Covariant derivative, action of vector to functions, coordinates

Question

Let $(\mathbb{e}_1,...,\mathbb{e}_n)$ be a frame of vector fields in a region of $U$. Let $\mathbb{X}=\mathbb{e}_jX^j.$

Suppose we have an expression for the covariant derivative in coordinates given as

$\nabla_X(\mathbb{e}_kv^k)=X^j\mathbb{e}_i\omega^i_{jk}v^k+X^j\mathbb{e}_j(v^k)\mathbb{e}_k$.

What is in coordinates and in words exactly $\mathbb{e}_j(v^k)$ here?

Answer 1

For any real value function $f : U \rightarrow \mathbb{R}$ defined on $U\subset M$, and a local vector field $X : U \rightarrow TU$ we can obtain new function $Xf : U \rightarrow \mathbb{R}$ defined by $$ (Xf)(p) = X_p f $$ I.e every (local) vector field $X : U \rightarrow TU$ is an operator on the space of real value function $C^{\infty}(U)$. That is $X : C^{\infty}(U) \rightarrow C^{\infty}(U)$. For your case, you have a local vector field $v : U \rightarrow TU $ with its expression in local frame is $v = v^ie_i$, where the components $v^i$ is just a real value function defined on $U$. The expression $\mathbf{e_j} (v^i)$ is just another real value function obtained by above construction $$ \mathbf{e_j} (v^i) : U \rightarrow \mathbb{R} $$

So when you evaluate the covariant derivatives at some point $p\in U$ you get the function $\mathbf{e_j} (v^i)$ evaluated at $p$, which gives you a real number $\mathbf{e_j} (v^i)(p) = \mathbf{e_j}|_p v^i \in \mathbb{R}$.

$\textbf{EDIT} :$

The local frame $\{\mathbf{e_j}\}$ is not always a coordinate frame $\{\partial_j\}$ so when we actually compute $\mathbf{e_j}|_p v^i$, its usually we have to expand $\{\mathbf{e_j}\}$ in terms of coordinate frame $\{\partial_j\}$. So e.g $\mathbf{e_j} = A_j^k \partial_k$ so $$ \mathbf{e_j} (v^i) = A_j^k \partial_k (v^i)= A^k_j \frac{\partial v^i}{\partial x^k} $$

Answer 2

Here ${\bf e}_j(v^k)$ is a shorthand notation for the derivative of the (smooth) function $v_k$ along the basis vector ${\bf e}_j$. This is what, 'in coordinates', is often denoted also by $\partial v_k/\partial x^j$, or $v_{k,j}$, or more accurately, $\langle dv_k,{\bf e}_j\rangle$.

More generally any vector field $X$ defined on $U$, such as $X={\bf e}_j$, gives rise to a first order differential operator acting on a smooth function $u\in C^{\infty}(U,\bf R)$, to give another smooth function $Xu$ (also denoted sometimes $\theta_X u$), defined by $Xu(x) = \langle du(x),X(x)\rangle$, $x\in U$. You must understand that $Xu$ here is a smooth function on $U$, not a single value defined on a point of $U$.