# Why does the covariant derivative of a $(p,q)$-tensor produce a $(p,q+1)$-tensor?

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:

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

• This question may be more appropriate for Mathematics, I will move it if so. – Charlie Feb 24 at 13:42
• It does map a $(p,q)$ tensor to a $(p,q+1)$ tensor by definition! – Mathphys meister Feb 24 at 14:46

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?

• I think the OP meant for $\nu$ to be an index, $\mu=0,1,2,3$. – J. Murray Feb 24 at 14:10
• I'll edit my answer to remove the reference to this index. – NDewolf Feb 24 at 14:13
• 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. – Charlie Feb 24 at 15:08
• To clarify, when I say "plus" above I don't mean addition. – Charlie Feb 24 at 15:09
• 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) – NDewolf Feb 24 at 15:34