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Hi,

this is again a question from me which did not get any answer at math.stackexchange (Link: http://math.stackexchange.com/questions/27366/)

This question is about how well one can choose a partition of unity.

So suppose $(M,g)$ is an open Riemannian manifold. I would like to have the following statement to be true:

"There exists an open cover $\{U_n\}$ of $M$ which is uniformly locally finite (i.e. each point of $p \in M$ lies in at most $q$ sets) and uniformly bounded (i.e. $\mathcal{U} := \bigcup U_n \times U_n$ is a controlled subset of $M \times M$, i.e. $\sup_{p \in \mathcal{U}} d(\pi_1(p), \pi_2(p)) < \infty$, where the $\pi_i$ are the projections on the first and second coordinate).

Furthermore, there exists an subordinate partition of unity $\{g_n\}$ with the properties that the functions $\{g_n^{1/2}\}$ are also smooth and for every $l \in \mathbb{N}$ the $i$-th derivatives $(g_n^{1/2})^{(i)}$ are uniformly bounded for all $i \le l$, i.e. $\parallel(g_n^{1/2})^{(i)}\parallel_\infty := \sup_{p \in M} |\nabla_{v_1, \ldots, v_i} g_n^{1/2}(p)| \le G_l$, where the $v_i$ are unit vectors at the point $p \in M$."

I think the most critical point is the uniform boundedness of the derivatives. Maybe someone knows a reference where it is proven that such a partition of unity always exists (or maybe just one, which has only the property of uniform boundedness of the derivatives)? Is this statement even true?

Thanks, Alex

edit: Though the metric $g$ on $M$ is fixed (in the application where I need this) I would be also happy about a solution of the form "There exists a metric $g$ on $M$, such that ... (the above holds)" or by giving some sufficient conditions on the metric, such that it is always possible.

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In what norm do you measure the derivatives? More precisely, what Riemannian metric do you use? Is it fixed? –  Orbicular Mar 22 '11 at 13:17
    
If your manifold is an open subset of a compact manifold, and $g$ is a restriction of a metric from this compact manifold, then it's true; but I suspect this isn't very useful to you. –  Łukasz Grabowski Mar 22 '11 at 13:30
    
I believe this can be done, if there is a uniform lower bound on either the injectivity radius or the volume of small geodesic balls, as well as the appropriate bounds on the Ricci or sectional curvature. You might want to look up the paper by Bemelmans, Min-Oo, Ruh on smoothing Riemannian metrics. The paper by Shi in JDG on the Ricci flow on a complete Riemannian manifold might also have some relevant results. –  Deane Yang Mar 22 '11 at 15:28
    
I thank you all for your help! –  AlexE Mar 23 '11 at 8:16

3 Answers 3

up vote 3 down vote accepted

As Deane Yang points out, you want bounds from below on the injectivity radius and from above on the curvature. You may want to consider a manifold of bounded geometry, which means that in addition all derivatives of the curvature tensor are bounded. This last condition is equivalent to having an atlas of coordinate charts such that all transition functions have uniformly bounded derivatives.

This topic is discussed in a very down-to-earth way in Appendix 1 of the paper "Spectral Theory Of Elliptic Operators On Non-Compact Manifolds" by Shubin. Your question is addressed in Lemma 1.2 on page 30 and Lemma 1.3 on page 31. Here is a link to the paper:

http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.27.4473

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No you can't have uniformly bounded second derivatives and diameters at the same time (unless you are allowed to change the Riemannian metric).

Consider a smoothened boundary of a tubular neighborhood of a 1-dimensional $\varepsilon$-grid in $\mathbb R^3$. This surface has lots of short closed geodesics (of length of order $\varepsilon$). If $f$ is a function whose second derivative is bounded by a constant $G_2$, then the maximum of the derivative of $f$ along such a geodesic is bounded above by $CG_2\varepsilon$ (where $C$ denotes a universal constant). Hence, at a point where two such geodesic intersect orthogonally, we have $\|df\|\le C G_2\varepsilon$. Since such points form an $\varepsilon$-net and $\|D^2f\|\le G_2$, we have $\|df\|\le C G_2\varepsilon$ everywhere. Thus the function must be $CG_2\varepsilon$-Lipschitz.

On the other hand, at each point some element of a partition of unity attains a value at least $1/q$. Since the function is $CG_2\varepsilon$-Lipschitz and vanishes away from the respective set $U_n$, the distance from this point to the boundary of $U_n$ is at least $1/qCG_2\varepsilon$.

Now consider a 2-manifold $M$ that contains arbitrary large pieces of such $\varepsilon$-grids with arbitrarily small $\varepsilon$'s. In each of these pieces, one of the covering sets has diameter at least $1/qCG_2\varepsilon$, but $C$, $q$ and $G_2$ are fixed, so the diameters go to infinity as $\varepsilon\to 0$.

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The simplest case is the following: If there exists $k, r, v > 0$ such that the volume of any geodesic ball of radius $r$ is bounded from below by $v$ and the absolute value of sectional curvature is bounded from above by $k$, then there exists a metric $C^0$ close to the original metric with all the desired properties listed in the question. You just combine the results proved by Stefan Peters or Greene-Wu on existence of harmonic co-ordinates with the theorem of Bemelmans-Min-Oo-Ruh on smoothing Riemannian metrics.

The curvature bound can be weakened if needed.

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