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2 votes
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82 views

Is isoperimetric hypersurface unique up to homeomorphism?

Is there a Riemannian structure on $\mathbb{R}^n $with two non homeomorphic compact hypersurfaces $M,N$ such that both satisfy the isoperimetric inequality. I precisely meanthe following: $$\...
Ali Taghavi's user avatar
0 votes
0 answers
183 views

Sufficient condition for existence of a closest-point projection from a neighborhood onto a subset of a Riemannian manifold

Let $(M,g)$ be a Riemannian manifold and let $N$ be a subset of $M$. On one hand, it is well known that if $N$ is an embedded submanifold of $M$, then it admits a tubular neighborhood, and, ...
gpr1's user avatar
  • 144
5 votes
2 answers
769 views

Checking that the image of a curve is not contained in a hyperplane

Let $\gamma : [0,1] \to \mathbb R^n$ be a smooth curve, $n \geq 2$. I would like to find an easy to check condition such that the image of $\gamma$ is not contained in an $n-1$ dimensional linear ...
J. Swail's user avatar
  • 437
2 votes
1 answer
256 views

Equidistant points on a compact Riemannian manifold

Let $(M,g)$ be a compact Riemannian manifold. To this Riemannian manifold, we associate a natural number $K(M,g)$ as follows: $K(M,g)$ is the maximum of all $n\in \mathbb{N}$ such that we have at ...
Ali Taghavi's user avatar
7 votes
1 answer
367 views

Does complexified isometry group act transitively on tangent bundle of compact Riemannian manifold?

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$Let $ g $ be the round metric on the sphere $ S^n $. Since $ S^...
Ian Gershon Teixeira's user avatar
7 votes
1 answer
246 views

Currents in sub-Riemannian geometry

Federer and Fleming's notion of "currents" is well established so far, and starting from the seminal work of Ambrosio and Kirchheim, the notion of metric currents is well studied also. The ...
Son Gohan's user avatar
  • 215
12 votes
1 answer
1k views

Smoothness of distance function to a compact set

Fix a non-empty compact subset $K\subseteq \mathbb{R}^n$ and let $d_K(x):=\min_{z \in K} \,\|z-x\|$ be the map sending any $x\in \mathbb{R}^n$ to its distance from $K$. Suppose that: $K$ is regular : ...
ABIM's user avatar
  • 5,405
15 votes
2 answers
1k views

When is a bi-Lipschitz homeomorphism smoothable?

Suppose I have a smooth Riemannian manifold $X$ with induced distance function $d$, and a bi-Lipschitz (with respect to $d$) homeomorphism $$\phi: X \to X.$$ Under what circumstances could $\phi$ be ...
Rohil Prasad's user avatar
  • 1,601
10 votes
2 answers
926 views

Is there a volume-preserving diffeomorphism of the disk with prescribed singular values?

This is a cross-post. While working on a variational problem, I have reached to the following question. Let $0<\sigma_1<\sigma_2$ satisfy $\sigma_1\sigma_2=1$, and let $D \subseteq \mathbb{R}^2$...
Asaf Shachar's user avatar
  • 6,741
1 vote
1 answer
237 views

A basic question about compact $C^1$ surfaces with boundary

Let $S \subset \mathbb{R}^3$ be a compact and locally $C^1$ simply-connected surface with a $C^1$ boundary with no self intersection. Is there a $C^1$ bijection $F: \overline{B(0,1)} \rightarrow \...
MathLearner's user avatar
2 votes
1 answer
210 views

$L^{2}$ Betti number

Let $\tilde{X}$ be a non-compact oriented, Riemannian manifold adimits a smooth metric $\tilde{g}$ on which a discrete group $\Gamma$ of orientation-preserving isometrics acts freely so that the ...
user94640's user avatar
5 votes
1 answer
375 views

A possible characterization of sphere or projective space

Is there a compact Riemanian manifold $M$ not diffeomorphic to sphere or real or complex or quaternion projective space which admit a diffeomorphism $f$ with the property that $$\forall x \in M, \...
Ali Taghavi's user avatar
6 votes
1 answer
148 views

Does a minimum area disk that is bounded by a cycle $C$ continuously deform in $R^3$ as $C$ moves in $R^3$?

Let $C_1=(v_1,v_2,\ldots,v_{i-1},v_i)$ and $C_2=(v_1,v_2,\ldots,v_{i-1},v'_i)$ be two cycles that are drawn in $R^3$ in the shape of an unknot (not knotted) with straight line segments as their edges (...
Hooman's user avatar
  • 415
5 votes
1 answer
1k views

Does every smooth manifold admit a metric with bounded geometry and uniform growth?

Let $M$ be a smooth manifold, $g_M$ a Riemannian metric, and consider for $x\in M$ the volume growth function, $gr_x$ that maps $r>0$ to the volume $vol_{g_M}(B(x,r))$. My interest is to see ...
user44172's user avatar
  • 541
4 votes
1 answer
487 views

Lorentzian metrics on the torus up to continuos deformations

Any two Riemannian metrics can easily be deformed into each other, only obtaining positive definite metrics in between. However, for metrics of other signatures this might not be possible. Which ...
Dominik's user avatar
  • 3,017
1 vote
1 answer
924 views

Normal tubular neighborhood theorem for semi(or pseudo)-riemannian manifolds

Suppose you have a manifold $M$ and a closed sub-manifold $A$, and let $g$ be a semi-riemannian metric,ie, $g_x$ defines a quadratic form on $T_xM$ such that $g_x(v,v)\ge0$, but $g_x(v,v)=0$ not ...
Javier Gargiulo's user avatar
2 votes
1 answer
453 views

Smooth maps transverse to a foliation

Let $M$ and $N$ be smooth manifolds and let $S$ be a submanifold of $N$ ($\dim S < \dim N$). Let $\mathfrak S$ be a foliation of $S$. We say that a map between $M$ and $N$ is transverse to $\...
Saurabh T's user avatar
  • 287
6 votes
0 answers
302 views

degenerating surface II

In degenerating surface, Robert Bryant give us an example of a sequence of minimal immersions which converges (in $C^2$- topology) to $z\mapsto z^{2k+1}$ on the unit disc $\mathbb{D}$. My question is ...
Paul's user avatar
  • 914
14 votes
4 answers
2k views

When do two holonomy maps determine flat bundles that are isomorphic as just bundles (w/o regard to the flat connections)?

Suppose we have a surface S (although the question might make as much sense in higher dimensions) and a topological group G. The data of a flat vector bundle on S (up to isomorphism) is the same as a ...
Ilya Grigoriev's user avatar