Question on covariant derivatives

I am reading I am reading Spacetime and Geometry : An Introduction to General Relativity -- by Sean M Carroll and have arrived at chapter 3 where he introduces the covariant derivative ##{\mathrm{\nabla }}_{\mu }##. He makes demands on this which are \begin{align}
\mathrm{1.\ Linearity:}\mathrm{\ }\mathrm{\nabla }\left(T+S\right)=\mathrm{\nabla }T+\mathrm{\nabla }S & \phantom {10000}(1) \\
\mathrm{2.\ Leibniz\ rule:}\mathrm{\nabla }\left(T\ \otimes \ \ S\right)=\left(\mathrm{\nabla }T\right)\ \ \otimes \ \ S+T\ \otimes \ \ \left(\mathrm{\nabla }S\right) & \phantom {10000}(2) \\
{\mathrm{3.\ Commutes\ with\ contractions:}\mathrm{\nabla }}_{\mu }\left(T^{\lambda }_{\ \ \ \lambda \rho }\right)={\left(\mathrm{\nabla }T\right)}^{\mathrm{\ \ \ }\lambda}_{\mu \ \ \lambda \rho } & \phantom {10000}(3) \\
{\mathrm{4.\ Reduces\ to\ partial\ derivative\ on\ scalars:}\mathrm{\nabla }}_{\mu }\phi ={\partial }_{\mu }\phi  & \phantom {10000}(4) \\
\end{align}1,2 and 4 seem reasonable but I cannot understand 3 and he does not seem to use it, even though he implies that he does.

The LHS of (3) seems straight forward\begin{align}
{\mathrm{\nabla }}_{\mu }\left(T^{\lambda }_{\ \ \ \lambda \rho }\right) & ={\partial }_{\mu }T^{\lambda }_{\ \ \ \lambda \rho }+{\mathrm{\Gamma }}^{\lambda }_{\mu \kappa }T^{\kappa }_{\ \ \ \lambda \rho }-{\mathrm{\Gamma }}^{\kappa }_{\mu \lambda }T^{\lambda }_{\ \ \ \kappa \rho }-{\mathrm{\Gamma }}^{\kappa }_{\mu \rho }T^{\lambda }_{\ \ \ \lambda k} & \phantom {10000}(5) \\
 & ={\partial }_{\mu }T^{\lambda }_{\ \ \ \lambda \rho }-{\mathrm{\Gamma }}^{\kappa }_{\mu \rho }T^{\lambda }_{\ \ \ \lambda k} & \phantom {10000}(6) \\
\end{align}Which is very like the rule for the covariant derivative of a (0,1) tensor.

I understand that the ##\mathrm{\nabla }T## in (1) and (2) means ##{\mathrm{\nabla }}_{\sigma}T## where ##T## is some tensor. So the RHS of (3) appears to be ##{\left({\mathrm{\nabla }}_{\sigma}T\right)}^{\mathrm{\ \ \ }\lambda}_{\mu \ \ \lambda \rho }## which leaves too many indices on the RHS. Otherwise the RHS is some kind of derivative with one contra- and three co-variant indices. What is that?

Help!
Posted on Physics Forums at https://www.physicsforums.com/threads/question-on-covariant-derivatives.967509/

Exercise 2.10 Maxwells equations in 2-dimensional spacetime

Question

Consider Maxwell's equations, dF = 0 , d*F = *J in 2-dimensional spacetime. Explain why one of the two equations can be discarded. Show that the electro-magnetic field can be expressed in terms of a scalar field. Write out the field equations for this scalar field in component form.

Answer 

"2-dimensional spacetime"? Is that one dimension of space and another of time or two space and one time? We assume the former and see what happens.

This question seemed too easy, especially after my labours on Maxwell's equation with three-forms. In one time and one space dimension, the first of Maxwell's equations becomes 0 = 0 and I think that the second produced the scalar field which I wrote down in almost one page: Ex 2.10 Maxwells equations in 2-dimensional spacetime.pdf.

Important Equations for General Relativity

Here are some important equations for General Relativity.

Babylonian equations

They are in Commentary Important Equations.pdf along with references and some notes.
Mathematics

• Definition of \partial\mu (Carroll 1.54)
• (Anti)symmetrisation operator (Carroll 1.79)
• Vector as derivative (Carroll 2.16)
• Commutator (Carroll 2.20/2.23)
• Tensor transformation equation (Carroll 2.30)
• Basis vectors (Physics Forums)
• Covariant derivative / Christoffel symbol (Carroll section 3.2)
• Torsion Tensor (C Eq 3.22)
• The Christoffel connection Γ  (C Eq 3.27)
• The geodesic equation (C Eq 3.44)
• Directional covariant derivative (C Eq 3.38)
• The parallel transport equation (C Eq 3.39, 3.40)
• Riemann tensor (c Eq 3.112/3.113)
• Bianchi identity (c Eq 3.140)
• Ricci tensor and scalar, Weyl tensor (c Eq 3.144-3.147)
• Einstein tensor and 'contravariant' derivative (c Eq 3.1452,2)
• Killing's equation, Killing vectors (c Eq3.174)
• Geodesic deviation equation (c Eq3.208)

Tensor tricks

• What tensor rank?
• Multi-dimensional Chain Rule
• Partial derivative of components gives Kronecker delta
• Coordinates
• Vectors (C1.152)
• Tensors
• Partial derivatives commute
• Metric is always symmetric (C section 2.5)
• Contracting with metric lowers / raises index
• You can lower or raise indices on a tensor equation
• Swap indices with metric or any similar tensor
• Inverses and determinants
• Inverse of a matrix
• The determinant of the inverse is reciprocal of the determinant
• Determinant of a tensor in terms of Levi-Civita symbol (C Eq 2.66)
• Inverse tensor
• A relationship for the derivative of the determinant
• Fully contracted symmetric × antisymmetric tensor vanishes
• Symmetrising a tensor equation
• Two formulas involving four-velocity
• Second formula
• The projection tensor on four-velocity (C Eq 1.21)

Physics

• Electro Magnetic Field Tensor (C Eq 1.69)
• Maxwell's equations (C Eq 1.96-1.98)
• Energy Momentum tensor for a perfect fluid (C Eq 3.93 and 1.114)
• Energy Momentum tensor for dust in SR (C Eq 1.110)
• Energy-momentum tensor from action for matter (C Eq 4.75)
• Energy-momentum conservation equation (C Eq 3.92 & 4.8)
• Einstein's equation x 3 for general relativity (C Eq2.44-4.46)
• Friedmann equations (C Eq 8.67)

More Maths

• Differential and Integration on Web
• Pullback / Pushforward operators (Carroll A.9, A.10)
• Levi-Civita symbol and tensor (Carroll section 2.8)
• p-forms (Carroll section 2.9)
• Exterior derivative (Carroll 2.76)
• Wedge product (Carroll 2.73)
• Hodge star operator (Carroll 2.82)
• Stokes's theorem (C Eq 3.35)
• Euler-Lagrange Equation

All in Commentary Important Equations.pdf along with references and some notes.

Image from Wikipedia Babylonian equations.

The rest of the fab four are

Commentary Tensor Tricks.pdf 

Commentary 2.5 The Metric.pdf 

Commentary Constants and conversion factors.pdf 

Carroll's (1.68) where the Levi-Civita symbol is defined says "the Levi-Civita symbol is a ##(0,4)## tensor. It is NOT a tensor as he reminds us elsewhere.

Maxwell's equation with three-forms

James Clerk Maxwell was one of the greatest theoretical physicists of all time. He was up there with Newton and Einstein. He published his famous equations in about 1865. Quite a bit was known about electricity and magnetism and the relationships between them at the time but he spotted a missing component and was able to show that electromagnetic waves must exist and moreover travel at the speed of light, which was already known. Therefore light had to be electromagnetic waves. It must have made physicists feel queasy. I remember being astonished when we did Maxwell's equations and proved it at university. Soon after his equations were known, radio waves (1887, Hertz) and  X-rays (1895, Roentgen ) were discovered. Then came Einstein with Special (1905) and General (1915) Relativity.

Carroll writes "The other one of Maxwell's equations, (1.96) can be expressed as an equation between three-forms:
\begin{align}
\mathrm{d}\left(*F\right)=*J & \phantom {10000}(1) \\
\end{align}where the current one-form ##J## is the current four-vector with index lowered. Filling in the details is left for you, as good practice converting from differential form notation to ordinary index notation." I need the practice .....

##J## was given as\begin{align}
J^{\mu }=\left(\rho ,\ J^x,J^y,J^z\right) & \phantom {10000}(2) \\
\end{align}where ##\rho ## is the classical charge density and ##J^i## the current.

His equation (1.96) ("the other one of Maxwell's equations") was
\begin{align}
{\partial }_{\mu }F^{\nu \mu }=J^{\nu } & \phantom {10000}(3) \\
\end{align}This equation is really four equations one for each value of ##\nu = 0,1,2,3## and on the Left Hand Side we might have four terms for the summation over ##\mu## but one is zero because of the antisymmetry of ##F## (in the next equation). The four equations are written out in (8)-(10).

We have a new form of the electromagnetic field strength tensor:
\begin{align}
F^{\mu \nu }=\left( \begin{array}{cccc}
0 & E^1 & E^2 & E^3 \\
-E^1 & 0 & B^3 & -B^2 \\
-E^2 & -B^3 & 0 & B^1 \\
-E^3 & B^2 & -B^1 & 0 \end{array}
\right)=-F^{\nu \mu } & \phantom {10000}(7) \\
\end{align}
I am not certain if the indices on the ##B##'s should be up or down.

We expand (3) and we get the two relevant Maxwell's equations (one in three parts):
\begin{align}
{\partial }_1F^{01}+{\partial }_2F^{02}+{\partial }_3F^{03} & =J^0 & \phantom {10000}(8) \\
{\partial }_0F^{10}+{\partial }_2F^{12}+{\partial }_3F^{13} & =J^1 & \phantom {10000}(9) \\
{\partial }_0F^{20}+{\partial }_1F^{21}+{\partial }_3F^{23} & =J^2 & \phantom {10000}(10) \\
{\partial }_0F^{30}+{\partial }_1F^{31}+{\partial }_2F^{32} & =J^3 & \phantom {10000}(11) \\
\end{align}
We now convert (1) into index notation. It becomes
\begin{align}
{\widehat{\varepsilon }}^{\ \ \ \ \ }_{\beta \gamma \delta \iota }{\partial }_{\alpha }F^{\delta \iota }+{\widehat{\varepsilon }}^{\ \ \ }_{\gamma \alpha \delta \iota }{\partial }_{\beta }F^{\delta \iota }+{\widehat{\varepsilon }}^{\ \ \ }_{\alpha \beta \delta \iota }{\partial }_{\gamma }F^{\delta \iota }=2{\widehat{\varepsilon }}_{\kappa \alpha \beta \gamma }J^{\kappa } & \phantom {10000}(29) \\
\end{align}On the face of it there the free indices ##\alpha ,\beta ,\gamma ## will give 64 components of the equation to check and the double summation on the LHS gives ##16\times 64=1,024## terms and the summation on the RHS a further 256. In total 1,280 terms which is rather more than the 16 terms in Maxwell's equations. Conceivably we might need to write out every combination. Fortunately the number of terms will be reduced by the antisymmetry of the Levi-Civita symbol ##\widehat{\varepsilon }## and the electromagnetic field tensor ##F##.

We prove that all components of (29) vanish except ones where ##\alpha ,\beta ,\gamma ## all differ. We only have 24 components (or 96 terms) to check! After plodding through all of them, I found six of each of Maxwell's equations (8)-(11) as predicted. The other forty are all 0 = 0. (You can see my notes calculating ##\widehat{\varepsilon }## in the previous post also on the Levi-Civita symbol.)

Having done all that I did discover the easy way to get there without all the plodding tables and paper work. But it was worth it, because as Carroll said, I needed practice converting from differential form notation to ordinary index notation and moreover practice with manipulating indexed equations. The plodding inspired me.

The full six page story is here: Commentary 2.9 Maxwell equation with 3 forms.pdf

The Levi-Civita Symbol

The Levi-Civita symbol, which is not a tensor, is defined as
\begin{align}
{\widetilde{\epsilon }}_{{\mu }_1{\mu }_2\dots {\mu }_n}=\left\{ \begin{array}{ll}
+1 & \mathrm{if\ }{\mu }_1{\mu }_2\dots {\mu }_n\mathrm{\ is\ an\ even\ permitation\ of}\ 01..(n-1)\  \\
-1 & \mathrm{if\ }{\mu }_1{\mu }_2\dots {\mu }_n\mathrm{\ is\ an\ odd\ permitation\ of}\ 01..\left(n-1\right) \\
0 & \mathrm{otherwise} \end{array}
\right. & \phantom {10000}(1) \\
\end{align}Instead of ##\widetilde{\epsilon }## it is also often written as ##\widehat{\varepsilon }##, which is what I usually do.

We usually only need it in three or four dimensions so it becomes much simpler:
\begin{align}
{\widehat{\varepsilon }}_{\alpha \beta \gamma \delta }=\left\{ \begin{array}{ll}
+1 & \mathrm{if\ }\alpha \beta \gamma \delta \mathrm{\ is\ an\ even\ permitation\ of}\ 0123\  \\
-1 & \mathrm{if\ }\alpha \beta \gamma \delta \mathrm{\ is\ an\ odd\ permitation\ of}\ 0123 \\
0 & \mathrm{otherwise} \end{array}
\right. & \phantom {10000}(2) \\
\end{align}or
\begin{align}
{\widehat{\varepsilon }}_{ijk}=\left\{ \begin{array}{ll}
+1 & \mathrm{if\ }ijk \mathrm{\ is\ an\ even\ permitation\ of}\ 123\  \\
-1 & \mathrm{if\ }ijk \mathrm{\ is\ an\ odd\ permitation\ of}\ 123 \\
0 & \mathrm{otherwise} \end{array}
\right. & \phantom {10000}(3) \\
\end{align}One simple thing is that if any two of the indices on ##\widehat{\varepsilon }## are the same, it vanishes.

Even in four dimensions it is fairly easy to calculate, for example
\begin{align}
{\widehat{\varepsilon }}_{1023}=-{\widehat{\varepsilon }}_{0123}=-1 & \phantom {10000}(4) \\
\end{align}It's easy to jot down the indices and determine the sign by walking indices in one direction or another. Here is an example from my real sad life

Lorentz transformation for velocity

There's an equation on Wikipedia for the Lorentz transformation for velocity at
https://en.wikipedia.org/wiki/Lorentz_transformation#Transformation_of_velocities
which states
\begin{align} \large
{\boldsymbol{\mathrm{u}}}^{\boldsymbol{\mathrm{{'}}}}\boldsymbol{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{1}\boldsymbol{\mathrm{-}}\frac{\boldsymbol{\mathrm{u}}\boldsymbol{\mathrm{\cdot }}\boldsymbol{\mathrm{v}}}{c^{\boldsymbol{\mathrm{2}}}}}\left[\frac{\boldsymbol{\mathrm{u}}}{{\gamma }_{\mathrm{v}}}\boldsymbol{\mathrm{-}}\boldsymbol{\mathrm{v}}\boldsymbol{\mathrm{+}}\frac{1}{c^2}\frac{{\gamma }_{\mathrm{v}}}{{\gamma }_{\mathrm{v}}+1}\left(\boldsymbol{\mathrm{u}}\boldsymbol{\mathrm{\cdot }}\boldsymbol{\mathrm{v}}\right)\boldsymbol{\mathrm{v}}\right]& \phantom {10000}(1)  \\
\end{align}
where the coordinate velocities and Lorentz factor are
\begin{align} \large
\boldsymbol{\mathrm{u}}=\frac{d\boldsymbol{\mathrm{r}}}{dt}\ ,\ \ \ {\boldsymbol{\mathrm{u}}}^{\boldsymbol{\mathrm{{'}}}}=\frac{d{\boldsymbol{\mathrm{r}}}^{\boldsymbol{\mathrm{{'}}}}}{dt^{'}}\ \ ,\ \ \ \ {\gamma }_{\mathrm{v}}=\frac{1}{\sqrt{1-\frac{\boldsymbol{\mathrm{v}}\boldsymbol{\mathrm{\cdot }}\boldsymbol{\mathrm{v}}}{c^2}}} & \phantom {10000}(2) \\
\end{align}
##\boldsymbol{\mathrm{v}}## is the relative velocity of the primed frame of reference. Instead of (1), it is sometimes useful to have the equation
\begin{align} \large
{\boldsymbol{\mathrm{u}}}^{\boldsymbol{\mathrm{{'}}}}\boldsymbol{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{1}\boldsymbol{\mathrm{-}}\frac{\boldsymbol{\mathrm{u}}\boldsymbol{\mathrm{\cdot }}\boldsymbol{\mathrm{v}}}{c^{\boldsymbol{\mathrm{2}}}}}\left[\frac{\boldsymbol{\mathrm{u}}}{{\gamma }_{\mathrm{v}}}\boldsymbol{\mathrm{-}}\boldsymbol{\mathrm{v}}\boldsymbol{\mathrm{+}}\frac{1}{v^2}\left(1-\frac{1}{{\gamma }_{\mathrm{v}}}\right)\left(\boldsymbol{\mathrm{u}}\boldsymbol{\mathrm{\cdot }}\boldsymbol{\mathrm{v}}\right)\boldsymbol{\mathrm{v}}\right] & \phantom {10000}(3) \\
\end{align}
where ##v^2=\boldsymbol{\mathrm{v}}\boldsymbol{\mathrm{\cdot }}\boldsymbol{\mathrm{v}}##.
This form of the equation was useful in the first exercise in Sean M Carroll's Spacetime and Geometry : An Introduction to General Relativity. To get it we just have to show that
\begin{align} \large
\frac{1}{c^2}\frac{{\gamma }_{\mathrm{v}}}{{\gamma }_{\mathrm{v}}+1}=\frac{1}{v^2}\left(1-\frac{1}{{\gamma }_{\mathrm{v}}}\right) & \phantom {10000}(4) \\
\end{align}
We use ##\gamma ## instead of ##{\gamma }_{\nu }## because have terms like ##{\gamma }^2## where the 2 is an exponent.
\begin{align} \large
\frac{1}{c^2}\frac{\gamma }{\gamma +1} & =\frac{1}{c^2}\frac{1}{\gamma +1}\left(\frac{{\gamma }^2}{\gamma }\right)=\frac{1}{c^2}\frac{1}{\gamma +1}\left(\frac{\frac{1}{1-{v^2}/{c^2}}}{\gamma }\right) & \phantom {10000}(6) \\
 & =\frac{1}{c^2}\frac{1}{\gamma +1}\left(\frac{\left[\frac{1-1+{v^2}/{c^2}}{1-{v^2}/{c^2}}\right]c^2}{\gamma v^2}\right) & \phantom {10000}(7) \\
 & =\frac{1}{c^2}\frac{1}{\gamma +1}\left(\frac{\left[\frac{1}{1-{v^2}/{c^2}}-1\right]c^2}{\gamma v^2}\right) & \phantom {10000}(8) \\
 & =\frac{1}{c^2}\frac{1}{\gamma +1}\left(\frac{\left[{\gamma }^2-1\right]c^2}{\gamma v^2}\right) & \phantom {10000}(9) \\
 & =\frac{1}{c^2}\frac{1}{\gamma +1}\left(\frac{\left[\gamma +1\right]\left[\gamma -1\right]c^2}{\gamma v^2}\right) & \phantom {10000}(10) \\
 & =\frac{\gamma -1}{\gamma v^2} & \phantom {10000}(11) \\
\therefore \frac{1}{c^2}\frac{\gamma }{\gamma +1} & =\frac{1}{v^2}\left(1-\frac{1}{\gamma }\right) & \phantom {10000}(12) \\
\end{align}

                                         QED

I tried to post this on the Wikimedia page, but it was not allowed :-(
This proof is also available as a pdf here.

The second exterior derivative always vanishes

We investigate the claim that second exterior derivative of any form A always vanishes:
$$\mathrm{d}(\mathrm{d}A)\mathrm{=0}$$This is Carroll's equation (2.80). He then tells us that this is due to the definition of the exterior derivative and the fact that partial derivatives commute, ##{\partial }_{\alpha }{\partial }_{\beta }={\partial }_{\beta }{\partial }_{\alpha }##. From the definition we get
$$\large\mathrm{d(dA)}=\left(p+1\right)p{\partial }_{[{\mu }_1}{\partial }_{[{\mu }_2}A_{{\mu }_3\dots {\mu }_{p+1}]]}$$This contains nested antisymmetrisation operators which we met in Exercise 2.08. The expansion of the equation contains ##p!\left(p+1\right)!## terms in total containing permutations of the indices ##{\mu }_1{\mu }_2{\mu }_3\dots {\mu }_{p+1}##. If ##p## was ##10## that would be 39,916,800 terms.

First I exercised my permutation skills with ##2, 3, 4## and ##n## indices to get the drift of the proof. I was then able to expand groups of ##p+1## terms of ##\mathrm{d}(\mathrm{d}A)##. Each one vanished due to ##{\partial }_{\alpha }{\partial }_{\beta }={\partial }_{\beta }{\partial }_{\alpha }##. It did not depend on the antisymmetry of ##A##. This is rather like the fact that the wedge product of a 2-form and a 1-form does not depend on the antisymmetry of the 2-form as we discovered in Commentary 2.9 Differential forms.pdf.

We also note that
$$\large A_{[{\mu }_1}B_{[{\mu }_2}C_{{\mu }_3\dots {\mu }_{p+1}]]}=0$$for any rank 1 tensors  ##A,B## and any tensor ##C## and the up/down position of any index in the tensors is immaterial.

Read all 5 pages at Commentary 2.9 Second exterior derivative.pdf.