The tag has no wiki summary.

learn more… | top users | synonyms

3
votes
2answers
48 views

Algebraic surface as a smooth manifold

Let $S$ be the set of points $x=(x_1,x_2,\ldots,x_9)\in \mathbb{R}^9$ which satisfy the following conditions: $$x_1^{2}+x_2^{2}+x_3^{2}=x_4^{2}+x_5^{2}+x_6^{2}=x_7^{2}+x_8^{2}+x_9^{2}=1$$ ...
1
vote
0answers
22 views

Standards in P.L. Topology

About a week ago, the reading course on PL topology I'm going to follow started. The aim of the reading course is to understand the basics of PL topology and have a reasonable to good understanding of ...
0
votes
0answers
10 views

fibered solid tori matched by a fiber preserving homeomorphism

how do I proof that two seifert fibered solid tori $V$ and $V'$ (not ordinary fibered) with the same fiber parameters matched together by a fiber preserving homeomorphism do not become a solid tori. ...
3
votes
1answer
59 views

Approach topological manifolds with smooth manifolds

Because I'm doing some problems that consider all the manifolds while the situation is really clear when considering only smooth manifolds. Thus my question is can we always appoint a topological ...
1
vote
1answer
38 views

Is $VV^T + D$ a submanifold?

If the positive definite matrix P forms a manifold, is that the subset that {P: P = V V^T + D} where V is a low rank matrix and D is a positive definite matrix a sub-manifold? This idea is ...
7
votes
2answers
90 views

Dimension of de Rham Cohomology groups?

Is there a simple way to prove that the de Rham cohomology groups of a compact manifold $M$ have finite dimension as $\mathbb{R}$-vector spaces?
1
vote
1answer
24 views

set of immersions on a manifold is an open set in the set of all mappings to $\mathbb{R}^{2m+1}$

I'm studying a proof of Takens' Delay Embedding Theorem. A key fact used is that the set of immersions is open in the set of all mappings (mappings from an $m$-dimensional manifold M to ...
3
votes
1answer
53 views

Prove that there are no convex functions on compact manifolds

This one seems intuitively obvious to me but I don't know how to prove it. Suppose you have a compact manifold $M$ with a function $f$ defined on it. Given two points $x$ and $y$ on the manifold, ...
1
vote
1answer
48 views

Invariant form on Lie algebra

Does anyone have a reference for the following fact? Let $G \subset GL(n)$ be a compact Lie group. Then the form $$f(A)=-Tr(A^2)$$ defined for $A \in T_e G \subset \mathfrak{gl}(n)$ is positive ...
2
votes
1answer
92 views

a theorem in topology

Is there anyone know there is a theorem in topology which states that a compact manifold "parallelizable" with N smooth independent vector fields. must be an N-torus? and why the vector field here is ...
4
votes
1answer
82 views

Schrödinger Kernels on manifolds

Let $M$ be a compact Riemannian manifold and $\Delta$ be the Laplace-Beltrami operator. It is well-known that the solution operator to the heat equation $e^{t \Delta}$ is smoothing for $t>0$ and ...
2
votes
1answer
95 views

Every compact orientable surface with $S^1\times\{0\}$ as its boundary intersects the $z$-axis

Let $M$ be a compact orientable surface (manifold in $\mathbb R^3$) with boundary $S^1\times\{0\}$.Show that $M$ intersects the $z$-axis. Some ideas: $1)$Since $M$ is a compact orientable manifold ...
5
votes
1answer
222 views

Uniformization Theorem for compact surface

Why in proof of proposition 6 of http://arxiv.org/abs/0909.1665, they claim that if a embedded surfaces $\Sigma^2 \subset (M^3,g)$ is homeomorphic to $\mathbb{RP}^2$, where $M$ is compact manifold, ...
0
votes
0answers
76 views

closed surface from a fundamental polygon.

In wikipedia about surface: Closed surfaces can be constructed from an oriented polygon with an even number of sides, called a fundamental polygon of the surface, by pairwise identification of its ...
1
vote
0answers
48 views

Geodesic On Compact Manifolds

Let $M$ be a compact Riemmanian manifold. Let $G$ denote the set of all geodesics of $M$. If $\gamma\in G$ let $l(\gamma)$ denote its length. Let $$S=\sup\{l(\gamma): \gamma\in G\}$$ Suppose ...