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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.

Hi,

this is again a question from me which did not get any answer at math.stackexchange (Link: https://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.

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)".

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.

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

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)".