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I am trying to understand a line in the proof of Lemma 3.6 on p. 18 of this book by Chow, Lu, and Ni.

Say we have a Riemannian manifold $M$ with Levi-Civita connection $\nabla$. Let $\Delta$ be the Laplacian on functions. Then the authors state that (in local cordinates) we have $$\Delta\nabla_i f = \nabla_j\nabla_i\nabla_j f=\nabla_i\nabla_j\nabla_j f-R_{jijk}\nabla_k f.\quad(*)$$ It seems that $\nabla_j\nabla_i\nabla_j$ and $\nabla_i\nabla_j\nabla_j$ are short for $$\sum_{j,k}g^{jk}\nabla_j\nabla_i\nabla_k,\quad\nabla_i\sum_{j,k}g^{jk}\nabla_j\nabla_k.$$ respectively. I tried to use that $\Delta=\sum_{j,k}g^{jk}\nabla_j\nabla_k$, together with $R_{ji}=[\nabla_j,\nabla_i]$ for coordinate vectors, but somehow I'm not able to work out the details.

Question 1: How do we prove $(*)$?

There is another part of the proof of that lemma that I don't understand:

Question 2: How do we show that $-R_{jijk}\nabla_k f=R_{ij}\nabla_j f$?

geometricK
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1 Answers1

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Repeated indices indicate summation. We know that $\nabla_i\nabla_j f=\nabla_j\nabla_i f$ is true for functions (Show this identity Ex. 1.14 in Chow's Book). Then $$\Delta\nabla_i f =(\nabla_j\nabla_j)\nabla_i f =\nabla_j(\nabla_j\nabla_i )f=\nabla_j(\nabla_i\nabla_j )f=(\nabla_j\nabla_i)(\nabla_jf)=\nabla_i\nabla_j\nabla_j f-R_{jijk}\nabla_k f=\nabla_i(\nabla_j\nabla_j f)-R_{jijk}\nabla_k f.$$

And note that $\nabla_jf$ is not a function so the action of $\nabla_j\nabla_i$ on it has a curvature term.

And in last note that $\nabla_j\nabla_j f=\Delta f$ and $R_{jijk}=-R_{ik}$ (trace of Riemann curvature is Ricci) by definition. So ...

C.F.G
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