I know of two ways of defining the (negative - depending on your convention) Laplace-Beltrami operator on the differential forms of a compact, orientable Riemannian manifold $M$.

  1. The Levi-Civita connection extends to a connection tensor bundles by Leibniz rule $\nabla(a\otimes b)=\nabla a\otimes b + (-1)^aa\otimes\nabla b$ (and similarly for wedges) and by defining it on $1$-forms by $$(\nabla\alpha)(X,Y) = \nabla_X\alpha(Y)-\nabla_Y\alpha(X)-\alpha([X,Y])$$ (is this correct?). In particular, we have a connection $\nabla:\Omega^k(M)\to\Gamma(M,T^*M\otimes\Lambda^kM)$ and another $\nabla:\Gamma(M,T^*M\otimes\Lambda^kM)\to\Gamma(M,T^*M^{\otimes 2}\otimes\Lambda^kM)$. We can now concatenate them and take the negative of the trace with respect to the metric $$\Delta=-tr_g(\nabla\nabla).$$
  2. Using Hodge theory, we can define $\Delta=-(dd^\star+d^\star d)$.

Is there an easy way to see whether these two definitions are equal (possibly without computing in coordinates)? A reference where it is done would be awesome!

  • 3
    $\begingroup$ They’re not the same. See en.wikipedia.org/wiki/Weitzenb%C3%B6ck_identity $\endgroup$
    – Deane
    Apr 1, 2023 at 21:58
  • $\begingroup$ @Deane Aha, that explains why I couldn't prove it... If you write your comment as an answer (if possible with a small description of the identity and a reference?) I will be very happy to accept it! $\endgroup$ Apr 2, 2023 at 10:23

1 Answer 1


The two Laplacians are not the same. One is the other plus a curvature term. The formula is known as the Weitzenböck formula. It is surprisingly difficult to calculate. It and its proof can be found in this note of Petersen.


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