Let $M$ be a compact manifold. Let $S^2 T^*M $ be the vector bundle of all symmetric $(0,2)$ tensors and $S_+^2 T^*M$ be the open subset of all positive definite ones. Does $S_+^2 T^*M $ have bounded geometry? It is understood that the metric on it is the tensor product of the induced metric on the cotangent bundle from some Riemannian metric on $M$.
$\begingroup$
$\endgroup$
12
-
$\begingroup$ see mathoverflow.net/questions/212713/… $\endgroup$– valeriFeb 18, 2016 at 11:05
-
1$\begingroup$ @Kaveh May you more explain on the structure of the metric on the bundle? $\endgroup$– Ali TaghaviFeb 18, 2016 at 11:25
-
$\begingroup$ I read mathoverflow.net/questions/212713/…. But I can not see the conclusion. $ s_+^2T^*M$ is a subbundle of $T^*M \otimes T^*M$, so the question is that is finite tensor product of cotangent bundle has bounded geometry. $\endgroup$– KavehFeb 18, 2016 at 11:30
-
$\begingroup$ About Riemannian metric, I think the natural metric is the tensor product of a Riemannian metric on the cotangent bundle. I dont not know yet what kind of metrics exist on cotangent bundles. $\endgroup$– KavehFeb 18, 2016 at 11:41
-
$\begingroup$ in the referred post $TM$ was considered with two types of metrics: Sasaki with flat fibers - not bounded geometry, and Cheeger-Gromoll, coming from submersion $GM \to TM$ of bounded geometry. So, the choice is yours. $\endgroup$– valeriFeb 18, 2016 at 11:51
|
Show 7 more comments