# General relativity tensors

1. Jul 28, 2013

### vidi

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!

Last edited by a moderator: Jul 29, 2013
2. Jul 28, 2013

### WannabeNewton

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.

3. Jul 29, 2013

### WannabeNewton

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 $\nabla_{b}\epsilon_{a_1...a_n} = 0$.

Last edited: Jul 29, 2013
4. Aug 1, 2013

### R136a1

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?

5. Aug 1, 2013

### WannabeNewton

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.