Let $\hat{g}$ and $\bar{g}$ be two smooth Riemannian metrics defined, say, on $\mathbb{R}^n.$ Consider a smooth function $\xi$ that acts as an interpolation function between the two metrics above on an annulus $B_{2R}(0)\backslash B_R(0).$
If we define $g=\xi\hat{g} + (1-\xi)\bar{g},$ which is clearly a smooth Riemannian metric, can we find write $Rm(g)$ in terms of $Rm(\hat{g})$ and $Rm(\bar{g})$ plus an error? More importantly, what would be the structure of this error?
After some Googling, the trick seems to be to consider the difference between the two metrics $ h = \tilde{g}-\bar{g}.$ Thus, $g= \bar{g} + \xi h.$ Using the notation above, we can write the Christoffel symbols of $g$ schematically as:
$\Gamma(g) = \Gamma(\bar{g}) + \frac{1}{2}g * \bar{\nabla}(\xi h),$
where $ h = \tilde{g}-\bar{g}.$ Thus,
$Rm(g)= Rm(\bar{g}) + g^{-1} *[ \bar{\nabla}^2\xi *h + \bar{\nabla}\xi*\bar{\nabla}h+ \xi^2\bar{\nabla}^2h] + g^{-2}[*\bar{\nabla}\xi*\bar{\nabla}\xi*h*h + \bar{\nabla}h*\bar{\nabla}\xi*\xi h +\xi^2*\bar{\nabla}h*\bar{\nabla}h],$
where $A*B$ denotes any linear combination of terms involving contractions of the tensor product $A\otimes B.$
As Professor Yang hinted above, the error involves up to second derivatives of the metric and is, in some sense, quadratic.