# A simple derivation involving the Hodge dual

1. Jul 2, 2007

### nrqed

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

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

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

$$(\,^* F)^{c_1...c_{n-k}} = \frac{1}{ \sqrt{g} k!} \epsilon^{c_1...c_{n-k}d_1...b_k} F_{d_1...d_k}$$
(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 $det(g^{ij}) = \frac{1}{g}$ 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 $g^{c_1 a_1} ....g^{c_{n-k} a_{n-k}}$ but it's obvious that this won't lead to a factor of $1/det(g)$ 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 $\epsilon_{a_1...a_{n-k}b_1...b_k} = \epsilon^{a_1...a_{n-k}b_1...b_k}$ 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

Last edited: Jul 2, 2007
2. Jul 2, 2007

### shoehorn

I don't have time at the moment to go through the derivation directly, but it might be helpful to look at the properties of the permutation tensor when you contract some indices with the metric.

For example, it should be immediately obvious that the following identity holds:

$$\epsilon^{\mu_1\mu_2\ldots\mu_m} = g^{\mu_1\nu_1}g^{\mu_2\nu_2}\ldots g^{\mu_m\nu_m} \epsilon_{\nu_1\nu_2\ldots\nu_m} = \det(g^{\mu\nu})\epsilon_{\mu_1\mu_2\ldots\mu_m}$$

(If this isn't obvious, try to see if you can convince yourself that it holds in two dimensions, then three. This should give you a feel for it and help to convince you that it holds in an arbitrary number of dimensions.) The above identity follows from

$$g^{1\mu_1}g^{2\mu_2}\ldots g^{m\mu_m} \epsilon_{\mu_1\mu_2\ldots\mu_m} = \det(g^{\mu\nu})$$

3. Jul 2, 2007

### nrqed

Perfect! Thanks!! this is exactly what I needed!!

I appreciate!

4. Jul 2, 2007

### nrqed

I am puzzled.

Just to be clear, the epsilon that appeared in my first post was the Levi-Civita symbol, not the Levi-Civita tensor (also called the volume form, I think?). And, if my understanding is correct, the Levi-Civita symbol is not a tensor so one can't use factors of the metric to raise or lower the indices. Actually, Felsager gives the following relation:
$$\epsilon^{a_1...a_n} = \epsilon_{a_1...a_n}$$

So I don't like the first part of the relation you gave (the lhs equal to the middle expression). It does not sound right to me.

However, I do like the second relation at the condition of raising the indices on the rhs, that is

$$g^{\mu_1\nu_1}g^{\mu_2\nu_2}\ldots g^{\mu_m\nu_m} \epsilon_{\nu_1\nu_2\ldots\nu_m} = \det(g^{\mu\nu})\epsilon^{\mu_1\mu_2\ldots\mu_m}$$

This would do the trick!

Patrick

Last edited: Jul 2, 2007
5. Jul 2, 2007

### nrqed

Might as well show the final proof, in case anyone would be interested!

A crucial point that was very instructive for me is to keep in mind that the Levi-Civita symbol is not a tensor so we may not use the metric to raise or lower indices. So in order to decide where to put the factors of the metric and with what indices, we must pay attention to the tensors in the problem. First, let's start with

$$(\,^* F)_{\gamma_1...\gamma_{n-k}} = \frac{\sqrt{g}}{k!} \epsilon_{\gamma_1..\gamma_{n-k} \beta_1...\beta_k} F^{\beta_1...\beta_k}$$

Clearly, to get its contravariant components we need to raise all the indices with factors of the metric:

$$(\,^* F)^{a_1...a_{n-k}} = g^{a_1 c_1} g^{a_2 c_2} \ldots g^{a_{n-k} c_{n-k}} (\,^* F)_{c_1...c_{n-k}}$$

which is equal to

$$(\,^* F)^{a_1...a_{n-k}}= \frac{\sqrt{g}}{k!} g^{a_1 c_1} \ldots g^{a_{n-k} c_{n-k}} \epsilon_{c_1..c_{n-k} d_1...d_k} F^{d_1...d_k}$$

Now a key point is that we should not look at the Levi-Civita symbol to decide what factors of the metric to put and with what indices and so on. We may only rely on the F. So the next step is to rewrite the F on the rhs as an expression with all the indices down times the appropriate factors of the metric (we do not lower the indices, we just rewrite the F in terms of its lower idinces, that's different!), giving:

$$(\,^* F)^{a_1...a_{n-k}}= \frac{\sqrt{g}}{k!} g^{a_1 c_1} \ldots g^{a_{n-k} c_{n-k}} \epsilon_{c_1..c_{n-k} d_1...d_k} g^{b_1 d_1} \ldots g^{b_k d_k} F_{b_1 ...b_k}$$

Notice how clever we were. We never used the epsilon to raise or lower indices.
Now, finally, we may use the identity given in the previous post to finally get

$$(\,^* F)^{a_1...a_{n-k}} = \frac{\sqrt{g}}{k!} det(g^{\mu \nu}) \epsilon^{a_1 ...a_{n-k} b_1...b_k} F_{b_1...b_k}$$

where the indices $\mu$ and $\nu$ in the determiant are only there to remind us that this is the determinant of the matrix $g^{\mu \nu}$ , with indices upstairs, so it's the inverse of the usual metric. Therfeore, we have $det(g^{\mu \nu}) = \frac{1}{(\sqrt{g})^2}$ . So finally, we have

$$(\,^* F)^{a_1...a_{n-k}} = \frac{1}{\sqrt{g} k!} \epsilon^{a_1 ...a_{n-k} b_1...b_k} F_{b_1...b_k}$$

which completes the proof!!

Last edited: Jul 2, 2007
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