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Suppose $f$ is a scalar function and $h$ a symmetric $(0,2)$ tensor on a closed Riemannian manifold $(M,g)$. Then is the following equality true? $$\int_Mf\nabla_i\nabla_jh_{ij}\mathsf{dvol}_g=\int_M(\nabla_i\nabla_j f)h_{ij}\ \mathsf{dvol}_g.$$

I know that $\nabla_i\nabla_jh_{ij}=\mathsf{div}(\nabla_jh_{ij})$ and $f\mathsf{div}(\nabla_jh_{ij})=\mathsf{div}(f\nabla_jh_{ij}) -\langle \nabla f , \nabla_jh_{ij} \rangle$ then integrating over $M$ and using divergence theorem gives: $$\int_Mf\nabla_i\nabla_jh_{ij}\ \mathsf{dvol}_g=\int_M-\langle \nabla f , \nabla_jh_{ij} \rangle\ \mathsf{dvol}_g$$ The other equation that can help is $\mathsf{div}(\nabla_jfh_{ij})=\nabla_i\nabla_j(fh_{ij})=(\nabla_i\nabla_jf)h_{ij}+f\nabla_i\nabla_jh_{ij}+2\nabla_if\nabla_j h_{ij}$ but I don't know how to relate this to the last equation and deduce the wanted equality.

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Hint: The divergence theorem, in abstract index notation, reads $$ \int_M\nabla_iX^idV_g=\int_{\partial M}N_iX^idV_{g|_{\partial M}} $$ This means that on a compact manifold without boundary, you can integrate by parts exactly as you would in single variable calculus, by moving a $\nabla_i$ from one term to another and reversing the sign, regardless of how indices are contracted.

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  • $\begingroup$ Can you elaborate "by moving a ∇i from one term to another"? Because one term is vector field and another term is a function in $fX$. $\endgroup$
    – C.F.G
    Commented Apr 28, 2021 at 23:32
  • $\begingroup$ The point is that it doesn't matter what type of tensors the terms are, since $\nabla$ distributes over all tensor products/contractions. In that case, it would be$$\nabla_i(fX^i)=(\nabla_if)X^i+f(\nabla_iX^i)$$and the integral of the l.h.s. vanishes due to divergence theorem. One could have arbitrarily complicated examples such as$$\nabla_i(T^{ij_1\cdots j_n}_{k_1\cdots k_m}U^{k_1\cdots k_m}_{j_1\cdots j_n})=(\nabla_iT^{ij_1\cdots j_n}_{k_1\cdots k_m})U^{k_1\cdots k_m}_{j_1\cdots j_n}+T^{ij_1\cdots j_n}_{k_1\cdots k_m}(\nabla_iU^{k_1\cdots k_m}_{j_1\cdots j_n})$$and do the same thing. $\endgroup$
    – Kajelad
    Commented Apr 28, 2021 at 23:41
  • $\begingroup$ The integrand is always a scalar, but the individual terms can be tensors of arbitrary type. $\endgroup$
    – Kajelad
    Commented Apr 28, 2021 at 23:55
  • $\begingroup$ So you meant $\int_M\nabla_if\nabla_j h_{ij}=-\int_M(\nabla_j\nabla_if) h_{ij}$? Well, but how to see this one explicitly? $\endgroup$
    – C.F.G
    Commented Apr 28, 2021 at 23:56
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    $\begingroup$ The intermediate step would be$$\int_M\nabla_if\nabla_jh_{ij}=\int_M\left[\nabla_j((\nabla_if)h_{ij})-(\nabla_j\nabla_if)h_{ij}\right]$$ and the first term integrates to $0$ since it is a divergence. $\endgroup$
    – Kajelad
    Commented Apr 29, 2021 at 0:00

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