#### nrqed

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I am trying to prove the following result (I am using the notation of Felsager, Geometry, Particles and Fields)

The Hodge dual of a k form "F" in n dimensions (here I work in Euclidean space) gives an n-k form with covariant components

[tex] (\,^* F)_{a_1...a_{n-k}} = \frac{1}{k!} \sqrt{g} \epsilon_{a_1...a_{n-k}b_1...b_k} F^{b_1...b_k} [/tex]

So far so good. Now the question is to prove that the contravariant components are given by

[tex](\,^* F)^{c_1...c_{n-k}} = \frac{1}{ \sqrt{g} k!} \epsilon^{c_1...c_{n-k}d_1...b_k} F_{d_1...d_k} [/tex]

(notice the key point that the sqrt(g) went from numerator to denominator. This means that a factor of 1/g must have been generated, in other words the determinant of the inverse metric [itex] det(g^{ij}) = \frac{1}{g} [/itex] was generated, somehow. This is the key of the difficulty.

I don't know how to get this. What I obviously through initially was to simply raise the "a" indices with n-k metrics [itex] g^{c_1 a_1} ....g^{c_{n-k} a_{n-k}} [/itex] but it's obvious that this won't lead to a factor of [itex] 1/det(g)[/itex] which I need to make it work.

Then I realized that the epsilon symbol does not need any metric to raise or lower the indices (is that correct?) so that, I think, I can set [itex] \epsilon_{a_1...a_{n-k}b_1...b_k} = \epsilon^{a_1...a_{n-k}b_1...b_k} [/itex] directly? So then, maybe I needed another set of metrics to raise the indices of F. Maybe after all I would end up with n factors of g and be able to get a determinant of g out of that. But I am stuck because if I simply change the indices of the epsilon symbol from downstairs to upstairs, I don't have any way to generate a determinant of the metric.

So I am stuck. The question boils down to : how to get from the epsilon tensor with all indices down to one with all indices up, and how to generate the determinant of g.

I hope someone can clarify this for me!!

Patrick

The Hodge dual of a k form "F" in n dimensions (here I work in Euclidean space) gives an n-k form with covariant components

[tex] (\,^* F)_{a_1...a_{n-k}} = \frac{1}{k!} \sqrt{g} \epsilon_{a_1...a_{n-k}b_1...b_k} F^{b_1...b_k} [/tex]

So far so good. Now the question is to prove that the contravariant components are given by

[tex](\,^* F)^{c_1...c_{n-k}} = \frac{1}{ \sqrt{g} k!} \epsilon^{c_1...c_{n-k}d_1...b_k} F_{d_1...d_k} [/tex]

(notice the key point that the sqrt(g) went from numerator to denominator. This means that a factor of 1/g must have been generated, in other words the determinant of the inverse metric [itex] det(g^{ij}) = \frac{1}{g} [/itex] was generated, somehow. This is the key of the difficulty.

I don't know how to get this. What I obviously through initially was to simply raise the "a" indices with n-k metrics [itex] g^{c_1 a_1} ....g^{c_{n-k} a_{n-k}} [/itex] but it's obvious that this won't lead to a factor of [itex] 1/det(g)[/itex] which I need to make it work.

Then I realized that the epsilon symbol does not need any metric to raise or lower the indices (is that correct?) so that, I think, I can set [itex] \epsilon_{a_1...a_{n-k}b_1...b_k} = \epsilon^{a_1...a_{n-k}b_1...b_k} [/itex] directly? So then, maybe I needed another set of metrics to raise the indices of F. Maybe after all I would end up with n factors of g and be able to get a determinant of g out of that. But I am stuck because if I simply change the indices of the epsilon symbol from downstairs to upstairs, I don't have any way to generate a determinant of the metric.

So I am stuck. The question boils down to : how to get from the epsilon tensor with all indices down to one with all indices up, and how to generate the determinant of g.

I hope someone can clarify this for me!!

Patrick

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