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I just came across [https://math.stackexchange.com/questions/1046370/heat-equation-on-manifold]. I do not really have knowledge in differential geometry but I at least want to get an intuition of how geometry influences equations like the heat equation. What is meant by operators like e.g. $\text{div}_g$ and $\nabla_g$ if $(M,g)$ is a Riemannian manifold? I tried to find references but they either give too much details which is too much for me as a beginner or they skip almost all details that I'm not able to really tell what's going there. Is it for example true that \begin{align} \text{div}_g f=0 \Leftrightarrow \text{div} f= g ? \end{align} Is there a reference which gives a lowkey overwiev (in simple words) to understand PDE's on manifolds?

user99432
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    No, $g$ is the Riemannian metric, and it is used to define covariant derivatives and curvature. You need the metric to go back and forth between vector fields and differential $1$-forms, the Hodge star operator, etc. So to define the divergence of a vector field will require this structure. What you wrote down makes no sense, since the metric $g$ is a $2$-tensor field on the manifold and the divergence of a vector field is a scalar function. – Ted Shifrin Jan 23 '23 at 19:14
  • What is then meant by $\text{div}_g$? – user99432 Jan 23 '23 at 20:01
  • It's the divergence (of a vector field) computed using the Riemannian metric $g$. For example, we start with a vector field $X$, use the metric to turn it into a $1$-form $\omega$, use the Hodge star operator to get the $(n-1)$-form $\star\omega$, then take the exterior derivative $d(\star\omega)$. This $n$-form is some scalar function $(\text{div}_g X)$ times the volume form $dV$. (It can also be defined without differential forms by computing the Lie derivative $\mathscr L_X(dV) = (\text{div}_g X) dV$. This is a bit simpler, but equivalent.) – Ted Shifrin Jan 23 '23 at 20:08
  • For the case of the heat equation, the metric is used to define the Laplacian. You can think of $\text{div}(\text{grad}f)$. This will turn out to be $\star d(\star df)$. That is, you the vector field $\text{grad}f$ corresponds (according to the metric) to the $1$-form $df$. – Ted Shifrin Jan 23 '23 at 20:12

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