# Proof that covariant derivative of contravariant components of metric vanish for metric compatabilit

In this excercise I want to show that $$\nabla_\rho g_{\mu \nu}=0$$ and $$\nabla_\rho g^{\mu \nu}=0$$

This should probably be very easy, but excuse me I'm completly new to GR.

So to do this I used that the covariant derivative of a rank 2 tensor is

$$\nabla_c t^{ab} = \partial_c t^{ab} + \Gamma^a_{dc}t^{db}+\Gamma^b_{dc}t^{ad}$$ $$\nabla_c t_{ab} = \partial_c t_{ab} - \Gamma^d_{ac}t_{db}-\Gamma^d_{bc}t_{ad}$$

And I rewrote the connection in therms of partial derivatives of the metric. So for the covariant components of the metric this worked out nicely, everything canceld and ended up to be zero.

Now with the contravariant components I'm running into trouble:

$$\nabla_\rho g^{\mu \nu} = \partial_\rho g^{\mu\nu} + \Gamma^\nu_{d\rho}g^{d\nu}+\Gamma^v_{d\rho}g^{\mu d}=\\ \partial_\rho g^{\mu\nu} + \frac{1}{2}g^{\mu e} (\partial_d g_{e \rho} + \partial_\rho g_{de} -\partial_e g_{d\rho})g^{d \nu}+ \frac{1}{2}g^{\nu e} (\partial_d g_{e \rho} + \partial_\rho g_{de} -\partial_e g_{d\rho})g^{\mu d}$$

Now the two parts inside the brakets are the same and the metrics outside replace the dummy indices d and e by $$\nu$$ and $$\mu$$. If I raie the indices d and e four of the six terms cancel each other but then I am left with:

$$\nabla_\rho g^{\mu \nu} = 2 \partial_\rho g^{\mu \nu}$$

This is clearly not zero for a general metric in non local coordinates. So I assume I'm messing something up completly, pleas ehelp me to understand this.

• You can’t raise the indices because of the partial derivative. Mar 4, 2020 at 15:03
• You choose the connection such that $\nabla_\rho g_{\mu \nu}=0$. It's not true in general for any connection. And there's no need to expand the Christoffel symbols $\Gamma$ in terms of the metric. Mar 4, 2020 at 21:03
• @Prahar Well technically I can for a connection whose covariant derivative vanishes. That's the whole point why one chooses metric compatibility in GR I suppose? Mar 4, 2020 at 21:08
• @CinaedSimson I know that this is not true for all and I know that this is an assumption or choice you put into the framework of classical GR. But the exercise explicitly asks me to do this... "Show that, if the affine connection is written in terms of the metric, ∇ ρ g μν = 0, ∇ ρ g μν = 0." So I think this is just a little practice to get comfortable with these kind of calculations not a real proof or anything... Mar 4, 2020 at 21:11
• @higgshunter - To raise or lower an index you must have precisely the following index structure $g^{\mu\nu} g_{\nu\alpha}$. What you have is $g^{\mu\nu} \partial_\rho g_{\nu\alpha}$. To bring it into the first form, you will need to move the derivative around, $g^{\mu\nu} \partial_\rho g_{\nu\alpha} = \partial_\rho ( g^{\mu\nu} g_{\nu\alpha} ) - (\partial_\rho g^{\mu\nu}) g_{\nu\alpha}$. You can now raise the index on the first term without issue. The SAME holds true even if I replace the partial derivative with a covariant derivative. Mar 4, 2020 at 21:11

Actually you were close but no cigar.

The purpose of this post to introduce you to the dual space connections.

In addition, I don't agree with the examples of covariant derivative of tensors which you provided in your question. For instance, taking the covariant derivative of a $$(r,s)$$ tensor changes it to a $$(r,s+1)$$ tensor.

I'm going to use $$D$$ to denote the covariant derivative (fewer keystrokes) - and I'm assuming the Levi-Civita connection (i.e., the connection is torsion free.)

The connections in the dual space are denoted as $$\omega^{j}_{i}=\Gamma^{j}_{ik} dx^{k}.$$

And I'm using the full metric tensor

$$g=g_{ij}dx^{i}\otimes dx^{j}$$

which are typically denoted by the tensor coefficients only, for example, $$g_{ij}.$$

The covariant derivative of the metric tensor

$$Dg=0$$

ensures the magnitudes of the tangent vectors and the angles between them remain invariant when the metric tensor is parallel transported.

Expanding the calculation of the covariant derivative

$$Dg =D(g_{ij})dx^{i}\otimes dx^{j}+g_{ij}D(dx^{i})\otimes dx^{j}+g_{ij}dx^{i} \otimes D(dx^{j})$$ yields $$Dg=(dg_{ij}-\omega^{k}_{i}g_{kj}-\omega^{k}_{j}g_{ik})\otimes dx^{i}\otimes dx^{j}=0$$

which implies

\begin{align} dg_{ij}=&\; \omega^{k}_{i}g_{kj}+\omega^{k}_{j}g_{ik}\\ \Gamma^{j}_{ik}=&\; \Gamma^{j}_{ki} \end{align}

where

\begin{align} D(g_{ij})\;=&\;dg_{ij}\\ D(dx^{i})=&-\omega^{k}_{i}g_{kj}dx^{i}\\ D(dx^{j})=&-\omega^{k}_{j}g_{ik}dx^{j}. \end{align}

At this point, I should probably also prove that when $$2$$ vector fields from the tangent space are parallel transported along a curve with respect to the metric then $$DG=0$$ - but I don't have time today - and it's a seperate question.

Also, it's possible to derive the Christoffel symbols from equation $$(1)$$ if one writes $$dg_{ij}$$ in terms of partials, the $$\omega^{i}_{j}$$ connections in the form $$\Gamma_{ijk}$$ and use the metric to raise the $$k$$ at the very end. But it can be a little tricky.

Ref: S.S. Chern, W.H. Chen, and K.S. Lam, $$Lectures\;on\;Differential\;Geometry$$, Series On University Mathematics, Vol. $$\mathbf{1}$$ (1998), 133-139.