4 votes
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Why the derivative of a vector field along a curve is not defined for a generic manifold?

Suppose $X$ is a vector a field on a smooth manifold, and $\gamma:I\rightarrow M$ a smooth curve. Let $(U,\phi$, and $(V,\psi)$ be coordinate charts with $U\cap V$, and $\gamma(I)\cap U\cap V$ not ...
Chris's user avatar
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3 votes
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Is this quotient space a manifold?

Let $X$ be a topological manifold, $G$ a Lie group, $\mu: G\times X\to X$ a continuous action. Definition. A submanifold $C\subset X$ is called a cross-section for the $G$-action on $X$ if each $G$-...
Moishe Kohan's user avatar
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3 votes
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Spivak's definition of a $k$-dimensional mainfold

Here in the expression $\mathbb R^k \times \{0\}$ the symbol $0$ denotes the zero vector in $\mathbb R^{n-k}$. Often the same symbol $0$ is used to mean different things. The description $\{ y \in V \...
littleO's user avatar
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3 votes

What structure do Smooth Maps and Diffeomorphism?

So first things first, every diffeomorphism is also a homeomorphism by definition, so diffeomorphisms automatically preserve every topological property about manifolds. Since Lee has already given you ...
Chris's user avatar
  • 2,717
3 votes

What structure do Smooth Maps and Diffeomorphism?

To answer question (1), it is the atlas itself which defines the smooth structure. Given a smooth manifold $M$ with its atlas that I will denote $\mathcal A$, it is quite possible that $M$ has another ...
Lee Mosher's user avatar
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3 votes
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Multiplying tangent vectors by positive numbers (Gullemin-Pollack 1.8.2)

Working locally is fine. However, you can simplify the proof using the following ingredients. The authors define smoothness for maps $f : X \to \mathbb R^m$ defined on any $X \subset \mathbb R^n$. ...
Paul Frost's user avatar
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2 votes

Why do derivations take only smooth inputs?

They indeed can be. But this would ask you to keep track of a lot of differentiability levels. You can have $C^k$-manifolds for every $k$, and every operation happens at the appropriate $k$. By simply ...
Trebor's user avatar
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2 votes
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When is an analytic function a first integral of a vector field?

Notice that $f$ is constant along the trajectories of $V$ if and only if $V$ is tangent to the level sets $f^{-1}(c)$, $c \in \Bbb R$. This observation implies that if $f$ has an isolated extremum at $...
Travis Willse's user avatar
1 vote
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Is the image of the projection of a smooth curve onto a subset of its coordinates a manifold if the projection is an injective immersion?

The answer is no. Suppose $M$ is the image of the embedding: $$f(t) =(\cos t,\cos t\sin t, t)$$ defined on $(-3\pi/2, \pi/2)$. This is an embedded submanifold of $\mathbf{R}^3$, but its projection ...
While I Am's user avatar
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1 vote
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Time dependent vector fields are complete on compact manifolds

Found the solution. One has to look at a maximal integral curve of the induced vector field $J\times M\to T(J\times M)$ and apply the Escape lemma.
Csaba Daniel Farkaš's user avatar
1 vote
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Prove that if every smooth function on a subset of a manifold can be extended to a smooth function on the whole manifold, then the subset is closed.

The statement in the title of this question is wrong. Consider $A=M=\mathbb{R}$. Then, clearly any smooth function on $A$ can be extended to one on $M$. But $A$ is not compact. Of course, $A$ is still ...
GraffL's user avatar
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1 vote
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Does the inner semidirect product of Lie groups need these two subgroups both be closed?

You're right. See the note about page 169 on my list of corrections.
Jack Lee's user avatar
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1 vote
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If $G$ is an abelian Lie group then the Lie algebra of $G$ is abelian

If $G$ is abelian, then $i$ is a Lie group homomorphism. So from Teorema 8.44 in Lee's book that $i_* : LieG \to LieG$ is a Lie algebra homomorphism. So for any $X \in LieG$, $i_*X \in LieG$ is ...
Kelvin Lois's user avatar
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1 vote
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Is the shear map of a smooth principal bundle always a diffeomorphism?

The shear map is a diffeomorphism: A local section $s:U\to P$ induces diffeomorphisms $$ U\times G\times G\to P_{|U}\times G,\;\;\;\;(x,h,g)\mapsto (s(x)h,g) $$ $$ U\times G\times G\to (P \times_{M} P)...
Claire's user avatar
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1 vote
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Understanding Tangent Space under Context of Lie Theory

You are thinking about tangent spaces on a too high level for this example. Since $GL(n,\mathbb R)$ is an open subset in the vector space $M_n(\mathbb R)$ of $n\times n$-matrices any tangent space can ...
Andreas Cap's user avatar
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1 vote

How Can I solve this problem on Hopf Map?

The Hopf map is given by $$\pi : S^3\to S^2,\quad \pi (z,w) = (2\Re(w\overline{z}),2\Im(w\overline{z}), |z|^2 - |w|^2 ) .$$ Thus for all $a \in [-1,1]$ $$\pi ^{-1}(Z_a) = \{(z,w) \in S^3 \mid|z|^2 - |...
Paul Frost's user avatar
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1 vote
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How Can I solve this problem on Hopf Map?

Haven't you answered both questions already? For (1), you even wrote the conditions out yourself. For (2), $|z|^2=\frac{1+a}2$ and $|w|^2=\frac{1-a}2$ define two circles and hence the solution set is ...
Three aggies's user avatar
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1 vote
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Relation between generators of $\mathfrak{sl}_2(\mathbb{R})$ and $SL_2(\mathbb{R})$

To conclude the discussion: the Cartan decomposition (aka the SVD) implies that already the 1-parameter subgroups of positive diagonal matrices and of rotations generate $SL(2, {\mathbb R})$. In ...
Moishe Kohan's user avatar
  • 95.8k
1 vote

Universal property of submanifold

Long comment but not a complete answer (it's been a long time since I looked at this carefully and unfortunately I can't remember any reference ): Regarding 3: First, $\iota^*$ maps $C^\infty(M)$ to $...
Deane's user avatar
  • 7,317
1 vote
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Universal property of submanifold

Consider a smooth injective map $i\colon \mathbb R\to \mathbb R^2$ given by $i(t)=(t^3, t^2)$. The image $i(\mathbb R)$ is the affine variety $\{(x, y)\in \mathbb R^2 \mid x^2-y^3=0 \}$, where $(0, 0)$...
Paweł Czyż's user avatar
  • 3,228

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