Timeline for Is Level set of Regular functions in Alexandrov spaces again an Alex. space?

9 events

when toggle format	what		by	license	comment
Mar 2, 2010 at 1:22	comment	added	J. GE		Ok, I see it converges to $A_2$ only because the ambient spaces are converging. Thanks!
Mar 2, 2010 at 1:12	comment	added	Anton Petrunin		The later is also true, but if I remember right we do not use it.
Mar 2, 2010 at 1:07	vote	accept	J. GE
Mar 2, 2010 at 1:06	comment	added	J. GE		oops, I meant the $\frac{1}{\theta_{n,1}\theta_{n,2}}M_{n,1}$ converge to $\mathbb R^k\times A_2$
Mar 2, 2010 at 1:03	comment	added	J. GE		OK, so it's the rescal of the whole manifold $\frac{1}{\theta_{n,2}}M_{n,1}$ converge to $\mathbb R^k\times A_2$, not the 'fiber' converges to $A_2$ as intrinsic metric spaces.
Mar 2, 2010 at 0:59	comment	added	Anton Petrunin		@Leonid. The simplest case is a distance map (i.e. each coordinate $x^i$ is a distance function) such that $dx^i(\xi)>0$ for some direction $\xi$ at each point.
Mar 2, 2010 at 0:53	comment	added	Anton Petrunin		No, first you pass to the limit space and then you note that it is splitting as $R^k\times A_2$.
Mar 2, 2010 at 0:42	comment	added	J. GE		Thanks Prof. Petrunin, In fact, I am reading your paper joint work with Kapovitch and TUSCHMANN. The 4.3 blow-up method, it seems the reason that $\frac{1}{\theta_{n,2}}M_{n,2}$ converges to some Alexandrov space $A_2$ requires that the level set $M_{n,2}$ also has curvature bounded below, right? Or did I miss something? (or should it be rescaling the whole manifold $M_n$ instead of the level set?
Mar 2, 2010 at 0:20	history	answered	Anton Petrunin	CC BY-SA 2.5