Thursday, 10 January 2019

Exercise 2.08 Exterior derivative and modified Leibniz rule

This exercise started in section 2.9 on which I wrote a seven page commentary before accepting Carroll's challenge to prove the modified Leibniz rule for the exterior derivative.

Carroll introduces differential forms: "A differential p-form is a completely antisymmetric (0,p) tensor. Thus scalars are automatically 0-forms and dual vectors (downstairs index) are one-forms. We also have the 4-form ##{\epsilon }_{\mu \nu \rho \sigma }##, the Levi-Civita tensor." I had a slight problem with 0- and 1-forms but this was resolved on Physics Forums.

First we meet the wedge product:

Given a##\ p##-form ##A## and a ##q##-form ##B## we can form a ##(p+q)##-form known as the wedge product ##A\wedge B## by taking the antisymmetrised tensor product:
$${\left(A\wedge B\right)}_{{\mu }_1\dots {\mu }_{p+q}}=\frac{\left(p+q\right)!}{p!q!}A_{[{\mu }_1\dots {\mu }_p}B_{{\mu }_{p+1}\dots {\mu }_{p+q}]}$$
I prove that the wedge product of an n-dimensional 2-form and 1-form is completely antisymmetric in any number of dimensions n ##\mathrm{\ge}## 2 and therefore a 3-form.

Then we meet the exterior derivative
$${\left(\mathrm{d}A\right)}_{{\mu }_1\dots {\mu }_{p+1}}=\left(p+1\right){\partial }_{[{\mu }_1}A_{{\mu }_2\dots {\mu }_{p+1}]}$$They both involve the ghastly total antisymmetrisation operation [] on indices. It is defined back in his equation (1.80) as
$$T_{[{\mu }_1\dots {\mu }_n]}=\frac{1}{n!}\left(T_{{\mu }_1\dots {\mu }_n}\pm \mathrm{sum\ over\ permutations\ of}\ {\mu }_1\dots {\mu }_n\right)$$
This led on to Exercise 2.08

Question

Verify (2.78): For an exterior derivative of a product of a p-form ω and a q-form η, we have the modified Leibnitz rule:
$$\mathrm{d}\left(\omega \wedge \eta \right)=\left(\mathrm{d}\omega \right)\wedge \eta +{\left(-1\right)}^p\omega \wedge \left(\mathrm{d}\eta \right)$$

Answer 

Here we have the ghastly total antisymmetrisation operation [] again and nested in itself. I had to invent a new notation
$$\sum_{\mp \mathrm{\circlearrowleft }}{A_{{\mu }_1\dots {\mu }_n}}\equiv \left(A_{{\mu }_1\dots {\mu }_n}\pm \mathrm{sum\ over\ permuta}\mathrm{tions\ of}\ {\mu }_1\dots {\mu }_n\ where\ we\ use\ -\ for\ odd\ permutations\ and\ +\ for\ even.\right)$$
because writing the stuff about the permutations every time would be stupid and does not fit on a line. I expanded each term in the question equation and reached expressions like
$$\frac{{\left(-1\right)}^{p\left(q+1\right)}}{\left(q+1\right)!p!q!}\sum_{\mp \mathrm{\circlearrowleft }}{\left(\sum_{\mp \mathrm{\circlearrowleft }}{{\mathrm{\partial }}_{{\mu }_1}{\eta }_{{\mu }_2\dots {\mu }_{q+1}}}\right){\omega }_{{\mu }_{q+2}\dots {\mu }_{p+q+1}}}$$
where you can see the nested expansions explicitly. Each term had a different variant of the nesting so the nesting had to be removed and I proved, for example and avoiding too many subscripts, that
$$\sum_{\mp \mathrm{\circlearrowleft }}{\left(\sum_{\mp \mathrm{\circlearrowleft }}{{\mathrm{\partial }}_a{\eta }_{c_1\dots c_q}}\right){\omega }_{b_1\dots b_p}}\mathrm{=}\left(q+1\right)!{\left(-1\right)}^{q(p+q)}\sum_{\mp \mathrm{\circlearrowleft }}{{\mathrm{\partial }}_a{\omega }_{b_1\dots b_p}{\eta }_{c_1\dots c_q}}$$
factorials cancelled beautifully but I was left with
$${\mathrm{d}\left(\omega \wedge \eta \right)}_{\ }=\left(\mathrm{d}\omega \right)\wedge \eta ={\left(-1\right)}^{\left(q+p\right)}\omega \wedge \left(\mathrm{d}\eta \right)$$
which is not the same as the modified Leibnitz rule, in other words, junk.

I also show that the equation does not work for two 1-forms. In fact I get$$2\mathrm{d}\left(\omega \wedge \eta \right)=\left(\mathrm{d}\omega \right)\wedge \eta +{\left(-1\right)}^p\omega \wedge \left(\mathrm{d}\eta \right)$$which agrees with my wrong answer. I am within spitting distance of the correct proof!  I hope to come back to this problem.

Read the full account
Commentary 2.9 Differential forms.pdf (7 pages)
Ex 2.08 Exterior derivative and modified Leibnitz rule.pdf (8 pages - I have long thought that he was Leibnitz. That is acceptable, but Leibniz is better.)

There is also an answer here from University of California, Santa Barbara (UCSB).
It starts from ##\mathrm{d}\left(\omega \wedge \eta \right)## and uses the ordinary Leibnitz rule to split that in two. It then manipulates the two parts to convert them into the RHS of the question equation. There are some rather 'hand-waving' steps such as the first line of text "We drop the inner set of antisymmetrization brackets in the second line because the terms are antisymmetrized in all of their indices". It works in the opposite direction to what I did.

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