# Exterior derivative of hodge dual

1. Aug 13, 2015

### JorisL

Hello all,

I'm having a minor annoyance in proving an identity.
The identity is the following

$$\star\text{d}\star A_p = \frac{(-)^{p(D-p+1)-1+t}}{(p-1)!}\nabla_\mu A^\mu_{\,\, \mu_1 \cdots \mu_{p-1}}\text{d}^\mu_1\wedge \cdots \wedge \text{d}^\mu_{p-1}$$

I'm stuck at the first step of proving this
$$\text{d}\star A_p = ?$$

Because I'm in arbitrary dimension it's difficult to write out the full expression.
Not in the least because this can be written as

$$\text{d}(\star A_p) = \text{d}\left(\frac{1}{p! (D-p)!}A_{\mu_1\cdots \mu_p}\epsilon_{\nu_1\cdots \nu_{D-p}}^{\quad\quad\mu_1\cdots\mu_p} \text{d}^\nu_1\wedge \cdots \wedge \text{d}^\nu_{D-p}\right)$$

If I immediately apply the definition of the exterior derivative to this I get a lot of ugliness.
I would need to introduce the metric $p$ times, once for each of the upper indices.
Then I need to use $\nabla_\alpha \epsilon_{\mu_1\cdots \mu_D} = 0$ to rewrite the partial in terms of the connection. The same happens for the metric factors.

Is there any way I can write this down in a clean way? If need be I can grind my way through but if there is an alternative I'm not going to do that (at first sight) useless exercise in index gymnastics.

Joris

PS.

The next step would be to write the identity as
$$\star\text{d}\star A_p = \frac{(-)^{p(D-p+1)-1+t}\partial_\mu\left(\sqrt{|g|}A^{\mu\nu_1\cdots\nu_{p-1}}\right)}{(p-1)!\sqrt{|g|}}g_{\nu_1\mu_1}\cdots g_{\nu_{p-1}\mu_{p-1}}\text{d}^\mu_1\wedge \cdots \wedge \text{d}^\mu_{p-1}$$

2. Aug 13, 2015

### stedwards

Could you clarify a few things?

In this equation,
$$\star\text{d}\star A_p = \frac{(-)^{p(D-p+1)-1+t}}{(p-1)!}\nabla_\mu A^\mu_{\,\, \mu_1 \cdots \mu_{p-1}}\text{d}^\mu_1\wedge \cdots \wedge \text{d}^\mu_{p-1}$$ is the five pointed star $\star$ the Hodge's duality operator, or the duality operation that acts on contravariant vectors introduced in Misner Thorn Wheeler, ff equation 15.15? Normally the Hodge star is rendered as a line centered asterisk $*$.

Is $A$ a one-form? On one side it is single indexed, and the other side multi-indexed. What's the relationship between the two $A$'s?

For background, do you have a reference source for your first equation?

3. Aug 13, 2015

### JorisL

It is indeed the hodge dual, I used the notation from my source. (didn't know about the other duality operation)
Unfortunately this source is my advisor's thesis so it's not accessible from what I've seen.

More notation; $A_p$ is a p-form in fact, the left side is written without indices.
The right side explicitly shows the components of $A_p$.

There aren't any free indices on either side btw.

What I've been doing, is getting used to working without any indices present.
To do that I need to have a "basis set" of properties from which I can then derive whatever I need without having to go back to index-notation.

4. Aug 13, 2015

### stedwards

Please use indices for this, at least to begin with.

And good, the indeed Hodge operator I know, the funny star is something of a mystery.

1. I still don't know the relationship between $A_p$, and multi-indexed $A$. Could you write it out explicitly?

2. What does $d_1^\mu$ mean? It looks like a basis vector with too many indices. I'm unfamiliar with this notation.

5. Aug 14, 2015

### JorisL

$$\star\text{d}\star A_p = \frac{(-)^{p(D-p+1)-1+t}}{(p-1)!}\nabla_\mu A^\mu_{\,\, \mu_1 \cdots \mu_{p-1}}\text{d}^{\mu_1}\wedge \cdots \wedge \text{d}^{\mu_{p-1}}$$

Woops a typo that got in. They indeed are the coordinate forms.

$$A_p = \frac{1}{p!}A_{\mu_1\cdots\mu_p}\text{d}^{\mu_1}\wedge \cdots \wedge \text{d}^{\mu_{p-1}}$$

6. Aug 14, 2015

### stedwards

Wow, what a lot of confusing LaTex.

I think you want this:
For $A$ a one-form,
$$*\text{d}*A_{\mu_p} = \frac{(-1)^{p(D-p+1)-1+t}}{(p-1)!}\nabla_\nu A^\nu_{\,\, \mu_1 \cdots \mu_{p-1}}\text{d}^{\mu_1}\wedge \cdots \wedge \text{d}^{\mu_{p-1}}.$$ It looks like $D$ might be the dimension of the manifold, and $t$ the number of negative signs in the diagonalized metric. Your equation defining the one-form $A$ in terms of the p-form $A$ seems to be correctly written as: $$A_{\mu_p} = \frac{1}{p!}A_{\mu_1\cdots\mu_p}\text{d}^{\mu_1}\otimes \cdots \otimes \text{d}^{\mu_{p-1}}$$ rather than $$A_{\mu_p} = \frac{1}{p!}A_{\mu_1\cdots\mu_p}\text{d}^{\mu_1}\wedge \cdots \wedge \text{d}^{\mu_{p-1}}.$$ The reasoning being that the the wedge products define a volume element. The volume element is not a tensor, but a tensor density, and apparently, both $A$'s are intended to be tensors. The product of a tensor and a tensor density is not a tensor.

Does this seem correct to you? Notice that I've changed $A_p$ to $A_{\mu_p}$.

Last edited: Aug 14, 2015
7. Aug 14, 2015

### stedwards

I've made a mistake.

Neither of these two equations produce a tensor with the volume element acting on a tensor: $$A_{\mu_p} = \frac{1}{p!}A_{\mu_1\cdots\mu_p}\text{d}^{\mu_1}\otimes \cdots \otimes \text{d}^{\mu_{p-1}},$$ $$A_{\mu_p} = \frac{1}{p!}A_{\mu_1\cdots\mu_p}\text{d}^{\mu_1}\wedge \cdots \wedge \text{d}^{\mu_{p-1}}.$$ So, at this point I think you need to consult your professor to find out what's what, and clear-up any transcription errors.

8. Aug 14, 2015

### Ben Niehoff

Stedwards, I think you've given rather confusing advice, and none of your formulas are grammatically correct.

To address the OP:

I think it's a bit clearer to write the coordinate basis 1-forms as $\mathrm{d} x^\mu$, because they are the exterior derivatives of the coordinate functions $x^\mu$.

Yep, pretty much. But because of antisymmetry of the wedge product, you actually only need to do this once, rather than $p$ times. Just differentiate the first one, and stick a factor $p$ in front, since the remaining $p-1$ of them have to give you the same result after reshuffling indices.

Personally, I find it less confusing to write things in terms of the epsilon symbol $\varepsilon_{\mu_1 \ldots \mu_D} = \pm 1$ which is always just a number. The volume form then gets an explicit $\sqrt{|g|}$.

It's tedious, but as I said, take advantage of symmetry. You can always do a simpler example, say with A a 2-form in 3 dimensions, just to get the hang of it before doing the completely general version.

9. Aug 15, 2015

### stedwards

Thank you very much. If you had bothered to read my posts, you would find thatI have attempted to correct a multiplicity of typographical/translational/grammatical errors in JirisL's posts. Maybe I have failed to find them all, but I wouldn't bother if you got off my back about it.

Given this, I am no longer comfortable with this problem, and am very happy that you would take it off my hands.

Would you please tell me which forumlas of mine you think are grammatically incorrect that I have not already addressed.

10. Aug 15, 2015

### JorisL

I figured as much. I'll just do it this way then.

Stedwards, thanks for the help mate.
I made some typo's. However I'm not sure what you meant with post #7.
When $A$ is a true p-form, both are the same. Because of the antisymmetry in the indices I can use wedges or tensor products all the same.

RE: $A_p$ vs $a_{\mu_p}$, my convention is that the subscript denotes that its a p-form. This because I'll be working on SUGRA-actions where multiple p-form field strengths are incorporated. e.g. a 2-form and 4-form in type IIA.

11. Aug 17, 2015

### Ben Niehoff

This isn't much worth arguing about, but people use both $\star$ and $\ast$ to denote the Hodge dual, it's just a matter of style. The Hodge dual is a linear map on the exterior algebra of a vector space; it is just as well defined on contravariant vectors as on covariant ones, although it's typically only used with forms (covariant vectors).

I'm not sure what you can possibly mean by these equations, since the indices are unbalanced on each side. How many indices does $A$ have?

Joris has been using a subscript $p$ (not an index) to indicate that $A$ is a p-form.