Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
I misunderstood your question, therefore I deleted my answer. – treble Mar 14 '12 at 2:17
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. – Yuri Vyatkin Mar 14 '12 at 5:45
Thanks Yuri, is there any good reference for this? I found most of the book use local coordinate instead of local frame. – Sun Mar 14 '12 at 12:37
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 $\nabla r$? – Willie Wong Mar 14 '12 at 15:27
@WillieWong, yes exactely. I will edit my post. – Sun Mar 14 '12 at 15:38

I don't think curvature terms should appear since $\nabla_{e_i} \nabla_{e_i} f = e_i \cdot e_i f$, where you think of the $e_i$ as first order differential operators. Then using your notation $$ r_{ikk} = e_k e_k e_i r = (e_k [e_k,e_i] + e_ke_i e_k) r = (e_k[e_k,e_i] + [e_k,e_i] e_k + e_ie_ke_k)r = (e_k[e_k,e_i] + [e_k,e_i] e_k)r + r_{kki}. $$ So $r_{ikk}$ differs from $r_{kki}$ by a second order term.

share|cite|improve this answer
Thanks Eric, I think exactely the samething. But in R.Schoen and S.T. Yau's book: Lectures on Differential Geometry. Prop2.2 they calculated $\Delta(|\nabla f|^2)$. Using local coordinate, they write: $f_{ij}=f_{ji}$ and $f_{jij}=f_{jji}+R_{ij}f_j$. Which confues me even more, cause the Lie bracket should be zero in the coordinate chart. – Sun Mar 14 '12 at 15:47
@SunParkJoe But this is different than what you asked. $\Delta f$ is not simply $\sum_i \nabla_{e_i}\nabla_{e_i} f$ since this is not invariantly defined. $\Delta$ is the Laplace-Beltrami operator and you can see a defintion here:… – Eric O. Korman Mar 14 '12 at 16:21
Yes, I know the difference. What I need to calculate is the one I asked in the post: $(|\nabla r|^2)_{kk}$ for a fixed $k$. Many papers states $\sum_i r_i r_{ikk}= \sum_i r_ir_{kki}+\sum_{i,j}R_{ikjk}r_ir_j$. Why this is true? – Sun Mar 14 '12 at 16:36
@SunParkJoe I see. Do you have any references available online? – Eric O. Korman Mar 14 '12 at 16:55
for example: P. Li and J. Wang Comparison theorem for Kähler manifolds and positivity of spectrum. On page 49 – Sun Mar 14 '12 at 16:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.