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No. The assumption is coordinate-independent (i.e., preserved by self-diffeomorphisms) but the desired conclusion is not. Begin with $R$ being the standard irrational flow and $\mathcal O$ a coordina …

Here is an explicit version of Ryan's example. Consider $S^3$ as the unit sphere in $\mathbb C^2$ and define a vector field $V$ on it by $V(z_1,z_2)=(iz_1,\sqrt 2 i z_2)$. Here $z_1,z_2\in \mathbb C$ …

These groups are not even quasi-isometric. (Two metric spaces are said to be quasi-isometric if they contain bi-Lipschitz equivalent nets. In the case of finitely generated groups, word metrics are as …

No, a generic Riemannian metric does not have totally geodesic 2-dimensional submanifolds at all. The property that you ask for is very rare. For example, it implies that $R(X,Y)Y$ belongs to the line …

The second identity is always true because both arguments of $d$ are smooth functions of $u$ and $t$ and they coincide when $u=0$. The first one holds true for all $u$ and $h$ only if the metric is f …

There is a couple of standard methods. First, one can embed $M$ into some $\mathbb R^N$ and fix a smooth neighborhood retraction onto the image. Then let $a$ be the composition of the weighted averag …

Yes. Fix $x\in\Omega$ and let $r=d_\Omega(x,\partial\Omega)$, then $\Omega$ contains the Euclidean $r$-ball centered at $x$. So it suffices to construct $\phi$ supported in this ball with $\|d\phi\|\l …

By Perelman's Stability Theorem, if a (compact) limit of $n$-dimensional Alexandrov spaces of curvature $\ge k$ has the same dimension, then the convergent spaces are eventually homeomorphic to the li …

It is a $C^\infty$ manifold if you define charts properly (e.g. using geodesics normal to the boundary as coordinate lines). The metric of the double is $C^2$ (but not always $C^3$). Indeed, since th …

As others mentioned, you have to remove the diagonal of $M\times M$ or square the distance function. Then, for a complete $M$, the answer is the following. The distance function is differentiable at …

This is an expansion of Anton Petrunin's comment. Let me describe how to perturb the standard metric of the plane so that the resulting metric is bi-Lipschitz to the original with Lipschitz constant …

First of all, the identity holds for any degree 1 map $F:\mathbb S^2\to\mathbb S^2$. Moreover, for any $F=(f,g,h):\mathbb S^2\to\mathbb S^2$, $$ \int_{\mathbb S^2} f\,dgdh = \frac43\pi \deg F. $$ Thi …

Here is a counter-example. Take a sequence of round 2-dimensional spheres $M_n$ of radii $r_n=n^{-1/2}$, $n=1,2,\dots$. Join them together into a long connected sum, namely connect each sphere to the …

I assume that by "all derivatives" you mean derivatives of every order. Suppose that all transitions between normal coordinates have uniformly bounded derivatives within some radius $r$. For every po …

The answer is yes. Suppose the contrary and rescale the picture so that $r=1$. We may assume that $P_1$ and $P_3$ are the endpoints of the graph. There must be points on the graph that are outside the …