From John Roe, Elliptic Operators, topology and asymptotic methods , page 99
Let $M$ be a manifold of dimension $n$ with fixed point $q$. Let a geodesic local coordinate system $x^{i}$ originate from $q$. Let $r^{2}=\sum(x^{i})^{2}=\sum g^{ij}x^{i}x^{j}$(is there a typo here?), so that $r$ is the geodesic distance from $q$ and $h$ is the function $$ (4\pi t)^{-n/2}e^{-\frac{r^{2}}{4t}} $$ Roe claimed the following: $$ \frac{\partial h}{\partial t}+\triangle h=\frac{rh}{4gt}\frac{\partial g}{\partial r},g=\det(g_{ij}) $$ I could not really follow his proof for this statement. After establishing the fact that $\nabla h=-\frac{h}{2t}r\frac{\partial}{\partial r}$, he wants to apply the formula $$ \nabla^{*}(fV)=f\nabla^{*}V-\langle \nabla f, V\rangle $$ where $f=-\frac{h}{2t}$, $V=r\frac{\partial}{\partial r}$ to $\nabla h$, thus compute the Laplacian. To compute the first part, he claimed that $$ \nabla^{*}(r\frac{\partial}{\partial r})=-\frac{1}{\sqrt{g}}\sum_{j}\frac{\partial}{\partial x^{j}}(x^{j}\sqrt{g})=-n-\frac{r}{2g}\frac{\partial g}{\partial r} $$ I am very puzzled with this computation. He used the formula that for a one form $\alpha=\sum A_{i}dx_{i}$ we have $$ d^{*}\alpha=-\frac{1}{\sqrt{g}}\sum_{i,j}\partial_{j}(A_{i}g^{ij}\sqrt{g}) $$ Thus since $rdr\leftrightarrow r\frac{\partial}{\partial r}$ by identifying $TM$ with $T^{*}M$, he get the above formula. Now I have some questions:
1) How do we exactly write down $r\frac{\partial}{\partial r}$ in local coordinates? Should it be $\sum (r\frac{\partial x^i}{\partial r})\frac{\partial}{\partial x^i}$? But then how to compute $\frac{\partial x^i}{\partial r}$ using the metric?
2) Assuming $A_{i}g^{ij}=x^{j}$ for now, how do we compute $$ -\frac{1}{\sqrt{g}}\sum_{j}x_{j}\frac{\partial}{\partial x_{j}}\sqrt{g}? $$ I remember Roe has computed $\frac{\partial}{\partial x_{j}}\log{\sqrt{g}}$ earlier using Cramer's rule, and the result is(see page 20, there may be a typo here) $$ \frac{\partial}{\partial x_{j}}\log{\sqrt{g}}=\sum_{\alpha}\Gamma^{\alpha}_{j\alpha} $$ But now all the Christoffel symbols has 'miraculously' disappeared, and it is replaced by $$ \frac{\partial g}{\partial r} $$ which I again do not know how to compute. So I need to ask for some help.
I think one piece of the puzzle you are missing is that $$ r \partial_r = \sum_i x^i \partial_i.$$ This means that $$ - \frac{1}{\sqrt{g}} \sum_i x^i \partial_i \sqrt{g} = -\frac{1}{\sqrt{g}} r \partial_r (\sqrt{g} ) = -\frac{r}{2g} \partial_r g,$$ as claimed.
Edit: I think there is some confusing notation at work here. It seems Roe uses $\nabla$ to mean two different things: for most of the book, he uses it to denote a connection on a bundle $S$, which is a map $$ \nabla: \Gamma(S) \to \Gamma(T^\ast M \otimes S).$$ In this context, on the other hand, he uses $\nabla$ to denote the gradient, which gives the direction of greatest increase of a function: $$ \nabla = \text{grad} : C^\infty(M) \to \Gamma(TM).$$ Each of these $\nabla$'s have a formal adjoint $\nabla^\ast$ going the other way; in this proof, it seems that $\nabla^\ast$ refers to the adjoint of the gradient operator, which is usually called the divergence (actually, I think the usual divergence is the negative of this). From now on in this answer, I will use the notation $\text{grad}$ for the gradient and $\text{div}$ for the divergence. In coordinates, these operators are given by \begin{align*} \text{grad} \, f &= \sum_{i,j} g^{ij} \partial_i f \, \partial_j \\ \text{div}(X^i \partial_i) &= -\frac{1}{\sqrt{g}} \sum_i \partial_i (X^i \sqrt{g}) \end{align*} (Note the sign on $\text{div}$, which is probably not standard. These formulas can be found in Jost, for example.)
The Laplacian of a function $f$ can be computed in (at least) two ways: $$\Delta f = d^\ast d f = \text{div} \, \text{grad} \, f.$$ It seems that Roe is computing $\Delta h$ as $\text{div} \, \text{grad} \, h$, although he uses the notation $\nabla^\ast \nabla h$. I think it is a mistake when he refers to formula 1.26, which is for $d^\ast$, not $\text{div}$ (perhaps in the first edition of the book, there was a formula for $\text{div}$).
Let's take a moment to compare $d^\ast$ and $\text{div}$, which are given by the very similar formulas \begin{align*} d^\ast (A_i dx^i) &= -\frac{1}{\sqrt{g}} \sum_{i,j} \partial_j (A_i g^{ij} \sqrt{g}) \\ \text{div}(X^i \partial_i) &= -\frac{1}{\sqrt{g}} \sum_i \partial_i (X^i \sqrt{g}) \end{align*} Note the appearance of the $g^{ij}$ in $d^\ast$. Interestingly, there is no nice "product rule" for $d^\ast$, whereas for $\text{div}$ we have (as Roe states) the following: $$\text{div}(fV) = f \, \text{div} \, V - \langle \text{grad} \, f, V \rangle .$$