10
votes
1answer
102 views

How can I understand the three-dimensional space forms?

Here is what I know: A space form is defined as a manifold admitting a Riemannian manifold of constant sectional curvature A classical result of Cartan states that a manifold is a space form if and ...
2
votes
2answers
80 views

Are these two 2-manifolds homeomorphic?

I have a 2-Sphere with a finite number $k$ of points removed (at least 3), and I want to equip it with a Riemannian metric of constant negative curvature. My first thought was to take a free ...
2
votes
1answer
38 views

Cohomology of volume forms

If g and h are Riemannian metrics on the same manifold, say both of volume 1, then it follows (I guess from Poincaré duality) that their volume forms dvol_g and dvol_h are cohomologous. Question: is ...
1
vote
1answer
65 views

Lifting problems existence

Let $g:\mathbb{R}^m \longrightarrow \mathbb{R}^m,g\in C^1(\mathbb{R}^m)$ such that: $\|g'(x)(v)\|\geq\|v\|,\forall v\in \mathbb{R}^m,\forall x \in \mathbb{R}^m$ show that any rectilinear path ...
5
votes
0answers
57 views

Automorphism on a kahlerian variety which acts as -id on $H^2(X,\mathbb{Z})$

I've read this proposition: "there is no automorphism (biholomorphic map) of a Kahlerian complex manifold $X$ which acts on $H^2(X,\mathbb{Z})$ as $-$identity" so I tried to prove this statement and ...
2
votes
0answers
88 views

Torsion of fundamental group is abelian

On Riemannian manifold with Ricci curvature bounded below (For example, flat torus. It has ${\bf Z}^2$ as fundamental group, which is nilpotent (=almost abelian). In fact Ricci curvature bouned ...
3
votes
1answer
112 views

Riemannian metric for surface of negative Euler characteristic

So, I want to equip a surface of negative Euler characteristic with a Riemannian metric of negative curvature. I know from the uniformization theorem, that a metric of constant curvature exists ...
3
votes
1answer
169 views

Existence of minimizing geodesic in each fixed-end-point homotopy class in a complete manifold?

This is intuitively clear, but I cannot solve this homework problem: 1) Let $(M,g)$ be a complete Riemannian manifold, let $c:[0,1]\to M$ be a continuous curve in $M$ such that $c(0)=p, c(1)=q$. Then ...
11
votes
1answer
247 views

Dolbeault Cohomology is invariant under homeomorphisms

If $X$ and $Y$ are two complex manifolds, which are homeomorphic but not necessarily diffeomorphic, must their Dolbeault cohomology groups be isomorphic? Here the Dolbeault cohomology groups ...
4
votes
0answers
118 views

Transforming the Dirac Operator on $S^1$

My goal is to understand as much as I can about the Dirac operator on $S^1$ where we give $S^1$ the spin structure given by the connected double cover of the frame bundle. The spinor bundle on $S^1$ ...
6
votes
1answer
265 views

Dirac Operators on $S^1$

I am trying to understand the Dirac operators associated to the 2 spinor bundles on $S^1.$ I have been getting very confused about why one bundle has nontrivial harmonic spinors and the other ...
2
votes
0answers
52 views

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

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 ...
2
votes
1answer
46 views

systole of space projective 3-dimensional

I'm study the papper "H. Bray, S. Brendle, M. Eichmair, and A. Neves, Area-minimizing projective planes in three-manifolds, Comm. Pure Appl. Math". (see http://arxiv.org/abs/0909.1665). Let be ...
5
votes
1answer
195 views

Parabolic elements correspond to punctures

In Mapping Class Group by Farb and Margalit page 22, they say: Let $S$ be a hyperbolic surface. If a non-trivial element of $\pi_1(S)$ is represented by a loop (up to homotopy) around a puncture, ...
1
vote
0answers
81 views

Manifold contains a totally geodesic closed hypersurface

Let $(M^n,g)$ be a closed simply-connected positively curved manifold. Show that if $M$ contains a totally geodesic closed hypersurface (i.e., the second fndamental form or shape operator is zero), ...
5
votes
1answer
122 views

Things related to the Preissman Theorem

I'm reading the proof of the Preissman Theorem, in Do Carmo's book of Riemannian Geometry. A crucial step in this demonstration is the following lema, Lema: Let $M$ be a compact riemannian ...
0
votes
1answer
66 views

Commutator map and the derived series

Let be $G$ a solvable group, let $$ G=G_0\supset G_1\supset\cdots\supset G_k=1$$ be the derived series for $G.$ Is clear that $G_ {k-1}$ is abelian. Now take $b\in G_{k-1}$ e $a\in G_{k-2}$ my ...
8
votes
2answers
228 views

Cartan Theorem.

Cartan Theorem: Let $M$ be a compact riemannian manifold. Let $\pi_1(M)$ be the set of all the classes of free homotopy of $M.$ Then in each non trival class there is a closed geodesic. (i.e a closed ...
3
votes
1answer
108 views

Question about immersions of $\mathbb{R}P^n$ into $\mathbb{R}^{n+1}$

I am currently reading a paper which takes for granted the following geometric fact: if $\mathbb{R}P^n$ can be immersed in $\mathbb{R}^{n+1}$ then for some $k$, $n=2^k-1$ or $n=2^k-2$. My initial ...
3
votes
1answer
341 views

Determining the “positivity” or “negativity” of Chern class (number?) of zero-sets of homogeneous polynomials

If $\Omega$ is the curvature 2-form on a $n-$manifold, then I would think that the Chern classes (forms), $c_k$ are defined as, $det(I + \frac{it\Omega}{2\pi}) = \sum c_k t^k$ I would like to ...