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Let $M$ be a Riemannian manifold and $f$ a smooth function on $M$. The Bochner formula proved in Schoen-Yau's book "lectures in Differential Geometry": (prop. 2.2) $$ \Delta |\nabla f|^2(p)=2\sum _{ij} |f_{ij}|^2 +2 R_{ij}f_i f_j +2 \sum f_i(\Delta f)_i $$ The proof assumes that ${x_i}$ are normal coordinates around $p$. and $f_i$ is the co variant differential of $f$ with respect to $\partial/\partial x_i$. $R_{ij}$ is the Ricci tensor.

Then the proof proceed as follows: Since $|\nabla f|^2=\sum f_i^2$. Hence at $p$ one has $$ \Delta |\nabla f|^2 =\sum_j(\sum f_i^2)_{jj}=..... $$

My question is actually the first claim $$|\nabla f|^2=\sum f_i^2$$ It seems for my that in general $$\nabla f=f_sg^{si}\frac{\partial}{\partial x_i}$$ In normal coordinate we can only get the expression $|\nabla f|^2=\sum f_i^2$ only holds at the point $p$, not other points. But clearly we need to take derivative, the value at the point $p$ is not enough.

It's most likely I miss some point here, but I can't figure out why. Anyone can help?

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By definition of Levi-Civita connection, the covariant derivative of the metric tensor vanishes. – achille hui Feb 4 '13 at 23:12
@achillehui how does that help? – Eric O. Korman Feb 4 '13 at 23:45
The $f_{ij}$ there are covariant derivatives of $f_i$. When you use Levi-Civita connection, the partial derivatives of $g^{st}$ in $\partial_j( g^{st}f_s f_t)$ will be stuffed into the $\Gamma_{ij}^k$ part of $f_{ij} = \partial_j f_i - \Gamma_{ji}^k f_k$. You no longer need to worry about the partial derivatives of $g_{st}$ explicitly. – achille hui Feb 5 '13 at 0:11
@achillehui I don't understand your answer. It seems that you agree with that $|\nabla f|^2 =g^{st}f_sf_t$ not $f_i^2$. My question is which one is THE expression of $|\nabla f|^2$? – user60933 Feb 5 '13 at 10:04
$|\nabla f|^2 = g^{st}f_sf_t$ in general. But at $p$, $g^{st} = g_{st} = \delta_{st}$ and the 'value' of $|\nabla f|^2$ reduces to $f_i^2$. Since we have used co-variant derivatives for bookkeeping the partial derivatives from $g^{st}$ and we only need numbers over a single point $p$, there is no need to explicitly include $g^{st}$ in the expression. – achille hui Feb 5 '13 at 10:37

If I am not mistaken, the point that achille hui was trying to make is that you can commute tracing with respect to the metric $g$ and covariant derivatives. In other words, you can take the terms $g^{ij}$ out of the covariant derivatives.

In general, you have $\left|\nabla f\right|^2=g^{ij}f_if_j$. Then,

$$\Delta\left(\left|\nabla f\right|^2\right)=tr_g\nabla^2\left(\left|\nabla f\right|^2\right)=g^{kl}\nabla^2_{k,l}(g^{ij}f_if_j)=g^{kl}g^{ij}\nabla^2_{k,l}(f_if_j).$$

So if you are working in geodesic normal coordinates centered at $p$, the RHS corresponds to the term $\sum\limits_j\left(\sum\limits_if_i^2\right)_{jj}$ in your question.

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