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Let $X^n$ be an Alexandrov space, and $f: X^n\to \mathbb R^k$ a regular map, does the level set necessary be an Alexandrov space?

In my mind, the intrinsic metric on the level set is 'comparable' to the ambient metric, but is it necessary an Alexandrov space?

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up vote 3 down vote accepted

The answer is "no" even for regular semiconcave function $f:X\to\mathbb R$

If $f:X\to \mathbb R$ is convex then it is a long-standing open problem.

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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? – John B Mar 2 '10 at 0:42
No, first you pass to the limit space and then you note that it is splitting as $R^k\times A_2$. – Anton Petrunin Mar 2 '10 at 0:53
@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. – Anton Petrunin Mar 2 '10 at 0:59
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. – John B Mar 2 '10 at 1:03
oops, I meant the $\frac{1}{\theta_{n,1}\theta_{n,2}}M_{n,1}$ converge to $\mathbb R^k\times A_2$ – John B Mar 2 '10 at 1:06

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