# Upper index covariant derivative $\nabla^\mu$

In the book Cosmology by Daniel Baumann, the author states that $$\nabla^\mu g_{\mu\nu}=0$$, where $$g_{\mu\nu}$$ is the metric tensor considered (usually the one associated to the Minkowski metric or to the FRW metric), and this is used to reason that we can introduce the cosmological constant in the form of $$\Lambda g_{\mu\nu}$$ in the left-hand side of the Einstein field equations, since its presence doesn't alter the conservation of the energy-momentum tensor, given by $$\nabla^\mu T_{\mu\nu}=0$$.

However, I don't understand what the covariant derivative with an upper index is supposed to mean, that is, what is $$\nabla^\mu$$? What I know is that the covariant derivative acts in the following way on four-vectors with upper and lower indices:

$$\nabla_\mu A^\nu=\partial_\mu A^\nu+\Gamma^\nu_{\mu\lambda}A^\lambda$$ $$\nabla_\mu B_\nu=\partial_\mu B_\nu-\Gamma^\lambda_{\mu\nu}B_\lambda$$

Should I just take $$\nabla^\mu=g^{\mu\nu}\nabla_\nu$$? My knowledge on general relativity is quite basic, so I'm unsure of how am I supposed to prove that $$\nabla^\mu g_{\mu\nu}=0$$. Is this valid in general or does it depend on the metric considered?

• Just expand Christoffel symbols in terms of metric
– KP99
Commented Feb 2, 2023 at 18:43
• Covariant derivatives of the metric vanish because the Levi-Civita connection is defined to make it vanish. Commented Feb 2, 2023 at 18:46
• You can use the metric to raise the index. Commented Feb 2, 2023 at 18:52
• Have you learned how to take the covariant derivative of tensors, using Christoffel symbols? Commented Feb 2, 2023 at 18:53
• As Ghoster mentioned, if you start with $\nabla_ag_{bc}=0$ , then you can express $\Gamma^a_{bc}$ in terms of $g_{ab}$. This is the Levi-Civita connection. What I said is basically the back-calculation: you substitute the levi-civita connection and get cov. der =0. And yes $\nabla^a=g^{ab}\nabla_b$
– KP99
Commented Feb 2, 2023 at 19:16

In General Relativity, the covariant derivative is defined in such a way that $$\nabla_\rho g_{\mu\nu} = 0$$. This can be seen by writing the covariant derivative explicitly in terms of Christoffel symbols, for example. One may then notice that $$\nabla^\mu g_{\mu\nu} = g^{\mu\rho} \nabla_{\rho} g_{\mu\nu} = 0$$.
"Should I just take $$\nabla^\mu = g^{\mu\nu}\nabla_\nu$$?" - Yes, you should, that is the definition of $$\nabla^\mu$$.
The covariant derivative is a map $$\nabla : \Gamma(TM)\times T^r_s(M)\to T^r_s(M)$$ taking a vector field $$X$$ and a tensor field $$T$$ and giving back a tensor field $$\nabla_X T$$. When you fix $$X = \partial_\mu$$, one of the coordinate basis vectors, you get the operator $$\nabla_\mu \equiv \nabla_{\partial_\mu}:T^r_s(M)\to T^r_s(M).$$ We then define $$\nabla^\mu = g^{\mu\nu}\nabla_\nu$$. This is, in particular, relevant in the case $$(r,s)=(0,0)$$ in which you are acting on scalars. Then $$\nabla_\mu f =\partial_\mu f$$ just gives you the exterior derivative $$df\in \Omega^1(M)$$, which is a one-form, while $$\nabla^\mu f$$ becomes the gradient $$\nabla f\in \Gamma(TM)$$, which is a vector field.
Regarding $$\nabla^\mu g_{\mu\nu}=0$$, it follows directly from the assumption that your connection $$\nabla$$ is metric-compatible, i.e. $$\nabla g=0$$. In fact, $$\nabla^\mu g_{\mu\nu} = g^{\rho\mu}\nabla_\rho g_{\mu\nu}$$ but $$\nabla_\rho g_{\mu\nu}=0$$ because it is exactly the coordinate version of $$\nabla g =0$$. This is a standard assumption in GR. In Differential Geometry it is perfectly possible to endow a manifold with a metric and a connection such that $$\nabla g \neq 0$$. Then your property simply doesn't hold.