# General relativity tensors

This is a question on the nitty-gritty bits of general relativity.

Would anybody mind teaching me how to work these indices?

**Definitions**:

Throughout the following, repeated indices are to be summed over.

Hodge dual of a p-form $X$:
$$(*X)_{a_1...a_{n-p}}\equiv \frac{1}{p!}\epsilon_{a_1...a_{n-p}b_1...b_p}X^{b_1...b_p}$$
Exterior derivative of p-form $X$: $$(dX)_{a_1...a_{p+1}}\equiv (p+1) \nabla_{[a_1}X_{a_2...a_{p+1}]}$$

Given the relation
$$\epsilon^{a_1...a_p c_{p+1}...c_n}\epsilon_{b_1...b_pc_{p+1}...c_n}\equiv p!(n-p)! \delta^{a_1}_{[b_1}...\delta^{a_p}_{b_p]}\,\,\,\,\,\,\,\,\,(\dagger)$$
where $\epsilon_{a_1...a_n}$ is an orientation of the manifold.

Why then is
$$(*d*X)_{a_1...a_{p-1}}=(-1)^{p(n-p)}\nabla^b X_{a_1...a_{p-1}b}$$?

Firstly, I believe $(*d*X)$ means $*(d(*X))$?
$$(d*X)_{c_1...c_{n-p+1}}=\frac{n-p+1}{p!}\nabla_{[c_1}\epsilon_{c_2...c_{n-p+1}]b_1...b_p}X^{b_1...b_p}$$
Then $$*(d*X)_{d_1...d_{p-1}}=\frac{n-p+1}{(n-p+1)!p!}\epsilon_{d_1...d_{p-1}c_1...c_{n-p+1}}\nabla^{[c_1}\epsilon^{c_2...c_{n-p+1}]b_1...b_p}X_{b_1...b_p}$$
$$=\frac{1}{(n-p)!p!}\epsilon_{d_1...d_{p-1}c_1...c_{n-p+1}}\nabla^{[c_1}\epsilon^{c_2...c_{n-p+1}]b_1...b_p}X_{b_1...b_p}$$

Now I know that I should apply $(\dagger)$ but I don't know how to given the antisymmetrisation brackets. Would someone mind explaining it to me please? Thank you!

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## Answers and Replies

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WannabeNewton
Science Advisor
Ok let's take it slow because you messed up in one of the steps. This is a bit messy (but not much). So we start off with ##d(*\alpha)_{c b_1...b_{n-p}} = (n - p + 1)\nabla_{[c}*\alpha_{b_1...b_{n-p}]}## hence ##*d(*\alpha)_{a_1...a_{p-1}} = \frac{n - p + 1}{(n-p + 1)!}\epsilon_{d_1...d_{n-p + 1}a_1...a_{p-1}}\nabla^{[d_1}*\alpha^{d_2...d_{n-p+1}]}##. Before proceeding, notice that ##\nabla^{[d_1}*\alpha^{d_2...d_{n-p+1}]} = \delta^{[d_1}_{e_1}...\delta^{d_{n-p +1}]}_{e_{{n-p+1}}}\nabla^{e_1}*\alpha^{e_2...e_{n-p+1}} =\\ \frac{1}{(p - 1)!(n - p +1)!}\epsilon^{f_1...f_{p-1}d_1...d_{n-p + 1}}\epsilon_{f_1...f_{p-1}e_{1}...e_{n-p+1}}\nabla^{e_1}*\alpha^{e_2...e_{n-p+1}}##

Now ##\epsilon^{f_1...f_{p-1}d_1...d_{n-p + 1}}\epsilon_{d_1...d_{n-p + 1}a_{1}...a_{p-1}} = (-1)^{(p-1)(n - p + 1)}(p - 1)!(n - p + 1)!\delta^{[f_1}_{a_1}...\delta^{f_{p-1}]}_{a_{p-1}}##
(the ##(-1)^{(p-1)(n - p + 1)}## comes from rearranging the indices on ##\epsilon^{f_1...f_{p-1}d_1...d_{n-p + 1}}## so that it takes the right form)
so ##*d(*\alpha)_{a_1...a_{p-1}} = \frac{(n - p + 1)(-1)^{(p -1)(n - p + 1)}}{(n - p +1)!}\epsilon_{a_1...a_{p-1}e_{1}...e_{n-p+1}}\nabla^{e_1}*\alpha^{e_2...e_{n-p+1}}##. Take it from there.

WannabeNewton
Science Advisor
Just to be on the safe side, let me know if you need more help and/or if the above isn't clear enough (thankfully there is very little you have to do beyond what I posted above). Also, I didn't mention this before because it isn't really important but for a Lorentzian space-time manifold, the relation you gave involving the orientation of the space-time has an overall negative sign. Finally, keep in mind that ##\nabla_{b}\epsilon_{a_1...a_n} = 0##.

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I've been following this thread because I'm interested in the solution, but sadly the OP never came back. Can somebody give it anyway?

WannabeNewton
Science Advisor
Sure, there's very little to do beyond what was in post #2 anyways. So we left off at ##*d(*\alpha)_{a_1...a_{p-1}} = \frac{(n - p + 1)(-1)^{(p -1)(n - p + 1)}}{(n - p +1)!}\epsilon_{a_1...a_{p-1}e_{1}...e_{n-p+1}}\nabla^{e_1}*\alpha^{e_2...e_{n-p+1}}##.

Now ##(-1)^{(p -1)(n - p + 1)}\epsilon_{a_1...a_{p-1}e_{1}...e_{n-p+1}}\nabla^{e_1}*\alpha^{e_2...e_{n-p+1}} \\ = \frac{(-1)^{p(n - p)}}{p!}\epsilon_{e_2...e_{n-p+1}a_1...a_{p-1}e_{1}}\epsilon^{e_2...e_{n-p + 1}d_1...d_p }\nabla^{e_1}\alpha_{d_1...d_p} \\ =(-1)^{p(n - p)}(n - p)! \nabla^{e_1}\alpha_{[a_1...a_{p-1}e_1]} ##

hence ##*d(*\alpha)_{a_1...a_{p-1}} = \frac{(-1)^{p(n - p)}(n - p + 1)(n - p)!}{(n - p + 1)!} \nabla^{e_1}\alpha_{[a_1...a_{p-1}e_1]} = (-1)^{p(n - p)}\nabla^{e_1}\alpha_{a_1...a_{p-1}e_1} ##
where ##\alpha_{[a_1...a_{p-1}e_1]} = \alpha_{a_1...a_{p-1}e_1}## by definition of a differential form. Thus we have the desired result.