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In the specific case of the covariant derivative acting on a scalar function:

$$\nabla_\nu f$$

it seems strange to me that this would return a covector. Am I wrong in thinking the covariant derivative is a generalisation of a directional derivative? If we take the directional derivative of some scalar field we get back a scalar:

$$\nabla f(x,y,z)\cdot\vec v$$

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    $\begingroup$ This question may be more appropriate for Mathematics, I will move it if so. $\endgroup$ – Charlie Feb 24 at 13:42
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    $\begingroup$ It does map a $(p,q)$ tensor to a $(p,q+1)$ tensor by definition! $\endgroup$ – Mathphys meister Feb 24 at 14:46
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Part of your intuition is correct: The directional derivative of a scalar function indeed again gives a scalar function and this caries over to the definition of a covariant derivative.

However you seem to be making a subtle mistake in the definition of the covariant derivative. It is the map $\nabla$ that maps a function $f$ to a covector $\nabla f$ and in general a $(p,q)$-tensor field to a $(p,q+1)$-tensor field. But if you plug in a vector $\vec{v}$ (which is what you did for the directional derivative) then you indeed obtain a new $(p,q)$-tensor field. So there is no mismatch between your intuition and the general definition.

Now, to address the part of your intuition that it would be strange if $\nabla_\mu f$ is not a scalar: Is it for example that strange that a partial derivative (i.e. the covariant derivative on a flat manifold) would be something more than a scalar function when it does not transform trivially under coordinate transformations?

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    $\begingroup$ I think the OP meant for $\nu$ to be an index, $\mu=0,1,2,3$. $\endgroup$ – J. Murray Feb 24 at 14:10
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    $\begingroup$ I'll edit my answer to remove the reference to this index. $\endgroup$ – NDewolf Feb 24 at 14:13
  • $\begingroup$ So the regular directional derivative is basically the covariant derivative (in flat space) plus the contraction of the result with a vector. Whereas the covariant derivative by itself stops a step before, other than that they are the same operation. $\endgroup$ – Charlie Feb 24 at 15:08
  • $\begingroup$ To clarify, when I say "plus" above I don't mean addition. $\endgroup$ – Charlie Feb 24 at 15:09
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    $\begingroup$ Yes exactly. In fact this is (up to minor details) part of the general definition of a linear connection (the mathematical notion of what physicists call a covariant derivative) $\endgroup$ – NDewolf Feb 24 at 15:34

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