# Hodge Duals and the Interior Product

1. Apr 24, 2005

### Kane O'Donnell

Hi everyone,

I'm having difficulty with an exercise I have to do in differential geometry this semester. Suppose that the interior product (also known as the interior derivative) is denoted by $$i_X$$. Then the exercise is to show that:

$$i_X\star\omega = \omega\wedge X^\flat$$​

where $$X^\flat$$ is the 1-form related to $$X$$ by the metric and the star is the Hodge Star dual operator.

The problem is I don't really know where to start. I've tried several approaches, for example taking the interior derivative of $$\phi\wedge\star\omega$$ where $$\phi$$ is a p-form, and using the defining property of the star operator, but I can't see it leading anywhere.

Any suggestions?

Kane O'Donnell

2. Apr 24, 2005

### matt grime

This isn't my area, so take this with a pinch of salt, but you may want to say what X is.

3. Apr 24, 2005

### mathwonk

not mine either but i bet if you'll tell us all the relevant definitions, we can help you, and probably by then you won't need any help. i.e. always start from the definition of all the symbols.

4. Apr 24, 2005

### Berislav

Here's my idea:

$$\star\omega = \frac{1}{(n-p)!}\epsilon^{i_1...i_p}_{i_{p+1}...i_n}\omega_{i_1...i_p}dx^{i_{p+1}}\wedge... \wedge dx^{i_n}$$

The interior derivative of this is:
$$\frac{1}{(n-p)!}X^i\omega_{ii_1...i_{p-1}}\epsilon^{i_1...i_p}_{i_{p+1}...i_n}dx^{i_{p+1}}\wedge... \wedge dx^{i_{n-1}}$$

$$\omega\wedge X^\flat=\omega\wedge i_X\omega=\omega_{i_1...i_p}}X^i\omega_{ii_1...i_{p-1}}dx^{i_{1}}\wedge...\wedge dx^{i_p}\wedge dx^{i_{1}}\wedge... \wedge dx^{i_{p-1}}$$

Now you could use "brute force" to permute the above expression to get the expression above that, through the use of the permutation tensor and taking into account the anti-commutativity of the wedge product. Not the most elegant of proofs, but I hope it's exceptable.

It's a vector field (I think).

5. Apr 24, 2005

### Kane O'Donnell

Yes, X is a vector field.

I don't see how $$\omega\wedge X^{\flat} = \omega\wedge i_{X}\omega$$. The formula for $$X^{\flat}$$ is simply:

$$X^{\flat} = g_{ac}X^{c}\theta^{a}$$​

where the thetas are just a basis of one-forms. Also, the interior derivative reduces the covariant degree of a form by one, so the equality in the last post doesn't actually add up.

Kane

Kane

6. Apr 25, 2005

### Berislav

My mistake. I was under the impression that we were dealing with a symplectic manifold, where $\flat$ is standardly defined to be the map $\flat: \chi(M) \rightarrow \chi^* (M)$

Yes. That's what (I think) I did.

7. Apr 25, 2005

### Kane O'Donnell

Nope, sorry about the confusion there. The flat notation is apparently fairly common and is a bit of a joke based on the similarity between flattening a note in music and lowering the index of a contravariant vector. Same kind of thing for raising the index of a covariant vector, it's often denoted by a sharp.

Is there any other notation/symbology that isn't immediate?

Kane

8. Apr 25, 2005

### matt grime

Well, I don't know what g_{ac} is, nor what ^c means, and I can only presume you're a physicist cos I think you're using summation convention, which, if I were a physicist as opposed to an algebraist, might let me guess what g_{ac} meant, but I don't think it's important you know what they "stand for" but "what they do".

That is to say, mathwonk's idea is the correct one: write out the definition of the * and put all the expressions into the formula, what drops out? (Like Berislav did but for a different manifold).

9. Apr 25, 2005

### Kane O'Donnell

g is the metric tensor. As I said in the first post, I *have* been doing the calculations, I'm just not seeing them lead anywhere. I was just hoping an experienced diff geometer would be familiar with the proof and be able to point me to the right starting point.

Don't worry, I will talk to my lecturer tomorrow.

I might as well post a quick review of what the hell I'm talking about. The interior derivative is an algebraic operator that reduces a p-form to a (p-1)-form. It's called a derivative because it has the 'Leibnitz-like' property:

$$i_X(\alpha\wedge\beta) = i_X(\alpha)\wedge\beta + (-1)^{a}\alpha\wedge i_X(\beta)$$​

where $$\alpha$$ is an a-form.

The interior derivative also has the property that if $$\alpha$$ is a one-form, then $$i_X(\alpha) = \alpha(X)$$. Remember X is a vector field here.

Now, recall that the space of p-forms and (n-p)-forms on an n-dimensional manifold have the same (finite) dimension and are therefore isomorphic. Usually there is no natural isomorphism. When there is a metric, however, there is such an isomorphism, called the Hodge dual map. The details are a bit tedious. It suffices to say here that the Hodge dual, $$\star\omega$$ of a p-form $$\omega$$ is defined to be the (n-p)-form such that:

$$\phi\wedge\star\omega = g_{p}(\phi, \omega)\Omega~\forall\phi\in A^p$$​
where A^p is the space of p-forms and $$\Omega$$ is the volume form, $$dx^1 \wedge .... \wedge dx^n$$. g here is the metric tensor again however the p subscript indicates that it is a metric tensor on p-forms not 1-forms.

On the basis of the above, I've tried taking the interior derivative of the wedge product $$\phi\wedge\star\omega$$ and using both the Leibnitz-like property of the interior derivative and the definition of star operator in terms of that wedge to get a fairly big expression in terms of a local basis. However, I'm drowning in a sea of wedge products of basis vectors. I agree it's something like what Berislav was saying in terms of manipulating the wedges until I get what I want, but it's not as straightforward as I'd have hoped.

Aaaaanyway, that's my rant for tonight. Enjoy your differential geometry, ladies and gents.

Kane O'Donnell

10. Apr 25, 2005

### matt grime

So, it seems like the meanings of all the terms make sense in context, probably explaining why I couldn't guess what X was.

There are two things that spring to mind. One is to use the fact that **w=+/-w on the LHS to make i_X act on w rather than *w, and the other is to do it for a couple of simple examples, such as R^3 where we can explicitly write *dx=dy/\dz and see what happens if we pick a nice g and a nice X.
If the proof is constructive, then this ought to help.

Last edited: Apr 25, 2005
11. Apr 25, 2005

### Doodle Bob

Are there specific dimensions given by this problem? The reason why I ask is because the two sides seem to give different dimensional forms.

If w is a p-form on an n-diml. manifold, then *w will be an (n-p)-form, so that $i_X *w$ is an (n-p-1)-form. But the right-hand side seems to give us a (p+1)-form.

12. Apr 25, 2005

### mathwonk

that was kind of you to try to bring us up to snuff, but it still sounds a bit imprecise to me. i.e. p forms and n-p forms are not really of finite dimension are they? how about some hypotheses, like smooth or harmonic or holomorphic, ....and I presume you want to assume compact connected and oriented for your manifold. and even then i think you want to assume you are discussing not the spaces of forms themselves, but the cohomology spaces of closed modulo exact forms.

and then for the hodge theory don't you want harmonic forms? i.e. doesn't hodge's theorem say that the quotient space of closed modulo exact forms on a compact connected oriented riemennian manifold, is isomorphic to the space of harmonic forms, (which is finite dimensional).

of course i may be out of my depth here as it has been 40 years since i played with these. i appreciate your patience.

some of us are perhaps not so knowledgable but we are logical, so if we have the data we think we can help.

Last edited: Apr 25, 2005
13. Apr 25, 2005

### Kane O'Donnell

Well the *spaces* of p-forms and (n-p)-forms are finite dimensional. That is, you can write down a certain number of linearly independent basis forms for a given p, and it's finite. In particular there are no p-forms on a manifold for p > n, and the spaces of 1 forms and n-forms are both 1-dimensional. The number of dimensions is simply $$\frac{n!}{(n-p)!p!}$$ which explains why the spaces of p-forms and (n-p)-forms are isomorphic - they're finite dimensional vector spaces of the same dimension.

As for orientation, the volume element $$\Omega$$ defines an orientation and the rules for computing Hodge duals ensures that the orientation is preserved across the isomorphism...ummm..I think...

I'm afraid I can't go into that kind of detail basically because we haven't in class - it's a course on general relativity and the lecturer has set this assignment which is supposed to show us some of the time-saving machinery of differential geometry, things that we haven't actually studied in class but will help with calculations.

Needless to say, they aren't yet helping with the calculations, because I'm bogged down in calculations .

Kane

14. Apr 25, 2005

### mathwonk

you seem to be confusing p forms on a vector space with p forms on a manifold. it is not likely true that the space of p forms on a manifold is finite dimensional.

i don't wish to be troublesome, but you seem not to have quite grasped what you are taking about here, or else, perhaps more likely i have not understood you properly.

i.e. maybe you are not working on manifolds at all, but are merely doing elementary linear algebra, at least then your dimensions would be correct.

i know, tell me your definition fo a p - form, that will help.

anyway it is an interesting subject for further discussion when daylight returns.

best wishes,

15. Apr 26, 2005

### dextercioby

Our LaTex code doesn't work with \$,but with [ itex ] ...[/itex] for inline formulas and [ tex] ... [/tex] for the other ones...(of course,without the spaces between the letters & the sq.brackets).

Daniel.

16. Apr 26, 2005

### Kane O'Donnell

A p-form is a (p, 0) totally antisymmetric tensor. That is, a multilinear map from p vector fields to the smooth functions on the manifold, where the map is antisymmetric with respect to the swap of any two vector fields. I am definitely working on manifolds.

Kane

17. Apr 26, 2005

### Kane O'Donnell

LOL!

It's ok everyone, the lecturer got the statement wrong.

The thing I'm supposed to be proving is $i_{X}\star\omega = \star(\omega\wedge X^{\flat})$

So, the dimensions on each side didn't match up after all!

Kane

18. Apr 26, 2005

### Doodle Bob

Sorry, I didn't mean "dimension" in terms of spatial dimension, but rather the "size" or "rank" of the forms, i.e. how many vectors they need to eat in order to produce a scalar, as described above.

The problem itself actually has nothing to do with manifolds and involves only pointwise operations, the Hodge star operator, the wedge product, and the interior product are all things that live in $\Lambda V^*$ for any given vector space $V.$

As I explained before, the two sides of the equation give forms of different rank. Assuming the w is a p-form and that dim V=n, then the LHS gives us a (n-p-1)-form, but the RHS gives a (p-1)-form.

So, the result isn't true in general. At the very least, n needs to be even.

The instructor must have specific numbers in mind, e.g. n=4 (which would make sense in a relativity setting) and, say, p=1. Even in that case I'm skeptical that it's true. However, in order to test this case, you can let $v_1, \dots, v_4$ be an orthonormal basis of $V$ and assume that $\omega= f{v_1}^*$ for some non-zero $f$ and let $X=a_1 v_1 + a_2v_2 + ... +a_4v_4$, and check both sides.

Actually, now that I think about it, you can assume also that $\omega={v_1}^*$ so that $*\omega={v_2}^*\wedge {v_3}^*\wedge {v_4}^*$ and $a_1=0$ for X.

19. Apr 26, 2005

### Doodle Bob

I knew it. Although a bit messy, you could use my proposed method for n=4 and p=1, to solve this for the general case.

Edit: actually it's even easier than that. You need really just to show that $(\omega\wedge X^{\flat})\wedge (i_{X}\star\omega) = volume\ form$, where the volume form is ${v_1}^*\wedge ... \wedge {v_n}^*$ is the dual basis to an orthonormal basis $v_1,..., v_n$. I would first show it's true for $X=v_1$ (so that $X^\flat = {v_1}^*$) and w a primitive p-form. Then by linearity, it's true in general.

-Doodle Bob

Last edited: Apr 26, 2005
20. Apr 26, 2005

### Kane O'Donnell

Thanks As soon as he said he'd made a mistake I was like, oh, well that shouldn't be hard then...just deal with the right hand side. Silly me!

Kane