# curvature of subset of Alexandrov spaces

If M is a Riemannian manifold with $Ric \ge - \left( {n - 1} \right)$, $$ds_M^2 = d{t^2} + \exp \left( {2t} \right)ds_N^2$$ N is a submanifold of M. Then by Gauss-equation, we can prove $Ric\left( N \right) \ge 0$. If M is an Alexandrov space with $Ric \ge - \left( {n - 1} \right)$ (or $\sec \ge - 1$) and a warped product like above. Can we get similar information about N?

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The case $\mathrm{sec}\ge -1$ is done in "Curvature bounds for warped products of metric spaces." by Alexander and Bishop. It should be possible to modify their proof so it will work for Ricci. – Anton Petrunin Aug 28 '13 at 13:46
@AntonPetrunin:The Ricci curvature I use is defined by Zhang-Zhu from your parallel transportation formula. I don't know whether the Ricci you mean is the same or equivalent. – jiangsaiyin Aug 28 '13 at 13:57