I'm struggling with proving that Bochner laplacian can be described by the following formula similar to the standard laplacian formula from calculus: $$\Delta = \sum_i \nabla_i^2,$$ where $\nabla_i = \nabla_{e_i}$ for some local orthonormal basis $e_i$ and $\nabla$ is a connection satisfying any reasonable conditions (Levi-Civita on $TM$, the induced connections on $\Omega^k(M)$).

We live on a compact $C^\infty$ Riemannian manifold without boundary and the laplace operator is defined as $\nabla^*\nabla$, where $\nabla$ is viewed as a map $\Gamma(E)\to \Gamma(T^*M \otimes E)$ between spaces with scalar product and thus the adjoint operator $\nabla^*\!\!:\Gamma(T^*M \otimes E) \to \Gamma(E)$ is well defined*.

I'm new to Levi-Civita, so please be understanding if the solution is trivial...

*Actually the spaces here are neither finite dimensional nor Hilbert, so it is not clear if the adjoint operator exists... The explanation will be highly appreciated, but it can be assumed that $\nabla^*$ exists.


This calculation is done in full detail (without the assumption that the basis is orthonormal) in example $10.1.32$ of Liviu Nicolaescu's wonderful book Lectures on the Geometry of Manifolds which is freely available as a pdf from his website.

  • $\begingroup$ Thanks a lot! But I must admit that I expected the calculation to be more straightforward :( $\endgroup$ – savick01 Sep 4 '13 at 22:44
  • 1
    $\begingroup$ It is more straightforward if the basis is orthonomal. In that case $g^{ij} = \delta^{ij}$ so things simplify quite a bit. $\endgroup$ – Michael Albanese Sep 4 '13 at 22:46
  • 2
    $\begingroup$ Also see page 210 of Petersen. $\endgroup$ – mdg Nov 14 '14 at 1:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.