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Let $M$ be a (closed if necessary) Riemannian manifold with Levi-Civita connection

Let $\nabla^*$ be the formal adjoint of $\nabla$ with respect to the $L^2$ inner product. Let $\Delta=\nabla^*\nabla$ denote the Laplacian.

Question 1: In general, do we have $\nabla\circ\Delta$=$\Delta\circ\nabla$, as operators $C^\infty(M)\to C^\infty(M)\otimes C^\infty(T^*M)$?

Question 2: If not, is this relation true if the metric is flat?

Comment: I feel that the answer to Q2 at least must be yes, but I am not good with these computations. So I would appreciate it if someone could work through the computation (or perhaps share a reference) for the commutator $[\nabla,\Delta]$ involving the curvature terms.

geometricK
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    In general, there will be some differences due to the curvature terms. For a special case, you can look up Bochner's formula. – User Mar 09 '22 at 18:41
  • If the metric is flat, the problem reduces to compute $(\sum_{j=1}^n \frac{\partial^2}{\partial x_j ^2} )\circ \frac{\partial}{\partial x_{j_0}}$ and Schwarz lemma gives the answer. – Didier Mar 10 '22 at 10:53
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    You may find this my answer useful for getting a more general perspective on this identity. – Yuri Vyatkin Mar 17 '22 at 10:23
  • @YuriVyatkin Thanks, indeed that looks useful (somehow it did not appear in my search the first time). Does your answer imply that there exist constants $C_1$ and $C_2$ such that $|\nabla\Delta A-\Delta\nabla A|\leq C_1|\nabla R||A|+C_2|R||\nabla A|$, where norms are induced from the Riemannian metric? – geometricK Mar 17 '22 at 14:22
  • Oh, I have no idea about the estimates. It looks like a totally new question. – Yuri Vyatkin Mar 17 '22 at 21:25

1 Answers1

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As it has been mentioned in comment, by Bochner's formula for function $f$ the answer is clear:

$$\Delta\nabla_if = \nabla_i\Delta f + \mbox{R}_{ij}\nabla_jf.$$

Obviously in flat metrics we have $\mbox{R}_{ij}=0$ and ..

C.F.G
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