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One characteristic of a CBA($K$) surface (a topological surface with an intrinsic metric of curvature $\leq K$ in the sense of Alexandrov) is that $\delta_K(T) \leq 0$, where $\delta_K(T)$ is the relative excess of a triangle T: $\delta_K(T)=\alpha+\beta+\gamma- \alpha_K+\beta_K+\delta_K$, where $\alpha, \beta, \gamma$ are the angles of the triangle (in the Alexandrov sense) and $\alpha_K, \beta_K, \delta_K$ are the angles of a comparison triangle in the model space of curvature $K$.

A surface with Bounded Integral Curvature (BIC) is a topological surface with an intrinsic metric with the property that any point as a neighborhood $U$ such that for any set $\mathcal{T}$ of "nice" triangles with disjoint interior in $U$, the sum of $\delta_0(T)$ over all $T\in \mathcal{T}$ is bounded by a constant $C$ depending only on $U$. See Reshetnyak, Two-dimensional manifolds of bounded curvature. Geometry, IV, Ecyclopaedia Math. Sci., or Aleksandrov, A. D.; Zalgaller, V. A. Intrinsic geometry of surfaces for more details.

If $K\leq 0$, then the CBA($K$) condition mentioned above obviously implies the BIC condition mentioned above with $C=0$, as $\delta_0(T)\leq 0$ for any triangle.

My question is: what is the argument if $K\geq 0$?

I saw the fact that CBA($K$) implies BIC noted a couple of times in the literature, so the argument is certainly quite stupid.

[Edit] :

Thomas Richard noted to me that what I am looking for follows from $$\delta_0(T) \leq Area(T) \;\;\; (*)$$

where $Area(T)$ is the 2-dimensional Hausdorff measure given by the CBA(1) metric.

Formula $(*)$ is stated (in a slightly more general form) as Lemma 5.3 in

[MO] Total Excess on Length Surfaces, Machigashira and Ohtsuka, Mathematische Annalen, 2001

No proof is provided, the authors say that the argument is similar to the the proof of the analog of (*) for curvature bounded from below (CBB) metrics, provided in

[M] The Gaussian curvature of Alexandrov surfaces, Machigashira, J. Math. Soc. Japan, 1998

The argument should rest on (3) of Lemma 5.1 of [MO], which says that, with suitable conditions, in a surface with a bounded from above metric, there are points with a neighborhood bi-Lipschitz to a Euclidean disc. There is no proof of this lemma in [MO], the authors refer to

[OT] The Riemannian Structure of Alexandrov Spaces, Otsu and Tanoue

which is a preprint, which seems currently unavailable (the analog result for CBB metrics is proved in the Burago-Gromov-Perelman paper).

If someone has any reference or arguments about those results (or a copy of [OT]), I would be highly interested.

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  • $\begingroup$ I guess that in the definition of BIC one should read "disjoint triangles" rather than "distinct", right? $\endgroup$ Jan 5, 2015 at 13:04
  • $\begingroup$ Thanks! I changed this and also I put the word "nice" for the triangles, as to be precise there are some conditions on them (e.g. convexity condition), but all details would be too long. I added references for interested people. $\endgroup$ Jan 6, 2015 at 9:35

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