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I am trying to calculate the following: \begin{equation} \int_{M}R_{ij}\nabla^{i}\varphi\nabla^{j}\varphi\; e^{-\varphi}d\mu=\int_{M}\nabla^{i}R_{ij}\nabla^{j}\varphi\;e^{-\varphi}d\mu+\int_{M}R_{ij}\nabla^{i}\nabla^{j}\varphi\;e^{-\varphi}d\mu. \end{equation}

I have this calculated: \begin{equation} e^{-\varphi}R_{ij}\nabla^{i}\varphi=e^{-\varphi}\nabla^{i}R_{ij}-\nabla^{i}(R_{ij}e^{-\varphi}). \end{equation} So i got that \begin{equation} \begin{split} \int_{M}R_{ij}\nabla^{i}\varphi\nabla^{j}\varphi\; e^{-\varphi}d\mu&=\int_{M}(e^{-\varphi}\nabla^{i}R_{ij}-\nabla^{i}(R_{ij}e^{-\varphi}))\nabla^{j}\varphi\;d\mu\\&=\int_{M}\nabla^{i}R_{ij}\nabla^{j}\varphi\;e^{-\varphi}d\mu-\int_{M}\nabla^{i}(R_{ij}e^{-\varphi})\nabla^{j}\varphi d\mu. \end{split} \end{equation} Now, I have the following: \begin{equation} \nabla^{i}(R_{ij}e^{-\varphi})\nabla^{j}\varphi=\nabla^{i}(R_{ij}e^{-\varphi}\nabla^{j}\varphi)-R_{ij}\nabla^{i}\nabla^{j}\varphi\;e^{-\varphi}. \end{equation} Then I have \begin{equation} \int_{M}R_{ij}\nabla^{i}\varphi\nabla^{j}\varphi\; e^{-\varphi}d\mu=\int_{M}\nabla^{i}R_{ij}\nabla^{j}\varphi\;e^{-\varphi}d\mu+\int_{M}R_{ij}\nabla^{i}\nabla^{j}\varphi\;e^{-\varphi}d\mu-\int_{M}\nabla^{i}(R_{ij}e^{-\varphi}\nabla^{j}\varphi)d\mu. \end{equation} I cannot remove the last term from the above equation. The only idea I have is to use the divergence theorem, but it doesn't appear in the equation.

Cal22
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    Is $M$ compact? If it is, then the last term is equals to zero, since the Divergence Theorem. – DiegoMath Jun 01 '21 at 00:44
  • Yes, $M$ is compact. But how do you use the divergence theorem in this case? I don't see it very clear. – Cal22 Jun 01 '21 at 00:46
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    $\nabla^i(T_i)$ is the divergent of the tensor $T_i=R_{ij}e^{-\phi}\nabla^j\phi$. – DiegoMath Jun 01 '21 at 00:50
  • $R_{ij}e^{-\varphi}\nabla^{j}\varphi$ is a escalar fuction, why is it a tensor? – Cal22 Jun 01 '21 at 01:07
  • It has i as an index, so it is a tensor. – Arctic Char Jun 01 '21 at 02:26
  • I am confused because the definition of divergence I understand is using the nabla operator as follows: $\nabla_{i} T^{i}$ when $T=T^{i}\frac{\partial}{\partial x^{i}}$. For me, this is the definition of divergence. I don't understand the meaning of $\nabla^{i}(T_{i})$. I made a mistake? – Cal22 Jun 01 '21 at 03:19
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    Is $M$ Riemannian and $\nabla$ the Levi-Civita connection? In that case you can just rise and lower indices and $\nabla^iT_i=\nabla_iT^i$. – nicrot000 Jun 01 '21 at 09:50
  • Could you recommend me literature to review these properties? – Cal22 Jun 01 '21 at 21:58
  • Do you understand the notation $\nabla^i \varphi$ and how it is different from $\nabla_i \varphi$? – Arctic Char Jun 02 '21 at 09:03
  • Yes. The first ones are the gradient components of a scalar function and the next one is the covariant derivative of a scalar function. – Cal22 Jun 02 '21 at 14:51

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