# On a family of $C^0$-convergent Riemann metrics

I am dealing with the following concrete situation that could be familiar to Riemannian geometers more experienced than myself.


A. We know that for any $p\in M$ there exists an open neighborhood $U\ni p$ and local coordinates $x^1,\dotsc, x^m$ on $U$ such that

$$g^\ve_{ij} \to g_{ij}$$

uniformly on the compacts of $U$, where

$$g^\ve =\sum_{i,j} g^\ve_{ij}dx^idx^j,\;\;g=\sum_{i,j}g_{ij}dx^idx^j.$$

B. (Edited following Deane Yang's inquiry.) The note by $Gr_2(TM)$ the bundle of Grassmanians of $2$-planes in the tangent bundle. The sectional curvature $K^\ve$ can then be viewed as a function $K^\ve: Gr_2(TM)\to\mathbb{R}$. We know that there exists a smooth function $K^0: Gr_2(TM)\to \mathbb{R}$ such that $K^\ve\to K^0$ uniformly.

Question. Can we conclude that the function $K^0$ in B is the sectional curvature of $g$?

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I just want to confirm that the sectional curvature converges only for co-ordinate 2-planes and not for other 2-planes? Others can answer your question better than me, but my experience has been that controlling the sectional curvature only along co-ordinate 2-planes (for a fixed set of co-ordinates) is not enough. – Deane Yang May 15 '12 at 20:46
It's a bit better than the uniform convergence mentioned in my question. let me edit the question. – Liviu Nicolaescu May 15 '12 at 22:14
Have you looked at the paper by Stefan Peters, "Convergence of riemannian manifolds", Compositio Math 62 (1987) 3-16? You can't conclude that the limiting sectional curvature is continuous but it is bounded almost everywhere. – Deane Yang May 15 '12 at 22:41
I have looked at the paper you mentioned. That paper uses only the assumption that $K^\varepsilon$ is uniformly bounded. In my case $K^\varepsilon$ is convergent to a smooth limit. The question is weather this limit is indeed the curvature of the limiting metric $g$ which we know is smooth. – Liviu Nicolaescu May 16 '12 at 0:10
You're absolutely right, and Anton Petrunin has a much better answer to your question. – Deane Yang May 16 '12 at 8:05

Yes, it is true. Here is a sketch of proof, but I am sure that there is a simper way to see it.

Note that by Gauss–Bonnet, this is true in dimension 2.

Assume we know it in dimension $n-1$.

There is a construction of smooth convex hypersurfaces $H^\varepsilon$ in $(M,g^\varepsilon)$ which converge nicely to a smooth hypersurfaces $H$ in $(M,g^\varepsilon)$. "Nicely" means that one can apply the induction hypothesis for $H^\varepsilon\to H$. Taking many hypersurfaces like that plus linear algebra finishes the proof.

The construction. To construct $H^\varepsilon$ near $p\in M$, choose n points $a_1,a_2,\dots a_n$ such that in all the metrics the angles $\angle a_ipa_j\approx\pi/2$ for all $i\ne j$. Then define $$H^\varepsilon=\{\\, x\in M\mid \sum\phi(|a_i-x|_{g^\varepsilon}-|a_i-p|_{g^\varepsilon})= 0 \\,\}$$ where $\phi(0)=0$, $\phi'(0)=1$ and $\phi''\ll -2$.

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@ Anton Thank you for your answer. I need to process the strategy that you outlined above. I have to admit that I do not see why $H^\varepsilon$ is smooth and what is the role of the angle condition that you mentioned. – Liviu Nicolaescu May 16 '12 at 9:35
@Liviu, sorry, it is very cryptic. Look at 4.3 in our "Extremal subsets in ..." and then look at Lemma B.5 in journals.tubitak.gov.tr/math/issues/mat-03-27-1/… . Hope it helps. – Anton Petrunin May 16 '12 at 10:26
@ Anton: Thanks for the ref. I'll check it out. – Liviu Nicolaescu May 16 '12 at 12:45
@Anton: Perhaps it is obvious, but why does it follow from Gauss-Bonnet in dimension 2? – Dan Lee May 16 '12 at 18:31
@Dan. Fix 3 very close points, let $\triangle^\varepsilon$ be geodesic triangles for $g^\varepsilon$ with the vertices at these points. Note that the angles and area of $\triangle^\varepsilon$ converge to the angles and area of $\triangle^0$. Apply Gauss–Bonnet. – Anton Petrunin May 17 '12 at 2:17