53 votes
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Are diffeomorphic smooth manifolds truly equivalent?

The counterexample just shows that two diffeomorphic smooth structures on the same set $X$ do not need to share a common atlas. However, in any case, two diffeomorphic structures cannot be ...
Francesco Polizzi's user avatar
36 votes
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Can every manifold be turned into a Lie group?

There is an easy counterexample: $S^2$ cannot be given a Lie group structure (this is a consequence of the hairy ball theorem). The problem with your construction is that it doesn't offer how to ...
kamills's user avatar
  • 2,097
34 votes

Are diffeomorphic smooth manifolds truly equivalent?

This is nothing specific to differentiable manifolds. If you have some set $X$ and you define some structure on $X$ (e.g. group, vector space, topological space, differentiable manifold, ...) there ...
M. Winter's user avatar
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31 votes
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The integral of a function on manifold and differential form

Differential forms are not introduced to answer the question "How do I integrate functions on manifolds?" Instead, they are introduced to answer a different question, namely "What are ...
Lee Mosher's user avatar
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30 votes
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Understanding the orientable double cover

Almost $2$ years later, I'll give a complete answer to my own question. Step 1 (Topology of $\widetilde{M}$): Take an atlas $\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in\Lambda}$ such that $\{U_\alpha\}_{\...
rmdmc89's user avatar
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29 votes
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Number of Differentiable Structures on a Smooth Manifold

The distinction to be made is that a differentiable structure is a choice of maximal smooth atlas $\mathcal A$, but two different choices $\mathcal A$ and $\mathcal A'$ can lead to isomorphic smooth ...
Pedro's user avatar
  • 122k
28 votes

Can every manifold be turned into a Lie group?

Lie groups as manifolds, are very special, owing to the group operations. Basically, "what happens at the identity" determines what happens everywhere. And this means that the tangent bundle $T G$ is ...
Matematleta's user avatar
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26 votes
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Show that $SL(n, \mathbb{R})$ is a $(n^2 -1)$ smooth submanifold of $M(n,\mathbb{R})$

You want to consider the smooth map $\det\colon M_n(\mathbb{R})\to\mathbb{R}$. If you can show $1$ is a regular value of $\det$, then $SL(n,\mathbb{R})=\det^{-1}(1)$ is a smooth manifold of dimension $...
Ben West's user avatar
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25 votes
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What is the structure group of the tangent bundle?

Structure group: Given a fiber bundle with total space $E$, base space $M$, model fiber $F$ and projection $\pi$, the local trivialization condition says that given any $u\in E$ such that $u$ ...
Bence Racskó's user avatar
24 votes

Can every manifold be turned into a Lie group?

To add to the previous answers, topological groups have abelian fundamental groups. $G$ is Topological $\implies$ $\pi_1(G,e)$ is Abelian Orientable surfaces of genus at least two are not ...
Moisés's user avatar
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22 votes
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Manifold has uncountable many smooth stuctures if it has one

It took me a while to understand the great idea proposed by Anthony Carapetis since I think that other people may have the same doubt that I had, I decided to write a more detailed answer using his ...
Matheus Manzatto's user avatar
21 votes
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Origins of Differential Geometry and the Notion of Manifold

[2016-07-25]: Section Differential Geometry added. Although OP narrowed down the post, there are still many more important historical facts which should be addressed to adequately answer the question,...
Markus Scheuer's user avatar
21 votes

Showing that $\bar{\mathbb{B}}^n$ is a manifold with boundary (Lee ITM Probelm 3-4)

I've been worked on this problem for some time, and i think i probably solved it based on the hint given on the book. Maybe this seems a little long, but it is really not. I tried my best to make this ...
Kelvin Lois's user avatar
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20 votes
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Understanding Takens' Embedding theorem

Practical meaning of Takens’ Theorem using your example The butterlfly-like structure traced out by the trajectories of the Lorenz system is the attractor of this dynamics. Its properties contain ...
Wrzlprmft's user avatar
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20 votes
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Showing that the composition of maps of constant rank does not have to be of constant rank

The map$f:\mathbb R\to \mathbb R^2:t \mapsto (t,t^2)\;$ has rank one everywhere. The map $g:\mathbb R^2 \to \mathbb R:(x,y)\mapsto y\;$ has rank one everywhere. Nevertheless the map $g\circ f:\mathbb ...
Georges Elencwajg's user avatar
19 votes
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Is the function ring $C^{\infty}(M)$ noetherian?

No, the ring $C^{\infty}(M)$ is never noetherian if $\dim M\gt 0$. Indeed consider a strictly decreasing sequence of closed subsets $$M=C_0\supsetneq C_1 \supsetneq C_2 \supsetneq C_3\supsetneq \...
Georges Elencwajg's user avatar
19 votes
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Definition of smooth manifold using sheaves.

Isomorphisms of locally ringed spaces are the same as isomorphisms of ringed spaces (which happen to be locally ringed), so you do not have to specify that $(M,\mathcal{O}_M)$ is locally ringed. That ...
Eric Wofsey's user avatar
19 votes

Since the Curvature tensor depends on a connection (not metric), is it the relevant quantity to characterize the curvature of Riemannian manifolds?

I think the point that should be understood is that there are three different levels of structure here: a Riemannian structure (a choice of metric) is richer than a choice of affine connection, which ...
Anthony Carapetis's user avatar
18 votes

Can we recover a compact smooth manifold from its ring of smooth functions?

[I assume all "smooth manifolds" are Hausdorff and paracompact.] Yes, you can recover $M$ as a smooth manifold from the ring $C^\infty(M)$. Here's a quick sketch. First, note that we can recover ...
Eric Wofsey's user avatar
18 votes

Number of Differentiable Structures on a Smooth Manifold

In the second statement, "unique" means unique up to diffeomorphism. If you have a manifold $M$ with a smooth structure $A$ and a homeomorphism $\varphi :M \rightarrow M$, which is not a ...
Lukas Heger's user avatar
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18 votes
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Is there a way to associate a Lie algebra to the group of diffeomorphisms?

This question certainly is too broad to be answered completely. Before talking about infinite dimensional Lie groups you need a concept of infinite dimensional manifolds. As you note correctly, you ...
Andreas Cap's user avatar
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18 votes
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Abstract Smooth Manifolds vs Embedded Smooth Manifolds

There's a reason that definition does not require that the map $\phi$ in a chart $(U,\phi)$ be a diffeomorphism: that would require knowing already that $M$ is a smooth manifold, but since that is ...
Lee Mosher's user avatar
  • 119k
18 votes
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Sign of codifferential

Okay. I've done it. I'll post it here. I think it may be of help to others. The whole purpose of the codifferential is to be the adjoint of the exterior derivative with respect to the Hodge inner ...
Jackozee Hakkiuz's user avatar
17 votes
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Relation between exterior derivative and Lie bracket

Eric, if you take $X=\partial/\partial x^i$, $Y=\partial/\partial x^j$, and $\omega = f dx^k$, then of course the formula gives you what you expect from the first two terms, and the bracket term ...
Ted Shifrin's user avatar
17 votes

Space of smooth structures

Let me answer your last question. Theorem. (Kirby, Siebenmann) Let $M^n$ be a closed $n$-dimensional topological manifold, where $n\ge 5$. Then the set of isomorphism classes of smooth structures ...
Moishe Kohan's user avatar
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17 votes

Are diffeomorphic smooth manifolds truly equivalent?

Keep in mind, the equivalence relation on differential manifolds is defined by the existence of a diffeomorphism between the two manifolds. This definition does not require that your favorite function ...
Lee Mosher's user avatar
  • 119k
17 votes

Continuous homomorphisms of Lie groups are smooth

Here is a better (more elegant) proof for both questions using Cartan-Von Neumann's theorem on closed subgroups. Let $H$ be the graph of $\gamma$, then $H$ is closed (and thus a Lie subgroup of $G\...
EternalBlood's user avatar
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16 votes
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The Pullback Bundle is an Embedded Submanifold of its Parent Space

$\newcommand{\M}{M}$ $\newcommand{\N}{N}$ $\newcommand{\brk}[1]{\left(#1\right)}$ $\newcommand{\be}{\beta}$ $\newcommand{\al}{\alpha}$ $\newcommand{\til}{\tilde}$ The pullback bundle is indeed an ...
Asaf Shachar's user avatar
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16 votes

Can we recover a compact smooth manifold from its ring of smooth functions?

I believe this is proven in Chapter 7 of Nestruev's Smooth Manifolds and Observables, but I haven't checked carefully. More precisely, the functor $M \to C^{\infty}(M)$ from smooth manifolds to the ...
Qiaochu Yuan's user avatar

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