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New answers tagged differential-topology

1 vote

Dimension of a smooth manifold

I believe you mean "$W$ neighbourhood of $x$", in general it is not possible to find a neighbourhood $W$ of $M$ such that $W \cap M$ is diffeomorphic to an open subset of $\mathbb R^m$. In ...
• 8,280
1 vote
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Dimension of a smooth manifold

In such situations it is always a good idea to look at examples. Let $k$ be arbitrary, and consider $$M=\{(x,0,\ldots,0)\in\mathbb{R}^k\ |\ x\in\mathbb{R}\}$$ You can verify very easily that $M$ is ...
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Is every connected surface locally symmetric?

Credit to Moishe Kohan https://math.stackexchange.com/a/3604136/724711: The Uniformization Theorem says that if $S$ is a connected Riemannian surface then it is conformally equivalent to a complete ...
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Example for Partition of Unity over circle

Here's a suggestion: Think of this on the unit interval $[0,1]$ — ultimately you will identify $0$ and $1$. You have two open subsets, $V_1=(0,1)$ and $V_2=[0,1/2)\cup (1/2,1]$. Take a smooth function ...
• 115k

In which points is $f: \mathbb{R}P^2 \rightarrow \mathbb{R}^3$ an inmersion.

Your chart $\phi_1$ is the one you would use if working with the projective space as the quotient $\Bbb{R}^3\setminus\{0\}\to\Bbb{RP}^2$, not $S^2\to\Bbb{RP}^2$. Thus your whole calculation is wrong. ...
• 56k
1 vote
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Extending Diffeomorphisms of Lie Manifolds

Not assuming $N$ is connected: Let $H \cong S^1 \times \mathbb Z_3$ which topologically is the disjoint union of $3$ circles embedded as $H$ concentrically into $M = G = \mathbb R^2$ in such a way ...
• 1,603
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Structure required for differentiability in topological context

In order to talk about differentiability you need the full structure of $\mathbb{R}^n$, i.e. its vector space structure together with the Euclidean norm. Because these are essential in defining ...
• 43.3k
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Reflection of the "Figure Eight" not diffeomorphic to the "Figure Eight"

A look into Boothby's book shows that he uses the concept of an imnmersed submanifold (see Definition 4.3). $N = G( \mathbb R)$ is regarded as such an object. This means that $N$ is not regarded as a ...
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1 vote

1 vote
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Prove that if a smooth manifold $M$ is contractible then every vector bundle over $M$ is trivial

You might also try Hussemoller's book "Fiber Bundles", Chapter 2, Corollary 4.8. Usually one proves, in the same breath, that if $f,g : X \to Y$ are two homotopic maps and if $B$ is a vector ...
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What are the orientable prime compact three-manifolds that can be embedded in $\mathbb{R}^4$?

Actually, regarding connected sums, it is much more subtle than you think. On one hand, if $M_1, M_2$ are 3-dimensional manifolds each of which embeds in $\mathbb R^4$, then their connected sum also ...
• 98k
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In complex geometry, is an holomorphic function continuous by definition?

Let $f : A \rightarrow \Bbb{C}$ be a complex function, with $A \subseteq \Bbb{C}$. Usually, to even say that $f$ is holomorphic, we require that $A$ be an open subset of $\Bbb{C}$, just like with ...
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Trivial examples can be obtained for non-connected $M$. Take e.g. $M = (0,1) \cup (1,2), N = \mathbb R$ and $f(x) = x$ for $x \in (0,1)$, $f(x) = 0$ for $x \in (1,2)$ Then the rank of $f$ is $1$ at \$x ...