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By a 1990 paper of Michael Anderson, the following is true:

Theorem. Let the metric space $(X,d,p)$ be a pointed Gromov-Hausdorff limit of a sequence of complete pointed Riemannian manifolds $(M_i,g_i, p_i)$ satisfying the uniform bounds

  • [two-sided Ricci] $|Ric(g_i)|\leq \lambda$
  • [$L^{n/2}$ norm of curvature] $\int\limits_M |Rm(g_i)|^{n/2} \leq \Lambda$
  • [non-collapsing] for some $r_0$, $\inf\limits_{q\in M_i, r\leq r_0} \ vol(B(q,r))/r^n\geq v$.

Then $(X,d,p)$ is (topologically) a smooth orbifold, whose singularities are of the form $\mathbb{R}^n/\Gamma$ for some finite $\Gamma\subseteq SO(n)$.

I'm trying to get a sense of how sharp this statement is. Here's a possible converse.

Question. Suppose I have a spherical space form $S^{n-1}/\Gamma$, and a $n$-manifold $M$ with boundary $\partial M = S^{n-1}/\Gamma$. Can I construct a sequence of complete Riemannian metrics on $M$'s interior, all satisfying the bounds above, such that some pointed Gromov-Hausdorff limit is homeomorphic to $\mathbb{R}^n/\Gamma$?

A few remarks on what I know: I am aware of some standard constructions for particular $\Gamma$, such as the ALE hyperkähler metrics for $\Gamma\subseteq SU(2)$. However, I don't know any good general ways to construct a family of metrics satisfying two-sided Ricci bounds.

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As far as I know, the only way to do this is to glue a small copy of an ALE Ricci-flat space (with bounded $L^{n/2}$ curvature and boundary at infinity equal to $S^{n-1}/\Gamma$) onto each orbifold singularity. So the answer is presumably yes for each $\Gamma$ for such an ALE space exists. Probably the best person to ask is Anderson himself or maybe his colleague Claude LeBrun. –  Deane Yang Feb 7 '12 at 17:45
    
Thanks for the comment. I agree the answer is yes for each manifold $M$ that admits an ALE Ricci-flat metric. (With the sequence of metrics just given by rescaling.) But presumably many (most?) manifolds with space form boundary don't admit ALE Ricci-flat metrics; I'm curious about them. –  macbeth Feb 7 '12 at 18:40
    
The manifold $M$ itself doesn't have to admit an ALE Ricci-flat metric. Just put an arbitrary smooth metric on it with bounded Ricci curvature, such that each orbifold singularity is metrically a point. It's only the small bubble that has to be Ricci-flat, so when you scale it down and perturb the metric (to do the gluing), the Ricci curvature doesn't blow up. –  Deane Yang Feb 7 '12 at 19:07
    
What I meant was, isn't it true that $M$ [in my notation not the "manifold itself," but the "part that gets collapsed"] doesn't itself need to admit an ALE Ricci-flat metric? For instance, it could have the topology of an ALE Ricci-flat orbifold surgery'd onto an ALE Ricci-flat manifold, corresponding to the situation of a "bubble on a bubble" in the limit. I guess I'm asking whether every manifold with space form boundary can be decomposed as a "tree" of ALE Ricci-flat orbifolds with manifold "leaves." –  macbeth Feb 7 '12 at 19:47
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CORRECTION: What I said in my last comment is not right (thanks to macbeth for correcting me on this). The bubble is not necessarily an ALE Ricci-flat manifold but an ALE Ricci-flat orbifold, where the singularities also arise from bubbles shrinking down to points. This sequence repeat, leading to a tree structure of ALE Ricci-flat orbifolds. I believe that very little is known about this tree structure except in special cases (e.g., where the original manifold is hyperkahler). The main problem is that there is no known "minimum energy" for an ALE Ricci-flat manifold. –  Deane Yang Feb 9 '12 at 10:33

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