Properties of Unit Tangent Bundle

Question

If $(M, g)$ is a Riemannian manifold with or without boundary, its unit tangent bundle $UTM := \{(p, v) \in TM \mid |v|_g = 1\}$ is a smooth, properly embedded codimension-1 submanifold with boundary in $TM$, with $\partial(UTM) = \pi^{-1}(\partial M)$ (where $\pi: UTM \rightarrow M$ is the canonical projection). The unit tangent bundle is connected iff $M$ is connected, and compact iff $M$ is compact.

I need to prove three things:

$UTM$ is a smooth, properly embeeded codimension-1 submanifold boundary in $TM$, and $\partial(UTM) = \pi^{-1}(\partial M)$.
$UTM$ is connected iff $M$ is connected.
$UTM$ is compact iff $M$ is compact.

I got stuck at 1. Here is my attempt so far: $g: TM \rightarrow \mathbb R$ is a smooth function, so $UTM = \{g = 1\}$ is a regular submanifold with codimension-1, i.e. an embedded manifold. However, I am not sure how to show $\partial(UTM) = \pi^{-1}(\partial M)$. Also, if $(p, v) \in \partial(UTM)$, what does it mean? By definition, if $(p, v)$ is a boundary point, then given a tangent vector $X_{(p,v)} \in T_{(p,v)}T_p M$, its "last" component is nonnegative. How does it lead to the fact that $X_{(p,v)} \in \pi^{-1}(\partial M)$?

2 is false as stated: $S^1$ is connected, but $UTS^1$ is disconnected. 2 is true for manifolds of dimension $2$ or more. — Jason DeVito - on hiatus, Jun 16 '21 at 01:52
@JasonDeVito Could you please elaborate why $UTS^1$ is disconnected? (edit: +1 and -1, so disconnected?) — James C, Jun 16 '21 at 01:56
Yes, $TS^1$ is isomorphic to $\mathbb{R}\times S^1$. Then $UTS^1$ is isomorphic to ${\pm 1}\times S^1$. The basic problem is that the unit sphere in $\mathbb{R}$ is disconnected. — Jason DeVito - on hiatus, Jun 16 '21 at 01:59

score 2 · Answer 1 · answered Feb 20 '25 at 01:36

Let $S$ be a subset of a smooth manifold $M$ (with or without boundary) having the following property: for each $p\in S$ there exists a neighborhood $U$ of $p$ in $M$ such that $U\cap S$ is an embedded submanifold (with or without boundary) of $U$. Then $S$ can be given a smooth structure making it an embedded submanifold (with or without boundary) of $M$. If $\partial M = \varnothing$ then this can be proved using slice charts. If $M$ has a non-empty boundary then we can reduce the problem to the boundaryless case by embedding $M$ in its double.

Let $(E_{1}, \dots, E_{n})$ be a local orthonormal frame on an open set $V$ in $M$ and let $\Phi : \pi^{-1}(V)\rightarrow V\times\mathbb{R}^{n}$ be the associated smooth local trivialization (where $\pi : TM\rightarrow M$ is the canonical projection). Since $V\times \mathbb{S}^{n-1}$ is an embedded submanifold (with boundary if $\partial V\neq \varnothing$) of $V\times\mathbb{R}^{n}$, its image $UTM\cap \pi^{-1}(V)$ under the diffeomorphism $\Phi^{-1}$ is embedded in $\pi^{-1}(V)$. Thus, we may use the result in the first paragraph to conclude that $UTM$ is embedded in $TM$.

Look here for connectedness and here for compactness.

Properties of Unit Tangent Bundle

1 Answers1

Linked