So the covariant derivative of a scalar function $f$ on a manifold $M$ w.r.t the vector $X$ is defined as

$$ \nabla_X f = X(f) .$$

From the very beginning of my course on general relativity, it has been stated that vectors in the tangent space are directional derivative operators. So $X$ is a map from the space of functions on the manifold $\mathcal{F}$ to the reals:

$$ X : \mathcal{F} \rightarrow \mathbf{R}. $$

However, I am fully aware that $\nabla_X f = X(f)$ defines a $(0,1)$ tensor, but this seems to contradict the definition that $X(f)$ is a scalar?

## Best Answer

It depends on how you read the symbol

$$\nabla_X f.$$

If you consider $X$ and $f$ fixed, i.e., you pick one specific vector field $X\in \Gamma(TM)$ and a specific smooth function $f\in C^\infty(M)$, then $\nabla_X f$ is just the specific function $X(f)$.

On the other hand, if you consider $f\in C^\infty(M)$ to be a fixed function and let $X$ be an arbitrary vector field, so that you are actually looking to the map $X\mapsto \nabla_X f$, then you have a $(0,1)$ tensor field usually denoted as $\nabla f$, whose action on $X$ is $\nabla f(X) = \nabla_X f$.

Why? Because of the properties of the covariant derivative. It is well known that $\nabla_X Y$ is defined to be $C^\infty(M)$-linear (or tensorial as some people prefer calling it) on the entry below.

This ensures that fixing $f$ you get $X\mapsto \nabla_X f$ a $C^\infty(M)$-linear mapping defined on vector fields, and hence a $(0,1)$ tensor field.

The thing is just that in one case $\nabla_X f$ is one specific calculation with a specific result and the other is actually one mapping, i.e., a function. This is the same as asking whether $f(x)$ is a function or a number. Some people read this with $x$ one arbitrary variable so that $f(x)$ is the "rule that defines $f$" and so is the function, but if $x\in \mathbb{R}$ is one specific number $f(x)$ is a number as well.