# Proof of $\int_Mf\nabla_i\nabla_jh_{ij}\mathsf{dvol}_g=\int_M(\nabla_i\nabla_j f)h_{ij}\ \mathsf{dvol}_g.$

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.

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.
• 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 $fX$. Commented Apr 28, 2021 at 23:32
• 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. Commented Apr 28, 2021 at 23:41
• 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? Commented Apr 28, 2021 at 23:56
• 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. Commented Apr 29, 2021 at 0:00