1. Feb 28, 2008

### HenryGomes

Usually the adjoint to the exterior derivative $$d^*$$ on a Riemannian manifold is derived using the inner product
$$\langle\langle\lambda_1,\lambda_2\rangle\rangle:=\int_M\langle\lambda_1,\lambda_2\rangle\mbox{vol}=\int_M\lambda_1\wedge*\lambda_2$$
where $$\lambda$$ are p-forms and $$*$$ is the Hodge duality operator taking p-forms to (n-p)-forms which is defined by the above equation where $$\langle\cdot,\cdot\rangle$$ is the canonical inner product induced on p-forms by the Riemannian metric g (it is just the tensor p-product of the inverse metric).
It is quite easy to derive $$d^*=*d*$$. But does anyone know how to do this without using Hodge star operator, through the g-induced inner product directly?

2. Mar 5, 2008

### Phrak

what text are you referencing?

3. Mar 6, 2008

### HenryGomes

Any good differential geometry book should have this. One I like, which is for physicist's is John Baes' "Knots, Gauge Theory and Gravity". You might also want to check out Bleecker's " Variational Principles in Gauge Theories". I re-read the post and it seemed a bit badly written. So just to spell out what I meant:
The Hodge star is defined by:
$$\langle \lambda_1,\lambda_2\rangle vol=\lambda_1\wedge*\lambda_2$$
Find the adjoint of d without using Hodge star, just the canonically induced metric $$\langle\cdot,\cdot\rangle$$

4. Mar 6, 2008

### Phrak

thanks, Henry.

I have R.W.R. Darling's "Differential Forms and Connections". Formal. Not written for physicists. And a 1985 dover reprint, Dominic G. B. Edelen, "Applied Exterior Calculus". Neither one have I read yet.

Perhaps you could tell me if either of these texts might be worth trying.

I don't want to be misleading. I find differential forms fascinating in their application to physics. Sadly, I'm not capable of responding to your past three posts, as yet, but I would surely like to get to that point. From what I've seen over the past 6 weeks, Hurkyl seems to talk of differential forms with some authority. You might try buttonholing him for some input.

-deCraig

Last edited: Mar 6, 2008
5. Mar 8, 2008

### Hurkyl

Staff Emeritus
Silly question -- if you want to express $* d *$ without the Hodge stars, what's wrong with simply replacing them with a formula that calculates them?

Or... maybe your question is more fundamental? You call it the adjoint, so I assume

$$\langle d^* f, g \rangle = \langle f, dg \rangle?$$

Was that what you wanted? Or maybe something like $d^*f$ is the transpose of the tangent multi-vector $\langle f, d \_\_\_ \rangle$?

Last edited: Mar 8, 2008
6. Mar 9, 2008

### HenryGomes

Because it is not really the answer I am looking for, but to apply a different method to obtain adjoints when we do not have the Hodge star. Thanks for the answer!