Question

Let $M$ be a compact subset of $\mathbb R^2$ with smooth boundary, and let $g$ be a Riemannian metric on $M$. If $g'$ is another Riemannian metric which is "close" to $g$, then they should have almost identical curvature profiles. I would like to prove a concrete estimate on the total difference of their curvatures in terms of the distance of $g'$ to $g$. Before I state the question precisely, I need to introduce some notation.

Write $\operatorname{Sym}$ for the space of symmetric $2\times 2$ real matrices, and let $\operatorname{SPD} \subseteq \operatorname{Sym}$ be those matrices which are also positive-definite. Consider the function space $\Omega = C^2(M, \operatorname{SPD})$. Denote partial derivatives of $g_{ij} \in \Omega$ by additional subscripts following a comma, so that $\tfrac{\partial}{\partial x^k} g_{ij} = g_{ij,k}$, et cetera. Endow the space $\Omega$ with the norm $$\|g\| = \sup_{x \in M} \max_{i,j,k,l} \left\{|g_{ij}(x)|, |g_{ij,k}(x)|, |g_{ij,kl}(x)| \right\},$$ so that it has the structure of an open cone within the Banach space $C^2(M, \operatorname{Sym})$.

Each $g \in \Omega$ defines a Riemannian structure on $M$ via the inner product $\langle v, g(x) v' \rangle$ for $v, v' \in T_x M$. Let $K(g,x)$ be the scalar curvature of the metric $g$ at the point $x \in M$.

What I want to prove: For each $g \in \Omega$, there exist constants $C$ and $\epsilon$ so that if $g' \in \Omega$ with $\|g - g'\| < \epsilon$, then $$\sup_{x \in M} \left| K(g,x) - K(g',x) \right| \le C\|g- g'\|.$$

My current approach to this is quite clunky, and involves calculating everything directly from the Christoffel symbols of the metrics. Is there a better, more geometric approach to this than brute force calculations?

I'm sure this type of lemma is well known to geometric analysts. Is a proof of a similar result written down somewhere?

Answer 1

This is a straightforward consequence of the fact that $K(x)$ is a continuous function of $g(x)$, $\partial g(x)$, and $\partial^2(g)$.

Answer 2

The coordinate-dependent approaches mentioned in other posts are the quickest way to proceed here, but if you want a more coordinate-independent approach, one can flow from one metric g to the next g' (e.g. by using the line segment $t \mapsto (1-t) g + t g'$) and using the standard formulae for the first variation of curvature, as can be found for instance in my blog post

http://terrytao.wordpress.com/2008/03/28/285g-lecture-1-ricci-flow/

Answer 3

It is a fact (due to Riemann I believe) that in normal coordinates, the Taylor expansion of $g_{ij}$ is $\delta_{ij}+\frac{1}{3}R_{ikjl}x^k x^l+O(||x||^3)$, where $R_{ikjl}$ are components of the $(4,0)$ curvature tensor. In dimension $2$ the tensor reduces to scalar curvature. Thus curvature is the second derivative of the metric in normal coordinates. In your setup you insist on global coordinates coming from the ambient Euclidean plane, so you need to take into account the coordinate change from the normal coordinates (defined locally) and global Euclidean coordinates. It seems to me that compactness of $M$ gives a bound on such coordinate change, a bound that depends on $g$.

EDIT: After seeing comments by Will Jagy and Willie Wong, I realized that I misread the question and it makes no sense as stated. Still I will leave my answer in the hope that it would help to the questioner.