begin{array}{cc}1\\1 \end{array} \right )$$
$$V^\alpha_{;\beta}\vec {e_\alpha}$$ - contravariant derivative

 

[Fredrik]

I don't understand the question. What do you mean by contravariant derivative? Do you mean covariant? What do you mean by those matrices? Are you saying that the components of T in the basis you're considering are (1,1)?

 

[rafal_]

I made a mistake I meant covariant. T is general tensor. I don't understand why
$$\frac{\partial \vec V}{\partial x^\beta}=V^\alpha_{;\beta} \vec e_{\alpha}$$
can be associated with a (1,1) tensor

 

[haushofer]

A covariant derivative is per definition a derivative which transforms as a tensor. Loosely, the covariant derivative of a tensor T adds another covariant index, so if T is type (k,l) then the covariant derivative of T is type (k+1, l) (I always forget the convention, but here the k is the number of covariant indices).

Here's a nice exercise: you know how the components of a vector V tranform:

$$V^{\mu'}(x') = \frac{\partial x^{\mu'}}{\partial x^{\nu}}V^{\nu}(x)$$

You also know how a partial derivative transforms:

$$\frac{\partial}{\partial x^{\lambda'}} = \frac{\partial x^{\alpha}}{\partial x^{\lambda '}}\frac{\partial}{\partial x^{\alpha}}$$

Now calculate the transformation the components of the partial derivative of a vector:

$$\frac{\partial}{\partial x^{\lambda'}}V^{\mu'}(x') = \frac{\partial x^{\alpha}}{\partial x^{\lambda '}}\frac{\partial}{\partial x^{\alpha}} [\frac{\partial x^{\mu'}}{\partial x^{\nu}}V^{\nu}(x)]$$

If you write this out, you'll see that this doesn't transform as a (1,1) tensor as you would naively expect on basis of the index structure of the expression. However, in derivatives you compare tensors at different points on the manifold and you need a certain prescription to perform this comparison: the connection $\Gamma$.

If you want to write things down with a basis, you can define the connection and the covariant derivative as

$$\nabla_{\mu}e_{\nu} \equiv \Gamma^{\lambda}_{\mu\nu}e_{\lambda}$$

Here $\nabla_{\mu}$ means covariant derivation with respect to the basis vector

$$e_{\mu} = \partial_{\mu}$$

This second-last formula states that the covariant derivative of the basis vector should be expressible in terms of your basis vectors. If you then write down a tensor in a certain coordinate basis, you can act on this tensor with the covariant derivative. For instance,

$$\nabla_{\mu} (V) = \nabla_{\mu}(V^{\alpha} e_{\alpha}) = [(\nabla_{\mu}V^{\alpha})e_{\alpha} + V^{\alpha}\nabla_{\mu}e_{\alpha}]$$

Demanding that the covariant derivative acting on scalar functions gives the same result as a partial derivative (and some basic rules such that it obeys Leibnitz), you arrive at the same answer as in a component-only treatment.

Hopefully this makes things a bit clear :)

 

[Fredrik]

I like the approach to covariant derivatives that starts with a connection. I'll quote myself (and fix a couple of mistakes at the same time):

Let V be the set of smooth vector fields on a manifold M. A connection is a map $\nabla:V\times V\rightarrow V$ such that

(i) $$\nabla_{fX+gY}Z=f\nabla_X Z+g\nabla_YZ$$

(ii) $$\nabla_X(Y+Z)=\nabla_XY+\nabla_XZ$$

(iii) $$\nabla_X(fY)=(Xf)Y+f\nabla_XY$$

$\nabla_XY$ is the covariant derivative of Y in the direction of X. The covariant derivative operator corresponding to a coordinate system x is

$$\nabla_{\frac{\partial}{\partial x^\mu}$$

The notation is often simplified to

$$\nabla_{\partial_\mu}$$

or just $\nabla_\mu$.

The above only defines the action of $\nabla_X$ on vector fields, but note that condition (iii) above suggests a way to extend the definition to scalar fields. If we define

$$\nabla_Xf=Xf$$

condition (iii) looks like the Leibnitz rule for derivatives:

$$\nabla_X(fY)=(\nabla_Xf)Y+f\nabla_XY$$

So we choose to define $\nabla_Xf$ that way. Can we do something similar for covector fields? It turns out we can. Suppose that $\omega$ is a covector field. The closest thing to a Leibnitz rule we can get is this:

$$\nabla_X(\omega(Y))=(\nabla_X\omega)(Y)+\omega(\nabla_XY)$$

so we choose to define $\nabla_X\omega$ by

$$(\nabla_X\omega)(Y)=\nabla_X(\omega(Y))-\omega(\nabla_XY)$$

for all Y. Note that this means that we define $\nabla_X\omega$ to be a covector field.

The same idea can be used to find the appropriate definition of $\nabla_X$ acting on an arbitrary tensor field. You can probably figure it out on your own.

Don't forget that the covariant derivative you're used to is the special case $X=\partial_\mu$.