Understanding the definition of the covariant derivative

Question

I'm currently working my way through the book "Mathematical Methods for Physics - An Introduction to Group Theory, Topology and Geometry" and I think I have a very fundamental misunderstanding when it comes to the covariant derivative of tensors. The book states $$ f_{;\mu\nu}=\nabla_\nu\nabla_\mu f=\partial_\nu\partial_\mu f-\Gamma_{\mu\nu}^\lambda\partial_\lambda f $$ for scalar functions $f$ over some smooth manifold with connection coefficients $\Gamma^\lambda$.

But, on the other hand, it is stated that (for generic vector fields $X$) $$ \nabla_X f=X[f]=X^\mu\partial_\mu f $$ and therefore $\nabla_\mu f=\partial_\mu f$, which would be a scalar. Applying this twice, I would obtain

$$ \nabla_\nu\nabla_\mu=\partial_\nu\partial_\mu f, $$ which clashes with the expression given in the book. So where and how do the connection coefficients come in?

I have to say, I'm already confused by the definition given for $\nabla_X f$. Shouldn't the covariant derivative be a map from $(n,m)$-tensor fields to $(n,m+1)$-tensor fields?

Answer 1

In the case of a scalar field, the covariant derivative equals the partial derivative.

In the case of a tensor field of higher rank, there are further terms involving connection coefficients: for rank $n$ there are $n$ such terms.

Starting from scalar $f$, one can obtain the first rank (i.e. 4-vector) field $$ A_a \equiv \nabla_a f = \partial_a f $$ The covariant derivative of this 4-vector field is then $$ \nabla_b A_a = \partial_b A_a - \Gamma^\lambda_{ba} A_\lambda = \partial_b \partial_a f - \Gamma^\lambda_{ba} \partial_\lambda f $$ In other words it is all consistent. In particular, $\nabla_a f$ is of first rank (as the index implies). And of course one would expect a gradient of a scalar field to be like a vector field.

Coming now to the question about $\nabla_X$, yes this is a bit more subtle but it is not too bad. To get some intuition, consider the concept of differentiation along a line in the more familiar territory of Euclidean space. In this case if you have a scalar field $f$ then its gradient $\partial_a f$ is a vector field (given here in terms of its covariant components; you can call it a one-form if you like). But, for any given path specified by some $X_a(u)$, where $u$ is a parameter along the path, one can also obtain $$ \frac{df}{du} = \frac{d X^\mu}{du} \partial_\mu f . $$ From the indices one would say this is a scalar not a vector. That is right: it is a single number at each point on the path. However when combined with the specification of the path $X_a(u)$ it can be used to construct a vector, at least for points along the path.

Answer 2

1. The covariant derivative of a covector field $V_\mu$ is:

$$ V_{\mu;\nu} = V_{\mu,\nu} - \Gamma^\lambda{}_{\mu\nu} V_{\lambda} $$

2. The covariant derivative of a scalar field is just its partial derivative, but the result is a covector field:

$$ f_{;\mu} = f_{,\mu} $$

Put 1 & 2 together, i.e. put $V_\mu = f_{;\mu}$, and I believe your confusion will be resolved.

3. Also note that when $X^\mu$ is a vector, the difference between $\nabla_X f$ and $\nabla_\mu f$ is very similar to the difference you know from standard vector calculus between $\vec{X}\cdot\nabla f$ and $\nabla f$. i.e observe that the analogous operations are:

$$ \vec{X}\cdot\nabla f =\vec{X}\cdot\big[(\hat{x}\partial_x+\hat{y}\partial_y+\hat{z}\partial_z) f\big] \Longleftrightarrow \nabla_X f = X^\mu f_{,\mu} \tag{1}$$

$$ \nabla f = (\hat{x}\partial_x+\hat{y}\partial_y+\hat{z}\partial_z) f \Longleftrightarrow \nabla_\mu f = f_{,\mu}\tag{2}$$

Namely, in $(1)$, you're taking the gradient with respect to some coordinate basis (with basis vectors $\hat{e}_\mu$) and immediately dotting the result with a certain vector field $\vec{X}$ written in that basis. Hence the result is indeed a scalar field.

In $(2)$, you're just taking the gradient of a scalar function $f$ with respect to some coordinate basis, which results in a covector field. The relation between the two operations then becomes clear from the above.