# [Physics] Covariant derivative of a function confusion

general-relativity

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.