Note: Laplacian Chain Rule
Steve Trettel
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Here are several chain rules for working with the Laplacian on a Riemannian manifold - these come up in several calculations so I want to preserve them somewhere.
A chain rule
Here we start with arbitrary functions $E\colon M\to\RR$ and $g\colon\RR\to\RR$ and compute the laplacian of the composite $g\circ E$.
We use the definition in terms of differential forms $\Delta =\star d\star d$ and work outwards applying one operator at a time.
(i) Differentiate:
$$d(g\circ E)= g^\prime(E)dE$$(ii) Apply Hodge Star:
$$\begin{align} d(g\circ E)&= d\left(g^\prime(E) dE\right)\\ &= g^\prime(E) \star dE \end{align}$$(iii) Differentiate:
$$\begin{align} d\star d(g\circ E)=d\left(g^\prime(E) \star dE\right)&\\ = d(g^\prime(E))\wedge \star dE &+ g^\prime(E)d\star dE\\ =g^{\prime\prime}dE\wedge \star dE &+ g^\prime(E)d\star dE \end{align}$$
We can simplify the first term using the definition of the hodge dual
$$dE\wedge\star dE = \langle dE,dE\rangle\mathrm{vol}$$where $\langle \cdot,\cdot\rangle$ is the Riemannian metric and $\mathrm{vol}$ is the Riemannian volume form. The metric on 1-forms is defined in terms of the metric isomorphism from $TM$ to $T^\ast M$, so
$$\langle dE, dE\rangle :=\langle (dE)^\sharp, (dE)^\sharp\rangle = \langle\nabla E,\nabla E\rangle$$where $\nabla E$ is the gradient vector. Putting this all together,
$$d\star d(g\circ E) = g^{\prime\prime}(E) \|\nabla E\|^2\mathrm{vol}+g^\prime(E)d\star dE$$- (iv) Apply Hodge Star: this pulls through scalars so immediately we get
The dual of the Riemannian volume is the constant function $1$ by definition, and $\star d \star d E$ is the Laplacian of $E$. Thus
$$\Delta (g\circ E)= g^{\prime\prime}(E)\|\nabla E\|^2 + g^\prime(E)\Delta E$$Two Variables
A similar computation holds when building a function out of multiple pieces: if $X\colon M\to\RR$ and $Y\colon M\to\RR$ are two functions and $g\colon\RR^2\to\RR$ we can build a single function $M\to\RR$ as their composite:
$$g(X,Y)\colon p\mapsto g(X(p),Y(p))$$The reason such a result might be useful is that it gives a way to build functions on $M$ out of simple pieces (for instance, maybe instead of the abstract functions $E$ and $R$ we work with $r$ and $z$ in cylindrical coordinates, etc). Similar reasoning to the above lets us calculate $\Delta g(X,Y)$ here:
If $X,Y\colon M\to\RR$ and $g\colon\RR^2\to\RR$ then
$$\Delta g(X,Y)=g_{xx}\|\nabla X\|^2+g_{yy}\|\nabla Y\|^2+ 2g_{xy}\langle \nabla X,\nabla Y\rangle +g_x \Delta X+g_y\Delta Y$$First, $dg(X,Y)=g_x(X,Y)dX+g_y(X,Y)dY$. The star passes right through the scalar partial derivatives, so
$$\star d g(X,Y)=g_x(X,Y)\star dX+g_y(X,Y)\star dY$$Differentiating this involves differentiating each of the two terms above; the calculations proceed similarly so we do one and then copy the analogous term:
$$\begin{align} d\left(g_x\star dX\right)&=d(g_x(X,Y))\wedge\star dX + g_x(X,Y)d\star dX\\ &=\left(g_{xx}(X,Y)dX+g_{xy}(X,Y)dY\right)\wedge\star dX + g_x(X,Y)d\star dX\\ &=g_{xx}(X,Y)dX\wedge\star dX + g_{xy}(X,Y)dY\wedge\star dX + g_x(X,Y)d\star dX \end{align}$$By definition of $\star$ we know
$$dX\wedge\star dX = \|dX\|^2\mathrm{vol}=\|\nabla X\|^2\mathrm{vol}$$$$ dY\wedge \star dX =\langle dY,dX\rangle \mathrm{vol}=\langle \nabla X,\nabla Y\rangle \mathrm{vol}$$Substituting these in (and supressing further copies of the input variables $(X,Y)$ for brevity)
$$d(g_x \star dX)=g_{xx}\|\nabla X\|^2\mathrm{vol}+g_{xy}\langle \nabla X,\nabla Y\rangle \mathrm{vol}+g_x d\star dX$$An analogous term occurs for $d(g_y\star dY)$, and putting them together we get the total differential:
$$\begin{align} d\star d g(X,Y)&=g_{xx}\|\nabla X\|^2\mathrm{vol}+g_{xy}\langle \nabla X,\nabla Y\rangle \mathrm{vol}+g_x d\star dX\\ &+g_{yy}\|\nabla Y\|^2\mathrm{vol}+g_{yx}\langle \nabla Y,\nabla X\rangle \mathrm{vol}+g_y d\star dY \end{align}$$Applying the star once more:
$$\begin{align} \star (d\star d g(X,Y))&=\star(g_{xx}\|\nabla X\|^2\mathrm{vol})+\star(g_{xy}\langle \nabla X,\nabla Y\rangle \mathrm{vol})+\star(g_x d\star dX)\\ &+\star(g_{yy}\|\nabla Y\|^2\mathrm{vol})+\star(g_{yx}\langle \nabla Y,\nabla X\rangle \mathrm{vol})+\star(g_y d\star dY)\\ &=g_{xx}\|\nabla X\|^2\star\mathrm{vol}+g_{xy}\langle \nabla X,\nabla Y\rangle \star\mathrm{vol}+g_x \star d\star dX\\ &+g_{yy}\|\nabla Y\|^2\star\mathrm{vol}+g_{yx}\langle \nabla Y,\nabla X\rangle \star\mathrm{vol}+g_y \star d\star dY\\ &=g_{xx}\|\nabla X\|^2+g_{xy}\langle \nabla X,\nabla Y\rangle +g_x \Delta X\\ &+g_{yy}\|\nabla Y\|^2+g_{yx}\langle \nabla Y,\nabla X\rangle +g_y\Delta Y \end{align}$$Finally, realzing the cross terms from each are equivalent (the metric and order of partial differentiation are symmetric) gives our final answer
$$\Delta g(X,Y)=g_{xx}\|\nabla X\|^2+g_{yy}\|\nabla Y\|^2+ 2g_{xy}\langle \nabla X,\nabla Y\rangle +g_x \Delta X+g_y\Delta Y$$This formula simplifes in the often-useful case that $X$ and $Y$ have orthogonal level sets (for instance, when choosing $X,Y$ from nice coordinate systems).
Coordinate Computations
The two variable chain rule generalizes directly to arbitrarily many input functions $X_i\colon M\to \RR$. One particularly useful case is when a collection ${X_i}$ forms a coordinate patch on $M$. Then the expression $g(X_1,\ldots, X_n)$ is just a ‘coordinate representation’ of our function, and the laplacian chain rule directly gives a formula for the laplacian in these coordinates. Here we record the often-useful case where the coordinates are orthogonal
Let $X_i\colon M\to\RR$ be a collection of functions for $i\in{1,\ldots,n}$ and $g\colon\RR^n\to \RR$ with pairwise orthogonal level sets. Then the composite function $g(X_1,\ldots X_n)\colon M\to\RR$ has laplacian
$$\Delta g(X_1,\ldots X_n)=\sum_{i=1}^n g_{x_ix_i}\|\nabla X_i\|^2+g_{x_i}\Delta X_i$$Example: Euclidean Cartesian
Consider the functions $X,Y$ on the plane $\RR^2$, where $X(x,y)=x$ and $Y(x,y)=y$. With respect to the Euclidean metric $\nabla X$ and $\nabla Y$ are both unit length and orthogonal, thus so are $dX$ and $dY$ (musical isometry). This implies $\star dX = dY$ and $\star dY = -dX$ as
$$\begin{align}dX \wedge \star dX = \langle dX, dX\rangle \mathrm{vol}=\mathrm{vol} &= dX\wedge dY \\ dY \wedge \star dY = \langle dY,dY\rangle \mathrm{vol}=\mathrm{vol} &= dX\wedge dY = dY\wedge(-dX) \end{align}$$This implies the laplacians of $X,Y$ vanish as $d^2=0$:
$$ \Delta X = \star d\star d X = \star d dY =0 $$So, plugging these facts into the chain rule we see
$$\begin{align} \Delta g(X,Y)&=g_{xx}\|\nabla X\|^2+g_{yy}\|\nabla Y\|^2+g_x\Delta X+g_y\Delta Y\\ &= g_{xx}\cdot(1)+g_{yy}\cdot(1)+g_x\cdot(0)+g_y\cdot(0)\\ &=g_{xx}+g_{yy} \end{align}$$Thus the laplacian in $X,Y$ is just the sum of the second derivatives with respect to each slot of $g$, as we well know!