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Suppose $\nabla$ is the Levi-Civita connection on Riemannian manifold $M$. $X$ be a vector fields on $M$ defined by $X=\nabla r$ where $r$ is the distance function to a fixed point in $M$. $\{e_1, \cdots, e_n\}$ be local orthnormal frame fields. We want to calculate $(|\nabla r|^2)_{kk}=\nabla_{e_k}\nabla_{e_k}|\nabla r|^2$. Let $$\nabla r=\sum r_i e_i$$ so $r_i=\nabla_{e_i}r$.

The standard calculation for tensor yields: $$(|X|^2)_{kk}=(\sum r_i^2)_{kk}\\ =2(\sum r_i r_{ik})_{k} \\ =2\sum r_{ik}r_{ik}+2\sum r_i r_{ikk} $$ My question is, how to switch the order of partial derivatives $r_{ikk}$ to $r_{kki}$. I know some curvature terms should apear, but I am very confused by this calculation.

My main concern is $r_i$ should be function, when exchange the partial derivatives Lie bracket will apear, how come the curvature term apears?

Anyone can help me with this basic calculations?

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    I misunderstood your question, therefore I deleted my answer.2012-03-14
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    Partial derivatives are defined w.r.t. a coordinate system, and you are talking about covariant derivative w.r.t an local orthonormal frame, that makes a big difference. The partial derivatives indeed commute unlike the covariant ones.2012-03-14
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    Thanks Yuri, is there any good reference for this? I found most of the book use local coordinate instead of local frame.2012-03-14
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    You seem to be interpreting $r_i$ as the i'th partial for some function $r$: are you defining your vector field $X$ as, in fact, the gradient field $\nabla r$?2012-03-14
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    @WillieWong, yes exactely. I will edit my post.2012-03-14
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    Is this in context of trying to derive [Bochner formula](http://en.wikipedia.org/wiki/Bochner%27s_formula)? If so, you don't want to define $f_{kk} = e_k(e_k(f))$, the correct term should be $e_k\cdot \nabla_{e_k}\nabla f$. (Since $\triangle = \mathrm{tr}\nabla^2$ and not $\sum_k \nabla_{e_k} \nabla_{e_k}$.)2012-03-14
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    @WillieWong, Yes. But if you choose o.n.b $e_i$, then it's the term I gave in the post.2012-03-14
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    You are not just assuming ONB. You are assuming that the Ricci rotation coefficients vanish suitably. (In particular you need that $\nabla_{e_i}e_i = 0$.)2012-03-15
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    @WillieWong, yes you are right, the Christofell symbole vanishes at the point $p$.2012-03-15

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