# Commutation formula for covariant derivative

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?

• 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.
• 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: en.wikipedia.org/wiki/… – 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
Commuting covariant derivatives leads to a term in the curvature tensor. Indeed, the Riemann curvature tensor is $$R(X,Y)T = \nabla_X\nabla_Y T - \nabla_Y\nabla_X T - \nabla_{[X,Y]} T$$ Often, this is written for $T = Z$, a vector field, but $T$ can be any tensor. If $T= f$ is a function, the curvature is zero on $f$, (assuming a zero torsion connection). However, for other kinds of tensors, the curvature is nonzero. Let me give you an example. Assume $J$ is a Jacobi field : $$[\partial_r,J] = 0$$ where $r$ is a distance function. Then, (here, the connection is Riemannian) $$[\partial_r,J] = \nabla_{\partial_r}J - \nabla_J \partial_r = 0$$ Taking a second covariant derivative gives $$\nabla^2_{\partial_r} J - \nabla_{\partial_r} \nabla_J \partial_r = 0$$ Now, commuting in terms of the curvature tensor, $$\nabla^2_{\partial_r} J + R(J,\partial_r)\partial_r - \nabla_J\nabla_{\partial_r}\partial_r = 0$$ The last term is zero because $\partial_r$ is self-parallel. So we have the Jacobi equation $$\nabla^2_{\partial_r} J + R(J,\partial_r)\partial_r = 0$$ ...can you point out to me the example you would like to work out...I don't have the book in question, but is it about the Bochner method, by any chance? -- Salem