Acting upon a vector say,(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# What valence tensor produced by the absolute derivative?

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