# Tag Info

5

We provide a careful proof of Theorem 5.1 from Bott & Tu. The main body of the proof is taken from this MO post. Let $M^n$ be a smooth manifold. An open cover $\{U_\alpha\}$ is good if each nonempty finite intersection $U_{\alpha_1}\cap\cdots\cap U_{\alpha_k}$ is diffeomorphic to $\Bbb R^n$. The goal here is to prove the following Theorem 1. Every ...

3

Pick any non-identity involution $i$ of $\Bbb{RP}^2$. It has two lifts to a map $S^2 \to S^2$; pick the one whose square is the antipodal map. Because the antipodal map has no fixed points and is order 2, your lift must have no fixed points and be order 4. Note that necessarily the involution $i$ downstairs must have fixed points; there is no surface with ...

2

I think your confusion is around the $k$ vs. $n$. If so, I think the point is that the dimension of a manifold is not the dimension of the Euclidean space of which it is a subset (if it even is a subset of some Euclidean space). For example, consider the Klein bottle $K$. $K$ can be viewed as a subset of $\mathbb{R}^4$, but is $2$-dimensional, and cannot be ...

2

Use the fact that $B^{n+1}\setminus\{p,q\}\cong S^n\times(0,1)$, and define your map $f\colon S^n\times(0,1)\to S^n$ to just be projection onto the first factor.

2

Even easier, note that $d\theta\wedge\theta = (dx\wedge dz)\wedge(dy-z\,dx) = -dx\wedge dy\wedge dz\ne 0$, so this distribution is not involutive. There are (even locally) no integral manifolds, as you said. (You can also see this by locally writing a putative such surface as a graph $y=\phi(x,z)$ and getting a contradiction by taking a small closed curve on ...

2

Yes, it's correct. (You should probably comment that $\mu_M$ is nowhere-zero because of the linear independence of the $v_i$. Can you check that?) You also have a trivialization of the normal bundle, so it's orientable, and that means the tangent bundle is orientable as well. (Any time two of three vector bundles in a short exact sequence are orientable, so ...

1

One example is translation in $\mathbb{R}^{d}$. Perhaps a slightly more interesting case is polar coordinates. Take $M=M'=\mathbb{R}^{2}$, $f=$ identity. On $\left(M,g\right)$, take polar coordinates, and on $\left(M',g'\right)$ the Euclidean chart. Fix a point $p\in M\backslash\left\{ 0\right\}$, $p=\left(r,\theta\right)=\left(x,y\right)$, with $x=r\... 1 If there is such an extension, then$H_n(f)$factors over$H_n(D^n)=0$. If$\deg(f)=0$, then$f$is zero in$\pi_n(S^n)$by Hurewicz, hence extends to a map on the disk. 1 Your conjecture is (mostly) correct, but perhaps stated badly: every connected compact surfaces-with-boundary admits an everywhere nonzero vector fields. On the other hand, there are compact surfaces-without-boundary, other than the sphere, that do not admit nonzero vector fields. A 2-holed torus, for example, does not. As for combing cats and dogs, it ... 1 The author is not claiming that such a diffeomorphism always exists. For instance, if$K = M = S^1$or$M = S^1\times S^1, K = S^1 \times \{0\}$, then it is clear that no neighborhood of$K\$ can be homeomorphic to an open ball.

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