# What valence tensor produced by the absolute derivative?

1. Apr 17, 2015

### binbagsss

Acting upon a vector say,
so it is defined as:
$\frac{d}{d\lambda}V^{u}+\Gamma^{u}_{op}\frac{dx^{o}}{d\lambda}V^{p}=\frac{DV^{u}}{D\lambda}$

And this can also be written in terms of the covariant derivative, $\bigtriangledown_{k}$ by $\frac{DV^{u}}{D\lambda}=\frac{d x^{k}}{d \lambda} \bigtriangledown_{k}V^{u}$ [1]

So I know that the covariant derivaite takes a tensor of valance $(p,q)$ to $(p,q+1)$

So using this and the fact that the RHS of [1] has a vector multiplied by a covariant derivative acting upon a vector I would conclude that it takes a $(1,0)$ to a $(1,1)$. Is this correct?

However my textbook states '$\frac{DV^{a}}{Du}$ transforms as a covariant vector, which I dont see,

Thanks.

2. Apr 17, 2015

### Orodruin

Staff Emeritus
$\nabla_k V^\mu$ is a type (1,1) tensor, but in $DV^\mu/D\lambda$ it is contracted with the vector $\dot x^k$ and therefore becomes a (1,0) tensor.

3. Apr 17, 2015

### binbagsss

Ahh I see thanks. And this is a contravariant not a covariant though? And typo in the book?

Last edited: Apr 17, 2015
4. Apr 17, 2015

### Orodruin

Staff Emeritus
Yes, it is contravariant. You can check this by checking the transformation properties.