I was feeling slightly ashamed at my inability to see the following, but Alpha's recent post encouraged me to seek help. What follows is rather specialized, so I will make no attempt to sugar the pill. So, earlier this year I found that the Laplacian in \(\mathbb{R}^3\) can be written in the language of differential forms as \(\nabla^2 = \ast d \ast d\), where \(d: \Lambda^p(V_n) \to \Lambda^{p-1}(V_n)\), and the Hodge operator \(\ast: \Lambda^p(V_n) \to \Lambda^{n-p}(V_n)\), and where \(n=3\) and for the p-forms \((\alpha, \beta)\) is taken to be an inner product, In short, we are in a metric space. Fine. Since, in these circumstances, the exterior derivative is an operator on a metric space, I may define its adjoint such that \((d \alpha, \beta) = (\alpha, d^{\dag}\beta)\), where, by definition, for some n-dimensional manifold \(M_n\), that \(d^{\dag}: \Lambda^p(M_n) \to \Lambda^{p-1}(M_n)\). Now my current text defines the Hodge-de Rham operator to be \(\mathcal{HD} \equiv d + d^{\dag}\), which is also fine (actually they use a strike-through of d, but I can't do this here). But they then go on to assert that the square of this operator \((\mathcal{HD})^2 = dd^{\dag}+ d^{\dag}d\) and that this is commonly referred to as the Laplacian; \((\mathcal{HD})^2 \equiv \nabla^2\)...Umm Now since \(\mathbb{R}^3\) is a vector space - specifically \(V_3\) in the present context, and also since \(\mathbb{R}^3\) is trivially the manifold \(M_3\), then it must be possible, in the specific case, to bring these definitions into register - that is \(\ast d\ast d \equiv d d^{\dag} + d^{\dag} d\) Know what? I cannot. Any help out there?
I think there's a few ways to view this, though I cannot think of a direct and short proof off the top of my head. The first is to consider the Dirac operator normally associated with Clifford algebras, ie \(D = \gamma^{\mu}\partial_{\mu}\). If you know the Dirac equation, you've seen this before and it's often written as D with a slash through it. Squaring this up you get \(\gamma^{\mu}\gamma^{\nu}\partial_{\mu}\partial_{\nu}\). Partial derivatives commute so \(\gamma^{\mu}\gamma^{\nu}\partial_{\mu}\partial_{\nu} = \frac{1}{2}(\gamma^{\mu}\gamma^{\nu}\partial_{\mu}\partial_{\nu} + \gamma^{\nu}\gamma^{\mu}\partial_{\nu}\partial_{\mu} ) = \frac{1}{2}\{ \gamma^{\mu},\gamma^{\nu}\}\partial_{\mu}\partial_{\nu} = g^{\mu\nu}\partial_{\mu}\partial_{\nu} = \partial^{2}\) The Dirac D is not an elliptic operator as the associated symbol doesn't have the right properties but if \(g^{ab} = \delta^{ab}\) then its positive definite and \(D^{2} = \nabla^{2}\). The second way is to use the symbol for the Laplacian : \(\sigma(\Delta,\xi)\omega = \sigma(dd^{\dag}+d^{\dag}d)\omega = -(\xi \wedge \iota_{\xi}(\omega} + \iota_{\xi}(\xi \wedge \omega)) = -\iota_{\xi}(\xi) \wedge \omega = -|\xi|^{2}\omega\) This is equivalent to saying \(\Delta \sim -\partial^{2}\) (or +, depending on sign conventions) and all you need to know to get that is the symbol for d and its adjoint. I'm not entirely sure how to do the \(\ast d \ast d = (d+d^{\dag})^{2}\) bit other than simply using the definition of the adjoint in terms of d's and Hodge stars. I also have a sneaking suspicion its not true in general, but in cases like Kahler manifolds they are equal, as I remember reading something where they explicitly stated "If..... then [something] = Laplacian = Kahler Laplacian = ....". There are definitely spaces where various definitions of Laplacians coincide. Unfortunately I returned the book I read it in to the library 2 days ago (3 months overdue....)
I think \(*d*d=dd^\dagger+d^\dagger d\) is only true for 0-forms since \(d^\dagger=*d*\) up to sign. The general one is called the Hodge Laplacian (probably also the Laplace-deRham operator) and the one acting on 0-forms is called the Laplace-Beltrami operator.
Yeah, having flicked a bit more through Nakahara it would seem there's various Laplacians, often related to one another by factors of 2 etc and the \(\ast d \ast d\) definitely seems like it's out by a factor of 2 from \((d+d^{\dag})^{2}\) so I'd wager it's only true on either particular manifolds or for particular value of p when applied to p-forms.
I thank you for you responses, and, in a masochistic sort of way, I am slightly encouraged by the evidence that I haven't missed a simple trick. However I have made some progress, not a lot, but some. temur: for the 0-form \(f\) we know that \(d^{\dag}f = 0\). Hence in the expression \(dd^{\dag}f+ d^{\dag}d f\) the first term vanishes. Now since \(df\) is a 1-form and that \(d^{\dag}df\) is then a 0-form by construction, we satisfy the Laplacian for a 0-form. Call it as \(f'\) We also know that for any 0-form, \(\ast d \ast d f\) is also a 0-form, call it \(f''\) (no derivatives implied). How do I know that \(f' = f''\)? I agree, though, on an arbitrary manifold, this probably doesn't extend to all p-forms. Alpha: My texts assure me that the equality \(\nabla^2 = dd^{\dag} + d^{\dag}d\) is valid only on a compact Riemann or pseudo-Riemann manifold, whereas, my earlier rubbish equating \(\nabla^2 = \ast d\ast d\) depended, I now see, on the manifold \(\mathbb{R}^3\) (Euclidean metric), which is neither Riemann nor compact. Umm... In my earlier thread you mentioned harmonic forms - I suspect there may be some clue here - lemme read yours again, together with my texts and get back, but one thing sticks out: our Laplacian is a self-adjoint differential operator. So, by my construction, I may have that, for arbitrary p-forms \((\nabla^2 \alpha, \nabla^2 \beta) = (d \alpha, d \beta) + (d^{\dag} \alpha, d^{\dag} \beta)\). So that if, say, \(\nabla^2\alpha =0\) (definition of an harmonic form), then the identity \((\nabla^2 \alpha, \nabla^2 \beta) = (d \alpha, d \beta) + (d^{\dag} \alpha, d^{\dag} \beta)\) collapses, since by a simple application of linear algebra we know that if, say, \(v=0 \Rightarrow (v,w) =0\)
because *d* = d+ I think it is Riemann. Noncompactness is not much of an issue if you consider only compactly supported entities.
You might want to check your algebra for that Laplacian identity you've given because I don't think that's right. The LHS involves second order derivatives on each form but the RHS is only 1st order derivatives on each form. Do you mean \((\nabla^{2}\alpha,\beta)\) instead, then you've got the right orders of d and d-dag everywhere.
Sorry, R^3 with its Euclidean metric is a Riemannian manifold, of course it is not a Riemann surface.
Then show me the proof, as this as at the heart of my question! (Alpha - back to you on your post, but now I have to run)
Yikes!! Did I really say that? It seems I did, but of course it is nonsense. It should be \( (\nabla^2 \alpha, \beta) = (d \alpha, d \beta)+(d^{\dag} \alpha, d^{\dag} \beta) = (\alpha,\nabla^2 \beta) \), since the operators \(d\) and \(d^{\dag}\) are adjoint, which implies that the Laplacian is self-adjoint. Logic all in a mess in my post. Sorry folks. Thank you for keeping me honest
