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.

$$\mathrm{d}(\mathrm{d}A)\mathrm{=0}$$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.

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