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Questions tagged [compact-manifolds]

For questions regarding the structure and properties of compact manifolds.

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Understanding of the term “compact support” in the proof of stokes theorem.

I am trying to understand the proof of Stokes theorem: $$ \int_M df = \int_{\partial M} f $$ for a differentiable Manifold $M$ with dimension $n$ and a differential $(n-1)$-form with compact support ...
Zaph's user avatar
  • 101
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23 views

Is there something like a compact surface classificator?

I'm currently studying compact surfaces and there are some exercises as the following: find a simple scheme equivalent to $abc, da^{-1}b,cef,e^{-1}f^{-1}d$ and classify the surface. After computations ...
Valere's user avatar
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1 vote
0 answers
30 views

Checking an error in a Differential Geometry problem on curvature and diffeomorphism types of compact surfaces

I am currently working through the following old exam problem. I have reached part (d)(i) but I believe the result required still is not guaranteed even with the new hypotheses. For example, if we ...
John Robertson's user avatar
3 votes
1 answer
94 views

Is it really true that $\mathcal{S}(\mathbb{R}^n)$ is identified with smooth functions on $S^n$ vanishing at a fixed point?

Let $\mathcal{S}(\mathbb{R}^n)$ be the Schwartz space on $\mathbb{R}^n$ and $C^\infty(S^n)$ be the space of smooth functions on $n$-sphere. Now fix a point $x \in S^n$ and define \begin{equation} C^\...
Keith's user avatar
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36 views

A regular, connected, compact surface with curvature on $[0,1]$

today was my final differential geometry exam and there was a problem that I partially solved, but I have some doubts. The problem asked to prove that there exists a regular, connected, compact ...
Nestor Bravo's user avatar
8 votes
1 answer
238 views

Does every finitely presented group have a finite index subgroup with free abelianisation?

Let $G$ be a finitely presented group. Does there exist a finite index subgroup $H$ such that its abelianisation $H^{\text{ab}} = H/[H, H]$ is free abelian? Note, if $G^{\text{ab}}$ is not already ...
Michael Albanese's user avatar
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0 answers
65 views

$Ric$ flat metric on $S^n$.

Suppose $\mathbb{R}\times S^{n}$ admits a complete riemannian metric $g$ such that $Ric_g = 0$. Prove that this metric $g$ induces a metric $\tilde g$ on $S^{n}$ such that $Rig_{\tilde g} = 0$. So far,...
M. Rubick's user avatar
2 votes
0 answers
38 views

Handles have the form $D^λ×D^{m−λ}$

I'm studying Matsumoto's An Introduction to Morse Theory. I want to solve a problem on page 76. Context: Let $M$ be a closed $m-$manifold and $f:M\rightarrow \mathbb{R}$ a Morse function. Let $c$ be a ...
3435's user avatar
  • 385
2 votes
0 answers
76 views

U(n) is compact and algebraic, but not abelian—why not a contradiction?

the subgroup of unitary matrices $\text{U}(n) \subset GL(n, \mathbb{C})$ is compact and definitely algebraic, with an algebraic group law; on the other hand, it's not abelian. why is this not a ...
BD107's user avatar
  • 611
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Lie subgroup of non-abelian compact Lie group is compact?

I need to decide if this statement is true or false: " Every Lie subgroup of non-abelian compact Lie group is compact." I think that it is false. I thought in a counterexample in which the ...
Kauê Orlando Pereira's user avatar
3 votes
1 answer
287 views

Prove that every compact manifold is homeomorphic to a subset of some Euclidean space.

I am trying to prove the following theorem: Theorem. Every compact manifold is homeomorphic to a subset of some Euclidean space. The manifolds I'm considering are the most general (without any ...
math-physicist's user avatar
2 votes
1 answer
98 views

Spectral triple on $\mathcal C(M)$ where $M$ is a compact Riemannian manifold, not necessarily spin

I have been reading Alain Connes' Compact metric spaces, Fredholm modules and hyperfiniteness. In proposition 1, it is mentioned that an unbounded Fredholm module (nowadays: spectral triple) over $C(M)...
ViktorStein's user avatar
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1 answer
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Holomorphic forms are closed on compact manifold $X$ if $\dim(X)=2$.

Let $X$ be a compact complex manifold and $\dim(X)=2$, $\eta$ is a holomorphic form on $X$. Prove that d$\eta=0$. I know when $X$ is a compact complex Kähler manifold, holomorphic forms are closed. In ...
save123's user avatar
  • 319
2 votes
1 answer
148 views

Choice in the non-smooth Whitney Embedding Theorem

Introduction In Munkres' Topology, he presents a precursor of what seems to be called the non-smooth Whitney Embedding Theorem in Section 50: Theorem 50.5 (The imbedding theorem). Every compact ...
Nick F's user avatar
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46 views

Motivation for the definition of a compact Riemann manifold with piecewise smooth boundary

I am asking for either motivation on the requirement regarding $f_i^{-1}(0)$ in the following definition, or better yet a reference to a book dealing with this subject. The following is the definition ...
Epsilon Away's user avatar
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153 views

Simplicial Homology Groups of Circle Wedge a Torus

Compute the simplicial homology groups of $S^1 \vee (S^1 \times S^1)$ in all dimensions. I'm trying to practice simplicial homology, and want to make sure I understand at a technical level what's ...
pyridoxal_trigeminus's user avatar
1 vote
1 answer
197 views

Fréchet derivative of the total variation norm for measures on a manifold

Let $\Theta$ be a compact $d$-dimensional Riemannian manifold without boundary and $M(\Theta)$ (resp. $M_+(\Theta)$) denote the set of signed (resp. nonnegative) finite Borel measures on $\Theta$. ...
ViktorStein's user avatar
  • 4,878
0 votes
1 answer
409 views

$S^1\times S^2$ embedded in $\mathbb{R}^4$ [duplicate]

It's easy to embed $S^1\times S^2$ in $\mathbb{R}^5$, since $S^1\subset \mathbb{R}^2$ and $S^2\subset\mathbb{R}^3$, but $S^1\times S^2$ lives also in $\mathbb{R}^4$. How can we write the embedding map ...
FreeFunctor's user avatar
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0 answers
329 views

Classify surface given by $abca^{-1}b^{-1}c^{-1}$

I'ven been solving problems from my Topology course, and don't understand something I saw while reading my solved examples. Here's a problem that will let me show my point: Given $X$ a compact ...
Alejandro Bergasa Alonso's user avatar
1 vote
1 answer
53 views

Proof that a compact surface has a tangent plane orthogonal to position

Given a compact surface $S$, is it true that $S$ has a tangent plane that is orthogonal to the position vector for at least one of its points? I believe that the statement is true, but I'm having ...
Joao Lima's user avatar
2 votes
1 answer
406 views

Subharmonic Functions on Closed Manifolds are Constant?

Apologies in advance if this question is trivial! Let M be a closed, connected, oriented, n-dimensional, manifold without boundary. Let $f$ be a smooth function on M where $\Delta f \geq 0$, where $\...
chiefdelta's user avatar
2 votes
1 answer
206 views

Regularity of higher order elliptic problem on compact smooth manifolds with boundary

I have trouble in finding a source in the literature for the following result: Let $\overline{M}$ be a compact smooth manifold of dimension $n \in \mathbb{N}$ with interior $M$ and non-empty boundary $...
Matthis Stresemann's user avatar
2 votes
0 answers
123 views

How to deal with boundary orientation in Exercise 16-5 from Lee's Introduction to Smooth Manifolds

16-5. Suppose $M$ and $N$ are oriented, compact, connected, smooth manifolds, and $F,G:M\to N$ are homotopic diffeomorphisms. Show that $F$ and $G$ are either both orientation-preserving or both ...
klein4's user avatar
  • 1,277
2 votes
0 answers
113 views

Extending an embedding with trivial normal bundle

Let $M^{m}$ and $N^{n}$ be $C^{\infty}$ compact manifold with boundary (eventually $\emptyset$) and let $j:M \to N$ be an embedding such that the normal bundle of the embedding $\nu(j)$ is trivial. ...
leobgg's user avatar
  • 193
18 votes
1 answer
580 views

Does there exist a compact submanifold of $\mathbb{R}^3$ whose fundamental group is $\mathbb{Z}^3$?

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 ...
Arshak Aivazian's user avatar
1 vote
1 answer
73 views

Can a connected closed manifold strictly contain a closed manifold of the same dimension?

A connected closed manifold can contain another one as a proper subset: for instance, the $1$-sphere (circle) is contained in the $2$-sphere. Is it possible with manifolds of the same dimension? ...
mathieu's user avatar
  • 125
3 votes
1 answer
84 views

Irreducible triangulations of manifolds

Does there exist a closed Riemann manifold $M$, two distinct irreducible triangulations $S_1$ and $S_2$ of $M$, and a triangulation $T$ of $M$ such that there exists a sequence of edge contractions on ...
Null Simplex's user avatar
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1 answer
288 views

Graph embeddings on nonorientable surfaces

If a finite graph $G$ can be embedded on an orientable surface of genus $n$, does this mean that it can be embedded on a nonorientable surface of genus $n$? Is the converse of this statement true?
Null Simplex's user avatar
2 votes
1 answer
119 views

Prove that if $f:M\rightarrow\Bbb R$ is a scalar function over a 1-manifold M without boundary then $\int_M df=0$

Well James Munkres in the text Analysis on Manifolds prove the general Stoke's theorem for $k$-form when $k>1$ and then he proves it for $k=1$ only when the bounary of the Manifold is not empty and ...
Antonio Maria Di Mauro's user avatar
2 votes
2 answers
624 views

Manifolds with Euler characteristic equal to $\pm 1$

A compact connected smooth surface has Euler characteristic equal to $\pm 1$ if and only it is homeomorphic to the real projective plane or the connected sum of $3$ real projective planes. What are ...
Eduardo Longa's user avatar
4 votes
1 answer
217 views

Can dimension of manifolds be understood similarly to dimension of schemes?

I’m only beginning to learn about schemes, but I know that at least in some cases, the dimension of a scheme (or variety) is 1 less than the length of the longest chain of irreducible closed subsets. ...
Jon H's user avatar
  • 187
1 vote
0 answers
63 views

Smooth identification of the complement of a disk

I was wondering, to costruct a map from a compact smooth manifold $M$ of dimesion $n$ to the sphere $\mathbb{S}^{n}$ of degree $1$, apparently, the most common idea is to wrap a disk $D$, neighborhood ...
jacopoburelli's user avatar
0 votes
0 answers
90 views

Intuitive explanation of degree

I do understand the for $f : \mathbb{S}^{1} \longmapsto \mathbb{S}^{1}$ the degree of a map (thinking $f$ as a closed curve $\gamma$ defined on $[0,1]$) can be seen as "how many times a closed ...
jacopoburelli's user avatar
2 votes
0 answers
370 views

degree of gauss map for genus $g$ torus in $\mathbb{R}^{3}$

I read that the degree of the gauss map for a $M$ compact orientable $2-$manifold (connected to use the fact that those are only the $g-$torus) in $\mathbb{R}^{3}$ should be $(1-g)$, which is $\frac{1}...
jacopoburelli's user avatar
13 votes
5 answers
991 views

Could exists a vector field on $\mathbb{S}^{2}$ with exactly $n$ zeroes?

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 ...
jacopoburelli's user avatar
4 votes
1 answer
410 views

Compact hyperbolic three-manifold - a question

In a recent paper the authors considered a spacetime described by $$AdS_4 \times \Sigma_3 \times \mathcal{I}_r \times S^2$$ where $\mathcal{I}_r$ is an interval of the $r$-coordinate and the two-...
user avatar
3 votes
2 answers
545 views

Homotopy invariance for compactly supported cohomology

I can't find any reference regarding homotopy invariance for compactly supported cohomology and I wonder under which conditions the homotopy invariance still holds for compactly supported cohomology. ...
Phi_24's user avatar
  • 1,114
2 votes
1 answer
159 views

Uniformization theorem for $C^k$ surfaces?

Does the uniformization theorem apply for surfaces that are $C^k$ ($k<\infty$)? I'm familiar with a couple of proofs of Uniformization (using Riemann-Roch, Ricci flow). But most of these proofs ...
sobol's user avatar
  • 622
1 vote
0 answers
39 views

Is there an example of reducible compact 3-manifold with boundary that has no embedded incompressible two-sided surface?

There is a theorem stating that for irreducible compact manifolds with non-empty boundary there always exists such an embedded surface and I'm trying to understand why the irreducibility condition ...
Haldot's user avatar
  • 830
2 votes
1 answer
60 views

Making intuition rigorous that integral of some positive function on set should be monotone in the Haar measure of the set

Let $\mathcal{M}$ be a compact Riemannian manifold with geodesic distance function $d$ and $\Omega$ its volume measure. Pick some $A,B\subseteq\mathcal{M}$ such that $\Omega(A)\ll\Omega(B)$, but: (1) $...
PPR's user avatar
  • 1,116
5 votes
1 answer
133 views

Covering $\Bbb RP^\text{odd}\longrightarrow X$, what can be said about $X$?

I am looking for any argument related to the following fact, which may or may not be true. Let $f:\Bbb RP^n\longrightarrow X$ be a covering space, where $n\geq 2$. Then, $X=\Bbb RP^n$. Now, for $n=\...
Sumanta's user avatar
  • 9,634
1 vote
1 answer
248 views

Compactification of log z Riemann Surface

I've been reading the 'Road to Reality' book of Roger Penrose and in the chapter on Riemann Surfaces, there is a note that we can compactify the log z Riemann Surface into a sphere. But I don't see ...
karkunow's user avatar
4 votes
2 answers
380 views

What is the "natural homomorphism" in the definition of an *essential manifold*?

The following definition of "essential manifold" is in this wiki page: A closed $n$-manifold $M$ is called essential if its fundamental class $[M]$ defines a nonzero element in the homology ...
Eduardo Longa's user avatar
1 vote
1 answer
340 views

Simplicial complexes embedded on a compact manifold

Every finite graph can be embedded on some compact surface of sufficient genus such that no two edges cross. If $S$ is a finite simplicial complex of dimension $n$, can $S$ be embedded in some ...
Null Simplex's user avatar
0 votes
1 answer
172 views

Can all connected graphs be embedded on a closed, compact 2-Manifold?

I know that there are spherical (planar) graphs such as $K_4$,and toroidal graphs such as $k_7$, but I was wondering if given any connected graph $G$, there exists a closed, compact 2-Manifold $M$ ...
Null Simplex's user avatar
4 votes
0 answers
488 views

An orientable surface that cannot be embedded into $\Bbb R^3$?

By Whitney's embedding theorem, every 2-dimensional manifold (aka. a surface) can be embedded into $\Bbb R^4$. Now, the Wikipedia page on that theorem states in this paragraph that we can even embedd ...
M. Rumpy's user avatar
  • 995
0 votes
0 answers
122 views

intersection number on the boundary of a manifold

Let $F: W \to N$ be a smooth map, where $W$ is a compact manifold with boundary, $Z \subset N$ is closed and all manifolds are oriented. Also $\partial F \pitchfork Z$ and $F^{-1}(Z)$ is a compact, ...
Dlmn's user avatar
  • 97
1 vote
1 answer
925 views

The Sobolev embedding inequality on manifolds

Let $(M,g)$ be a (smooth) compact Riemanian manifold of dimension $n$. I expect that the following inequality is true for any smooth function $f$: $$(\int_{M} |f|^{\beta})^{1/\beta} \leq C \;(\int_{M}...
Hana's user avatar
  • 79
2 votes
0 answers
175 views

square root of a Riemannian metric

Let $(M,g)$ be a compact Riemannian manifold. Taking a smooth vector field $X$, there is an associated smooth function $$g(X,X):M\longrightarrow \mathbb{R}^+, \quad p\longmapsto g_p(X_p,X_p)\geq 0.$$ ...
AmSa's user avatar
  • 139
0 votes
0 answers
171 views

Proof of classification theorem for compact surfaces

I am reading Massey's 'A basic coruse in Algebraic Topology'. In first chapter, he proved classification theorem for compact surfaces (compact connected 2-manifold). This theorem classifies compact ...
ZWJ's user avatar
  • 427

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