For questions regarding the structure and properties of compact manifolds.

Compact manifolds are manifolds which are compact as topological spaces.

For questions regarding the structure and properties of compact manifolds.

Compact manifolds are manifolds which are compact as topological spaces.

235 questions

votes

I am having some trouble understanding Takens' embedding theorem, and was hoping that someone with greater knowledge could help out.
Formally, the theorem goes as follows:
Let $M$ be a compact manifold of dimension $m$. For pairs $(\phi,y)$, where…

Astrid

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Does there exist a compact submanifold of $\mathbb{R}^3$ whose fundamental group is $\mathbb{Z}^3$ ?
The question in the title is a generalization of the question that really interests me:
Does there exist a connected finite set of unit cubes of…

Aivazian Arshak

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What are some examples of manifolds that do not have boundaries and are not boundaries of higher dimensional manifolds?
Is any $n$-dimensional closed manifold a boundary of some $(n+1)$-dimensional manifold?

Xiaoyi Jing

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Let $M$ be a compact, connected, oriented $n$-dimensional manifold without boundary. Suppose that $M\#M\cong M$. Does it imply that $M \cong S^n$?
Sorry if this is a naive question. This is not my area, and I have very few examples of higher…

Bruno Joyal

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Consider the following problem:
Suppose $(M, g_{ij})$ is a compact Riemannian manifold. Assume $u$ is a smooth, nonnegative function which satisfies the differential inequality
$$\Delta u \geq -cu$$
where $c$ is a constant.
Show…

user100463

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I just started to learn index theory of tangent vector fields. I'm aware of two examples on the sphere $\mathbb{S}^{2}$ with exactly one zero, which, which are $F(x,y) = (1-x^2-y^2)\partial x$ thought on $\mathbb{D}^2$ and then indentify the…

jacopoburelli

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Can someone please remind me how this goes?
Here's the idea of proof I'm trying to recall: let $S$ be a closed surface (connected, compact, without boundary) embedded in $\mathbb{R}^3$. Then one can define the "outward-pointing normal unit vector"…

Seub

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This was a past exam question: Let $M$ be a compact connected orientable topological $n$-manifold with boundary $\partial M$ so that $H_*(\partial M;\mathbb{Q}) \cong H_*(S^{n-1};\mathbb{Q})$. If $n \equiv 2$ mod $4$, show that the Euler…

xy15

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Let $M$ be a compact, connected $n$-manifold. Consider the homology groups $H_n(M)$ with coefficients in $\mathbb{Z}$.
It is well known that if $M$ is not $\mathbb{Z}$-orientable, then we have $H_n(M) =0$ and $H_{n-1}(M) = \mathbb{Z}/2 \oplus…

KarlPeter

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Let G be a finitely generated abelian group and M a compact manifold, I want to prove that $H_r(M,G)$ is finitely generated for $r\ge 0 $.
First I was thinking if I could do induction over $r$ because for $r=0$, $H_0(M,G) \cong \bigoplus_{i=1}^nG$…

jandrew

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Reading some books on diferential geometry, a found that $S^{2n}$ (with $ n>1$) are not symplectic manifolds. They say it's because the de Rham cohomology of this spheres are R, but I do not understand this argument. It will be helpfull if anyone…

Edgar Ortiz

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Perhaps this is obvious and I am overlooking something, but why are the homology groups of compact manifolds finitely generated?

Holdsworth88

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I want to find a differentiable $n$-dimensional compact manifold $M$ which can be endowed with an affine structure but cannot be endowed with a euclidean structure.
An affine (resp. euclidean) structure is a geometric structure with $X=\Bbb R^n$…

Adam Chalumeau

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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?

Heitor Fontana

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I am trying to classify all compact 1-manifolds. I believe I can do it once I can show every 1-manifold is orientable. I have tried to show prove this a bunch of ways, but I can't get anywhere.
Please help,
Note, I am NOT assuming that I already…

Bates

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