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Let $\nabla$ be a Riemannian connection, $\Delta$ be the Laplacian defined as $\text{tr}_g \nabla^2$. The equation I want to show is $$ \Delta\nabla_i f = \nabla_i\Delta f + \sum_{j} Ric_{ij} \nabla_jf $$ where $Ric$ is the Ricci curvature tensor, $f$ is a smooth function. The proof my text gives is (they use the convention that basically $g_{ij} = \delta_{ij}$ which I am not using)

$$ \Delta \nabla_i f = \nabla_j\nabla_i\nabla_j f = \nabla_i\nabla_j\nabla_j f - R_{jijk} \nabla_k f $$

...which I understand 0% about it. My guess is that the first interchange $$ \begin{align*} \Delta\nabla_i f &= \sum_{j,k}g^{jk}\nabla_j\nabla_k\nabla_i f = \sum_{j,k}g^{jk}\nabla_j\nabla_i\nabla_k f \end{align*} $$ is free since $\nabla_i\nabla_j f = \nabla_j \nabla_if$ and $\nabla$ has no torsion. But what about the second one (i.e. how do we interchange $\nabla_j$ and $\nabla_i$?)

Edit: so I've found and use the Ricci identity for 1-forms

$$ (\nabla_i \nabla_j - \nabla_j \nabla_i) w(\partial_k) = w(R(\partial_j,\partial_i)\partial_k) = \sum_{l} w^l R_{jik}^l $$ put $w = \nabla f$ says $$ (\nabla_i \nabla_j - \nabla_j \nabla_i) \nabla f(\partial_k) = (\nabla_i \nabla_j - \nabla_j \nabla_i) \nabla_k f = \sum_{l} (\nabla f)^l R_{jik}^l $$ and I don't know how to proceed to get $Ric$.

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  • $\begingroup$ The second equality follows from the definition of curvature $R(X, Y) Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X, Y]} Z$ for vector fields $X, Y, Z$. $\endgroup$ Commented Apr 8, 2022 at 18:28
  • $\begingroup$ @TravisWillse I'm not sure why that directly follows. I can take $X = \partial_i, Y = \partial_j$, but what should $Z$ be? $\endgroup$
    – macton
    Commented Apr 9, 2022 at 9:07
  • $\begingroup$ You can take $Z = (\nabla f)^\sharp$ and use the symmetries of the Riemannian curvature tensor. In any case, from your last line you can use the metric to raise and lower indices to raise an unpaired index, then use the symmetries of the Riemann curvature tensor and form the appropriate trace to compute an expression in terms of the Ricci tensor. $\endgroup$ Commented Apr 10, 2022 at 1:45
  • $\begingroup$ @TravisWillse My idea is that $$\sum_{j,k,l}g^{jk}\nabla_lf R_{jik}^l = \sum_{j,k,l,m}g^{jk}g^{lm}\nabla_lf R_{jikm} = -\sum_{j,k,l,m}g^{jk}g^{lm}\nabla_lf R_{jimk} = -\sum_{l,m}g^{lm}\nabla_lf Ric_{im} \overset{???}{=} -\sum_{l}\nabla_lf Ric_{il}$$ but I don't think the last equality holds in any possible cases. Are there something that I am missing? $\endgroup$
    – macton
    Commented Apr 10, 2022 at 12:44
  • $\begingroup$ math.stackexchange.com/q/4399574/272127 $\endgroup$
    – C.F.G
    Commented Apr 14, 2022 at 1:44

1 Answer 1

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You're almost there. Using the Einstein summation convention, so that we sum over repeated indices, we have

$$\Delta \nabla_i f = \nabla^j \nabla_j \nabla_i f = \nabla^j \nabla_i \nabla_j f = - R^j{}_i{}^k{}_j \nabla_k f +\nabla_i \nabla^j \nabla_j f = g^{jk} \operatorname{Ric}_{ij} \nabla_k f + \nabla_i \Delta f.$$

The third equality follows from applying the Ricci identity for $1$-forms that you mention in your edit to $\nabla_j f$, and the fourth from the symmetries of the curvature tensor $R$ and the definitions of $\operatorname{Ric}$ and $\Delta$.

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