# Tag Info

8

This is correct, and a fun idea. Here's why. I'll talk about smooth manifolds first, but it's true in full generality. All my manifolds have dimension $n \geq 3$. First off, we need to classify $n$-dimensional vector bundles over the circle. There are precisely two: one is trivial, the other is non-orientable, with total space homeomorphic to the (open) ...

7

If you could embed two copies of $\Bbb{RP}^2$ such that they don't intersect, you could demand that the second one be outside a (sufficiently small) tubular neighborhood of the first. The tubular neighborhood of any embedded $\Bbb{RP}^2$ is nontrivial by Stiefel-Whitney number manipulations. The boundary of such a tubular neighborhood is a 2-sphere. ...

7

Let me explain two ways to derive the parametric equations: Coordinate transformations: Think that you are in an amusement park and you are represented by a stick which rotates with respect to 2 axes at the same time. With respect to the big pole in the center of the rotating machine, and with respect to a center attached to the basket holding you. The ...

5

You can find some discussion and references on page 136 of the book by Lee "Introduction to smooth manifolds". In particular one of the result quoted is that Wall showed in 1965 that every 3-manifold can be embedded in$\mathbb{R}^5$. It also says that while the optimal immersion dimension is known for each $n$ and given by $2n-a(n)$, where $a(n)$ is the ...

4

At least for locally-flat topological embeddings (which are necessary for the spheres to be attaching spheres of some handle decomposition), your generalization is false. Lee and Wilczynski [LW] actually compute topological genus functions (the function sending a $\Bbb Z$-homology class to the minimum genus of an embedded surface representing it) for some ...

3

As the other answer suggests, there is an obstruction in the form of properness. If $f$ is a closed (synonymously, proper) embedding, then so is $\phi \circ f$. But even restricting to closed embeddings this is not true. Consider the case of long knots. That is, embeddings of $\mathbb R^1 \hookrightarrow \mathbb R^3$ that are the standard embedding outside ...

2

To be a manifold, a space $X$ has to be covered in coordinate maps $\varphi: A \subseteq \Bbb R^n \to X$. This is part of the definition. In particular, consider $\varphi$ to be a coordinate map onto a neighborhood of your point $x$. If $X$ is embedded into $R^m$ by some other map $\psi$, then $\psi \circ \varphi$ is a smooth map that could be considered a ...

1

For 1, positive mass theorem can be one of the examples. Here is some information about positive mass theorem. Schoen and Yau used the minimal surface to prove the positive mass theorem, which is originally a problem in general relativity.

1

Note that $E^k \setminus \{p\}$ deformation retracts to $S^{k-1}$ by $r : E^k \setminus \{p\} \to S^{k-1}, \; x \mapsto \frac x {\| x \|}$. Therefore, $\mathrm{id} \sqcup _f r : X \sqcup _f (E^k \setminus \{ p \}) = V \to X \sqcup _f S^{k-1} \simeq X$ is a retract of $V$ to $X$. To clarify why $X \sqcup _f S^{k-1} \simeq X$ let us show that, in general, if ...

1

There is a more general notion of a mapping cone. That is, an adjunction space corresponding to $f:S^{k-1} \to X$ is the same as a mapping cone $C_f$. For those you find plenty of answers which use the same strategy, hence you can build your intuition and see formulas by considering e.g. https://en.wikipedia.org/wiki/Mapping_cone_(topology), Homology of ...

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