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Let $M$ to be a manifold $m$-dimensional with a smooth hermitian metric $g$. The tangent bundle of $M$ is given by $TM= \bigcup_{p\in M} T_{p}M$, and the unit tangent bundle is given by

$S=\{x \in TM: \|x\|_{g(p)} =1$ where $x \in T_{p}M\} $.

(1)

In If M is a compact manifold, how to prove that the unitary tangent bundle is also compact?, the solutions assumed that given $x$ in $M$, you can obtain $U$ open neighborhood of $x$ such that $\pi^{-1}(U) \cap S \cong U \times S^{n-1}$. But this step is very important, and they do not show this homeomorphism.

(2)

In "The unit ball fibration in a tangent bundle", Pete L. Clark showed that $S$ is a compact set of $TM$ if $M$ is a compact manifold $m$-dimensional:

"1) Do the problem in the special case that the ball fibration is trivial, i.e., isomorphic to a product.

2) Convince yourself that the ball fibration is locally trivial.

3) Since the base is compact, you can find a finite open cover $\{U_i\}$ such that the restriction of the fibration to each $U_i$ is trivial. Now use the fact that a finite union of compact sets is compact."

I was trying to understand this answer, but I failed to follow this steps.

First of all, we consider $\pi: TM \to M$ the canonical projection. Using that $M$ is a compact manifold, there exists a finite charts $(\phi_{i}, U_{i})_{i=1}^{n}$ that covers $M$.

We note that if $M$ is a compact manifold then $TM$ is a trivial bundle, just consider $\Phi: TM \to M \times \mathbb{R}^{m}$ given by $x \mapsto (\pi(x), \theta_{\pi(x)}^{(U_{i}, \varphi_{i})} (x)) $ where $\theta_{\pi(x)}^{(U_{i}, \varphi_{i})}: T_{\pi(x)} M \to \mathbb{R}^{m}$ is a isomorphism linear and $i = \min\{j : \pi(x) \in U_{j}\}$.

But I'm stuck in this part. Can some one give some hint?

Thank you


Recall the definition:

A vector bundle of rank $m$ over $M$ is a topological space $E$ together with a surjective continuous map $\pi:E \to M$ satisfying the following conditions:

($1$) For each $p \in M$, the fiber $E_{p}= \pi^{-1}(p)$ over $p$ is endowed with the structure of a $k$-dimensional real vector space.

($2$) For each $p$ in $M$, there exists a neighborhood U of p in M and a homeomorphism $\Phi: \pi^{-1}(U) \to U \times \mathbb{R}^{m}$ (called a local trivialization of $E$ over $U$), satisfying the following conditions:

(2.1) $\pi_{U} \circ \Phi = \pi$ (where $\pi_{U}:U\times \mathbb{R}^{m} \to U$ is the projection);

(2.2) for each $q$ in $U$, the restriction of $\Phi$ to $E_{q}$ is a vector space isomorphism from $E_{q}$ to $\{q\}\times\mathbb{R}^{m}$.


  • Or someone know where I can find this proof? Any help would be appreciated. – Pedro do Norte Apr 12 '18 at 19:37
  • For your problem with (1), the identification is provided by any orthonormal frame $e : U \times \mathbb R^m \to TM:$ the unit sphere bundle is exactly the image of $U \times S^{m-1}.$ – Anthony Carapetis Apr 14 '18 at 06:11
  • I do not understand this argument. For consider a local frame from $U\times \mathbb{R}^{m}$ to $TM$, we will have to define our projection as $\widehat{\pi}: TM \to M \times \mathbb{R}^{m}$ (where $m$ is the dimension of manifold $M$), here $\widehat{E} =M \times \mathbb{R}^{m}$. So $U \times \mathbb{R}^{m}$ is an open set in $M \times \mathbb{R}^{m}$. – Pedro do Norte Apr 14 '18 at 13:07
  • But in this case, for a point $(p,v)$ in $U \times \mathbb{R}^{m}$, we have that $\widehat{E}{(p,v)} = (\widehat\pi)^{-1}(p,v)= v$ where $v \in T{p}M$, and then ($\widehat{\pi}, TM , M \times \mathbb{R}^{m}$) is a vector bundle of rank 1 over $M \times \mathbb{R}^{m}$. In this case, the local frame $TM$ over $U \times \mathbb{R}^{m}$ consists of one function and therefore it do not span $T_{p}M$. – Pedro do Norte Apr 14 '18 at 13:07
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    I think you misinterpreted me. I simply mean to take an orthonormal frame $e_1,\ldots,e_m$ (where each $e_i$ is a vector field on $U$) and consider it as a map $e : U \times \mathbb R^m \to TM$ by sending the $k^{\rm th}$ standard basis vector to $e_k$; i.e. $e(p,\delta_k) = e_k(p)$ where $\delta_k$ is the standard (or any orthonormal) basis of $\mathbb R^m.$ – Anthony Carapetis Apr 14 '18 at 13:12
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    But where I can find the "Gram-Schmit Algorthm for frames"? I know only in the context that $U$ is a subset of $\mathbb{R}^{n}$ for $E = T\mathbb{R}^{n}$. – Pedro do Norte Apr 14 '18 at 14:58
  • I was looking the proof of Gram-Schmit Algorthm for frames in Lemma 8.13 of Lee's "Introduction to Smooth Manifolds", and apparently, we can use the same argument to prove Gram-Schmit for frames in general manifolds. – Pedro do Norte Apr 14 '18 at 23:31
  • It's not true, that for $M$ compact $TM$ is trivial. – Mathy Jan 29 '20 at 10:49
  • @Mathy Cover $M$ by finitely many open neighbourhoods ${\mathcal{O}i}{i = 1}^N$ that admit normal frames on their compact closure. Choose a sequence $(x_i, v_i) \in TM$ where $v_i \in T_{x_i}M$, and $|v_i| = 1$. We can find $(x_{i_j}, v_{i_j})$ which converge $(x, v)$. A sub-sequence of $x_i$ converges to some $x$. Inifinitely many such x_i fall into the same neighbourhood from above, say $\mathcal{O}$. Using the compactness of $\overline{\mathcal{O}}$, we can refine the subsequence of $x_i$ such that the corresponding $v_i$ also converge to some $v$, also of unit norm due to continuity. – Master.AKA Jul 29 '24 at 15:22
  • @AnthonyCarapetis Why does $(M, g)$ necessarily admit an orthonormal local frame? Is it because $g$ is Hermitian? – Master.AKA Jul 29 '24 at 15:30
  • @Master.AKA: Any positive-definite bundle metric admits a local orthonormal frame - just apply Gram-Schmidt to the local frame provided by a local trivialization. – Anthony Carapetis Jul 31 '24 at 01:06
  • @Master.AKA But that does not mean that $TM$ is trivial. E.g., for compact $M= S^2$, $TS^2$ is non-trivial by the hairy ball theorem. – Mathy Jul 31 '24 at 10:14
  • @AnthonyCarapetis I see. I was thinking about normal frames. Those imply that the metric is flat. But I am still confused about something. Because an orthonormal frame means that you can always diagonalize the metric. Right? But it feels like you can then scale the vectors so that the metric becomes the usual Euclidean metric in a local coordinate. Isn't this a non-trivial question? I'm clearly missing something I guess. – Master.AKA Aug 08 '24 at 10:39
  • @Master.AKA: If there's an orthonormal coordinate frame (aka holonomic frame) this implies the metric is flat; but coordinate frames are a very special type of frame. – Anthony Carapetis Aug 09 '24 at 11:52

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