A clarification regarding analytic perturbation of metrics and Laplacian

This question is in reference to the following Mathoverflow question and the accepted answer to it. It seems to me that it is taken for granted that if the metric $g_t$ perturbs real analytically in time, so does the Laplacian $\Delta_t$ associated to $g_t$. I am trying to prove this step by step by trying to demonstrate a Taylor series expansion for $\Delta_{a + h}$ in terms of powers of $h$ and $t$ derivatives of $\Delta$ evaluated at $a$. In doing this, I have started from the local coordinate expression of $\Delta_{a + h}$ in terms of $g_{a + h}$, which is given by $\Delta_{a + h}f = \frac{1}{\sqrt {|g_{a + h}|}} \partial_i \left(\sqrt{|g_{a + h}|} g_{a + h}^{ij} \partial_j f \right)$. But trying to use the power series expansion for $g_{a+ h}$ (which is given by the fact that $g_t$ is varying real analytically) in the above expression for the Laplacian is becoming real messy real quick.

So, my question is, is this the proper way to see that the Laplacian is real analytic? Or there is some slick coordinate-invariant, or other methods of seeing this? Sorry if the question is too basic, I just wish to make sure that I understand all the details.

• I guess a coordinate approach would look a bit clearer, but the key points would be the same. For example, write $\Delta=\operatorname{div}\circ\operatorname{grad}$, then check that both operators depend analytically on $g$. The gradient of $f$ is simply $g^{-1}df$. For the divergence, it's just a bit more work. – Sebastian Goette Nov 23 '15 at 16:54
• If you don't need an explicit formula for the expansion, then the analyticity follows directly by the fact that the formula for the coefficients of the Laplacian in terms of the metric and its derivatives is a composition of analytic functions. The square root is analytic, because its input is always strictly positive. – Deane Yang Nov 23 '15 at 18:10