# Tag Info

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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 ...
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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 $G$-...
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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 ...
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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.
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 ...
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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.
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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 ...
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• 74.9k
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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 ...
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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 ...
• 95.8k
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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 $... • 7,317 1 vote Accepted ### Universal property of submanifold Consider a smooth injective map$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)\$...
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