## How does curvature change under perturbations of a Riemannian metric?

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?

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Your target inequality, as typed, seems to imply that $$K(g,x)$$ is constant in $x.$ Indeed, given fixed points $x,y$ take a conformal perturbation on a very small disc that avoids both points and let the perturbation shrink. The right hand side goes to $0,$ and as $g' = g$ near $y$ we get $$K(g,x) = K(g,y).$$ – Will Jagy Aug 21 2010 at 17:37
Well, along with wondering if the inequality is really a correctly typed version of what you wish to ask about, let me mention that there is plenty of relevant material under the name "Yamabe Problem" when the perturbations are restricted to conformal changes (at each point $g'$ is a positive scalar multiple of $g,$ often written $g' = e^f g.$ Probably anything you can think of on more general perturbations is included in Gromov's works. – Will Jagy Aug 21 2010 at 18:59
@Will: why even perturb? Let $g$ be a metric such that the scalar curvature is non constant. Let $g = g'$, the RHS vanishes, the LHS is manifestly non-zero. So the inequality as written is definitely fishy. – Willie Wong Aug 21 2010 at 22:56
Willie, yes, you are right. At some point Tom will notice this and perhaps be able to fix it. – Will Jagy Aug 22 2010 at 1:25
Willie and Will, thanks for your comments. You are correct: the statement as I'd written it was nonsense. I've edited the question to something less trivial. – Tom LaGatta Aug 22 2010 at 17:56

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

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Better yet, if you use the fact that $K(x)$ is a polynomial function of $g^{-1}$, $\partial g$, and $\partial^2 g$, you see that the constant $C$ is polynomial in the $C^2$ norm of $g$ and the $C^0$ norm of $g^{-1}$. – Willie Wong Aug 22 2010 at 18:24
Deane and Willie, thank you for your comments. I ended up proving the result directly based on your suggestions. – Tom LaGatta Aug 29 2010 at 0:42

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/

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Although I agree with the sentiment here (in the long run you really want to understand things and work as much as possible using a co-ordinate-free point of view), I think that a student should first understand how to derive the inequality above in the most direct way possible with minimal machinery. The inequality is rather basic, and, in this particular case, the co-ordinates don't get in the way at all (there are none of the difficulties you face when you try to prove an inequality in the opposite direction). – Deane Yang Aug 23 2010 at 22:07

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.

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 Igor, I've edited the question to something that does make sense. I think your approach is right: rather than estimate the scalar curvature directly by brute force, one should switch into normal coordinates and estimate the coordinate transformation. – Tom LaGatta Aug 22 2010 at 17:59 I have to say that I don't understand the need to use normal co-ordinates to derive the desired inequality. – Deane Yang Aug 23 2010 at 21:57