Covariant derivative - do the link between 2 expressions

Question

From the answer of @levap on the following link, on Difference between Covariant derivative notations, I try to understand the subtilities of covariant derivative.

levap wrote : " $\nabla_{\frac{\partial}{\partial x^i}} \frac{\partial}{\partial x^j} = \Gamma_{ij}^k \frac{\partial}{\partial x^k}$ "

Then If I apply this equation to the vector position $\vec{OM}$, I know this definition which makes appear the basis vector of curvilinear coordinates, i.e :

$\nabla_{\frac{\partial}{\partial x^i}} \frac{\partial \vec{OM}}{\partial x^j} = \Gamma_{ij}^k \frac{\partial\vec{OM}}{\partial x^k}$

that I can write also (with $\vec{e_j}$ and $\vec{e_k}$ local curvilinear basis vectors) :

$\nabla_{\frac{\partial}{\partial x^i}} \vec{e_j} = \Gamma_{ij}^k \vec{e_k}$ (equation 1)

I try to do the link between what I know, i.e I can make appear Christoffel's symbols like this :

$d\vec{e_i}=w_i^j \vec{e_j} = \Gamma_{ik}^{j} dx^{k} \vec{e_j}$ (equation 2)

How to do the link between expressions (equation 1)and (equation 2) ?

Any help is welcome, thanks

UPDATE 1 :

As I said in my comment, (equation 2) just represents the classic differential of a vector $$\vec{e_i}$$ like for example : $\text{d}\vec{e_r}=\text{d}\theta\,\vec{e_\theta}$ in polar coordinates. So in this case, I would have :

$\Gamma_{r\theta}^{\theta}=\Gamma_{\theta r}^{\theta}=1$

Concerning your answer, I need precisions about the notations that you used in your demonstration. I think that you assimilate $\{X_{a}\}$ basis to my notation $\text{d}x^{a}$ and its dual base $\{e^{b}\}$ to my notation $\{\dfrac{\partial}{\partial x^{b}}\}$ : by doing scalar product between 2 vectors of each of these basis, I get :

$$X_{a}\,e^{b} = \dfrac{\partial(\text{d}x^{a})}{\partial x^{b}}=\delta^{ab}$$

Is this equality above correct ?

Secondly, I have a problem with the following definition :

$$\nabla_{X_{a}}X_{b} \,=\, \Gamma_{ab}^{c}X_{c} \,=\, \omega^{\,c}_{\,\;b}(X_{a})X_{c}\quad\quad\text{(equation 3)}$$

With my notations, Covariant derivative is defined by :

$$\nabla_i{V_j}=\partial_i V_j -\Gamma_{ij}^{k}V_k = \dfrac{\partial V_j}{\partial x^{i}} - \Gamma_{ij}^{k}V_k\quad\quad (equation 4)$$

So one term into (equation 3) is missing relatively to my (equation 4) : this the term $\partial_i V_j = \dfrac{\partial V_j}{\partial x^{i}}$

Have you deliberately ommit this term or is it contained in others terms in your expression, i.e (equation 3).

Concerning the property (by swapping the dual and normal basis components into covariant derivative definition) :

$$\nabla_{X_{b}}e^{a} \,=\, -\omega^{\,a}_{\,\;c}(X_{b})e^{c}$$

I didn't know it : is there a quick way to proove it ?

One last question concerning the equality :

$$\text{d} \,\equiv\, e^{b} \wedge \nabla_{X_{b}}$$

Is this corresponding to the definition of total differential ?? , that I know under the form of :

$$\text{D}\,V_{i} \,\equiv\, \nabla_b\,V^{i}\,\,\text{d}x^{b}$$

with $\text{D}$ the total differential operator and $\text{d}$ the classical differential.

UPDATE 2 :

I am going to begin by boring you but I still have issues with some details.

I calculate the details of $\nabla_{X_{i}}V$ expression:

You wrote :

$$\begin{align} \nabla_{X_{i}}V &\,=\, \nabla_{X_{i}}(V_{a}e^{a}) \,=\, (\nabla_{X_{i}}V_{a})e^{a} + V_{a} (\nabla_{X_{i}}e^{a}) \\[0.1cm] & \,=\, (X_{i}V_{a})e^{a} - V_{a} (\Gamma^{a}_{ij}e^{j}) \\[0.1cm] & \,=\, \left(X_{i}V_{j} - V_{a}\Gamma^{a}_{ij}\right) e^{j} \end{align}.$$

Here my calculation, I start by the following expression (equation 5) :

$$\begin{align} \nabla_{X_{i}}V &\,=\, \nabla_{X_{i}}(V_{a}e^{a}) \,=\, (\nabla_{X_{i}}V_{a})e^{a} + V_{a} (\nabla_{X_{i}}e^{a}) \\[0.1cm] \end{align}.$$

Up to here, we agree.

Then, for the first term of right member into (equation 5):

$$\begin{align} (\nabla_{X_{i}}V_{a})\,=\,\dfrac{\partial V_{a}}{\partial x^{i}}-V_{c}\Gamma_{ia}^{c} \end{align}$$

So if I add factor $e^{a}$, we get

$$\begin{align} (\nabla_{X_{i}}V_{a})e^{a} \,=\, \dfrac{\partial V_{a}}{\partial x^{i}}\,\text{d}x^{a}-(V_{c}\Gamma_{ia}^{c})\,\text{d}x^{a} \end{align}$$

this yields :

$$\begin{align} \nabla_{X_{i}}(V_{a})e^{a} &\,=\, \delta^{a}_{i}\,V_{a}-(V_{c}\Gamma_{ia}^{c})\,e^{a}\\[0.1cm] &\,=\, V_{i} - V_{c}\Gamma_{ia}^{c}\,\text{d}x^{a}\\[0.1cm] \end{align}$$

Then, for the second term of right member into (equation 5):

$$\begin{align} V_{a} (\nabla_{X_{i}}e^{a}) &\,=\, V_{a}\,\delta^{a}_{i} + V_{a}\,\Gamma_{ic}^{a}\,e^{c}\\[0.1cm] &\,=\, V_{i}+V_{a}\,\Gamma_{ic}^{a}\,\text{d}x^{c} \end{align}$$

So If I add the 2 terms, I get :

$$\begin{align} \nabla_{X_{i}}V &\,=\, \nabla_{X_{i}}(V_{a}e^{a})++ V_{a} (\nabla_{X_{i}}e^{a})\\[0.1cm] &\,=\,2V_{i} \end{align}$$

This result is not good because it sould be equal to (like you proove it) :

$$\nabla_{X_{i}}V = \left(X_{i}V_{j} - V_{a}\Gamma^{a}_{ij}\right) e^{j}$$

At first sight, could you see where is my error in this calculation ?

Secondly, could I write :

$$\begin{align} (\nabla_{X_{j}}\,e^{a}) &\,=\, \dfrac{\partial \text{d}x^a}{\partial x^{i}}\,+\,\Gamma_{jc}^{a}\,e^{c}\\[0.1cm] &\,=\, \delta_{j}^{a}+\Gamma_{jc}^{a}\,e^{c} \end{align}$$

???

Thanks a lot for your help, this is precious.

Answer 1

I have a problem with equation 2 as you seem to be taking the exterior derivative of a vector. If I interpret your question correctly, you are trying to understand a relation between the covariant derivative and the exterior derivative, so let me try to answer that.

Let $g$ be a metric on some Riemannian manifold $\mathcal{M}$ coordinated by $(x^{a})$:

$$g \,=\, g^{ab}\frac{\partial}{\partial x^{a}} \otimes \frac{\partial}{\partial x^{b}} \,=\, \,\delta^{ab}\,X_{a}\otimes X_{b}$$

where $g^{ab}$ are the components of the metric, $\delta^{ab}$ is the Kronecker $\delta$ function and $\{X_{a}\}$ denotes a $g-$orthonormal frame with dual $g$-orthonormal coframe $\{e^{a}\}\;$ (i.e. $e^{a}(X_{b})=\delta^{a}_{b}$). Let $\nabla$ be the (unique) Levi-Civita connection on $\mathcal{M}$:

$$\nabla_{X_{a}}X_{b} \,=\, \Gamma_{ab}^{c}X_{c} \,=\, \omega^{\,c}_{\,\;b}(X_{a})X_{c}$$

where $\{\Gamma^{c}_{ab}\}$ are the connection coefficients (in a co-ordinate frame these are the Christoffel symbols) and $\{\omega^{\,a}_{\,\;b}=\Gamma^{a}_{cb}\,e^{c}\}$ are the connection $1-$forms. It follows from the definition above, the duality of $\{e^{a}\}$ and $\{X_{a}\}$, and the fact that $\nabla$ commutes with contractions that:

$$\nabla_{X_{b}}e^{a} \,=\, -\omega^{\,a}_{\,\;c}(X_{b})e^{c}.$$

Then from this it follows

$$e^{b} \wedge \nabla_{X_{b}}e^{a} \,=\, -\omega^{\,a}_{\,\;c} \wedge e^{c}$$

and hence from Cartan's first structure equation:

$$T^{a} \,=\,de^{a} + \omega^{\,a}_{\,\;c} \wedge e^{c} \,=\, de^{a} - e^{b} \wedge \nabla_{X_{b}}e^{a} \,=\, (d - e^{b} \wedge \nabla_{X_{b}})e^{a}$$

where $\{T^{a}\}$ are the torsion $2-$forms. Since $\nabla$ is Levi-Civita (and hence torsion-free: $T^{a}=0$), this yields:

$$d \,\equiv\, e^{b} \wedge \nabla_{X_{b}}$$

which is a relation between the Levi-Civita connection $\nabla$ and exterior derivative $d$ in terms of the dual $g-$orthonormal frames $\{X_{a}\}$ and $\{e^{a}\}$.

In fact, we only really require the connection to be torsion-free and not necessarily the Levi-Civita connection. We may even relax that assumption and would find by a similar argument that for any linear connection:

$$d \,\equiv\, e^{b} \wedge \nabla_{X_{b}} + T^{b} \wedge i_{X_{b}}$$

in terms of the interior derivative $i_{X_{b}}$. I hope this helps with some of your questions.

RESPONSE TO UPDATE:

I feel I may be confusing matters, but let me try to explain. Let's take the case you mention: spherical polar coordinates $(r,\theta,\phi)$. The metric is:

$$g\;=\; dr\otimes dr + r^{2}d\theta \otimes d\theta + r^{2}\sin^{2}(\theta)\,d\phi\otimes d\phi.$$

I can define the $g-$orthonormal coframe:

$$e^{1} \,=\, dr \quad e^{2}\,=\, r\,d\theta, \quad e^{3}\,=\, r\sin(\theta)\,d\phi$$

so that the metric then becomes:

$$g\;=\; e^{1} \otimes e^{1} + e^{2} \otimes e^{2} + e^{3}\otimes e^{3}$$.

Essentially all the metric components have now been "hidden away" in the definition of this coframe. Similar I would define

$$X_{1} \,=\, \frac{\partial}{\partial r}, \quad X_{2} \,=\, \frac{1}{r}\frac{\partial}{\partial \theta}, \quad \text{and the third :}\,\,\, X_{3} \,=\, \frac{1}{r\sin(\theta)}\frac{\partial}{\partial \phi}$$

to be the $g-$orthonomal frame. So the $\{e^{a}\}$ are 1-forms and $\{X_{a}\}$ are vector fields. Then, for example:

$$e^{1}(X_{2}) \,=\, dr\left(\frac{1}{r}\frac{\partial}{\partial \theta}\right) \,=\, \frac{1}{r} \cdot dr\left(\frac{\partial}{\partial \theta}\right) \,=\, 0$$

whereas

$$e^{2}(X_{2}) \,=\, r\,d\theta\left(\frac{1}{r}\frac{\partial}{\partial \theta}\right) \,=\, r\cdot\frac{1}{r} \cdot d\theta\left(\frac{\partial}{\partial \theta}\right) \,=\, 1.$$

The equality $e^{a}(X_{b})=\delta^{a}_{b}$ is correct, but it is more general as it holds for any $g$-orthonormal duals $\{e^{a}\}$ and $\{X_{a}\}$, not just the inertial frames $\{e^{a}=dx^{a}\}$ and $\{X_{a}=\frac{\partial}{\partial x^{a}}\}$.

Also, there is nothing missing in my definition. Notice that what is really meant by $\nabla_{i}V_{j}$ is $(\nabla_{X_{i}}V)_{j}$: the $j-$th component of the covariant derivative of the 1-form $V$. Writing $V=V_{a}e^{a}$:

$$\begin{align} \nabla_{X_{i}}V &\,=\, \nabla_{X_{i}}(V_{a}e^{a}) \,=\, (\nabla_{X_{i}}V_{a})e^{a} + V_{a} (\nabla_{X_{i}}e^{a}) \\[0.1cm] & \,=\, (X_{i}V_{a})e^{a} - V_{a} (\Gamma^{a}_{ij}e^{j}) \\[0.1cm] & \,=\, \left(X_{i}V_{j} - V_{a}\Gamma^{a}_{ij}\right) e^{j} \end{align}.$$

Then the $j-$th component is easily seen to be $X_{i}V_{j} - V_{a}\Gamma^{a}_{ij}$, just as you as $X_{i}V_{j}=\partial_{i}V_{j}$ in the case of intertial coordinates with $\{X_{i}=\partial_{i}\}$.

The way of proving the formula for the covariant derivative of the frames $\{e^{a}\}$ is from duality relation and $\nabla_{X_{a}}X_{b}=\omega^{\,c}_{\,\;b}(X_{a})X_{c}$:

$$\begin{align} e^{a}(X_{b})\,=\,\delta^{a}_{b} \quad&\Longrightarrow\quad \nabla_{X_{j}}\left[e^{a}(X_{b})\right] \,=\, \nabla_{X_{j}}\delta^{a}_{b} \,=\, 0 \\[0.1cm] &\Longrightarrow\quad (\nabla_{X_{j}}e^{a})(X_{b}) + e^{a}(\nabla_{X_{j}}X_{b}) \,=\, 0\\[0.1cm] &\Longrightarrow\quad (\nabla_{X_{j}}e^{a})(X_{b}) + e^{a}(\omega^{\,c}_{\,\;b}(X_{j})X_{c}) \,=\, 0 \\[0.1cm] &\Longrightarrow\quad(\nabla_{X_{j}}e^{a})(X_{b}) + \omega^{\,c}_{\,\;b}(X_{j})\,e^{a}(X_{c}) \,=\, 0 \\[0.1cm] &\Longrightarrow\quad(\nabla_{X_{j}}e^{a})(X_{b}) + \omega^{\,c}_{\,\;b}(X_{j})\,\delta^{a}_{c} \,=\, 0 \\[0.1cm] &\Longrightarrow\quad(\nabla_{X_{j}}e^{a})(X_{b}) + \omega^{\,a}_{\,\;b}(X_{j}) \,=\, 0. \end{align}$$

Then since $\alpha(X_{a})e^{a}=\alpha$ for any $1-$form $\alpha$, we find

$$\begin{align} (\nabla_{X_{j}}e^{a})(X_{b})e^{b} + \omega^{\,a}_{\,\;b}(X_{j})e^{b} \,=\, \nabla_{X_{j}}e^{a} + \omega^{\,a}_{\,\;b}(X_{j})e^{b} \;=\; 0 \end{align}$$

and the result follows.

And finally... the $d$ I use is called the exterior derivative and is very much related to the total derivative $D$ which you mention. They are the same for functions but $d$ extends the notion of a total derivative to higher degree differential forms; something perhaps for later study or another time.

RESPONSE TO UPDATE 2:

The key to understand your problem is to realise what people actually mean by $\nabla_{X_{i}}V_{j}$. It is actually shorthand for:

$$\nabla_{X_{i}}V_{j} \;\equiv\; (\nabla_{X_{i}}V)_{j}$$

so when I calculated $\nabla_{X_{i}}V=\left(X_{i}V_{j} - V_{a}\Gamma^{a}_{ij}\right) e^{j}$ it is easy to note what the $j-$th component is:

$$ (\nabla_{X_{i}}V)_{j} \;=\; X_{i}V_{j} - V_{a}\Gamma^{a}_{ij}.$$

If by $\nabla_{X_{i}}V_{j}$ we meant "take the covariant derivative of the components $V_{j}$ of the vector $V$", then this would simply be $X_{i}V_{j}$ as the components are just smooth functions. Understanding this should get you over your issues and you'll find everything works out.