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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 …

No, such functions do not exist. More precisely, let $f:\mathbb R\to\mathbb R$ be an arbitrary function, $\Sigma$ is the set of $x\in\mathbb R$ such that $f'(x)$ exists and equals 0. Then $f(\Sigma)$ …

Consider a two-dimensional sphere with a Riemannian metric of total area $4\pi$. Does there exist a subset whose area equals $2\pi$ and whose boundary has length no greater than $2\pi$? (To avoid tec …

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 …

The answer is no, assuming that you seek an orientation preserving square root. (I see unknown's answer appeared while I'm typing. I don't quite understand it at the moment but the construction looks …

The geodesics between points are not unique in both cases. Moreover the following is true: if $M$ is a universal cover of a compact Riemannian manifold whose fundamental group is virtually solvable bu …

It is rarely possible. First of all, if the manifold is simply connected, then there are no nowhere vanishing closed 1-forms at all. Indeed, every closed 1-form on such a manifold is a derivative of …

Consider co-dimension 0. In this case, bundles are covering maps, with all the goodies that they bring. And submersions are just local homeomorphisms - not very exciting compared to coverings.

Here is a self-contained proof not using any classification. With some effort, it can be made to work under weaker assumptions: the curvature is nonnegative outside a compact set and is bounded from a …

Yes. In fact, $\mathbb R^3$ could be any 3-manifold and $f(\mathbb R^1)$ any countable union of embedded segments. Lemma 1. Let $U$ be an open ball in $\mathbb R^3$ and $f:I\to\mathbb R^3$ an embeddi …

Let $M^n$ be a smooth manifold equipped with a nondegenerate Lagrangian $L:TM\to\mathbb R$, $L=L(x,y)$, $x\in M$, $y\in T_xM$. The stationary points of the corresponding integral functional on curves …

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 …

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 answer is yes. Moreover you don't need to assume that $B$ is nonpostively curved. (And, if you are not interested in the equality case or can afford a convex boundary, the nonpositive curvature of …

Given three points $a,b,c$ in a (geodesic) metric space $X$, one defines a comparison angle $\angle(a,b,c)$ by the cosine law: $$ \angle(a,b,c) = \arccos \frac{|ab|^2 + |ac|^2 - |bc|^2}{2\cdot|ab|\cd …