# Covariant derivative of a function confusion

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?

• $\nabla_Xf$ does not define a (0,1) tensor. $\nabla f$ does. $\nabla_Xf$ is the scalar you get when you act on $X$ with $\nabla f$. – AccidentalFourierTransform Dec 4 '17 at 22:33
• Note that different sources introduce define $\nabla$ to increase the rank of the tensor by one; others define $\nabla_X$ first, which keeps the rank of the tensor the same. Unfortunately both types of sources will call their thing 'the covariant derivative'. – knzhou Dec 4 '17 at 22:35
• @AccidentalFourierTransform But in order to evaluate the second order covariant derivative of a function $\nabla _a \nabla _b f$, I have to treat $\nabla _b f$ as a tensor. – Matt0410 Dec 4 '17 at 22:35
• Don't confuse $\nabla_a f$ (which really means $(\nabla f)_a$), which is a $(0, 1)$ tensor, with $\nabla_X f$, which is a scalar! These are the two different types of covariant derivative mentioned in my comment. – knzhou Dec 4 '17 at 22:36
• What's the actual question? – Phoenix87 Dec 4 '17 at 22:45

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 association $X\mapsto \nabla_X f$ 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.

• Hope you would notice my comment on this old post. As far as I could understand, when $f$ is a fixed function and $X$ is treated as an arbitrary vector field then $X \to \nabla_X f$ is a $C^\infty(M)$ linear map defined on vector fields $X$. (i) How do we know that this will be a $(0,1)$ tensor field? (ii) In the symbol $\nabla_X f$ in this case, the lower index $X$ only indicates that this is associated to the vector field $X$ and the symbols $\nabla_X f$ and $\nabla f$ are equivalent as long as $X$ is not specified. Am I right? – damaihati May 2 '19 at 9:40

On a smooth manifold $M$, the tangent vector fields $v\in \Gamma_M(TM)$ may be defined as a collection of paths $\gamma_x : \mathbb R \to M$ with $\gamma(0) = x$ for all points $x\in M$, being equivalent if the derivative $v_x$ coincides.

If we have a scalar function $f : M \to \mathbb R$ defined on the manifold, we can naturally find a map from $\mathbb R$ to $\mathbb R$ by defining $f \circ\gamma _x$. Then the rate of change of $f$ along $\gamma$ at $x$ is given by,

$$(D_vf) (x):= \mathrm d(f \circ \gamma)_0.$$

This motivates a derivation $D_v : C^\infty(M) \to C^\infty(M)$ which we know as $v^\mu\partial_\mu$ when acting on a scalar and it can be proven every derivation can be shown to arise from a tangent vector field. We can also interpret $D_v$ as the Lie derivative along $v$ for a scalar.

Notice that explicitly in components,

$$v^\mu \nabla_\mu f = v^\mu \partial_\mu f = \sum_{i=1}^d v^i \partial_i f$$

which is clearly a scalar, since there are no free indices left; we have only one pair which have been contracted giving the sum. If on the other hand we write $\nabla_\mu f$, this is a $(0,1)$ tensor since we have a free index $\mu$, uncontracted. $\nabla_\mu f$ is just the covariant derivative then, unrelated to any vector field.

Explicit Example

Take a vector field $V^\mu$ and a function $\phi$ on four-dimensional space time, endowed with Cartesian coordinates. Then we have that,

$$\mathcal L_V \phi = V^\mu \nabla_\mu \phi = V^\mu \partial_\mu \phi = V^t\frac{\partial}{\partial t}\phi + V^x\frac{\partial}{\partial x}\phi+V^y\frac{\partial}{\partial y}\phi+V^z\frac{\partial}{\partial z}\phi$$

which is clearly a scalar since components of a vector are scalar, and derivatives of scalars are scalar. On the other hand,

$$\nabla_\mu \phi = \partial_\mu \phi = \begin{pmatrix} \partial_t\phi \\ \partial_x\phi \\ \partial_y \phi \\ \partial_z \phi \end{pmatrix}^T$$

which is clearly a $(0,1)$ tensor.

• Thank you for your answer. I thought $\nabla_\mu f = \nabla_{e_\mu} f = e_\mu (f)$? So must be a scalar? – Matt0410 Dec 4 '17 at 22:51
• @Matt0410 See the updated answer. – JamalS Dec 4 '17 at 22:58
• @Matt0410 $e_\mu (f)$ is a covector, or rather the components of one. – Javier Dec 4 '17 at 23:03