de Rham-Laplace operator

[Xerxes]

Otherwise known as the Laplacian.

 

So I should apologize for the length of this post and for taking no prisoners here, but it would take me WEEKS to flesh out the background, so I dive in......

 

Given a finite [math]n[/math]-dimensional vector space [math]V_n[/math], define a space of [math]p[/math]-vectors by [math]\Lambda^p(V_n)[/math].

 

Now define the exterior derivative operator by [math]d:\Lambda^p(V_n) \to \Lambda^{p+1}(V_n)[/math].

 

It seems that to this operator one may assign an adjoint provided only one has an inner product on the space of [math]p[/math] -vectors and that these vectors, as differential forms, are exact. Fine.

 

Thus, for [math]\alpha,\,\beta \in \Lambda^p(V_n) [/math] and writing [math](\alpha,\beta)[/math] for the inner product, one may have that [math](d \alpha.\beta) = (\alpha, d^{\dagger} \beta)[/math].

 

Which is the usual way of expressing the adjoint of an operator EXCEPT it seems that, whereas [math]d:\Lambda^p(V_n) \to \Lambda^{p+1}[/math] one has that [math]d^{\dagger}: \Lambda^p(V_n) \to \Lambda^{p-1}(V_n)[/math]. I am having some difficulty making sense of these switching of dimensions.

 

Whatever.

 

The Hodge-de Rham operator is now defined as [math]\not{d} \equiv d + d^{\dagger}[/math]. Obviously the domain of this operator is [math]\Lambda^p(V_n)[/math], but from the above I cannot find the codomain. Is it simply [math]\Lambda^{p+1-1}(V_n)[/math]? This seems to fly in the face of the algebra of exponents, doesn't it?

 

Anyway, finally the square of the Hodge-de Rham operator is called the Laplacian by my text, ie it is what in "ordinary" vector calculus is a second-order differential operator that sends scalar and vector fields to scalar and vector fields, respectively. Simple algebra and the Lemma of Poincare gives that [math]\Delta = \not{d}^2 = (d +d^{\dagger})(d +d^{\dagger}) = dd^{\dagger} + d^{\dagger}d[/math], but I cannot equate this to any expression of the Laplacian with which I am familiar.

 

Specifically, and sticking with differential forms, it is not hard (and is in fact a fun exercise) to show that [math]\Delta = \ast d \ast d[/math] where the Hodge operator [math] \ast:\Lambda^p(V_n) \to \Lambda^{n-p}(V_n)[/math] and the exterior derivative is as above. More specifically, I want that [math]\ast d \ast d = dd^{\dagger}+ d^{\dagger}d[/math].

 

Know what? I cannot show this. Please help

[DrRocket]

 

Know what? I cannot show this. Please help

 

I am not expert in Hodge theory, but it appears that you have things rather contorted, too contorted to attempt to work out here. A couple of suggestions:

 

1. Get a good book that treats Hodge Theory. Principles of Algebraic Geometry by Griffiths and Harris or Foundations of Differentiable Manifolds and Lie Groups by Frank Warner perhaps.

 

2. Learn to walk before running. This is rather sophisticated stuff and more than one quantum jump up from material in your earlier posts which you had not mastered.

[Xerxes]

Points taken, thanks for your efforts to help me anyway