Skip to main content

All Questions

Filter by
Sorted by
Tagged with
1 vote
2 answers
290 views

Examples on small cut radius of totally convex set in non-negatively curved manifold

Suppose $M^n$ is an open complete nonnegatively curved Riemannian manifold. In Cheeger-Gromoll's proof of the soul theorem. They need an estimate on the cut radius of a totally convex set $C$. By a ...
Ralph's user avatar
  • 283
12 votes
1 answer
720 views

Is a manifold with flat ends of bounded geometry?

A Riemannian manifold $(M,g)$ is said to have flat ends if the curvature tensor of $g$ vanishes outside a compact set $K$. I was wondering if such manifolds are of bounded geometry. Recall that a ...
Thomas Rot's user avatar
  • 7,583
7 votes
2 answers
807 views

Asymptotic expansion of the Schrödinger kernel?

My stackexchange post was somewhat unsatisfactory (also because I may not have stated clear enough what my interest was). So here it goes! Let $M$ be a compact Riemannian manifold and $\Delta$ be the ...
Matthias Ludewig's user avatar
4 votes
2 answers
626 views

Uniqueness of Kähler form with same volume

Hallo, Let $M$ be a compact real-analytic Riemannian manifold with Riemannian metric $g$. Let $U \subset T^{*}M$ be a open neighbourhood of the zero section. On $U$ there exists a complex structure $...
hapchiu's user avatar
  • 339
3 votes
0 answers
242 views

What is known about analogous results of Kazdan and Warner in higher dimensions?

First let me state a Theorem due to Kazdan and Warner: ``Let M be a compact two dimensional orientable manifold. Let $f: M \rightarrow \mathbb{R}$ be a function that has the same sign as $\chi(M)$,...
Ritwik's user avatar
  • 3,245
26 votes
5 answers
7k views

Intuition for mean curvature

I would like to get some intuitive feeling for the mean curvature. The mean curvature of a hypersurface in a Riemannian manifold by definition is the trace of the second fundamental form. Is there ...
nicolas's user avatar
  • 693
9 votes
3 answers
479 views

twisted Poisson structures, degenerate metrics and integrability properties of (2,0)-tensors

Given a regular (constant rank) bi-vector $\Pi \in \Gamma(\bigwedge^2TM)$ on a smooth manifold $M$ the necessary and sufficient condition for the image of $\Pi^\sharp:T^*M\to TM$ to be an integrable ...
issoroloap's user avatar
6 votes
2 answers
633 views

Rigorous solution to Ricci Flow on dumbbell $S^3$

To begin a small interest in Ricci Flow and similar tools, I am starting with Hamilton's expository paper The Formation of Singularities in the Ricci Flow. This was posted in 1995, so I am wondering ...
Chris Gerig's user avatar
  • 17.5k
1 vote
1 answer
218 views

Isometric embedding of a neighbourhood of a totally real submanifold in a Kähler manifold

Hallo, Let $(M,J,\omega)$ be a real-analytic Kähler manifold. Let furthermore $A \subset M$ be a real analytic, totally real, Lagrangian submanifold and set $g := h|_{A}$. Where $h$ is the Kähler ...
hapchiu's user avatar
  • 339
8 votes
1 answer
458 views

Different complexifications of a real analytic Riemannian manifold

I have a question concerning the complexification of a real analytic Riemannian manifold. Let $(M,g)$ be a compact Riemannian manifold. It is a well known fact that in a neighbourhood $U$ of the zero ...
Dmitri's user avatar
  • 101
3 votes
2 answers
669 views

Real analytic submanifolds of $\mathbb{R}^{n}$

Hallo, Let $(M,g)$ be a Riemannian $k$-dim real analytic submanifold of $\mathbb{R}^{n}$. Is it true that $M$ in $\mathbb{R}^{n}$ looks locally (in a small neigbourhood around some point in $M$) as ...
hapchiu's user avatar
  • 339
2 votes
1 answer
283 views

Isometric embedding of a compact Lie Group in $M(n,\mathbb{C})$

Greetings, Let $G$ be a compact Lie group with a bi-invariant inner product $h$ on it. Can one embedd $G$ in $M(n,\mathbb{C})$ isometrically for some $n \in \mathbb{N}$. By isometrically I mean that ...
Dmitri's user avatar
  • 101
3 votes
1 answer
388 views

Isometric embedding of a real-analytic Riemannian manifold in a compact Kähler manifold

Hallo, It is a known fact that any real-analytic Riemannian manifold $M$ admits a isometric embedding in a Kähler manifold $\Omega$, where $M$ is totally real in $\Omega$. Of $\Omega$ can be taught ...
hapchiu's user avatar
  • 339
15 votes
0 answers
637 views

"Homogeneity" of the Hopf fibration $S^7\to S^{15}\to S^8$ [closed]

My question has to do with an apparent contradiction I get regarding the Hopf fibration $S^7\to S^{15}\to S^8$. Namely, the two following statements cannot be true at the same time (but I do not see ...
Renato G. Bettiol's user avatar
5 votes
1 answer
339 views

Are there countably many diffeomorphism classes of finite radius balls of complete Riemannian manifolds?

Suppose $M$ is a smooth complete Riemannian manifold and $x$ is a point in $M$. For any positive radius $r$ we consider the open ball $B(x,r)$ centered at $x$ with radius $r$. If we ignore the ...
Pablo Lessa's user avatar
  • 4,304
2 votes
0 answers
179 views

About Thom Theorem (representation submanifold for $H_{n-2}(M^n)$)

Recall Thom theorem : If $M^n$ is a smooth orientable closed manifold then any homology class in $H_{n-2}(M)$ is represented by the fundamental class of a smooth submanifold. And in the Harper and ...
Hee Kwon Lee's user avatar
  • 1,100
13 votes
2 answers
3k views

Intuition for Levi-Civita connection?

Levi-Civita connection is usually defined as the unique connection which is torsion free and preserves metric. Question Is there some intuitively transparent constructive way to define it (or ...
Alexander Chervov's user avatar
7 votes
1 answer
332 views

volume of exceptional group orbits

Assume that $G$ is a compact group acting by isometries on a (compact) Riemannian manifold (M,g), with principal orbits of dimension $d>0$. For $x\in M$, let $G(x)$ denote the $G$-orbit of $x$, by $...
Paolo Piccione's user avatar
0 votes
1 answer
339 views

Polarisation in a neighbourhood of a Lagrangian submanifold

Let $(X, \omega)$ be a symplectic manifold of dimension $2n$ and $\omega$ is an exact symplectic form i.e. $\omega = -d\alpha$. Let furthermore $M \subset X$ be a compact Lagrangian submanifold such ...
hapchiu's user avatar
  • 339
3 votes
3 answers
1k views

Rotation in Hyperkähler manifolds

Any Hyperkähler manifold has 3 complex structures $I_{1}, I_{2}, I_{3}$. Assume that there is an additional complex structure $J$. Can this be written as $J = aI_{1} + bI_{2} + cI_{3}$, where $(a,b,c) ...
hapchiu's user avatar
  • 339
4 votes
2 answers
575 views

Do transvers foliations induce complex structure?

Hallo, I have the following question: Let $M$ smooth analytic manifold of dimension 4n. Assume furthermore that $M$ admits two foliations $A$, $B$, both with leaves of dimension 2n such that the ...
Marin's user avatar
  • 41
3 votes
0 answers
585 views

Differentiation of Logarithm Map in Riemannian Geometry

I have a simple question regarding the differentiation of the logarithm mapping in Riemannian manifolds: Assume that $M$ is a compact Riemannian manifold, isometrically embedded into $\mathbb{R}^n$. ...
pil's user avatar
  • 233
14 votes
2 answers
964 views

A riemannian manifold with finitely many closed contractible geodesics

By a closed geodesic, I mean a smooth periodic geodesic $\mathbb{R} \rightarrow (M,g)$. I will consider them up to geometric distinction. This means that any two closed geodesics are equivalent if ...
Malte's user avatar
  • 847
6 votes
1 answer
1k views

Holonomy of a Kähler manifold

Hi, I have the following question: Let $(M,J, \omega)$ be a Kähler manifold (not necessary compact). We know that the holonomy group is a subgroup of $U_{n}$. Let $\Omega$ be a constant ($\nabla \...
Mina's user avatar
  • 93
7 votes
1 answer
502 views

Fundamental groups of compact manifolds with non-negative Ricci curvature.

I would like to find an appropriate reference for the following statement: Statement. Let $M$ be a compact Riemannian manifold with non-negative Ricci curvature. Then $\pi_1(M)$ is virtually abelian. ...
aglearner's user avatar
  • 14.3k
4 votes
1 answer
1k views

Curvature as metric invariant

This is quite well-known: the ONLY metric invariants are curvature, its higher derivatives, and any possible contractions between them. The meaning of an invariant is, to put it simply, a tensor ...
Caramba's user avatar
  • 133
7 votes
1 answer
496 views

Open problems about CMC hypersurfaces with symmetries?

Recently, Andrews and Li announced a complete classification of CMC ($H=const.$) tori in $S^3$, confirming a conjecture of Pinkall and Sterling. Their main result is that any such torus is ...
Renato G. Bettiol's user avatar
8 votes
0 answers
422 views

Exhaustion of an open manifold of bounded curvature and finite volume

In the Cheeger-Gromov paper "On the Characteristic Numbers of Complete Manifolds of Bounded Curvature and Finite Volume", http://www.maths.ed.ac.uk/~aar/papers/cheegergr1.pdf, the authors make the ...
J. GE's user avatar
  • 2,623
3 votes
1 answer
745 views

Star-shaped domain in a space form

Let $M$ be either $\mathbb R^n$, $\mathbb H^n$ or $\mathbb S^n$ and $p\in M$, by a star-shaped domain w.r.t $p$ I mean a connected open subset $\Omega$ in $M$ containing $p$ such that its boundary is ...
Kwong's user avatar
  • 103
13 votes
3 answers
851 views

Are negatively pinched manifold locally conformally flat?

One knows that hyperbolic manifolds are locally conformally flat. How about those negatively pinched manifolds, i.e. the sectional curvature $K$ satisfy: $$ -\Lambda \le K \le -\lambda$$ for $\Lambda&...
J. GE's user avatar
  • 2,623
9 votes
2 answers
2k views

$J$-holomorphic curve as a minimal surface

The following is a part of the proof of Gromov nonsqueezing theorem. The existence of a $J$-holomorphic curve gives an upper bound for the radius of a symplectically embedded ball. Let $\psi: B(r) \...
Hwang's user avatar
  • 1,398
8 votes
2 answers
2k views

Almost constant bump function

I ran into the following situation and it turned out to be more subtle than it looked. I have a complete Riemannian manifold $M$ and my objective is to construct a sequence of functions $f:M \to [0,1]...
Pablo Lessa's user avatar
  • 4,304
2 votes
1 answer
807 views

a result of soul theorem,right?

$X$ is an $n$-dim positively curved manifold and $Y$ is a totally geodesic submanifold of codimension 1. Then cutting along $Y$ we get $n$-dim positively curved manifolds without boundary, by soul ...
jiangsaiyin's user avatar
3 votes
1 answer
5k views

Relationship between sectional curvature, bisectional curvature and conjugate points

Tian has defined bisectional curvature for unit and perpendicular tangent vectors $X,Y$ as follow $$R(X,Y,X,Y)+R(X,JY,X,JY).$$ If bisectional curvature be constant, is there any relationship between ...
Reza's user avatar
  • 105
7 votes
3 answers
736 views

Nearly constant curvature implies "nearly isometric" to a space form?

It is well known a Riemannian manifold with constant sectional curvature is a quotient of the Euclidean space, hyperbolic space or sphere. In particular we know how their metric looks like locally. ...
Kwong's user avatar
  • 103
2 votes
1 answer
484 views

Sobolev norm of distance function on Riemannian manifold

$\DeclareMathOperator\SL{SL}$Suppose $M$ is a Riemannian manifold with distance function $d:M\times M \rightarrow [0,\infty)$. If it helps let $M$ be a Lie group with finite Haar measure $\mu$ and ...
Florek's user avatar
  • 23
1 vote
1 answer
423 views

Extension of groups in Bieberbach's theorem

I am reading de la Harpe's book "Topics in Geometric Group Theory". On page 145, there is a theorem: Let $V$ be a complete $n$-Riemannian manifold with sectional curvature satisfying $K\ge 0$. Then ...
J. GE's user avatar
  • 1,101
2 votes
1 answer
425 views

holomorphic extension of forms

hallo, I have the following question: Let $M$ be a $n-$dimensional complex manifold and $X \subset M$ be a compact $n-$dimensional totally real analytic Riemannian submanifold. Let furthermore $\...
bruno's user avatar
  • 29
9 votes
2 answers
638 views

Curvature of the Cayley projective plane

The Cayley projective plane can be realized as the compact homogeneous space $F_4/\mathrm{Spin}(9)$. In this way one can compute the curvature of this symmetric space in terms of a suitable ...
atreyee's user avatar
  • 403
19 votes
1 answer
2k views

Does this Banach manifold admit a Riemannian metric?

First, the question; after, the motivation. Consider 27.6 (pdf pp. 262-263) in The convenient setting of global analysis (AMS, 1997), and, in particular, the example given at the end of it, which ...
Benjamin Dickman's user avatar
4 votes
1 answer
201 views

estimate over simply-connected Riemannian manifold with non-positive sectional curvature

Let $M$ be a Complete simply-connected $n$-dimensional Riemannian manifold with non-positive curvature,$\Omega $ is a open subset of $M$ ,If $n\geq 4$ Is anyone give a estimate of $\frac{Vol_{n}\...
喻yuwei's user avatar
  • 119
9 votes
2 answers
367 views

Is compact flat manifold cusp cross-sections of a complete finite volume hyperbolic manifold?

Let $M^{n-1}$ be a closed flat manifold. Is it true that there exists a hyperbolic manifold $N^n$ with finite volume such that $M$ is a cusp cross-section of $N$? It was proved in "On the geometric ...
J. GE's user avatar
  • 2,623
7 votes
1 answer
554 views

Minimal distance spheres in complex projective spaces

My question has to do with distance spheres in $\mathbb CP^{n+1}$. I am interested in knowing what is the radius $r$ of a distance sphere $S(r)$ around a point that makes it a minimal submanifold of $\...
Renato G. Bettiol's user avatar
4 votes
1 answer
2k views

Totally geodesic submanifold of round sphere

Let $S^n$ be the $n$-dimensional round sphere (i.e. with Riemannian metric of constant curvature +1). Is there any classification result of totally geodesic embedded submanifolds? Are they all round ...
J. GE's user avatar
  • 2,623
0 votes
1 answer
314 views

G-structures and complete riemannian manifolds

what are possible fundamental and introductory texts about G-structures ? and where i can find the proof of this proposition: if G(group) acts properly discontinuously on a space X , then G is a ...
DAVID's user avatar
  • 165
9 votes
1 answer
2k views

Isoperimetry and Poincaré Inequality

What are the known relations between isoperimetric and Poincaré inequalities on manifolds? For example, for manifolds with a lower bound on Ricci curvature, the Cheeger-Buser inequality relates the ...
ThiKu's user avatar
  • 10.4k
10 votes
1 answer
820 views

A strange question about closed geodesics on a closed manifold

I'm studying a particular kind of curve evolution on Riemannian manifolds. It would help me to know the answer to the following kinda weird question: Does there exist a closed Riemannian manifold $M$ ...
Dmitri Gekhtman's user avatar
2 votes
1 answer
551 views

Heisenberg group: research themes

I am currently studying the Heisenberg group from the Riemannian geometry point of view, particularly focusing on its Gromov boundary and more generally its metric properties. I would like to know ...
Maria's user avatar
  • 31
3 votes
2 answers
845 views

How to solve the $C^\alpha$ Poisson equation on closed Riemannian manifolds?

To be specific, suppose $M$ is a closed oriented manifold, $g$ is a Riemannian metric of $M$. Let $\Delta_g$ be the Laplace-Beltrami operator w.r.t. $g$. Prove: Suppose $f\in C^\alpha(M)$ satisfies ...
Yuchen Liu's user avatar
  • 1,083
5 votes
2 answers
1k views

Kähler potentials that depend only on geodesic distance

Hermitian symmetric spaces of constant curvature have the property that the potential for their Kähler metric can be expresed as some function of the geodesic distance. Does anyone know if there are ...
Oliver Jones's user avatar
  • 1,378

1
34 35
36
37 38
40