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?

  • 2
    $\begingroup$ $\nabla_Xf$ does not define a (0,1) tensor. $\nabla f$ does. $\nabla_Xf$ is the scalar you get when you act on $X$ with $\nabla f$. $\endgroup$ – AccidentalFourierTransform Dec 4 '17 at 22:33
  • $\begingroup$ 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'. $\endgroup$ – knzhou Dec 4 '17 at 22:35
  • $\begingroup$ @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. $\endgroup$ – Matt0410 Dec 4 '17 at 22:35
  • 2
    $\begingroup$ 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. $\endgroup$ – knzhou Dec 4 '17 at 22:36
  • $\begingroup$ What's the actual question? $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.