1
$\begingroup$

Let $(M,g)$ be a compact manifold with a metric $g$ (not necessarily Riemannian one). Suppose that $(M,g)$ is a Lipschitz limit of a sequence of Riemannian manifolds $(M_n,g_n)$ with the sectional curvature bounded between $-1$ and $1$.

Question. Is it true that $(M,g)$ is an Alexandrov space with curvature bounded below by $-1$?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Lipschitz convergence implies Gromov-Hausdorff convergence, so the answer is yes.: $(M,g)$ is an Alexandrov space with curvature at least $-1$. The upper curvature bound on $g_n$ is not needed for this conclusion. $\endgroup$
    – Igor Belegradek
    Commented Dec 9, 2019 at 16:33
  • $\begingroup$ Thanks a lot Igor! Do you think there is a reference that proves that the Gromov-Hausdorff limit of manifolds with curvature $\ge -1$ are Alexandrov spaces of curvature $-1$? $\endgroup$
    – aglearner
    Commented Dec 9, 2019 at 16:45
  • 1
    $\begingroup$ You meant to say "of curvature at least $-1$. A reference is "A Course in Metric Geometry" by Burago, Burago, Ivanov, which is easy to find online. $\endgroup$
    – Igor Belegradek
    Commented Dec 9, 2019 at 16:52
  • $\begingroup$ Igor, I looked in the book but can't find the statement so far. If by any chance you remember where approximately it was written, could you please let me know? $\endgroup$
    – aglearner
    Commented Dec 9, 2019 at 23:20
  • 1
    $\begingroup$ In the book Example 7.4.3 states that Lipschitz implies GH, and Proposition 10.7.1 shows that lower curvature bounds are inherited by GH limits. $\endgroup$
    – Igor Belegradek
    Commented Dec 10, 2019 at 1:36

0

Reset to default

You must log in to answer this question.