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Picture below is from Li and Yau's Estimates of eigenvalues of a compact Riemannian manifold. How to get the (1.8) ? The main difficult is how to get the term with curvature. I guess it is because $u_{jii}$ become $u_{iij}$, but $u$ just a function, exchange the partial derivative will not creat term with curvature.

Besides, I found I am puny to calculate exchange partial derivative, whether there are some suitable example or exercise.

enter image description here

enter image description here

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    Higher *covariant* derivatives of scalars do not commute, since the derivative of a scalar is a one-form. $u_{jii} -u_{iij}= u_{iji}-u_{iij} = [\nabla_i, \nabla_j]u_i = R_{ijik}u_k = R_{kj} u_k.$ – 2017-01-01
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    @AnthonyCarapetis I find a formular $\nabla_i\nabla_j T_{b_1...b_s}^{a_1...a_r} - \nabla_j\nabla_i T_{b_1...b_s}^{a_1...a_r} =-R_{kji}^{a_n} T_{b_1...b_s}^{a_1...k...a_r} +R_{b_mji}^k T^{a_1...a_r}_{b_1...k...b_s} $. I use it get the result is $-R_{kj}u_k$. Could you give a likely formular ? – 2017-01-01

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