# Covariant derivative of the Levi-Civita symbol

In general relativity, the Levi-Civita symbol is defined by for example in spacetime with dimension 2+1 $$$$\varepsilon^{abc}=\frac{\epsilon^{abc}}{\sqrt{-g}},~~\epsilon^{abc}=0,\pm 1.$$$$ Here $$\varepsilon^{abc}$$ is a pesudotensor. My question is why the covariant derivative of this tensor vanishes: $$$$\nabla_d \varepsilon^{abc}=0?$$$$

• $\epsilon^{abc}$ is constant and and $\nabla_a \sqrt{-g}=0$ Commented Dec 13, 2019 at 16:28
• @jinawee why does $\nabla_a \sqrt{-g} =0$? Commented Feb 22, 2020 at 4:09
• @Greg.Paul See physics.stackexchange.com/q/249961 Commented Feb 22, 2020 at 12:22
• There seems to be some confusion about which is the tensor and which is the symbol. Since the Levi-Civita tensor is on the left, it is the Levi-Civita tensor which is being defined in terms of the Levi-Civita symbol and the determinant of the metric on the right. Commented Dec 6, 2020 at 22:05

Before answering your question, let me clear up some confusions that seem rather omnipresent.

I will use $$\tilde{\epsilon}_{\mu\nu\rho\sigma}$$ to denote the Levi-Civita symbol and $$\epsilon_{\mu\nu\rho\sigma} = \sqrt{-g} \tilde{\epsilon}_{\mu\nu\rho\sigma}$$ to denote the Levi-Civita tensor. Note that while $$\epsilon_{\mu\nu\rho\sigma}$$ defines a legitimate tensor field, neither $$\sqrt{-g}$$ nor $$\tilde{\epsilon}_{\mu\nu\rho\sigma}$$ does.

(Covariant derivatives only make sense when they operate on a tensor field. To qualify for a tensor field, the value at each point must be geometric in nature and be independent of coordinates. $$\sqrt{-g}$$, as a number, clearly depends on your particular choice of coordinates.)

So one should never write down non-sensical equations like $$\nabla_\alpha \epsilon_{\mu\nu\rho\sigma} = \sqrt{-g} \nabla_\alpha \tilde{\epsilon}_{\mu\nu\rho\sigma} + \tilde{\epsilon}_{\mu\nu\rho\sigma} \nabla_\alpha \sqrt{-g},$$

because on the right-hand side things like $$\nabla_\alpha \sqrt{-g}$$ is not well defined.

To define covariant derivatives, all that we need is connection coefficients, usually written as $$\Gamma^\rho_{\mu\sigma}$$. We can write $$\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu\lambda} V^\lambda.$$

To make the concept of covariant derivatives useful, we want it to satisfy the following minimal set of criteria.

1. linearity: $$\nabla (T + S) = \nabla T + \nabla S$$;
2. Leibniz (product) rule: $$\nabla(T \otimes S) = (\nabla T)\otimes S + T \otimes(\nabla S)$$.
3. commutes with contractions: $$\nabla_\mu ({T^\lambda}_{\lambda\rho}) = {{(\nabla T)_\mu}^\lambda}_{\lambda\rho}$$.
4. reduces to the partial derivative on scalars: $$\nabla_\mu \phi = \partial_\mu \phi$$.

These four requirements put certain constraints on $$\Gamma$$, so that we also, by necessity, have $$\nabla_\mu \omega_\nu = \partial_\mu \omega_\nu - \Gamma^\lambda_{\mu\nu}\omega_\lambda.$$

This can be then easily extended to rules for covariant derivatives of tensors of arbitrary ranks such as $${R^\mu}_{\nu\rho\sigma}$$. We can, e.g., easily derive $$\nabla_\mu \delta^\rho_\sigma = 0,$$ which makes perfect sense because $$\delta^\rho_\sigma$$ is simply the identity operator and should not be change from point to point.

N.B. The four criteria above will not give us the usual formula to calculate Christoffel symbols. Neither will they give us $$\nabla_\lambda \epsilon_{\mu\nu\rho\sigma}$$. For these, we need to add some more criteria.

In general relativity, we usually also add two more requirements.

• torsion-free: $$\Gamma^\lambda_{\mu\nu} = \Gamma^\lambda_{\nu\mu}$$.
• metric compatibility: $$\nabla_\rho g_{\mu\nu} = 0$$.

With these two additional requirements, we can get the usual formula to calculate Christoffel symbols from the metric itself.

But even with only metric compatibility, the following are also true. $$\nabla_\rho g^{\mu\nu} = 0$$ $$\nabla_\lambda \epsilon_{\mu\nu\rho\sigma} = 0$$

The first can be shown by having

$$0 = \nabla_\rho \delta^\mu_\sigma = \nabla_\rho (g^{\mu\nu} g_{\nu\sigma}) = g_{\nu\sigma} \nabla_\rho g^{\mu\nu} + g^{\mu\nu} \nabla_\rho g_{\nu\sigma} = g_{\nu\sigma} \nabla_\rho g^{\mu\nu},$$

and then multiplying both sides by $$g^{\sigma\nu'}$$.

The second can be shown by having first

$$0 = \nabla_\rho (\epsilon_{\mu\nu\rho\sigma} \epsilon^{\mu'\nu'\rho'\sigma'}) = \epsilon^{\mu'\nu'\rho'\sigma'} \nabla_\rho \epsilon_{\mu\nu\rho\sigma} + \epsilon_{\mu\nu\rho\sigma} \nabla_\rho \epsilon^{\mu'\nu'\rho'\sigma'},$$ where the first equality is because $$\epsilon_{\mu\nu\rho\sigma} \epsilon^{\mu'\nu'\rho'\sigma'}$$ is equal to the sum of a bunch of products of delta (i.e. identity) operators.

Now multiply both sides by $$\epsilon_{\mu'\nu'\rho'\sigma'}$$. The LHS remains 0. The first term on RHS becomes $$4! \nabla_\rho \epsilon_{\mu\nu\rho\sigma}$$, while the second becomes $$\epsilon_{\mu'\nu'\rho'\sigma'} \epsilon_{\mu\nu\rho\sigma} \nabla_\rho \epsilon^{\mu'\nu'\rho'\sigma'} = \epsilon^{\mu'\nu'\rho'\sigma'} \epsilon_{\mu\nu\rho\sigma} \nabla_\rho \epsilon_{\mu'\nu'\rho'\sigma'} = 4! (\delta^{\mu'}_\mu \delta^{\nu'}_\nu \delta^{\rho'}_\rho \delta^{\sigma'}_\sigma) \nabla_\rho \epsilon_{\mu'\nu'\rho'\sigma'} = 4! \nabla_\rho \epsilon_{\mu\nu\rho\sigma},$$ where we exchanged downstairs and upstairs for dummy variables within the same pair, even though one is within and one is outside the derivative. This is because $$\nabla_\rho g^{\mu\nu} = 0$$, which we just proved earlier.

This gives us the desired the result $$\nabla_\rho \epsilon_{\mu\nu\rho\sigma} = 0.$$

To reiterate, for this result to be valid, we only need metric compatibility, but not torsion-free, of covariant derivatives.

• Hi, thanks for your great answer! However, I am not totally understand your following description, “ To qualify for a tensor field, the value at each point must be geometric in nature and be independent of coordinates”. Why the tensor field should be independent of coordinates? We can operate $\nabla_\mu$ on $V^\lambda$, and I think $V^\lambda$ can depend on coordinates. Could you please elaborate more on this? Commented Apr 2, 2023 at 2:36
• The numerical value $V^\lambda$ obviously depends on coordinates. But $V$, as a vector, does not. The freezing point of water is 0 in Celsius and 32 degrees in Fahrenheit. So it is numerically dependent on the grade that you use. But its essence is independent thereof. Commented Apr 3, 2023 at 10:10
• Oh, that's a great explanation! So the $V$ has real physical meaning, despite its value may different in different coordinates. $\nabla_\mu V^\mu$ is Lorentz invariant in Minkowski spacetime. However, $g$ is a pure metric measure, it's a tensor density of weight -2. Commented Apr 4, 2023 at 13:13

To clarify a little, the Levi-Civita symbol $$\epsilon_{abc}$$ is a number, the one you wrote down, and not a tensor. It is a generalisation of the Kronecker delta. You can use it to build a tensor, the Levi-Civita tensor $$\varepsilon$$, whose components you wrote down. So okay, maybe it's a pseudo-tensor. It is the canonical volume form on the (pseudo-)Riemannian manifold.

To even talk about covariant differentiation, there must be an implicit connexion and hence Christoffel symbols. Well, you have assumed a metric $$g_{ij}$$, so okay, there is an important theorem that says that even though there are many different possible connexions, there is only one connexion on a (pseudo-)Riemannian manifold that:

(a) has no torsion, so it is symmetric, and

(b) it is consistent with the metric in that the covariant derivative (using that connexion) of the metric tensor is zero.

Another basic theorem of Riemannian geometry says that there is, at any one point $$p$$ that you decide on (and then do not get to change), a coordinate system such that all the Christoffel symbols vanish (but only at that point, they do vary in a neighbourhood of p, no matter how small) and all the first derivatives of the metric tensor's components $$g_{ij}$$ vanish at $$p$$.

Now we are ready to answer your question, and since the question is covariant, independent of the coordinates, our answer will generally be true for any coordinate system even though we are doing our calculation in a very special coordinate system.

The volume form is unique so even in these coordinates, it is $$\sqrt{-g}\wedge dy \wedge dz$$. And since the Christoffel symbols are all zero at $$p$$, the covariant derivative in any direction, e.g. the three coordinate directions, is given by the usual partial derivative of these coordinates.

With this set up, the ideas that other posters have attempted to communicate now work. Consider any directional derivative of $$\sqrt{-g} dx \wedge dy \wedge dz$$, e.g., $$\frac \partial {\partial x} (\sqrt{-g}dx \wedge dy \wedge dz)$$. By the Leibniz rule, for covariant derivatives, it is $$(\frac \partial {\partial x} \sqrt{-g}) dx \wedge dy \wedge dz + \sqrt{-g}\frac \partial {\partial x}(dx \wedge dy \wedge dz)$$.
As I remarked, these derivatives can be either the $$x$$-component of the covariant derivative of the tensor or the ordinary partial derivative with respect to $$x$$ of the component of the tensor, because the Christoffel symbols are zero.

But the partial of $$g$$ is zero since we assumed all first derivatives of the $$g_{ij}$$ in our coordinate system vanished. (It is also true that the covariant derivative of the metric tensor, and hence any function of it, also vanishes, in any coordinate system, but this is why.) So the first summand vanishes.

The second summand vanishes since all its components vanish: the coefficients of $$dx \wedge dy \wedge dz$$ are constant, so their partials vanish. This proves your formula.

• On the one hand, I have relied on a very deep theorem, that such a system of coordinates exists, but on the other hand, the technique of simplifying a calculation by choosing a special coordinate system is a very useful technique, as long as you are sure you have set up the question to be a covariant question Commented Dec 7, 2020 at 6:04