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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 [itex]X[/itex]:

[tex](*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}[/tex]

Exterior derivative of p-form [itex]X[/itex]: [tex](dX)_{a_1...a_{p+1}}\equiv (p+1) \nabla_{[a_1}X_{a_2...a_{p+1}]}[/tex]

Given the relation

[tex]\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)[/tex]

where [itex]\epsilon_{a_1...a_n}[/itex] is an orientation of the manifold.

Why then is

[tex](*d*X)_{a_1...a_{p-1}}=(-1)^{p(n-p)}\nabla^b X_{a_1...a_{p-1}b}[/tex]?

Firstly, I believe [itex](*d*X)[/itex] means [itex]*(d(*X))[/itex]?

[tex](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}[/tex]

Then [tex]*(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}[/tex]

[tex]=\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}[/tex]

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

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 [itex]X[/itex]:

[tex](*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}[/tex]

Exterior derivative of p-form [itex]X[/itex]: [tex](dX)_{a_1...a_{p+1}}\equiv (p+1) \nabla_{[a_1}X_{a_2...a_{p+1}]}[/tex]

Given the relation

[tex]\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)[/tex]

where [itex]\epsilon_{a_1...a_n}[/itex] is an orientation of the manifold.

Why then is

[tex](*d*X)_{a_1...a_{p-1}}=(-1)^{p(n-p)}\nabla^b X_{a_1...a_{p-1}b}[/tex]?

Firstly, I believe [itex](*d*X)[/itex] means [itex]*(d(*X))[/itex]?

[tex](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}[/tex]

Then [tex]*(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}[/tex]

[tex]=\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}[/tex]

Now I know that I should apply [itex](\dagger)[/itex] 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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