# Covariant derivative acting on vector field

I'm following the lecture here, and this question is with respect to the content on the whiteboard at the timestamp provided.

Consider Newtonian spacetime $(\mathcal{M}, \mathcal{O}, \mathcal{A}, \nabla, t)$ where $(\mathcal{M}, \mathcal{O}, \mathcal{A})$ is a smooth 4-dimensional manifold where all charts $(\mathcal{U}, x)$ are of the form $$x^0 : \mathcal{U} \to \mathbb{R}\\ x^1 : \mathcal{U} \to \mathbb{R}\\ \vdots\\ x^3 : \mathcal{U} \to \mathbb{R}$$ where $x^0 = t|_\mathcal{U}$ is the restriction of the absolute time function $t$ to the chart domain $\mathcal{U}$ and $\nabla$ is a prescribed covariant derivative operator with given connection coeffitient functions.

I'm trying to reevaluate the claim on the right, namely $$0 = \nabla dt.$$ As I understand it, one does so by picking any direction $\frac{\partial}{\partial x^a}$ from a chart-induced basis for $a=0,1, \ldots, 3$, then act on this vectorfield with the prescribed covariant derivative operator and apply the result to the $(0,1)$-tensor field $dt$, which is the gradient of the 0th component function: $d(x^0)$. The action of the covariant derivative along a vectorfield on a $(p, q)$-tensor field yields again a $(p, q)$-tensor field, hence we can look at the resulting $(0,1)$-tensor-field, i.e. a vector field, component-wise for components $b=0, \ldots, 3$. Formally, we apply the rules for covariant derivatives $$\left(\nabla_{\frac{\partial}{\partial x^a}} d(x^0)\right)_b = \frac{\partial}{\partial x^a}\left(d(x^0)_b\right) - \Gamma_{b m}^n d(x^0)_n \left(\frac{\partial}{\partial x^a}\right)^m$$ According to the notes on the blackboard, the first term should vanish entirely, while the second one should reduce to $\Gamma_{ba}^0$. I'm havin trouble to see this.

Let's start with the first term: $d(x^0)$ is a covector, so $d(x^0)_b$ is the 0th component function. $\left(\frac{\partial}{\partial x^a}\right)$ is a vector field, which can be applied to a fucntion to yield another function. Alright. Still I can't see how I need to proceed there.

Secondly, I'm stuck with the rightmost term. Somehow, $\left(\frac{\partial}{\partial x^a}\right)^m$ must reduce to $\delta_a^m$ and $d(x^0)_n$ must reduce to $\delta_n^0$, however I don't see how.

Basically, as far as I understand, the whole expression needs to be the zero function in the end.

Can anyone explain where I seem to go wrong? Or help me understand the step missing?

## Edit 1

After the commetnt by @willie-wong, I think I can answer the first part. First I get the component functions of $dx^0$ by means of $$(df)_j := (df) \left(\frac{\partial}{\partial x^j}\right) = \left(\frac{\partial}{\partial x^j}\right)(f).$$ In the case I presented this would result in $$(dx^0)_b = \frac{\partial x^0}{\partial x^b} =\delta_b^0,$$ which results in the colletion of components $(1, 0, 0, 0)$ and $$\frac{\partial}{\partial x^a}\delta_b^0 = \partial_a(\delta_b^0 \circ x^{-1}) = 0,$$ since the output of $\delta$ is constantly 0 or one, depending on $b$ with respect to any index $a$. Makes sense to me.

## Edit 2

As in edit 1, the term $d(x^0)_n$ becomes $\delta_n^0$ following the same calculation. In regards to the second term, I may interpret $$\left( \frac{\partial}{\partial x^a}\right)^m = \left( \frac{\partial x^m}{\partial x^a}\right),$$ i.e. as the $a$-th basis vector applied to the $m$-th component function. Hence the result doubtlessly is the $\delta_a^m$ function, and hence the second term is nonvanishing only for $n=0$ and $m=a$, resulting in the "time flows uniformly"-equation presented by Prof. Schuller.

• relative the the coordinates $x^0, \ldots x^3$, the components of the one form $d(x^0)$ are $(1, 0, 0, 0)$. Can you see why? [This answers your first question.] – Willie Wong Sep 19 '18 at 13:38
• for your second question: the vectors $\partial / \partial x^a$ form a basis of the tangent space. Relative to this basis, what is $\partial / \partial x^i$, for any given $i$? – Willie Wong Sep 19 '18 at 13:39
• @willie-wong, I inserted two edits in regards to your comments. Is this what you wanted to point me at? – marc Sep 19 '18 at 15:22
• The coefficients $v^m$ of a vector field $v$ is defined to be such that $$v = \sum v^m (\partial / \partial x^m)$$ So yes, if you take $v(x^k)$ you will get $\sum_{m} v^m (\partial x^k / \partial x^m) = v^k$. And yes, you correctly applied this to compute the $m$th coefficient of $\partial/\partial x^a$. – Willie Wong Sep 19 '18 at 15:50