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On a Riemannian Manifold $M^n$, the Hodge Laplacian is defined on k-forms by $\Delta\omega=\operatorname{d}\operatorname{d}^*\omega+\operatorname{d}^*\operatorname{d}\omega$. For 0-forms, e.i. smooth functions, one can easily see that w.r.t. some local coordinates with inverse metric tensor $g^{ij}$ it holds:$$\Delta\omega=-\partial_j(g^{ij}\partial_i\omega)$$ This can be used to apply results of 1-dimensional elliptic regularity to $\Delta$. My goal is to understand elliptic regularity on all k-forms, $\Delta:\Omega^k(M)\rightarrow\Omega^k(M)$. My question is: How does the term of highest order (2nd order in derivatives) look like in a local trivialisation of k-forms? (e.g. $\omega=\sum_{I=(i_1,...,i_k)\\0≤i_1≤...≤i_k≤n}\omega_I \operatorname{d}x^{i_{1}}\wedge...\wedge\operatorname{d}x^{i_{k}}$ or alternatively using some local orthogonal frames).

Is it true or wrong that its highest order is diagonal?$$(\Delta\omega)_I=A^{I,ij}\partial_i\partial_j\omega_I+\sum_JB^{I,J,i}\partial_i\omega_J+\sum_J C^{I,J}\omega_J$$ Is it maybe even true that the $A^{I,ij}=-g^{ij}$ as in the k=0 case?

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  • $\begingroup$ By hand, I could also veryfy all assertions for k=1 $\endgroup$
    – Felix Schlag
    Commented May 19, 2018 at 13:19
  • $\begingroup$ I think I proofed everything now, but my proof involves very shady business with the metric tensor. $\endgroup$
    – Felix Schlag
    Commented May 19, 2018 at 14:17
  • $\begingroup$ Very good question. Actually I have been thinking about it recently since I need it in my research. $\endgroup$
    – Piotr Hajlasz
    Commented May 22, 2018 at 5:54
  • $\begingroup$ If there is interest, I will post the proof, but these days I am too busy. $\endgroup$
    – Felix Schlag
    Commented May 23, 2018 at 20:37
  • $\begingroup$ Note that, if all you want is the top order term, then you can ignore all terms involving the derivatives of the metric tensor (equivalently, the Christoffel symbols and their derivatives). You can also do the calculation at a single point and assume that $g_{ij} = \delta_{ij}$ at that point. This simplifies the calculations. $\endgroup$
    – Deane Yang
    Commented Jul 11, 2018 at 22:00

1 Answer 1

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Your formula is true.

The Weitzenböck formula states that the Hodge Laplacian on $k$-forms satisfies $$ \Delta\omega=(d\delta+\delta d)\omega=\nabla^*\nabla\omega +\operatorname{Ric}(\omega), $$ where $\nabla^*\nabla$ is the Bochner Laplacian. For a proof, see Theorem 9.4.1 in:

P. Petersen, Riemannian geometry. Third edition. Graduate Texts in Mathematics, 171. Springer, Cham, 2016.

Finally, the Bochner Laplacian in suitable local coordinates can be represented as $$ \nabla^*\nabla=-\sum_{k,j} \big\{\ g^{kj}\nabla_k\nabla_j+\frac{1}{\sqrt{|g|}}\partial_{x^k}\big(\sqrt{|g|}g^{kj}\big)\cdot\nabla_j\ \big\}. $$ This is proved in Example 10.1.32 in

L. I. Nicolaescu, Lectures on Geometry of Manifolds.

Beautiful Nicolaescu's lectures are freely available on his homepage.

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    $\begingroup$ Yes, it is correct. $\endgroup$
    – Liviu Nicolaescu
    Commented Jul 11, 2018 at 20:36
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    $\begingroup$ Note that the curvature term in the Weitzenböck formula is only the Ricci curvature when $\omega$ is a $1$-form. In general it involves the Riemannian curvature tensor. However, it always acts on $\omega$ as a zeroth-order operator, so this has no effect on the principal symbol. $\endgroup$
    – Jeffrey Case
    Commented Jul 12, 2018 at 12:41

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